jeestudymaterialocean.files.wordpress.comβ¬Β¦Β Β· 3 | p a g e topic- relations and functions π...
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Study Material
Class XII - Mathematics
1 & 2
MARKS QUESTIONS
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TABLE OF CONTENTS
TOPIC- RELATIONS AND FUNCTIONS ................................................................................................................................ 3
TOPIC-: INVERSE TRIGONOMETRIC FUNCTIONS ............................................................................................................... 8
TOPIC-: MATRICES ........................................................................................................................................................... 16
TOPIC-: DETERMINANTS ................................................................................................................................................. 21
TOPIC: CONTINUITY ........................................................................................................................................................ 34
TOPIC- DIFFERENTIABILITY .............................................................................................................................................. 38
TOPIC-: APPLICATION OF DERIVATIVES .......................................................................................................................... 43
TOPIC-: INDEFINITE AND DEFINITE ................................................................................................................................ 48
TOPIC-: DIFFERENTIAL EQUATION ................................................................................................................................. 52
TOPIC-: VECTOR ALGEBRA .............................................................................................................................................. 56
TOPIC-: 3 β DIMENSIONAL GEOMETRY ........................................................................................................................... 60
TOPIC-: LINEAR PROGRAMMING .................................................................................................................................... 71
TOPIC-: PROBABILITY ...................................................................................................................................................... 83
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TOPIC- RELATIONS AND FUNCTIONS πΊπππππ πΈππππππππ ππ πππ π»ππππ πΉππππππππ πππ πππππππππ (π π΄πππ πΈππππππππ)
1 πΉπππ π‘βπ ππ’ππππ ππ πππ β πππ ππ’πππ‘ππππ ππππ π πππππ‘π π ππ‘ π΄ π‘π π΄.
2 πΏππ‘ π΄ = {4, 5, 0}. πΉπππ π‘βπ ππ’ππππ ππ ππππππ¦ ππππππ‘ππππ π‘βππ‘ πππ ππ πππππππ ππ π΄.
3 Let f: R βR be a function defined by π(π₯) = x2β3π₯ + 4, for all x β R, find the value of πβ1(2)
4 πΏππ‘ β ππ π ππππππ¦ ππππππ‘πππ ππ π πππππππ ππ¦ π β π = ππ , ππππ π‘βπ π£πππ’π ππ (3 β 2) β 1.
5 πΉπππ π‘βπ ππ’ππππ ππ πππ‘π ππ’πππ‘ππππ ππππ π‘βπ π ππ‘ {1, 2, 3, 4β¦π} π‘π ππ‘π πππ.
6 πΌπ π: π β π ππ πππ£ππ ππ¦ π(π₯) = (3 β π₯3)1/3, π‘βππ ππππ π(π(π₯)).
7 πΉπππ πππ(π₯), ππ π(π₯) = |π₯| πππ π(π₯) = |5π₯ β 2|
8 πΌπ π(π₯) = π₯ + 7 πππ π(π₯) = π₯ β 7, π₯ β π π‘βππ ππππ πππ (7).
9 πΏππ‘ β ππ π‘βπ ππππππ¦ ππππππ‘πππ ππ π πππ£ππ ππ¦ π β π = πΏπΆπ (π, π) πππ πππ π, π β π πΉπππ 5 β 7.
10 πβπ ππππππ¦ ππππππ‘πππ β : π π₯π β π ππ πππππππ ππ π β π = 2π + π. ππππ (2 β 3) β 4.
11 πβππ‘ ππ π‘βπ πππππ ππ π‘βπ ππ’πππ‘πππ π(π₯) =
|π₯ β 1|
(π₯ β 1) ?
12 ππππ‘π πππ ππ π: π β π πππ π: π β π πππ πππ£ππ ππ¦ π(π₯) = 8π₯3 πππ π(π₯) = π₯1/3
13 If X and Y are two sets having 2 and 3 elements respectively then find the number of functions
from X to Y.
14 πΏππ‘ β ππ π ππππππ¦ ππππππ‘πππ ππ π πππ£ππ ππ¦ π β π = π»πΆπΉ ππ π, π π€βπππ π, π β
π π€πππ‘π π‘βπ π£πππ’π ππ 22 β 4.
15 πΌπ π: π β π πππππππ ππ¦ π(π₯) = 3π₯ + 2 ππππππ π (π(π₯)).
16 πΏππ‘ π΄ = {1,2,3} , π΅ = {4,5,6,7} πππ πππ‘ π = {(1,4), (2,5), (3,6)} ππ π ππ’πππ‘πππ ππππ
π€βππ‘βππ π ππ πππ β πππ ππ πππ‘.
17 If the binary operation β ππ π‘βπ π ππ‘ π of integers is defined ππ¦ π β π = π + π β 5, then write the
identity element for the operation β ππ π.
18 πΌπ π = {(π₯, π¦): π₯ + 2π¦ = 8} ππ π πππππ‘πππ ππ π π€πππ‘π π‘βπ πππππ ππ π .
19 πΉπππ π‘βπ πππ ππ π: π β π πππ π: π β π πππ πππ£ππ ππ¦ π(π₯) = πππ π₯ πππ π(π₯) = 3π₯2.
20 πΌπ πππ ππ πππ‘π π‘βππ ππ ππ‘ πππππ π πππ¦ π‘βππ‘ π πππ π πππ‘β πππ‘π? ππ π€βππ‘?
21 πΊππ£π ππ₯πππππ ππ π‘π€π ππ’πππ‘ππππ π: π β π πππ π: π β π π π’πβ π‘βππ‘ πππ ππ ππππππ‘ππ£π ππ’π‘ π ππ
πππ‘ ππππππ‘ππ£π.
22 πΊππ£π ππ₯πππππ ππ π‘βπ ππ’πππ‘ππππ π: π β π πππ π: π β π π π’πβ π‘βππ‘ πππ ππ πππ‘π ππ’π‘ π ππ πππ‘
πππ‘π.
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23 πΏππ‘ π ππ π‘βπ ππ₯ππππππ‘πππ ππ’πππ‘πππ πππ π ππ π‘βπ πππππππ‘βπππ ππ’πππ‘πππ.πβππ‘ ππ πππ (1)?
24 πΆπππ ππππ π‘βπ ππππππ¦ ππππππ‘πππ β ππ π‘βπ π ππ‘ {1, 2, 3, 4, 5} πππππππ ππ¦ π β π = πππ {π, π}.
ππππ‘π π‘βπ ππππππ‘πππ π‘ππππ.
25 ππ‘ππ‘π π‘βππ‘ πππ£ππ ππ’πππ‘πππ ππ π π’πππππ‘ππ£π ππ πππ‘ , π: π β π, πππ£ππ ππ¦ π(π₯) = π₯2
πΊπππππ πΈππππππππ ππ πππ π»ππππ πΉππππππππ πππ ππππππππ (π π΄ππππ πΈππππππππ)
1 ππ‘ππ‘π π‘βπ ππππ ππ π€βπ¦ π‘βπ πππππ‘πππ π = {(π, π): π β€ π2} ππ π‘βπ π ππ‘ π ππ ππππ ππ. ππ πππ‘ ππππππ₯ππ£π.
2 πΌπ β ππ π ππππππ¦ ππππππ‘πππ ππ π‘βπ π ππ‘ π ππ ππππ ππ’πππππ . πππππππ ππ¦ π β π = π + π β 2, then
ππππ π‘βπ πππππ‘ππ‘π¦ πππππππ‘ ππ π‘βπ ππππππ¦ ππππππ‘πππ β
3 πΉπππ πππ π€βππ π: π β π πππ π: π β π πππ πππππππ ππ¦ π(π₯) =
2π₯ + 1
3 , π(π₯) = π₯ + 1.
4 πΌπ π‘βπ ππππππ¦ ππππππ‘πππ β ππ πππππππ ππ π ππ π β π = 2π + π + ππ, πππ πππ π, π β π ππππ
π‘βπ π£πππ’π ππ 3 β 4 πππ 2 β 5
5 πβππ€ π‘βππ‘ π: π β π π·ππππππ ππ¦ π(π₯) = π πππ₯ ππ ππππ‘βππ πππ πππ πππ πππ‘π.
6 πΌπ π‘βπ ππππππ¦ ππππππ‘πππ β πππππππ ππ π‘βπ π ππ π β π = π + ππ π‘βππ ππππ π€βππ‘βππ ππππππ¦ ππ
πππππ’π‘ππ‘ππ£π ππ πππ‘.
7 πΆβπππ π€βππ‘βππ π‘βπ π πππ’π‘πππ π ππ π πππππππ ππ π = (π, π): π β€ π3) ππ ππππππ₯ππ£π.
8 πΏππ‘ β ππ π ππππππ¦ ππππππ‘πππ ππ π‘βπ π ππ‘ ππ ππππ ππ’πππππ . πΌπ π β π = π + π β ππ, 2 β (3 β
π₯) = β7, ππππ π‘βπ π£πππ’π ππ π₯.
9 πβππ€ π‘βππ‘ π‘βπ ππ’πππ‘πππ π: π β π πππ£ππ: π(π₯) = ππ₯, ππ πππ πππ ππ’π‘ πππ‘ πππ‘π.
10 πΌπ π(π₯) = π₯2+3, π(π₯) = 3π₯ + 2. πΉπππ πππ.
11 πβππ€ π‘βππ‘ π‘βπ πππππππ‘βπππ ππ’πππ‘πππ π: π + β π πππ£ππ ππ¦ π(π₯) = ππππ₯, π₯ > 0 ππ πππ πππ.
12 π·ππ‘ππππππ π€βππ‘βππ π‘βπ ππ’πππ‘πππ β πππππππ ππ¦ π β π = π β π ππ ππππππ¦ ππ πππ‘ ππ π
13 π·ππ‘ππππππ π€βππ‘βππ π‘βπ πππππ‘πππ π ππ π = {π₯, π¦; π₯ β π¦ ππ ππ πππ‘ππππ} ππ ππππππ₯ππ£π
14 πβππ€ π‘βππ‘ π‘βπ ππ’πππ‘πππ π: π β π πππππππ ππ¦ π(π₯) = 2π₯ β 6 ππ πππ πππ.
15 πΏππ‘ π΄ = {1, 2,3} πππ πππππππ π = {(π, π): π β π = 12}, πβππ€ π‘βππ‘ π ππ ππ ππππ‘π¦ πππππ‘πππ ππ
π ππ‘ π΄.
16 π·ππ‘ππππππ π€βππ‘βππ π‘βπ ππ’πππ‘πππ π(π₯) = |π₯| + π₯
π βΆ π β π ππ (π) πππ β πππ (π) πππ‘π
17 πΆβπππ π€βππ‘βππ π‘βπ πππππ‘πππ π ππ π‘βπ π ππ‘ {1,2,3} πππ£ππ ππ¦ π = {(1,2), (2,1)} ππ π‘ππππ ππ‘ππ£π
18 πΌπ π(π₯) ππ ππ πππ£πππ‘ππππ ππ’πππ‘πππ, π‘βππ ππππ π‘βπ πππ£πππ π ππ π(π₯) =
3π₯ β 2
5.
19 πβππ€ π‘βππ‘ π(π₯) = 2π₯ ππ πππ β πππ πππ πππ‘ππππ π₯ β π
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20 πΌπ π(π₯) = log |
π₯ β 1
π₯ + 1| , β1 < π₯ < 1 π‘βππ π βππ€ π‘βππ‘ π(βπ₯) = π(π₯)
21 πΌπ π(π₯) = [π₯] πππ π(π₯) = |π₯| π‘βππ ππ£πππ’ππ‘π πππ (5
2) β πππ (
β5
2).
22 πΏππ‘ π: π: π β π ππ πππππππ ππ¦ π(π₯) = 3π₯ + 4 ππππ π‘βπ πππ£πππ π ππ π ππ ππ₯ππ π‘.
23 πβππ€ π‘βππ‘ β: π Γ π β π πππ£ππ ππ¦ π β π = π + 2π ππ πππ‘ ππ π πππππ‘ππ£π.
24 πβππ€ π‘βππ‘ π π’ππ‘ππππ‘πππ πππ πππ£ππ πππ πππ πππ‘ ππππππ¦ ππ π.
25 πβππ€ π‘βππ‘ ππ πππ‘π ππ’πππ‘πππ π: {1, 2, 3} β {1,2,3}ππ πππ€ππ¦π πππ β πππ.
πΉπ¬π³π¨π»π°πΆπ΅πΊ π¨π΅π« ππΌπ΅πͺπ»π°πΆπ΅πΊ (π π΄π¨πΉπ² πΈπΌπ¬πΊπ»π°πΆπ΅πΊ π¨π΅πΊπΎπ¬πΉ)
1 π!, where n(A)=n
2 πβπ ππ’ππππ ππ ππππππ¦ ππππππ‘ππππ ππ π‘βπ π ππ‘ ππππ ππ π‘πππ π πππππππ‘π ππ πππ£ππ ππ¦ ππ2 ,
π‘βπππππππ ππππ’ππππ ππππππ‘ππππ πππ 39
3 1 or 2
4 (3 β 2) β 1=9
5 π!
6 π(π₯) = (3 β π₯3)13β then find π(π(π₯))
π [(3 β π₯3)13β ] = [3 β ((3 β π₯3)
13β )3
]
13= (π₯3)
13β = π₯
7 πππ(π₯) = π(π(π₯)) = π(|5π₯ β 2|) = |5π₯ β 2|
8 πππ(7) = π(π(7)) = π(7 β 7) = π(0) = 0 + 7 = 7
9 5 β 7 = πΏπΆπ ππ 5 πππ 7 = 35
10 As π β π = 2π + π. find (2 β 3) β 4=(2 Γ 2 + 3) β 4 = 7 β 4 = 2 Γ 7 + 4 = 18
11 π(π₯) =|π₯β1|
(π₯β1)
|π₯ β 1| = π(π₯) = {(π₯ β 1)ππ π₯ β 1 > 0 ππ π₯ > 1
β(π₯ β 1)ππ π₯ β 1 < 0 ππ π₯ < 1 ,
πΉππ π₯ > 1, π(π₯) = 1, πΉππ π₯ < 1, π(π₯) = β1
π ππππ ππ π(π₯) = {β1,1}
12 πππ(π₯) = π(π(π₯)) = π (π₯1
3) =8(π₯1/3 )3 = 8π₯
13 ππ’ππππ ππ ππ’πππ‘ππππ ππππ π π‘π π ππ 32 = 9
14 As π β π = π»πΆπΉ ππ π, π, π‘βππ 22 β 4 = π»πΆπΉ ππ 22 πππ 4 ππ = 2
15 π(π(π₯) = π(3π₯ + 2) = 3(3π₯ + 2) + 2 = 9π₯ + 8
16 π ππ πππ β πππ πππππ’π π π(1) = 4, π(2) = 5, π(3) = 6 , ππ π‘π€π πππππππ‘π ππ π΄ βππ£π π πππ π image.
17 πΏππ‘ π β π ππ π‘βπ ππππ’ππππ πππππ‘ππ‘π¦
π β π = π β π + π β 5 = π β π = 5
18 {1,2,3}
19 πππ(π₯) = 3πππ 2π₯
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20 π ππ πππ‘ πππππ π πππππ¦ ππ πππ‘π ππ’π‘ π ππ’π π‘ ππ πππ‘π
21 π(π₯) = π₯, π(π₯) = |π₯|
22 π(π₯) = π₯ + 1, π(π₯) = {
π₯ β 1 ππ π₯ > 11 ππ π₯ = 1
23 1
24
* 1 2 3 4 5
1 1 1 1 1 1
2 1 2 2 2 2
3 1 2 3 3 3
4 1 2 3 4 4
5 1 2 3 4 5
25 πππ‘ π π’πππππ‘ππ£π
RELATIONS AND FUNCTIONS (2 MARKS QUESTIONS ANSWER)
1 1
2> (
1
2)2
β(1
2,1
2) βR , Hence R is not reflexive
2 e=2
3 πππ =2π₯+3
3.
4 (i) 3 β 4 = 2 Γ 3 + 4 + 3 Γ 4 = 22, (ππ) 2 β 5 = 2 Γ 2 + 5 + 2 Γ 5 = 19
5 Sππππ π(0) = π(π) ππ’π‘ 0 β π πβπππππππ π ππ πππ‘ πππ πππ
πππ π ππππ ππ π = [β1,1] β π π‘βπππππππ π ππ πππ‘ πππ‘π.
6 π β π = π + ππ πππ π β π = π + ππ, πβπππππππ π β π β π β π, π π ππ‘ ππ πππ‘ πππππ’π‘ππ‘ππ£π
7 Since π < π3 ππ πππ‘ π‘ππ’π πππ π =1
2, (π, π) β π
8 π₯ = β3
9 πΏππ‘ π, π β π , π(π) = π(π), ππ = ππ β π = π, π π ππ‘ ππ πππ β πππ
π ππππ ππ π = π + β ππππππππ, ππ ππ‘ ππ πππ‘ πππ‘π
10 3π₯2 + 11
11 πβππ€ πππ πππ πππ , πππ‘ π, π β π + β π(π) = π(π) β log π = log π β π = π
12 π β π = π β π, 5,2 β π, ππ’π‘ 2 β 5 = β3 β N so is not binary.
13 πΉππ (π, π) β π, π β π = 0 ππ ππ πππ‘ππππ . ππ ππ‘ ππ ππππππ₯ππ£π.
14 πΏππ‘ π₯1, π₯2 β π, π(π₯1) = π( π₯2) β 2π₯1 β 6 = 2 π₯2 β 6 β π₯1 = π₯2, π π ππ‘ ππ πππ β πππ.
15 πβπππ ππππ πππ‘ ππ₯ππ π‘ πππ¦ πππππππ‘π ππ π΄ πππ π€βππβ π β π = 12 π π ππ‘ ππ ππ ππππ‘π¦ πππππ‘πππ.
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ππ. |1 β 1| = 0 β 12 β (1, 1) β π ,
|1 β 2| = 1 β 12 β (1, 2), (2, 1) β π ,
|1 β 3| = 2 β 12 β (1, 3), (3, 1) β π ,
|2 β 3| = 1 β 12 β (3, 2), (2, 3) β π ,
16 ππππ‘βππ πππ β πππ πππ πππ‘π
17 π ππ πππ‘ π‘ππππ ππ‘ππ£π ππ (1,2) β π , (2,1) β π ππ’π‘ (1,1) β π ππ ππ‘ ππ πππ‘ π‘ππππ ππ‘ππ£π
18 πΏππ‘ π¦ = π(π₯) = 3π₯β2
5., π‘βππ π·π = π , π π = π
5y=3x-2 , π₯ =5π¦+2
3, π₯, π¦ β π β fβ1(y) =
5π¦+2
3
19 πΏππ‘ π, π β π , π(π) = π(π) β 2π = 2π β π = π, ππ ππ‘ ππ πππ β πππ
πΏππ‘ π¦ β π , π¦ = 2π₯ β π₯ =π¦
2β π (ππππππ), ππ π ππ πππ‘π.
20 π(βπ₯) =
πππ|1 + π₯|
πππ |1 β π₯|= β
πππ |1 β π₯|
πππ |1 + π₯| = β π(π₯)
21. πππ(5
2) = 2, πππ(
β5
2) = 3, π΄ππ ππ β 1
22 πβππ€ πππ β πππ πππ πππ‘π π‘βππ π-1(π₯) = π₯β4
3
23 ππππ 3, 5, 8 ππ ππππ ππ. π‘βππ(3 β 5) β 8 = 29 πππ 3 β (5 β 9) = 45, βππππ βis not ππ π πππππ‘ππ£π
24 β : π Γ π β π ππ πππ£ππ ππ¦ π₯ β π¦ = π₯ β π¦ π€βππβ ππ πππ‘ π ππππππ¦ ππππππ‘πππ ππ π₯ < π¦
(πππ ππ. π₯ = 4, π¦ = 6)
πππππππππ¦ πππ πππ£ππ πππ (π₯, π¦) = π₯ Γ· π¦ πππ¦ πππ‘ ππ π πππ‘π’πππ ππ. (πππ ππ. π₯ = 4, π¦ = 6)
25 ππ’ππππ π π ππ πππ‘ πππ πππ . πβππ π‘βπππ ππ₯ππ π‘π π‘π€π πππππππ‘π π ππ¦ 1 πππ 2 . π΄ππ π π‘βπ πππππ ππ 3
π’ππππ π πππ ππ ππππ¦ πππ πππππππ‘. πβπππππππ π‘βπ πππππ π ππ‘ πππ βππ£π ππ‘ πππ π‘ π‘π€π πππππππ‘π ππ π‘βπ ππππππππ {1, 2, 3}. π βππ€πππ π‘βππ‘ π ππ πππ‘ πππ‘π π, ππππ‘ππππππ‘πππ. π»ππππ π ππ’π π‘ ππ πππ β
πππ.
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8 | P a g e
TOPIC-: INVERSE TRIGONOMETRIC FUNCTIONS Q.NO 1 MARK PROBLEMS
1. If π¦ = sinβ1 π₯ then write the range of π¦
2. Find the principal value of sinβ1 (β1
2) + sinβ1 (
1
β2)
3. Find the principal value of cosβ1 (β3
2) + cosβ1 (
β1
β2)
4. Find the principal value of tanβ1(β3) β cotβ1(ββ3)
5. Find the principal value of tanβ1(β1) + secβ1(β2)
6. Find the principal value of cosecβ1(2) β cosecβ1(ββ2)
7. Find the principal value of secβ1 (2
β3)
8. Find the principal value of cotβ1 (β1
β3)
9. Find the value of tanβ1(1) + cosβ1 (β1
2) + sinβ1 (
β1
2)
10. Find the value of cos (π
2β sinβ1
3
7)
11. Find the value of sinβ1 (sin3π
5)
12. Find the value of cosβ1 (cos13π
6)
13. Find the value of cosβ1 (cos7π
6)
14. Find the value of tanβ1 (tan7π
6)
15. Find the principal value of cosβ1(cos(β680Β°))
16. Find the value of tanβ1 (tan3π
4)
17. Find the value of sinβ1 (cos (43π
5))
18. If secβ1(2) + cosecβ1 π¦ =π
2 π‘βππ ππππ π‘βπ π£πππ’π ππ π¦.
19. If tanβ1 β3 + cotβ1 π₯ =π
2 π‘βππ ππππ π‘βπ π£πππ’π ππ π₯
20. Find the value of sin (π
3 β sinβ1 (
β1
2))
21. If sin (sinβ11
5+ cosβ1 π₯) = 1, then find the value of π₯
22. If tanβ1 π₯ + tanβ1 π¦ = π
4 ; π€βπππ π₯π¦ <1,find the value of π₯ + π¦ + π₯π¦
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9 | P a g e
23. Find the value of sinβ1 (ββ3
2) + tanβ1(ββ3)
24. Find the value of tanβ1 [2 cos (2 sinβ11
2)]
25. Find the value of tan (2 tanβ1 (1
5))
INVERSE TRIGONOMETRIC FUNCTIONS
Q.NO 2 MARK PROBLEMS
1. Show that tanβ1 (1
2) + tanβ1 (
2
11) = tanβ1 (
3
4)
2. Show that tanβ1 (2
11) β tanβ1 (
7
24) = tanβ1 (
β29
278)
3. Show that 2 tanβ1 (1
2) + tanβ1 (
1
7) =tanβ1 (
31
17)
4. Show that 2 sinβ1 (3
5) = tanβ1 (
24
7)
5. Show that sinβ1(2π₯β1 β π₯2) = 2 sinβ1 π₯ = 2 cosβ1 π₯
6. Prove that 3 sinβ1 π₯ = sinβ1(3π₯ β 4π₯3), π₯ β [β1
2 ,1
2]
7. Prove that 3 cosβ1 π₯ = cosβ1(4π₯3 β 3π₯) , π₯ β [β1
2 , 1]
8. Prove that tanβ1(π₯) + tanβ1 (2π₯
1β π₯2) =tanβ1 (
3π₯ β π₯3
1β3π₯2), |π₯| <
1
β3
9. Prove that tanβ1 (π₯
π¦) β tanβ1 (
π₯βπ¦
π₯+π¦) =
π
4
10. Prove that tanβ1 βπ₯ =1
2cosβ1 (
1βπ₯
1+π₯) , π₯ β [0,1]
11. Find the value of tan (sinβ13
5+ cotβ1
3
2)
12. Find the value of tan1
2[sinβ1
2π₯
1+ π₯2 + cosβ1
1βπ¦2
1+π¦2] , |π₯| < 1, π¦ > 0 πππ π₯π¦ < 1
13. Simplify tanβ1 (β1+ π₯2β1
π₯) , π₯ β 0
14. Simplify tanβ1 (π₯
βπ2βπ₯2) , |π₯| < π
15. Simplify tanβ1 (3π2π₯ βπ₯3
π3β3ππ₯2) , π > 0;
βπ
β3< π₯ <
π
β3
16. Simplify cotβ1 (1
βπ₯2 β1) , π₯ > 1
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10 | P a g e
17. Simplify tanβ1 (β
1βcosπ₯
1+cosπ₯) , 0 < π₯ < π
18. Simplify tanβ1 (cosπ₯
1β sinπ₯) ,
β3π
2< π₯ <
π
2
19. Simplify tanβ1 (cosπ₯ β sinπ₯
cosπ₯+sinπ₯) , 0 < π₯ < π
20. Simplify tanβ1 (π cosπ₯ βπ sinπ₯
π cosπ₯ +π sinπ₯) ,if
π
πtan π₯ > β1
21. Solve 2 tanβ1(cos π₯) = tanβ1(2 cosec π₯)
22. Solve 2tanβ1(sin π₯) = tanβ1(2 sec π₯)
23. Solve tanβ1(2π₯) + tanβ1(3π₯) = π
4
24. Solve cos(tanβ1 π₯) = sin (cotβ1 (3
4))
25. If tanβ1 (π₯β1
π₯β2) + tanβ1 (
π₯ +1
π₯ + 2) =
π
4 ,then find the value of π₯
INVERSE TRIGONOMETRIC FUNCTIONS
Q.NO SOLUTIONS FOR 1 MARK QUESTIONS
1. βπ
2β€ π¦ β€
π
2 OR π¦ β [
βπ
2,π
2]
2. sinβ1 (β1
2) + sinβ1 (
1
β2) = sinβ1 (β sin
π
6) + sinβ1 (sin
π
4) =
βπ
6+π
4=
π
12
3. cosβ1 (
β3
2) + cosβ1 (
β1
β2) = cosβ1 (cos
π
6) + cosβ1 (β cos
π
4) =
π
6+3π
4=11π
12
4. tanβ1(β3) β cotβ1(ββ3) = tanβ1 (tan
π
3) β cotβ1 (βcot (
π
6)) =
π
3β5π
6=βπ
2
5. tanβ1(β1) + secβ1(β2) =
βπ
4+2π
3=5π
12
6. cosecβ1(2) β cosecβ1(ββ2) =
π
6+π
4=5π
12
7. secβ1 (
2
β3) =
π
6
8. cotβ1 (
β1
β3) = cotβ1 (β cot
π
3) = cotβ1 (cot (π β
π
3)) =
2π
3
9. tanβ1(1) + cosβ1 (
β1
2) + sinβ1 (
β1
2) =
π
4+2π
3βπ
6=3π
4
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11 | P a g e
10. cos (
π
2β sinβ1
3
7) = cos (cosβ1
3
7) =
3
7
11. sinβ1 (sin
3π
5) = sinβ1 (sin (π β
3π
5)) =
2π
5
12. cosβ1 (cos
13π
6) = cosβ1 (cos (2π +
π
6)) =
π
6
13. cosβ1 (cos
7π
6) = cosβ1 (cos (2π β
5π
6)) =
5π
6
14. tanβ1 (tan
7π
6) = tanβ1 (tan (π +
π
6)) =
π
6
15. cosβ1(cos(β680Β°)) = cosβ1(cos(680Β°)) = cosβ1(cos(4π β 40Β°))=40Β°
16. tanβ1 (tan
3π
4) = tanβ1 (tan (π β
π
4)) = tanβ1 (tan (
βπ
4)) =
βπ
4
17. sinβ1 (cos (
43π
5)) = sinβ1 (cos ((8π +
3π
5)))
= sinβ1 (cos (3π
5)) = sinβ1 (sin (
π
2β3π
5)) =
βπ
10
18. If secβ1(2) + cosecβ1 π¦ =π
2 π‘βππ π¦ = 2
19. If tanβ1 β3 + cotβ1 π₯ =π
2 π‘βππ π₯ = β3
20. sin (
π
3 β sinβ1 (
β1
2)) = sin (
π
3+π
6) = sin
π
2= 1
21. If sin (sinβ11
5+ cosβ1 π₯) = 1,then π₯ =
1
5
22. tanβ1 π₯ + tanβ1 π¦ = π
4 βΉ tanβ1 (
π₯+π¦
1βπ₯π¦) =
π
4 βΉ
π₯+π¦
1βπ₯π¦= 1βΉ π₯ + π₯π¦ + π¦ = 1
23. sinβ1 (
ββ3
2) + tanβ1(ββ3) = β
π
3βπ
3= β
2π
3
24. tanβ1 [2 cos (2 sinβ1
1
2)] =
π
4
25. tan (2 tanβ1 (
1
5)) = tan (2 tanβ1 (
1
5)) = tan π π€βπππ
π
2= tanβ1
1
5
Now tan π =2 tan
π
2
1βtan2π
2
=5
12
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12 | P a g e
Q.NO SOLUTIONS FOR 2 MARKS QUESTIONS
1.
L. H. S = tanβ1 (1
2) + tanβ1 (
2
11) = tanβ1(
12+211
1 β12 Γ
211
) = tanβ1 (3
4) = π .π». π
(πππππ’ππ: tanβ1 π₯ + tanβ1 π¦ = tanβ1 (π₯ + π¦
1 β π₯π¦) )
2.
tanβ1 (2
11) β tanβ1 (
7
24) = tanβ1(
211β724
1 +211 Γ
724
) = tanβ1 (β29
278) = π .π». π
(πππππ’ππ: tanβ1 π₯ β tanβ1 π¦ = tanβ1 (π₯ β π¦
1 + π₯π¦)
3. L.H.S = 2 tanβ1 (1
2) + tanβ1 (
1
7)
= tanβ1 (2Γ
1
2
1β(1
2)2) + tanβ1 (
1
7) apply formula: 2 tanβ1(π₯) = tanβ1 (
2π₯
1βπ₯2)
tanβ1 (4
3) + tanβ1 (
1
7)
=tanβ1 (31
17) apply formula tanβ1 π₯ + tanβ1 π¦ = tanβ1 (
π₯+π¦
1βπ₯π¦)
=R.H.S
4.
2 sinβ1 (3
5) = sinβ1(2 Γ
3
5β1 β (
3
5)2
) = sinβ1 (24
25) = tanβ1 (
24
7)
( formula 2 sinβ1(π₯) = sinβ1 (2π₯β1 β (π₯)2) )
5. πΏππ‘ π₯ = sinπ π‘βππ π = sinβ1 π₯
πππ€ sinβ1 (2π₯β1 β π₯2) = sinβ1 (2 sin πβ1 β sin2 π) =
sinβ1(2 sin π cos π) = sinβ1(sin 2π) = 2π = 2 sinβ1 π₯
Similarly By taking π₯ = cosπ , π€π πππ‘ sinβ1(2π₯β1 β π₯2) = 2 cosβ1 π₯
6. As above problem by taking π₯ = sinπ π€π π€πππ πππ‘ 3 sinβ1 π₯ = sinβ1(3π₯ β 4π₯3)
Use formula sin 3π = 3 sinπ β 4 sin3 π
7. As above problem by taking π₯ = cos π π€π π€πππ πππ‘ 3 cosβ1 π₯ = cosβ1(4π₯3 β 3π₯)
Use formula cos 3π = 4 cos3 π β 3 cos π
8. πππ‘ π₯ = tanπ π‘βππ π = tanβ1 π₯
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13 | P a g e
tanβ1 (3π₯ β π₯3
1 β 3π₯2) = tanβ1 (
3 tan π β tan3 π
1 β 3 tan2 π) = tanβ1(tan 3π) = 3π = 3 tanβ1 π₯
9. Apply the formula tanβ1 π₯ β tanβ1 π¦ = tanβ1 (π₯βπ¦
1+π₯π¦) and then simplify
10. π .π». π =
1
2cosβ1 (
1 β π₯
1 + π₯) =
1
2cosβ1 (
1 β (βπ₯)2
1 + (βπ₯)2) =
1
2Γ 2 tanβ1 π₯ = tanβ1 π₯ = πΏπ»π
11. tan (sinβ1
3
5+ cotβ1
3
2) = tan (tanβ1
3
4+ tanβ1
2
3) = tan (tanβ1
17
6) =
17
6
12.
tan1
2[sinβ1
2π₯
1 + π₯2 + cosβ1
1 β π¦2
1 + π¦2]
=tan1
2[2 tanβ1 π₯ + 2 tanβ1 π¦]
= tan[tanβ1 π₯ + tanβ1 π¦]
= tan (tanβ1 (π₯+π¦
1βπ₯π¦))
=π₯+π¦
1βπ₯π¦
13. πππ‘ π₯ = tan π π‘βππ π = tanβ1 π₯
tanβ1 (β1 + π₯2 β 1
π₯) = tanβ1 (
β1 + tan2 π β 1
tanπ) = tanβ1
1 β cos π
sin π= tanβ1 (tan
π
2)
=π
2=1
2tanβ1 π₯
14. Let π₯ = π sinπ π‘βππ π = sinβ1π₯
π
tanβ1 (π₯
βπ2 β π₯2) = tanβ1 (
π sin π
βπ2 β π2 sin2 π)
= tanβ1 (sin π
cos π) = tanβ1(tan π) = ΞΈ = sinβ1
π₯
π
15. ANS: 3 tanβ1π₯
π
Hint:By taking π₯ = π tan π π‘βππ π = tanβ1 (π₯
π) and then substitute and simplify as
above problem
16. Ans: secβ1 π₯
By taking π₯ = sec π π‘βππ π = secβ1 π₯ and then substitute and simplify
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14 | P a g e
17.
tanβ1(β1 β cosπ₯
1 + cosπ₯) = tanβ1(β
2sin2π₯2
2 cos2π₯2
) = tanβ1 (tanπ₯
2) =
π₯
2
18.
tanβ1 (cos π₯
1 β sin π₯) = tanβ1(
sin (π2 β π₯)
1 β cos (π2β π₯)
) = tanβ1 (tan (π
4+π₯
2)) =
π
4+π₯
2
19. tanβ1 (
cos π₯ β sin π₯
cos π₯ + sin π₯)
by dividing numerator denominator with cos π₯
tanβ1 (1βtanπ₯
1+tanπ₯)
= tanβ1 (tan (π
4β π₯))
=π
4β π₯
20. tanβ1 (
π cos π₯ β π sin π₯
π cos π₯ + π sin π₯)
Divide with π cos π₯ both numerator and denominator we get
tanβ1 (π
πβtanπ₯
1+π
πtanπ₯
) πππππ’ππ: tanβ1 π₯ + tanβ1 π¦ = tanβ1 (π₯+π¦
1βπ₯π¦) )
=tanβ1π
πβ tanβ1(tan π₯)
=tanβ1π
πβ π₯
21. 2 tanβ1(cos π₯) = tanβ1(2 cosec π₯)
tanβ1 (2 cos π₯
1 β cos2 π₯) = tanβ1(2 csc π₯)
tanβ1 (2 cos π₯
sin2 π₯) = tanβ1 (
2
sin π₯)
cot π₯ = 1
π₯ =π
4
22. 2tanβ1(sin π₯) = tanβ1(2 sec π₯)
tanβ1 (2 sin π₯
1 β sin2 π₯) = tanβ1 (
2
cosπ₯)
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15 | P a g e
tan π₯ = 1 implies π₯ =π
4
23. tanβ1(2π₯) + tanβ1(3π₯) = π
4
tanβ1 (5π₯
1β6π₯2) =
π
4 πππππ’ππ: tanβ1 π₯ + tanβ1 π¦ = tanβ1 (
π₯+π¦
1βπ₯π¦) )
By simplifying π₯ =1
6 ππ π₯ = β1
π₯ = β1 ππππ πππ‘ π ππ‘ππ ππ¦πππ π‘βπ πππ£ππ πππ’ππ‘πππ
Therefore π₯ =1
6
24. cos(tanβ1 π₯) = sin (cotβ1 (
3
4))
tanβ1 π₯ =π
2β cotβ1 (
3
4)
π
2β cotβ1 π₯ =
π
2β cotβ1 (
3
4)
π₯ =3
4
25. tanβ1 (
π₯ β 1
π₯ β 2) + tanβ1 (
π₯ + 1
π₯ + 2) =
π
4
πππππ’ππ: tanβ1 π₯ + tanβ1 π¦ = tanβ1 (π₯ + π¦
1 β π₯π¦)
After applying formula and simplify then we get
2π₯2 β 4 = β3
2π₯2 = 1
π₯ = Β±1
β2
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16 | P a g e
TOPIC-: MATRICES
1 mark questions:
1. πΌπ π΄ ππ π‘βπ π ππ’πππ πππ‘πππ₯ ππ πππππ 3 πππ |2π΄| = π |π΄| . πΉπππ π‘βπ π£πππ’π ππ πΎ ?
2. Write the element a12 of the matrix A = [πππ]2x2 Whose elements πππ are given
ππ¦ πππ = π2ππ₯. π ππππ₯ .
3. πΌπ π΄ |π΄πππ΄| = [
7 0 00 7 00 0 7
] π‘βππ ππππ |π΄πππ΄|
4. If A is an invertible matrix of order 3x3 and ΣAΣ =9. Then find π΄ππ|π΄πππ΄|
5. If a matrix has 18 elements, what are the possible orders it can have?
6. If [3 42 π₯
] [π₯1] = [
1925] , find the value of x.
7. Find the value of x and y that makes the following pair of matrices equal.
[π₯ + 3π¦ π¦7 β π₯ 4
] = [4 β10 4
]
8. πΌπ 2 [3 45 π₯
] + [1 π¦0 1
] = [7 010 5
] , π‘βππ ππππ π₯ β π¦.
9. If A is a 3x3 matrix, whose elements are given by aij = 1
3 I-3i + jI, then write the value of
a23.
10. If matrix A = [1 2 3] write AAβ. Where Aβ is the transpose of matrix A.
11. If A= [
2 3 β51 4 90 7 β2
] and B= [2 1 β1β3 4 41 5 2
] then find a22 +b21.
12. If A =[aij] is a 2x2 matrix such that aij = i2 β j2
13. Let A =[2 43 2
] and C = [β2 53 4
] find 3A β C.
14. If [1 0 00 β1 00 0 1
] [π₯π¦π§] = [
101] , find x,y,z.
15. If AT = [β2 31 2
] and B = [β1 01 2
], find ( A + 2B )T
16. If A = [2 35 7
] , find A + AT
17. If A = [π ππ π
] ,Verify that (A - AT)is a skew symmetric matric.
18. If A = [1 0β1 7
], find k such that A2 = 8A + KI
19. If A = [3 4β4 β3
], find f(A), where f(x) = x2 -5x + 7
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17 | P a g e
20. If A = [
2 0 12 1 31 β1 0
] then find the value of A2 β 3A + 2I
Answers of 1 mark questions
1. k = 8.
2. a12 = e2x sin2x
3. A |π΄πππ΄| = 7I
4. |π΄πππ΄| = IAIn-1 = 93-1 = 81.
5. 6.5.18x1, 9x2, 6x3, 3x6, 2x9, 1x18.
6. x=5
7. x= 7, y = -1.
8. 10
9. a23 =1.
10. AAβ = [1 2 3] [
123] = [1 4 9]
11. a22 +b21 = 4-3=1.
12. [0 β33 0
]
13. 3[2 43 2
] - [β2 53 4
] = [10 76 2
]
14. [π₯βπ¦π§] = [
101] Implies x=1, y =0, z= 1.
15. A + 2B = [β2 13 2
] + 2[β1 01 2
] = [β4 15 6
]
( A + 2B )T =[β4 51 6
].
16. A + AT = [2 35 7
] + [2 53 7
] = [4 88 14
]
17. Let (A - AT) = B, show that BT = -B.
18. K = -7.
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18 | P a g e
19. [β15 β2020 15
]
20. [2 0 12 1 31 β1 0
]
MATRICES
2 mark questions:
1. Shows that skew symmetric matrix of odd order is always singular.
2. If A = [πππ π₯ βπ πππ₯π πππ₯ πππ π₯
] then for what value of x , A is an identity matrix.
3. Simplify tanx [π πππ₯ βπ‘πππ₯π‘πππ₯ βπ πππ₯
] +secx [βπ‘πππ₯ βπ πππ₯βπ πππ₯ π‘πππ₯
]
4. For what value of x, is the matrix A= [β
0 1 β21 0 3π₯ β3 0
] a skew symmetric matrix?
5. Construct a 2 Γ 2 matrix where, aij = I -2i +3j I.
6. If [
π₯ 3π₯ β π¦2π₯ + π§ 3π¦ β π€
] = [3 24 7
], find x,y,z,w.
7. If A and B are symmetric matrices, show that AB is symmetric, if AB = BA.
8. Using elementary transformations, find the inverse of the matrix A= [6 55 4
]
9. Construct a 2x2 matrix, A = [aij] Whose elements are given by aij = ( i + j)2
2πππ
10. If A = [β3 00 β3
], find A4
11. If A is a square matrix such that A2 = A then write the value of (I + A )3 β 7A .
12. If A = [2 0 21 0 β1
] and B = [β3 1 02 0 1
] , find the matrix C such that A + B + C
is a null matrix.
13. Express the matrix A = [
4 2 β13 5 71 β2 1
] as the sum of a symmetric and a skew
symmetric matrix.
14. If A =[0 3β2 5
], find k, so that KA2 β 5A β 6I2=0
15. Given matrix A = [
β245] and B = [1 3 β6] verify that (AB)T = BTAT
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19 | P a g e
16. Given matrix A =[1 23 4
], find f (A), if f(x) = 2x2 - 3x +5.
17. Find the matrix X such that [
2 β10 1β2 4
] X = [β1 β8 β103 4 010 20 10
]
18. If A = [cos Ρ² π π ππΡ²π π ππΡ² cos Ρ²
] where i = ββ1 , By the principle of Mathematical induction
prove that An =[cosnΡ² π π πππΡ²π π πππΡ² cosnΡ²
].
19. Construct a matrix A = [πππ]2x2 whose elements aij are given by aij are given by aij
π2ππ₯π ππππ₯.
20. If A = [2 β π 21 3 β π
] is a singular matrix, then find the value of 5k β k2 ?
Answers of 2 mark questions
1. Let A be a skew-symmetric matrix of order n
Then, Aβ = -A
IAβI =I βAI
IAI = (-1)n IAI
IAI = -IAI => 2IAI = 0
IAI=0, Hence A is a singular matrix.
2. If A = [πππ π₯ βπ πππ₯π πππ₯ πππ π₯
] = I
If A = [πππ π₯ βπ πππ₯π πππ₯ πππ π₯
] = = [1 00 1
]
πΆππ π₯ = 1 , π πππ₯ = 0 X= 0π.
3. [0 β1β1 0
]
4. πΉππ π π πππ€ π π¦ππππ‘πππ πππ‘πππ₯, πππ = βπππ ,
π₯ = β(β2) = 2.
5. Required matrix = [1 41 2
]
6. x=3, y=7, z=-2, w=14.
7. Aβ =A, Bβ =B and if AB is symmetric, then (AB)β = AB --------------(i)
Also, (AB)β = BβAβ = BA -----------------------------------------------(ii)
From (i) and (ii), AB= BA
8. Using A = IA
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20 | P a g e
A-1 = [β4 55 β6
]
9 a11 = (1+1)2/2 = 2, a12 = (1+2)2/2 = 9/2, a21= (2+1)2/2 =9/2, a22 = (2+2)2/2 = 8
[2 9/29/2 8
]
10. A2 = [β3 00 β3
] [β3 00 β3
] = [9 00 9
]
A4 = [9 00 9
] [9 00 9
] = [81 00 81
]
11. ( I + A )3 β 7A = I3 + 3I2A + 3IA2+ A3 β 7A = I +3A + 3A + A β 7A = I
12. C = O β ( A + B ) = [0 0 00 0 0
] - [β1 1 23 0 0
] = [1 β1 β2β3 0 0
]
13. Symmetric part = [
4 5/2 05/2 5 5/20 5/2 1
], Skew-symmetric part = [
0 β1/2 β11/2 0 9/21 9/2 0
]
14. K = 1
15. Show that LHS = RHS.
16. [16 1421 37
]
17. X = [1 β2 β53 4 0
]
18. By using principle of mathematical induction get the result of An.
19. A= [π2π₯π πππ₯ π2π₯π ππ2π₯π4π₯π πππ₯ π4π₯π ππ2π₯
]
20 5k- k2 =4.
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21 | P a g e
TOPIC-: DETERMINANTS
ONE MARK QUESTIONS:
1. Evaluate the following determinants: (one mark each)
(i) |2 4β1 2
|
(ii) |π πππ βπππ ππππ π π πππ
|
(iii) |πππ 15Β° π ππ15Β°π ππ75Β° πππ 75Β°
|
(iv) |πππ 80Β° βπππ 10Β°π ππ80Β° π ππ10Β°
|
(v) |π₯ π₯ + 1
π₯ β 1 π₯|
(vi) |π + ππ π + ππβπ + ππ π β ππ
|
(vii) |2 34 β5
|2
(viii) |2 β1 β20 2 β13 β5 0
|
(ix) |2 β1 31 0 14 2 β1
|
2
(x) |0 π πππΌ βπππ πΌ
βπ πππΌ 0 π πππ½πππ πΌ βπ πππ½ 0
|
2. πΌπ |2 45 1
| = |2π₯ 46 π₯
|, Then find the value(s) of x.
3. πΌπ |π₯ β 2 β33π₯ 2π₯
| = 3, Then find the integral value(s) of x.
4. If |2π₯ + 5 35π₯ + 2 9
| = 0, find x.
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22 | P a g e
5. Find the value of x for which|3 π₯π₯ 1
| = |3 24 1
|.
6. If|π₯ π₯1 π₯
| = |3 41 2
|, write the positive value of x.
7. Write the minor and cofactor of the element a21 of [1 β24 3
]
8. Find the value of a11A11 + a12A12 of [2 β3 56 0 41 5 β7
]
9. Check whether the following matrix is singular.
[2 β1 β20 2 β13 β5 0
]
10. Find the value(s) of x for which the matrix A = [π₯ 1 21 0 35 β1 4
] is singular.
11. For what value of x, the matrix A = [1 β2 31 2 1π₯ 2 β3
] is singular?
12. Using determinants, find the area of the triangle whose vertices are (0, 0), (4, 3) and (8, 0).
13. Show that the points (1, 0), (6, 0) and (0, 0) are collinear.
14. Let A be square matrix of order 3 X 3. Write the value of|2π΄|, where |π΄| = 4.
15. Let A is square matrix of order 3, |π΄| β 0 πππ |3π΄| = π|π΄|, then write the value of k.
16. If A and B are square matrices of order 3 such that |π΄| = β1 and |π΅| = 3, then find the value
of|7π΄π΅|.
17. If a matrix A of order 3 X 3 has determinant 2, then find the value of|π΄ (8πΌ)|.
18. If A is a skew-symmetric matrix of order 3, write the value of πππ‘(π΄).
19. If A is a square matrix and |π΄| = 2, then write the value of|π΄π΄β²|, where Aβ is the transpose of matrix A.
20. If A is a non-singular matrix such that |π΄| = 5, then find|π΄β1|.
21. If A is an invertible matrix of order 3 X 3 and |π΄| = 5, then find|πππ π΄|.
22. If A is an invertible matrix of order 3 and |πππ π΄| = 64, then find|π΄|.
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23 | P a g e
23. If A is a square matrix such that π΄(πππ π΄) = 8πΌ, where I denotes the identity matrix of the same order, then find the value of|π΄|.
24. If A is a square matrix of order 3 such that |π΄| = 3, then find|π΄(ππππ΄)|.
25. If A is an invertible matrix of order 3 X 3 and |π΄| = 7, then find adj (adjA).
26. If A is an invertible matrix of order 3 X 3 and |π΄| = 4, then find |πππ(ππππ΄)|
27. Use matrix method to examine the given system of equations for consistency or inconsistency.
5x + 2y = 2, 3x + 2y = 5
28. Using properties of determinants, prove that |1 3 52 6 1031 11 38
| = 0
29. Using properties of determinants, prove that |8 2 712 3 516 4 3
| = 0
30. Using properties of determinants, prove that |2 3 713 17 515 20 12
| = 0
ANSWERS: (1 MARK QUESTIONS) 1.
(i) |2 4β1 2
| = (2)(2) β (β1)(4) = 8
(ii) |π πππ βπππ ππππ π π πππ
| = (π πππ)(π πππ) β (πππ π)(βπππ π) = sin2ΞΈ + cos2ΞΈ = 1.
(iii) |πππ 15Β° π ππ15Β°π ππ75Β° πππ 75Β°
| = (πππ 15Β°)(πππ 75Β°) β (π ππ15Β°)(π ππ75Β°) = πππ (15Β° + 75Β°) =
πππ 90Β° = 0.
(iv) |πππ 80Β° βπππ 10Β°π ππ80Β° π ππ10Β°
| = (π ππ10Β°)(πππ 80Β°) + (π ππ80Β°)(πππ 10Β°) = π ππ(10Β° + 80Β°) = sin90Β° = 1.
(v) |π₯ π₯ + 1
π₯ β 1 π₯| = (x)(x) β (x β 1)(x β 1) = x2 β (x2 β 1) = 1
(vi) |π + ππ π + ππβπ + ππ π β ππ
| = (π + ππ)(π β ππ) β (c + id)( βc + id) = (a2 β i2b2) β (βc2 + i2d2)
= (a2 + b2) β (βc2 β d2) = a2 + b2 + c2 + d2.
(vii) |2 34 β5
|2
= [(2)(β5) β (4)(3)]2 = (β22)2 = 484
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24 | P a g e
(viii) |2 β1 β20 2 β13 β5 0
| = 2(0 β 5) + 1 (0 + 3) β 2 (0 β 6) = 5
(ix) |2 β1 31 0 14 2 β1
|
2
= [2 (0 β 2) + 1 (β1 β 4) + 3 ( 2 β 0)]2 = (β3)2 = 9.
(x) |0 π πππΌ βπππ πΌ
βπ πππΌ 0 π πππ½πππ πΌ βπ πππ½ 0
| = 0 β π πππΌ (0 β πππ πΌ π πππ½) β πππ πΌ (π πππΌ π πππ½ β 0)
= π πππΌ πππ πΌ π πππ½ β πππ πΌ π πππΌ π πππ½ = 0.
2. Given, |2 45 1
| = |2π₯ 46 π₯
|
β β18 = 2x2 β 24 β 2x2 = 6
β x = Β± β3
3. Given, |π₯ β 2 β33π₯ 2π₯
| = 3
β 2x2 β 4x + 9x = 3 β2x2 + 5x β 3 = 0 β (2π₯ β 1) ( π₯ + 3) = 0
βπ₯ = 1
2, β3
4. Given, |2π₯ + 5 35π₯ + 2 9
| = 0
β (9)(2π₯ + 5) β 3(5π₯ + 2) = 0 β 18x + 45 β 15x β 6 = 0 β 3x + 39 = 0 β x = β13
5. As |3 π₯π₯ 1
| = |3 24 1
|
β 3 β x2 = 3 β 8 β π₯2 = 8
β x = Β±2β2
6. As |π₯ π₯1 π₯
| = |3 41 2
|
β π₯2 β π₯ = 6 β 4 βx2 β x β 2 = 0 β(π₯ β 2)(π₯ + 1) = 0 β x = 2, - 1 So, x = 2 (as we are asked only the positive value)
7. Minor of a21 = M21 = β2 Cofactor of a21 = A21 = (β1)2+1M21 = 2.
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25 | P a g e
8. a11A11 + a12A12 = (2)(β20) + (β3)(46) = β40 β 138 = β178
9. let A = [2 β1 β20 2 β13 β5 0
]
then |π΄| = |2 β1 β20 2 β13 β5 0
| = 2(0 β 5) + 1(0 + 3) β 2( 0 β 6) = 5 β 0 . Therefore given matrix is non-
singular.
10. Given that A = [π₯ 1 21 0 35 β1 4
] is singular.
β΄ |π΄| = 0
β |π₯ 1 21 0 35 β1 4
| = 0
β π₯(0 + 3) β 1(4 β 15) + 2(β1 β 0) = 0 β 3x + 9 = 0 β x = β3
11. Given that A = [1 β2 31 2 1π₯ 2 β3
] singular
β΄ |π΄| = 0
βΉ |1 β2 31 2 1π₯ 2 β3
| = 0
βΉ 1(β6 β 2) + 2(β3 β x) + 3(2 β2x) = 0 βΉ β8 β 6 β 2x + 6 β 6x = 0 βΉ β8x = 8 βΉ x = β1
12. Given vertices of triangle are (0,0), (4, 3) and (8, 0)
Then, β = 1
2 |0 0 14 3 18 0 1
| = 1
2 [0 β 0 + 1(0 β 24)] = β12
Hence, area of triangle = |β| = 12 sq units
13. Given points are (1, 0), (6, 0) and (0, 0)
Then, β = 1
2 |1 0 16 0 10 0 1
| = 1
2 [1( 0 β 0) β 0 + 1 (0 β 0)] = 0
Hence, given points are collinear.
14. Given that A is a matrix of order 3 X 3 and |π΄| = 4. Then, |2π΄| = 23|π΄| [β΅ |ππ΄| = πππ΄, where A is n X n matrix] = (8)(4) = 32.
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26 | P a g e
15. Given that A is a matrix of order 3 X 3 and |3π΄| = k |π΄|
We have, |3π΄| = k |π΄| β 33|π΄| = k |π΄| [β΅ |ππ΄| = πππ΄,π€βπππ π΄ ππ π π π πππ‘πππ₯] β 27 |π΄| = k |π΄| β k = 27
16. Given that A and B are square matrices of order 3 such that |π΄| = β1 and |π΅| = 3
Then, |7π΄π΅| = 73 |π΄π΅| [β΅ |ππ΄| = πππ΄, π€βπππ π΄ ππ π π π πππ‘πππ₯] = 73 |π΄||π΅| = (343) (β1) (3) = β1029
17. Given that A is a square matrix of 3 X 3 and |π΄| = 2
Then, |π΄(8πΌ)| = |8π΄πΌ| = |8π΄| = 83|π΄| = (512) (2) = 1024
18. Given that A is a skew symmetric matrix of order 3. β΄ AT = βA β |π΄π| = |βπ΄| β |π΄| = (β1)3 |π΄| [β΅|π΄π| = |π΄| ] β |π΄| = β|π΄| β 2|π΄| = 0 β |π΄| = 0
19. Given that A is a square matrix such that |π΄| = 2
Then, |π΄β²π΄| = |π΄β²||π΄| = |π΄||π΄| = (2) (2) = 4
20. |π΄β1| = 1
|π΄| = 1
5
21. Given, A is an invertible matrix of order 3 X 3 and |π΄| = 5
Now, |ππππ΄| = |π΄|πβ1, where n = 3 β = 52 β = 25
22. Given, A is an invertible matrix of order 3 X 3 and |ππππ΄| = 64
Now, |ππππ΄| = |π΄|πβ1, where n = 3 β 64 = |π΄|2 β |π΄| = Β± 8
23. Given A is a square matrix of order 3 X 3 such that A (adjA) = 8I
Then, |π΄|I = 8I [β΅ π΄(ππππ΄) = |π΄| πΌ] β |π΄| = 8
24. Given A is a square matrix of order 3 X 3 such that |π΄| = 3
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27 | P a g e
Now, |π΄ (πππ π΄)| =|π΄|π, where n = 3. = (3)3 = 27.
25. Given, A is an invertible matrix of order 3 X 3 and |π΄| = 7
Now, πππ (πππ π΄) = |π΄|πβ2A, where n = 3 = (7)(3 β 2) A = 7A.
26. Given, A is an invertible matrix of order 3 X 3 and |π΄| = 4
Now, |πππ (ππππ΄)| = |π΄|(πβ1)2, where n = 3
= (4)(3β1)2 = 44 = 256
27. Given system of equation is
5x + 2y = 2 3x + 2y = 5
Which can be written in the matrix form as AX = B where A =[5 23 2
], X = [π₯π¦] and B = [
25]
Now, |π΄| = |5 23 2
| = 4 β 0. So, the given system of equations is consistent.
28. L.H.S. = |1 3 52 6 1031 11 38
| = 2 |1 3 51 3 531 11 38
| (On taking 2 common from R2)
=2(0) [as R1 and R2 are identical] = 0 = R.H.S.
29. L.H.S. = |8 2 712 3 516 4 3
| = 4 |2 2 73 3 54 4 3
| (On taking 4 common from C1)
= 4 (0) [as C1 and C2 are identical] = 0 = R.H.S.
30. L.H.S. = |2 3 713 17 515 20 12
|
= |15 20 1213 17 515 20 12
| [On applying R1 βΆ R1 + R2]
= 0 [as R1 and R3 are identical].
TWO MARKS QUESTIONS:
1. Without expanding, show that |π π π
π + 2π₯ π + 2π¦ π + 2π§π₯ π¦ π§
| = 0.
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28 | P a g e
2. Prove that the determinant |π₯ π πππ πππ π
βπ πππ βπ₯ 1πππ π 1 π₯
| is independent ofπ.
3. If |β3 π¦π₯ β1
| = |1 42 3
|, then find the possible values of x and y, where x, y β β
4. Using determinants, show that the points (π, π + π), (π, π + π) πππ (π, π + π) are collinear.
5. Using determinants, find βaβ so that points (a, 2), (1, 5) and (2, 4) are collinear.
6. Using determinants, prove that πΌ + π½ = πΌπ½ so that points(πΌ, 0), (0, π½) and (1, 1) are collinear. 7. Using determinants, find the equation of the line joining the points (1, 2) and (3, 6).
8. If A and B are square matrices of same order such that |π΄| = 6 and AB = I , then find the value of |π΅|
9. πΌπ π΄ ππ π π ππ’πππ πππ‘πππ₯ π π’πβ π‘βππ‘ π΄ππ΄ = πΌ, write the value of|π΄|.
10. Ifπ΄ = [π₯ 22 π₯
] πππ |π΄4| = 625, find the value(s) of x.
11. Ifβ = |1 βπ πππ 1
βπ πππ 1 π πππβ1 βπ πππ 1
|, then prove that2 β€ β β€ 4, πππ πππ π.
12. πΉπππ π‘βπ πππππππ‘ ππ π‘βπ πππ‘πππ₯ [2 β14 3
].
13. Write π΄β1 πππ π΄ = [2 51 3
].
14. If A is a matrix of order 3 X 3 and π΄(ππππ΄) = [5 0 00 5 00 0 5
] , π‘βππ ππππ |ππππ΄|
15. If A is a square matrix of order 3 X 3 such that πππ(4π΄) = π ππππ΄, then find k.
16. If A is a non-singular matrix such that π΄β1 = [15 3β2 51
], then find (AT)β1
17. Use matrix method to examine the given system of equations for consistency or inconsistency:
x+ 3y = 5 ; 2x + 6y = 8
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29 | P a g e
18. Without expanding prove that |41 1 579 7 929 5 3
| = 0
19. Using properties of determinants, prove that |π₯ + π¦ π¦ + π§ π§ + π₯π§ π₯ π¦1 1 1
| = 0
20. Find the area of the triangle formed by (2, - 6), (5, 4) and (12, 4).
ANSWERS (TWO MARKS QUESTIONS):
1. L.H.S. = |π π π
π + 2π₯ π + 2π¦ π + 2π§π₯ π¦ π§
|
On applying R2 βΆ R2 β R1
= |π π π2π₯ 2π¦ 2π§π₯ π¦ π§
|
Taking 2 common from R2.
= 2 |π π ππ₯ π¦ π§π₯ π¦ π§
|
= 2(0) [as R2 and R3 are identical]
2. We have, |π₯ π πππ πππ π
βπ πππ βπ₯ 1πππ π 1 π₯
| = x (x2 β 1) β sinΞΈ (βx sinΞΈ β cosΞΈ) + cosΞΈ (βsinΞΈ + xcosΞΈ)
= βx3 β x + xsin2ΞΈ + sinΞΈ cosΞΈ β sinΞΈ cosΞΈ + x cos2ΞΈ = β x3 β x + x (π ππ2π + πππ 2π) = β x3 β x + x = β x3, which is independent of π.
3. Given, |β3 π¦π₯ β1
| = |1 42 3
|
βΉ 3 β xy = 3 β 8 βΉ xy = 8 Also, given that x and y are natural numbers. β΄ Possible values of x and y are x = 1, y = 8 or x = 2, y = 4 or x = 4, y = 2 or x = 8, y = 1.
4. Given points are (π, π + π), (π, π + π) πππ (π, π + π)
Then, β = 1
2 |π π + π 1π π + π 1π π + π 1
|
= 1
2 [π(π + π β π β π) β (π + π)(π β π) + 1(ππ + π2 β π2 β ππ)]
= 1
2[ac β ab β b2 + c2 + ab + b2 β c2 β ac]
= 0 Hence, given points are collinear.
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30 | P a g e
5. Given points (a, 2), (1, 5) and (2, 4) are collinear
Then, β = 1
2 |π 2 11 5 12 4 1
| = 0
βΉ 1
2 [π( 5 β 4) β 2( 1 β 2) + 1( 4 β 10)] = 0
βΉ β4 + a = 0 βΉ a = 4
6. Given points(πΌ, 0), (0, π½) and (1, 1) are collinear.
Then, β = 1
2 |πΌ 0 10 π½ 11 1 1
| = 0
βΉ 1
2 [πΌ(π½ β 1) β 0 + 1(0 β π½)] = 0
βΉ [πΌπ½ β πΌ β π½]= 0 βΉ πΌ + π½ = πΌπ½
7. Let (x, y) be any point on the line joining (1, 2) and (3, 6)
So the points (x, y), (1, 2) and (3, 6) are collinear.
Hence β = 1
2 |π₯ π¦ 11 2 13 6 1
| = 0
βΉ 1
2 [x (2 β 6) β y (1 β 3) + 1
(6 β 6)] = 0 βΉ [β4x + 2y] = 0 βΉ 2x β y = 0 is the equation of the line joining (1, 2) and (3, 6).
8. Given A and B are square matrices of same order such that |π΄| = 6 and AB = I.
Now, AB = I βΉ |π΄π΅| = |πΌ| βΉ |π΄||π΅| = 1 [β΅ |πΌ|= 1] β 6 |π΅| = 1
β |π΅| = 1
6
9. Given that A is a square matrix such that π΄ππ΄ = πΌ
Now, π΄ππ΄ = πΌ βΉ |π΄ππ΄| = |πΌ| βΉ |π΄π||π΄| = 1 [As |πΌ| = 1] βΉ |π΄||π΄| = 1 [As|π΄π| = |π΄|] βΉ |π΄|2 = 1 βΉ |π΄| = Β±1
10. Given that, π΄ = [π₯ 22 π₯
] πππ |π΄4| = 625
βΉ |π΄4| = 625 βΉ |π΄|4 = 625 βΉ |π΄| = Β±5
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31 | P a g e
βΉ [π₯ 22 π₯
] = Β±5
βΉ π₯2 β 4 = Β±5 βΉ π₯2 β 4 = 5 ππ π₯2 β 4 = β5 βΉ π₯2 = 9 ππ π₯2 = β1 βΉ π₯ = Β±3
11. Given, β = |1 π πππ 1
βπ πππ 1 π πππβ1 βπ πππ 1
| = 1(1 + π ππ2π) β π πππ (βπ πππ + π πππ) + 1(π ππ2π + 1)
= 1 + π ππ2π β 0 + π ππ2π + 1 = 2 + 2 π ππ2π
We know that β1 β€ π πππ β€ 1 β π
βΉ 0 β€ π ππ2π β€ 1 β π βΉ 1 β€ 1 + π ππ2π β€ 2 β π βΉ 2 β€ 2 + 2π ππ2π β€ 4 β π βΉ 2 β€ β β€ 4 β π
12. Let π΄ = [2 β14 3
]
A11 = 3, A12 = β4, A21 = 1, A22 = 2
Hence πππ π΄ = [3 1β4 2
]
13. As π΄ = [2 51 3
]
Then, |π΄| = |2 51 3
| = 6 β 5 = 1
Also, A11 = 3, A12 = β1, A21 = β5, A22 = 2
So, πππ π΄ = [3 β5β1 2
]
Hence π΄β1 = 1
|π΄| (πππ π΄) = [
3 β5β1 2
]
14. Given, π΄(ππππ΄) = [5 0 00 5 00 0 5
]
Then, π΄(ππππ΄) = 5πΌ βΉ |π΄| πΌ = 5πΌ [β΅ π΄(ππππ΄) = |π΄| πΌ] βΉ |π΄| = 5 πππ€, |ππππ΄| = |π΄|πβ1 βΉ |πππ π΄| = 52 = 25
15. Given that A is a square matrix of order 3 X 3 such that πππ(4π΄) = π ππππ΄
βΉ πππ(4π΄) = π πππ π΄ βΉ 4πβ1ππππ΄ = π ππππ΄ β΅ πππ(ππ΄) = ππβ1ππππ΄ βΉ 4πβ1 = π βΉ 43β1 = π βΉ π = 16
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32 | P a g e
16. Given, π΄β1 = [15 3β2 51
]
Then, (π΄π)β1 = πβ1π= [
15 3β2 51
]π
= [15 β23 51
]
17. Given system of equation is
x + 3y = 5 2x + 6y = 8
Which can be written in the matrix form as AX = B where A =[1 32 6
], X = [π₯π¦] and B = [
58]
Now, |π΄| = |1 32 6
| = 0.
So, we will calculate (ππππ΄)π΅. A11 = 6, A12 = β2, A21 = β3, A22 = 1
So, ππππ΄ = [6 β3β2 1
]
Now, (ππππ΄)π΅ = [6 β3β2 1
] [58] = [
6β2] β π
Therefore, given system of equations is inconsistent and has no solution.
18. L.H.S. = |41 1 579 7 929 5 3
|
C1 βΆ C1 β C2
= |40 1 572 7 924 5 3
|
Taking 8 common from C1
= 8|5 1 59 7 93 5 3
|
= 8(0) [as C1 and C3 are identical] = 0 = R.H.S.
19. L.H.S. = |π₯ + π¦ π¦ + π§ π§ + π₯π§ π₯ π¦1 1 1
|
On applying R1 βΆ R1 + R2
= |π₯ + π¦ + π§ π₯ + π¦ + π§ π₯ + π¦ + π§
π§ π₯ π¦1 1 1
|
Taking (x + y + z) common from R1
= (x + y + z) |1 1 1π§ π₯ π¦1 1 1
|
= (x + y + z) (0) [as R1 and R3 are identical] = 0
20. Let the given points are A (2, - 6), B (5, 4) and C (12, 4).
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33 | P a g e
So, ππ(βπ΄π΅πΆ) = 1
2|2 β6 15 4 112 4 1
|
= 1
2[2(4 β 4) + 6(5 β 12) + 1(20 β 48)]
= 1
2[0 β 42 β 28]
= 1
2[β70]
= [β35] Since area of a triangle canβt be negative
So ππ(βπ΄π΅πΆ) = 35 sq. units
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34 | P a g e
TOPIC: CONTINUITY SAMPLE QUESTION OF 1 MARK
1 Prove that the functionf(x) = 5x β 3 is continuous at x = 5.
2. Find the point of discontinuity of the function defined by f(x) =x2β25
x+5
3. Check the continuity of the function f(x) = xn is at x = n where n is a positive
integer.
4. For what value of k is the function f(x) = {tan2x
xif x β 0
k ifx = 0 continuous at x = 0.
5. A function f is defined by : f(x) = {
1βcosx
x2, when x β 0
A, when x = 0
Find the value of A.
6. A function f is defined by f(x) = {
sinβ1 x
x+ ex when x β 0
2 when x = 0
Show that f is continuous at x = 0
7. Find all points of discontinuity of f, where f is defined by:f(x) = {
x
|x|if x < 0
β1 if x β₯ 0
8. Find all points of discontinuity of f, where f is defined by: f(x) = {x3 β 3 if x β€ 2x2 + 1 if x > 2
9. Find all points of discontinuity of f, where f is defined by:f(x) = {|x|
xif x β 0
0 if x = 0
10. Find all points of discontinuity of f, where f is defined by:f(x) = {x10 β 1 if x β€ 1x2 if x β₯ 1
11. Is the defined by: f(x) = {2x if x < 22 if x = 2x2 ifx > 2
discontinuous at x = 2.
12. Is the function continuous at x = 1defined by f(x) = 2x + 3
13. Is the function continuous at x = 0 defined by f(x) = x2
14. Is the function continuous at x = 0 defined by f(x) = |x |
15. Is the function defined by f(x)=x3 + x2 β 1 for all R.
16. What is the interval over which f(x) = sin x is continuous.
17. At what point function is discontinuous;f(x) = {3x β 2 x β€ 0x + 1 x > 0
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35 | P a g e
18. Is the function continuous at x = 1: f(x) = {5x β 4 if x β€ 14x2 β 3x if x > 1
19. Is the function continuous at x = 3: f(x) = {2x2 β 5x β 3 x β 3
0 x = 3
20. Is the function continuous at x = 0: f(x) = {x x β€ 0x2 x β₯ 0
ANSWERS OF 1 MARK QUESTIONS
1 LHL=RHL= π(π₯)=22
2 Function is continuous at each real number other than(β5)
3 Continuous at π₯ = π
4 K=2
5 A=1/2
6 LHL=RHL= π(0)=2
7 No point of discontinuity
8 No point of discontinuity
9 Discontinuous at π₯ = 0
10 Discontinuous at π₯ = 1
11 Discontinuous at π₯ = 2
12 Continuous at π₯ = 1
13 Continuous at π₯ = 0
14 Continuous at π₯ = 0
15 Function is continuous
16. (ββ,β)
17. π΄π‘ π₯ = 0
18. Yes, continuous at π₯ = 1
19. Yes, continuous at π₯ = 3
20. Yes, continuous at π₯ = 0
SAMPLE QUESTION OF 2 MARKS:
1 Is the function defined by π(π₯) = π₯2 β π πππ₯ + 5 ππππ‘πππ’ππ’π ππ‘ π₯ = π ?
2 Discuss the continuity of the function π(π₯) = sin π₯ + cos π₯
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36 | P a g e
3 Discuss the continuity of the function π(π₯) = sin π₯ β cos π₯
4 Discuss the continuity of the function π(π₯) = sin π₯. cos π₯
5 If π(π₯) = {π₯ + π π₯ < 34 π₯ = 3
3π₯ β 5 π₯ > 3 is continuous at π₯ = 3, π‘βππ ππππ π‘βπ π£πππ’π ππ π.
6 If π(π₯) = {π₯2β9
π₯β3ππ π₯ β 3
2π₯ + π ππ‘βπππ€ππ π is continuous at x=3, then find the value of k.
7 Examine the continuity of the function: π(π₯) = {sinπ₯
π₯π₯ β 0
1 π₯ = 0 at π₯ = 0
8 Check the continuity of the functionπ(π₯) = |π₯ β 5| at π₯ = 5
9 Find the value of π, so that π(π₯) = {ππ₯ + 5 πππ₯ β€ 2π₯ β 1 ππ π₯ > 2
ππ ππππ‘πππ’ππ’π ππ‘ π₯ = 2.
10 Find the point of discontinuity of the function f defined by f(x)= {π₯ + 2 ππ π₯ β€ 1π₯ β 2 ππ π₯ > 1
11 For π(π₯) = {2|π₯|+π₯2
π₯π₯ β 0
0 π₯ = 0 examine the continuity of f(x) at x=0.
12
Discuss the continuity of the function at the point π defined by
π(π₯) =
{
π₯3
π20 < π₯ < π
π π₯ = π
2π βπ3
π₯2π₯ > π
13 If π(π₯) = {1βcosπ₯
π₯2π₯ β 0
π π₯ = 0 is continuous at π₯ = 0, then find the value of π
14 Is the function defined by π(π₯) = {
π₯ + 5 ππ π₯ β€ 1π₯ β 5 πππ₯ > 1
a continuous function? If not,
find the point(s) of discontinuity.
15 Examine the continuity of the function π(π₯) = π₯2 + 5 ππ‘ π₯ = β1
16 Examine the continuity of the function π(π₯) =1
π₯+3, π₯ β π .
17 Give an example of the function which is continuous at π₯ =
1, ππ’π‘ πππ‘ πππππππππ‘πππππ ππ‘ π₯ = 1
18 State the points of discontinuity for the function π(π₯) = [π₯] , ππ β 3 < π₯ < 3.
19 Is the function π(π₯) = 2π₯ β |π₯| ππ ππππ‘πππ’ππ’π ππ‘ π₯ = 0
20. For what value of k is the function π(π₯) = {π₯2 β 3 ππ π₯ β 3π ππ π₯ = 3
ππππ‘πππ’ππ’π ππ‘ π₯ = 3
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37 | P a g e
ANSWERS OF 2 MARKS QUESTIONS:
1 Yes, function is continuous at π₯ = π
2 Function is continuous for all R.
3 Function is continuous for all R.
4 Function is continuous for all R.
5 K=1
6 K=0
7 Function is discontinuous at π₯ = 0
8 Function is continuous at π₯ = 5. LHL=RHL=π(5) = 0,
9 LHL=RHL= π(2), 2π + 5 = 1 = 2π + 5 π‘βπππππππ π = (β2)
10 π₯ = 1 is the only point of discontinuity of f.
11 Function is discontinuous at π₯ = 0.
12 LHL=RHL= π(π)=π Hence π(π₯)ππ ππππ‘πππ’ππ’π ππ‘ π₯ = π.
13 π =
1
2
14 Function is not continuous at x=1.
15 Function is continuous.
16 For π₯ = β3 ππ’πππ‘πππ ππ πππ‘ πππππππ. Hence , not continuous for π₯ β π .
17 Absolute value function π(π₯) = |π₯ β 1| ππ ππππ‘πππ’ππ’π ππ‘ π₯ =
1 ππ’π‘ πππ‘ πππππππππ‘πππππ ππ‘ π₯ = 1.
18 π(π₯) = [π₯] ππ πππ‘ ππππ‘πππ’ππ’π πππ πππ‘πππππ . π»ππππ , πππ‘ ππππ‘πππ’ππ’π ππ‘ π₯ = Β±2,Β±1,0
19 LHL=RHL= π(0), function is continuous at π₯ = 0.
20 π = 6
**********************************************************************************
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38 | P a g e
TOPIC- DIFFERENTIABILITY 1 Marks Questions
S.No. Question
1. Differentiate: πππππ₯2 with respect to x.
2. Differentiate: 5sinπ₯ w.r.to x
3. For what value of x, f(x) = |2x β 7| is not derivable?
4. Write the derivative of logπ₯ 5, x > 0.
5.
Write the derivative of Sin (π₯7 +1) w.r. to x.
6. Differentiate Sin (log x), w. r. to x.
7. Differentiate cosβ1 βπ₯, w.r. to x.
8. Differentiate ππtanβ1 π₯, w.r. to x.
9. Differentiate πβπ₯+3, w.r. to x.
10. Differentiate πππ π₯, w. r. to ππ₯
11. Differentiate log7(ππππ₯), w.r. to x.
12. Write the Second derivative of log x, with respect of x.
13. For what value of x, f (x) = |x| is not differentiable?
14. For what value of x in (0, 2), f (x) = [π₯] is not differentiable?
15. Write an example of a function which is continuous at any point but not
differentiable at same point.
16. Write the statement of Rolleβs Theorem.
17. Write the statement of Mean Value Theorem.
18. Differentiate cosβ1(sin π₯), w.r. to x.
19. Find ππ¦
ππ₯ if x=π‘2 , and y= π‘3
20. What is the derivative of sin x w. r. t. Cos x?
21. If π¦ = cos(ππππ₯ + ππ₯) find ππ¦
ππ₯
22. If π(π₯) = |cos π₯|, ππππ πβ² (3π
4)
23. If π(π₯) = |cos π₯ β π πππ₯|, ππππ πβ² (π
6)
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39 | P a g e
24. If π₯2 + π¦2 = π2 then write the value of ππ¦
ππ₯
25. If π¦ = tanβπ₯ then find its derivative.
ANSWERS FOR 1 MARK SAMPLE PROBLEMS
1. 2x 2. 5sinπ₯ logπ 5 πππ π₯ 3. X = 7
2 4.
βπππ5
π₯(ππππ₯)2
5. 7π₯2πππ (π₯7 + 1) 6. cos(ππππ₯)
π₯ 7.
β1
2βπ₯ β1βπ₯ 8.
πππtanβ1 π₯
1+π₯2 9.
πβπ₯+3
2βπ₯+3
10 βπβπ₯Sinx 11. 1
π₯ πππ7ππππ₯ 12. β1
π₯2β
13. X=0 14. X=1 15. F(x) = |π₯| at x=0 18. -1 19. 3
2 t 20. β cot π₯ 21.
ππ¦
ππ₯= β(
1
π₯+ ππ₯) sin(ππππ₯ + ππ₯)
22.1
β2 23. β
1
2(1 + β3) 24.
ππ¦
ππ₯= β
π₯
π¦ 25.
ππ¦
ππ₯=
π ππ2βπ₯
2βπ₯
2 Marks Questions
1 If y= sinβ1[π₯β1 β π₯ β βπ₯ β1 β π₯2] and 0<x<1, then findππ¦
ππ₯.
2 Differentiate , 2πππ 2π₯ w.r.t. x
3 Differentiate , sinβ1(1
βπ₯+1)w.r.t. x
4 Find ππ¦
ππ₯ when x and y are connected by the relation given as tanβ1(π₯2 + π¦2) = a.
5 Find ππ¦
ππ₯ of a function expressed in parametric form
Sin x= 2π‘
1+ π‘2 , tan y=
2π‘
1β π‘2
6 If x= ππ₯π¦β , Prove that
ππ¦
ππ₯ =
π₯βπ¦
π₯ logπ₯
7 If y =tanβ1 π₯ , find π2π¦
ππ₯2 in terms of y alone.
8 If x= 4t and y = 4
π‘ then find
ππ¦
ππ₯ .
9 Differentiate 99π₯ w .r. t. x
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40 | P a g e
10 Differentiate cosβ1β
1βcosπ₯
2 w.r.t. x
11 Differentiate cotβ1 (π+π₯
πβπ₯) w.r.t. x
12 Differentiate tanβ1 (β1+π₯2 β1
π₯) w.r.t. x
13 Differentiate tanβ1 (βπ₯ +βπ
1ββππ₯) w.r.t. x
14 Differentiate tanβ1 (4βπ₯
1β4π₯) w.r.t. x
15 If y= tanβ15π₯
1β6π₯2 , β1
β6 <x<
1
β6 then prove that
ππ¦
ππ₯ =
2
1+4π₯2 +
3
1+9π₯2
16 Find second order derivative of ππ₯sin 5x
17 If y= A sin x + B Cos x , then find the value of π2π¦
ππ₯2 + y
18 Find ππ¦
ππ₯ for the function 2x + 3y = sin π¦
19 If π¦ = βtan π₯ + βtan π₯ + βtan π₯ +β¦β¦β¦β then find
ππ¦
ππ₯
20 Examine the applicability of Rolleβs Theorem for the function π(π₯) = π₯2 β 4 on [β2,2]
21 If π₯ = π cos π πππ π¦ = π sin π find ππ¦
ππ₯ ππ‘ π =
π
3
22 If ππ₯ + ππ¦ = ππ₯+π¦ Prove that ππ¦
ππ₯= βππ¦βπ₯
23 Find the value of C in the Rolleβs Theorem for the function f(x) = π₯3 β 3π₯ ππ π₯ β
[0, β3]
24 For the function π(π₯) = π₯ +1
π₯ ππ π₯ β [1,3].Find the value of C for Mean Value
Theorem.
25 For the function π(π₯) = π₯3 β 5π₯2 β 3π₯ ππ π₯ β [1,3].Find the value of C for Mean
Value Theorem.
Answers for 2 Marks Questions
1 Y= sinβ1 π₯- sinβ1 βπ₯
ππ¦
ππ₯=
1
β1βπ₯2 -
1
2βπ₯β1βπ₯
2 ππ¦
ππ₯= -2πππ
2π₯sin2x log2
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41 | P a g e
3 ππ¦
ππ₯= β
1
2βπ₯(1 + π₯)
4 ππ¦
ππ₯= β
π₯
π¦
5 Hint; Put t= tan π then findππ₯
ππ‘ πππ
ππ¦
ππ‘.
ππ¦
ππ₯= 1
6 Take log on both sides then it will be π¦ =π₯
ππππ₯ then diff w. r. t. x
7 π2π¦
ππ₯2= β2 sin π¦πππ 3π¦
8 ππ¦
ππ‘ =
β4
π‘2 ,
ππ₯
ππ‘ =4 then
ππ¦
ππ₯ =β1
π‘2
9 ππ¦
ππ₯= 99
π₯9π₯(log 9)2
10 ππ¦
ππ₯= β1
2
11 ππ¦
ππ₯=
βπ
π2+π₯2 Hint; cotβ1 (
π+π₯
πβπ₯) = tanβ1 (
πβπ₯
π+π₯) = tanβ1 1 β tanβ1
π₯
π
12 ππ¦
ππ₯=
1
2(1+π₯2) Hint; Put x= tan π then simplify
13 ππ¦
ππ₯=
1
2βπ₯(1+π₯) Hint; tanβ1 (
βπ₯ +βπ
1ββππ₯) = tanβ1 βπ₯ + tanβ1 βπ
14 ππ¦
ππ₯=
2
βπ₯(1+4π₯) Hint; tanβ1 (
4βπ₯
1β4π₯) =
2 tanβ1 2βπ₯ π‘βππ πππππππππ‘πππ‘π
15 Hint y= tanβ1 3π₯+tanβ1 2π₯ then prove it
16 π2π¦
ππ₯2=2ππ₯(5cos 5x- 12 sin 5x)
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42 | P a g e
17 0
18 ππ¦
ππ₯=
2
cosπ¦β3
19 ππ¦
ππ₯ =
π ππ2π¦
2π¦β1
20 Show all the three conditions
21 ππ¦
ππ₯=βπ
β3π
23 C = 1
24 C = β3
25 C = 7
3
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43 | P a g e
TOPIC-: APPLICATION OF DERIVATIVES
1 MARK QUESTION:
1 The side of a square is increasing at a rate of 0.2 cm/sec. Find the rate of increase
of perimeter of the square.
2 The radius of the circle is increasing at the rate of 0.7 cm/sec. What is the rate of
increase of its circumference?
3 If the radius of a soap bubble is increasing at the rate of 1
2 cm / sec. At what rate
its volume increasing when the radius is 1 cm.
4 A stone is dropped into a quiet lake and waves move in circles at a speed of 4
cm/sec. At the instant when the radius of the circular wave is 10 cm, how fast is
the enclosed area increasing?
5 The total revenue in rupees received from the sale of x units of a product is given by R(x) = 13x2 + 26x + 15. Find the marginal revenue when x = 7.
6 If a manufacturerβs total cost function is C(x) = 1000 + 40x + x2, where x is the output, find the marginal cost for producing 20 units.
7 If the rate of change of volume of a sphere is equal to the rate of change of its radius, find the radius of the sphere.
8 Find the rate of change of the area of a circle with respect to its radius βrβ when r = 4cm.
9 Find the value of βkβ for which the function f(x) = x2 β kx + 6, x > 0 is strictly increasing.
10 Write the interval for which the function f(x) = cos x, 0 x 2Ο is decreasing.
11 What is the interval on which the function f(x) = log x
x is increasing?
12 For which values of x, the functions y = x4 β 4
3x3 is increasing?
13 Write the interval for which the function f(x) = 1
x is strictly decreasing.
14 Find the sub-interval of the interval (0,Ο
2) in which the function f(x) = sin 3x is
increasing.
15 It is given that at x = 1, the function f(x) = x4 β 62 x2 + kx + 9 attains its maximum value is the interval [0, 2]. Find the value of k.
16 What is the slope of the tangent to the curve f = x3 β 5x + 3 at the point whose x co-ordinate is 2?
17 At what point on the curve y = x2 does the tangent make an angle of 45Β° with the x-axis?
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44 | P a g e
18 Find the equation of the normal to curve x2 = 4y which passes through the point (1, 2).
19 What is the slope of the normal to the curve y = 5x2 β 4 sin x at x = 0.
20 If the curves y = 2ex and y = aeβx intersect orthogonally (cut at right angles).
What is the value of a?
21 Find the slope of the normal to the curve y = 8x2 β 3 at x =1
4 .
22 For the curve y = (2x + 1)3 find the rate of change of slope as x = 1.
23 Find the slope of the tangent to the curve y = 3x4 β 4x at x=4.
24 If y = loge x , then find βy when x = 3and βx = 0.03.
25 Find the approximate change in the volume V of a cube of side x meters caused by increasing the side by 2%.
1 MARK QUESTION ANSWER:
1 0.8 cm/sec.
2 4.4 cm/sec.
3 2 cm3/sec.
4 80 cm2/sec.
5 Rs. 208.
6 Rs. 80
7 1
2βπ
8 8πcm2/cm
9 k β€ 0
10 (0, π]
11 (0, π]
12 π₯ β₯ 1
13 R
14 (0,π
6)
15 k= 120
16 7
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45 | P a g e
17 (1
2,1
4)
18 x + y -3 = 0
19 1
4
20 1
2
21 β1
4
22 72
23 764
24 0.01
25 0.06 x3 m3
2 MARKS QUESTION:
1 Find the rate of change of the total surface area of a cylinder of radius r and height h
with respect to radius when height is equal to the radius of the base of cylinder.
2 Find the rate of change of the area of a circle with respect to its radius. How fast is the
area changing w.r.t. its radius when its radius is 3 cm?
3 An edge of a variable cube is increasing at the rate of 5 cm per second. How fast is the volume increasing when the side is 15cm?
4 Find the least value of ΞΌ such that the function x2 + ΞΌx + 1 is increasing on [1, 2].
5 Find the maximum and minimum values of function f(x) = sin 2x + 5.
6 Find the maximum and minimum values if any of the function f(x) = β |x β 1| + 7,β x β
R.
7 Without using derivatives, find the maximum and minimum value of y = |3 sin x + 1|.
8 Find a for which f(x) = (x + sin x) + a is increasing.
9 If y = a log x + bx2 + x has its extreme values at x = -1 and x = 2, then find a and b.
10 Find the maximum and minimum values of f, if any , of the function given by f(x) = |x|, xβ R
11 Find the local minimum value of the function f given by f(x) = 3 + |x| , x β R
12 What is the maximum value of the function sin x + cos x?
13 Show that the function g(x) = log x do not have maximum or minima.
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46 | P a g e
14 Find the absolute maximum and the absolute minimum value of f(x) = 4x - 1
2 x2, x β
[β2,9
2]
15 Find the point on the curve y = 3x2 β 12x Γ 9 at which the tangent is parallel to x-axis.
16 Find the point on the curve y = 3x2 + 4 at which the tangent is perpendicular to the line
with slope β1
6 .
17 Find the point on the curve y = x2 where the slope of the tangent is equal to the y co-ordinate.
18 Find the slope of the normal to the curve x = 1 β a sinΞΈ; y = bcos2ΞΈ at x =Ο
2 .
19 At what points on the curve x2 + y2 β 2x β 4y + 1 = 0, the tangents are parallel to y-axis?
20 Find a point on the curvey = x3 β 3x, where tangents is parallel to the chord joining (1, -2) and (2, 7).
21 Find the value of m for which the line y = mx + 1 is a tangent to the curve y2 = 4x.
22 Find the points on the curve 9y2 = x3, where the normal to the curve makes equal intercepts with the axes.
23 Using differentiations, find the approximate value ofβ0.082.
24 Using differentiations, find the approximate value ofβ0.48.
25 Use differential to approximate(25)1
3.
26 Find the approximate value of f (3.02), where f(x) = 3 x2 + 5x + 3.
27 If the radius of a sphere is measured as 9cm with an error of 0.03 cm, then find the approximate error in calculating its volume.
2 MARKS QUESTION ANSWER:
1 8ππ
2 6π ππ2/ππ
3 3375 ππ3/π ππ
4 -2
5 Minimum value = 4, maximum value = 6.
6 Maximum value = 7, minimum value does not exist.
7 Maximum value = 4, minimum valve = 0.
8 π > 0
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47 | P a g e
9 a = 2, b = 1
2
10 Minimum value 0, no maximum value in R
11 3.
12 β2
14 Absolute minimum value = -10, absolute maximum value = 8
15 (2, β 3)
16 ( 1, 7)
17 ( 0, 0), (2, 4)
18 βπ
2π
19 (-1, 2) and (3, 2)
20 (2, 2) and (-2, -2)
21 1
22 (4, Β± 8
3)
23 0.2867
24 0.693
25 2.926
26 45.46
27 9.72π cm3
gsgsgsgsggsgsggsgsgsgsgsggsgsgsgsgsgsggsgsgsgsgsggsgsgsgsgsgsgsggsgsgsgsgsggsgsggsgsgsgsgsgsgsggsg
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48 | P a g e
TOPIC-: INDEFINITE AND DEFINITE SAMPLE QUESTION OF 1 MARK
1. Evaluate : β«(π ππβ1 βπ₯ +πππ β1βπ₯) ππ₯
2. Evaluate :β« πππ πππ₯(πππ πππ₯ + πππ‘π₯)ππ₯
3. Evaluate : β«1
1βπ ππ2π₯ ππ₯
4. Evaluate :β«1β π πππ₯
πππ 2π₯ dx
5. Evaluate : β«(8π₯ + π₯8)dx
6. Find the anti derivatives of (βπ₯+1
βπ₯)
7. Find the Integration : β«1
β1+4π₯2ππ₯
8. Evaluate :β« πππππ π₯ + ππ₯πππ π ππ₯
9. Find the Integration :β«
πtanβ1 π₯
1+π₯2ππ₯
10. Evaluate : β«πππ 2π₯+2π ππ2π₯
πππ 2π₯ππ₯
11. Evaluate : β«1
π ππ2π₯πππ 2π₯ππ₯
12. Evaluate βΆ β« πβπππ2π₯ ππ₯
13. Evaluate βΆ β«(1+log π₯)2
π₯ππ₯
14. Find β« ππππ₯ ππ₯
15. Evaluate βΆ β« ex (tanβ1 x +
1
1 + x2)dx
16. Evaluate βΆ β« x(x β 1) dx
4
1
17. Evaluate βΆ β«
1
4 + 9x2dx
23
0
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49 | P a g e
18. Evaluate βΆ β« sinx dx
Ο4
βΟ4
19. πΌπ β« (3 π₯2 + 2π₯ + π)ππ₯ = 0 , π‘βππ ππππ π‘βπ π£πππ’π ππ πΎ1
0.
20. Evaluate: β«1
π₯ππππ₯ππ₯
π2
π
21. Evaluate βΆ β« π ππ3π₯
π4
βπ4
ππ₯
22. Write the value of β«π₯+πππ 6π₯
3π₯2+π ππ6π₯ππ₯
23. Evaluate βΆ β«π ππ2βπ₯
βπ₯dx
24. Evaluate βΆ β«π2π₯βπβ2π₯
π2π₯+πβ2π₯dx
25. Evaluate βΆ β« π₯ π ππβ1π₯ ππ₯
SAMPLE QUESTION OF 2 MARKS
26. Evaluate: β«π₯3sin (tanβ1 π₯4)
1+ π₯8 ππ₯
27. Evaluate: β«1
β2π₯βπ₯2ππ₯
28. Find β«1
(π₯+1)(π₯+2) ππ₯
29. Find β« ππ₯π πππ₯ ππ₯
30. Evaluate : β« sec π₯ ππ₯
31. Evaluate : β«1
π₯ββπ₯ dx
32. Evaluate : β«1
exβ1 ππ₯
33. Evaluate : β« π ππβ1(2π₯
1+π₯2) ππ₯
34. Evaluate :β« π ππ7π₯ ππ₯
π
2
βπ
2
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50 | P a g e
35. Evaluate :β«β
1βπππ π₯
1+πππ π₯ ππ₯
36. Evaluate : β«1
π₯2+5π₯+6ππ₯
37. If β«1
1+4π₯2ππ₯
πΎ
0= π
8 then find the value of K.
38. Evaluate : β« πππ β1 (π πππ₯)ππ₯
39. Evaluate : β«(1 β π₯)βπ₯ ππ₯
40. Evaluate : β«π₯3βπ₯2+π₯β1
π₯β1 ππ₯
41. Evaluate :β«1
π2+(ππ₯)2ππ₯
β
0
42. Evaluate:β« β1 + sin π₯ π
20
ππ₯
43. β« [π₯2]ππ₯ β2
0Where [ ] is greatest integer function.
44. Find β« π[log(π₯+1)βlog π₯]
ππ₯
45. Evaluate : β«1
π₯(1+π₯7)ππ₯
46. Evaluate : β«π₯
βπ2βπ₯2ππ₯
π
βπ
47. Evaluate : β« π₯2π πππ₯π
20
ππ₯
48. Evaluate : β«1
β4βπ₯2ππ₯
49. Evaluate: β«log (π πππ₯)
π‘πππ₯ ππ₯
50. Evaluate : β«ππ₯
ππ₯ππ₯
ANSWER OF 1 MARK
1. π π₯
2+ π 2. βπππ‘π₯ β πππ πππ₯ + π 3. π‘πππ₯ + c 4. π‘πππ₯ β π πππ₯+ C
5. 8π₯
πππ8+
π₯9
9+ πΆ 6.
π
πππ
π + πππ
π + C 7. 1
2πππ|2π₯ + β1 + 4π₯2|+ C
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51 | P a g e
8. π₯π+1
π+1+
ππ₯
logπ+ πΆ 9. πtan
β1 π₯ + πΆ 10. π‘πππ₯ + c 11. π‘πππ₯ β πππ‘ π₯ + C
12. 1
2 ππππ₯ + C 13.
1
3(1 + log π₯)3 + πΆ 14. xlogx β x + C 15. ex tanβ1 x + C
16. 27
2 17.
π
24 18. 0 19. K= -2 20. πππ 2 21. 0 22.
1
6[log{3π₯2 + π ππ6π₯}] + πΆ
23. 2 tan βπ₯ +C 24. 1
2πππ|π2π₯ + πβ2π₯| + πΆ 25.
1
4(2π₯2 β 1)π ππβ1π₯ +
π₯β1βπ₯2
4 + C
ANSWER OF 2 MARKS
26. β1
4cos (tanβ1 π₯4) + πΆ 27. sinβ1(π₯ β 1) + πΆ 28. πππ |
π₯+1
π₯+2| + πΆ
29. ππ₯
2(π πππ₯ β πππ π₯) + πΆ 30. πππ |π ππ π₯ + π‘πππ₯|+ C 31. 2log|βπ₯ β 1| + π
32. πππ |ππ₯β1
ππ₯|+ πΆ 33. 2π₯ tanβ1 π₯ β log(1 + π₯2) + π 34. 0
35. 2 πππ(π πππ₯
2) + C 36. πππ |
π₯+2
π₯+3| + πΆ 37.
1
2 38.
ππ₯
2βπ₯2
2 + C
39. 2
3π₯3
2 β2
5π₯5
2 + C 40. π₯3
3+ π₯ + πΆ 41.
π
2ππ 42. 2
43. β2 β 1 44. π₯ + ππππ₯ + C 45. 1
7πππ |
π₯7
1+π₯7| + C 46. 0 47. π β 2
48. sinβ1 π₯/2 + π 49. [log(π πππ₯)]2
2 + C 50.
( π
π)π₯
ππππ
π
+ C
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52 | P a g e
TOPIC-: DIFFERENTIAL EQUATION SAMPLE QUESTIONS OF 1 MARK
1. Find the degree and order of the differential equation π₯π¦π2π¦
ππ₯2+ π₯(
ππ¦
ππ₯)2 β π¦
ππ¦
ππ₯= 0.
2. Find the degree and order of the differential equation(π2π¦
ππ₯2)2 + cos (
ππ¦
ππ₯) = 0.
3. Find the degree and order of the differential equation (π¦β²β²β²)2 + (π¦β²β²)3 + (π¦β²)4 + π¦5 = 0.
4. Verify that the function π¦ = acos π₯ + ππ πππ₯, where π, π β π is a solution of the differential equation: π2π¦
ππ₯2+ π¦ = 0.
5. Verify that the function π₯ + π¦ = π‘ππβ1π¦ is a solution of the differential equation:π¦2π¦β² + π¦2 + 1 = 0.
6. Verify that the function π¦ = β1 + π₯2 is a solution of the differential equation: π¦β² =π₯π¦
1+π₯2.
7. Form the differential equation representing the family of curvesπ¦ = ππ₯, where, m is arbitrary
constant.
8. Form the differential equation representing the family of curves π¦ = asin (π₯ + π), where, a,b is
arbitrary constant.
9. Form the differential equation of the family of parabolas having vertex at origin and axis along
positive y-axis.
10. Find general solution of the differential equation: ππ¦
ππ₯=
π₯+1
2βπ¦, (π¦ β 2).
11. Find general solution of the differential equation: ππ¦
ππ₯=
1+π¦2
1+π₯2.
12. Find general solution of the differential equation: ππ¦
ππ₯=
1βπππ π₯
1+πππ π₯.
13. Find general solution of the differential equation: ππ¦
ππ₯+ π¦ = 1 (π¦ β 1).
14. Find the integrating factor of the differential equation: ππ¦
ππ₯β 2π₯π¦ = π₯.
15. Find the integrating factor of the differential equation: π₯ππ¦
ππ₯+ 2π¦ = π₯2 log π₯.
16. Find the integrating factor of the differential equation: π₯ log π₯ππ¦
ππ₯+ π¦ =
2
π₯log π₯.
17. Find the integrating factor of the differential equation: (π₯ + π¦)ππ¦
ππ₯= 1.
18. Write the solution of the given differential: ππ¦
ππ₯+ ππ₯ = π.
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53 | P a g e
SAMPLE QUESTIONS OF 2 MARKS
1. Find the degree and order of the differential equation: (1 + 3ππ¦
ππ₯)2
3 = (4π3π¦
ππ₯3).
2. Find the degree and order of the differential equation: π¦β²β² + 2π¦β² + π πππ¦ = 0.
3. Find the degree and order of the differential equation: (π¦β²β²)3 + (π¦β²)2 + sin(π¦β²) + 1 = 0.
4. Verify that the function π¦ = π₯π πππ₯ is a solution of the differential equation
π₯π¦β² = π¦ + βπ₯2 β π¦2, (π₯ β 0 πππ π₯ > π¦ ππ π₯ < βπ¦).
5. Verify that the function π¦ β πππ π¦ = π₯ is a solution of the differential equation
(π¦π πππ¦ + πππ π¦ + π₯)π¦β² = π¦.
6. Form the differential equation representing the family of curves π¦ = asin(π₯ + π), where, π, π are
arbitrary constants.
7. Form the differential equation representing the family of curves π¦ = π2π₯(π + ππ₯), where, π, π are
arbitrary constants.
8. Form the differential equation of the family of circles touching the y-axis at origin.
9. Form the differential equation of the family of ellipses having foci on y-axis and centre at origin.
10. Find the equation of the curve passing through the point (1, 1) and whose differential equation is
π₯ππ¦ = (2π₯2 + 1)ππ₯ (π₯ β 0).
11. Solve the differential equationππ₯π‘πππ¦ππ₯ + (1 β ππ₯)π ππ2π¦ππ¦ = 0.
12. Solve the differential equation cos (ππ¦
ππ₯) = π, (π β π ); π¦ = 1 π€βππ π₯ = 0.
19. Find general solution of the differential equation(π₯ β π¦)(ππ₯ + ππ¦) = ππ₯ β ππ¦.
20. Write order and degree of the differential equation: β1 +
ππ¦
ππ₯= (
π2π¦
ππ₯2)1
3.
21. Solve ππ¦
ππ₯+ πππ π₯πππ π¦ = 0.
22. Find the differential equation of the family of curves given by π₯2 + π¦2 + 2ππ₯ = 0.
23. Verify that xy=c is a solution of the differential equationπ¦ππ₯βπ₯ππ¦
π¦= 0.
24. Find the degree and order of the differential equation: π4π¦
ππ₯4β log (
π3π¦
ππ₯3) = 0.
25. Find the integrating factor of the differential equation (1 + π¦2)ππ₯
ππ¦+ π¦π₯ = ππ¦, (β1 < π¦ < 1).
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54 | P a g e
13. Find the equation of the curve passing through the point (0, -2) given that at any point (x, y) on the
curve, the product of the slope of its tangent and y coordinate of the point is equal to the x
coordinate of the point.
14. Solve the differential equation π₯πππ (π¦
π₯)ππ¦
ππ₯= π¦πππ (
π¦
π₯) + π₯.
15. Solve the differential equation π₯ππ¦ β π¦ππ₯ = βπ₯2 + π¦2 ππ₯.
16. Solve the differential equation π₯ππ¦
ππ₯β π¦ + π₯π ππ (
π¦
π₯) = 0.
17. Solve the differential equation 2π₯π¦ + π¦2 β 2π₯2ππ¦
ππ₯= 0; π¦ = 2π€βππ π₯ = 1.
18. Find the general solution of the differential equation ππ¦
ππ₯β π¦ = πππ π₯.
19. Find the particular solution of the differential equation log(ππ¦
ππ₯) = 3π₯ + 4π¦, given that π¦ =
0 π€βππ π₯ = 0.
20. Find the general solution of the differential equation π¦ππ₯ + (π₯ β π¦2)ππ¦ = 0.
21. Find the equation of a curve passing through the origin given that the slope of the tangent to the
curve at any point (x, y) is equal to the sum of the coordinates of the point.
22. Find the particular solution of the differential equation(π₯ + 1)ππ¦
ππ₯= 2πβπ¦ β 1, given that y=0 when
x=0.
23. A population grows at the rate of 8% per year. How long does it takes for the population to
double?
24. Solve π₯ππ¦
ππ₯= π¦(log π¦ β log π₯ + 1)
25. Solve (π₯2 + 1)ππ¦
ππ₯+ 2π₯π¦ = βπ₯2 + 4.
ANSWERS OF 1 MARK QUESTIONS
1. Deg=1, order=2 2. Deg=not defined, order=2 3. Deg=2,
order=3 7.π₯ππ¦
ππ₯β π¦ = 08.
π2π¦
ππ₯2+ π¦ = 0 9. π₯π¦β² β 2π¦ = 0
10. π₯2 + π¦2 + 2π₯ β 4π¦ + π = 0 11. π‘ππβ1π¦ = π‘ππβ1π₯ + π 12.π¦ =
2π‘πππ₯
2β π₯ + π 13. π¦ = 1 + π΄πβπ₯ 14. πβπ₯
2
15. π₯2 16. log π₯ 17.πβπ¦
18. π¦. πΌπΉ = β«π. πΌπΉππ₯ + π , π€βπππ πΌπΉ = πβ«πππ₯
19. π₯ + π¦ = log(π₯ β π¦) + π 20. Ord=2, deg=2
21. log(π πππ¦ + π‘πππ¦) = βπ πππ₯ + π 22. π¦2 β π₯2 β 2π₯π¦π¦β² = 0
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24. Ord=4, deg=not defined 25. πΌπΉ = β1 + π¦2
Answers of 2 marks questions
1. Order=3, deg=3, 2. Order=2, deg=1
3.order=2, deg=not defined 6. π¦β²β² + π¦ = 0.
7. π¦β²β² β 4π¦β² + 4π¦ = 0 8. 2π₯π¦π¦β² + π₯2 = π¦2
9. π₯π¦π¦β²β² + π₯(π¦β²)2 β π¦π¦β² = 0. 10. π¦ = π₯2 + log |π₯|
11. π‘πππ¦ = πΆ(1 β ππ₯) 12. cos (π¦β2
π₯) = π.
13. π¦2 β π₯2 = 4. 14. sin (π¦
π₯) = log |ππ₯|
15.π¦ + βπ₯2 + π¦2 = ππ₯2 16. π₯ [1 β cos (π¦
π₯)] = ππ ππ(
π¦
π₯)
17. π¦ =2π₯
1βlog |π₯|, π₯ β 0, π₯ β π. 18. π¦ =
1
2(π πππ₯ β πππ π₯) + πππ₯
19.π3π₯
3+πβ4π¦
4=
7
12. 20.π₯ =
π¦2
3+
π
π¦.
21. π₯ + π¦ + 1 = ππ₯. 22. 1
2βππ¦= (π₯ + 1)ππ π¦ = log |
2π₯+1
π₯+1| , π₯ β β1.
23. π‘ =25
2log 2. 24. π¦ = π΄π₯ππ₯.
25.π¦(π₯2 + 1) =π₯
2βπ₯2 + 4 + 2 log |π₯ + βπ₯2 + 4 | + π
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TOPIC-: VECTOR ALGEBRA EXPECTED TOPICS IN
1 Mark
Algebra of Vectors ( Direct Questions on Addition, Subtraction, Dot Product, Cross Product)
Projection of Vector on another vector.
Section Formula.
Unit Vector
2 Marks
Area of Triangle
Area of Parallelogram
Unit vector perpendicular to two given vectors.
Find cross product when dot product is given and vice versa.
Direct problems based on Scalar Triple Product.
To prove Coplanarity of points.
To find value of an unknown constant when the given points are Coplanar.
Volume of parallelepiped.
Sample problems on 1Mark Questions
1. Find the values of π₯ and π¦ so that the vectors 2πΜ + 3πΜ β π¦οΏ½ΜοΏ½ and π₯πΜ + 3π Μ+5οΏ½ΜοΏ½ are equal.
2. For given vectors οΏ½βοΏ½ = (2πΜ β πΜ + 2οΏ½ΜοΏ½) and οΏ½ββοΏ½ = πΜ + πΜ β οΏ½ΜοΏ½ , ππππ π unit vector in the direction of
the vector οΏ½βοΏ½ + οΏ½ββοΏ½ .
3. Find π vector in the direction of vector 5πΜ β πΜ + 2οΏ½ΜοΏ½ which has magnitude 8 units.
4. Find the direction cosines of the vector joining the points A(1, 2, β3) and B(β1,β2, 1)
directed from A to B.
5. Find the position vector of a point R which divides the line joining two points with position
vectors (πΜ + 2πΜ β οΏ½ΜοΏ½) and (βπΜ + πΜ + οΏ½ΜοΏ½) in the ratio 2: 1
(π) Internally (ππ) externally.
6. Find the angle between two vectors οΏ½βοΏ½ and οΏ½ββοΏ½ with magnitude β3 and 2 respectively, having
οΏ½βοΏ½ . οΏ½ββοΏ½ = β6 .
7. Find the projection of the vector πΜ β πΜ on the vector πΜ + πΜ.
8. Find the magnitude of vectors οΏ½βοΏ½ and οΏ½ββοΏ½ having same magnitude such that angle between them is 60Β° and
their scalar product is 1
2 .
9. Find, |οΏ½βοΏ½ Γ οΏ½ββοΏ½|, if οΏ½βοΏ½ = πΜ β 7πΜ + 7οΏ½ΜοΏ½ and οΏ½ββοΏ½ = 3πΜ β 2πΜ + 2οΏ½ΜοΏ½ .
10. Find Ξ» and ΞΌ if (2πΜ + 6πΜ + 27οΏ½ΜοΏ½) Γ (πΜ + ππΜ + ΞΌοΏ½ΜοΏ½) = 0ββ
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11. If οΏ½βοΏ½ = πΜ + πΜ , οΏ½ββοΏ½ = πΜ + οΏ½ΜοΏ½ , π = οΏ½ΜοΏ½ + π,Μ find a unit vector in the direction of οΏ½βοΏ½ + οΏ½ββοΏ½ + π .
12. Write the value of πΜ β (πΜ Γ οΏ½ΜοΏ½) + πΜ β (οΏ½ΜοΏ½ Γ π)Μ + οΏ½ΜοΏ½ β (πΜ Γ π)Μ.
13. Is it possible that |οΏ½βοΏ½ + οΏ½ββοΏ½| = |οΏ½βοΏ½| + |οΏ½ββοΏ½| ? If yes, when?
14. Find the magnitude of (3οΏ½ΜοΏ½ + 4πΜ) Γ (πΜ + πΜ β οΏ½ΜοΏ½) .
15. Write the value of π for which, οΏ½βοΏ½ = 3πΜ + 2πΜ + 9οΏ½ΜοΏ½ and οΏ½ββοΏ½ = πΜ + ππΜ + 3οΏ½ΜοΏ½ are parallel vectors.
16. Write the value of π for which, οΏ½βοΏ½ = 3πΜ β 5πΜ + 2οΏ½ΜοΏ½ and οΏ½ββοΏ½ = πΜ + ππΜ + οΏ½ΜοΏ½ are orthogonal vectors.
17. Find the value of π if (1πΜ β 6πΜ + 7οΏ½ΜοΏ½) Γ (πΜ + 3πΜ + ποΏ½ΜοΏ½) = 0ββ.
18. Show that the vector πΜ + πΜ + οΏ½ΜοΏ½ is equally inclined to the axes OX, OY, and OZ.
19. Find |οΏ½βοΏ½|and |οΏ½ββοΏ½| if (οΏ½βοΏ½ + οΏ½ββοΏ½) . (οΏ½βοΏ½ β οΏ½ββοΏ½) = 8 andv |οΏ½βοΏ½| = 8|οΏ½ββοΏ½|.
20. Let οΏ½βοΏ½ and οΏ½ββοΏ½ be two vectors such that |οΏ½βοΏ½| = 3 and |οΏ½ββοΏ½| =β2
3 and οΏ½βοΏ½ Γ οΏ½ββοΏ½ is a unit vector. Then what
is the angle between οΏ½βοΏ½ and οΏ½ββοΏ½ ?
Sample problems on 2Marks Questions
1. If οΏ½βοΏ½ is perpendicular to both οΏ½ββοΏ½ and π, such that |οΏ½βοΏ½| = 2, |οΏ½ββοΏ½| = 3, |π| = 4 and the angle between οΏ½ββοΏ½ and π is 2π
3, then prove that οΏ½βοΏ½ β (οΏ½ββοΏ½ Γ π ) = 12β3 .
2. Let ΞΌββ = πΜ + πΜ, οΏ½βοΏ½ = πΜ β πΜ and οΏ½βββοΏ½ = πΜ + 2πΜ + 3οΏ½ΜοΏ½. If οΏ½ββοΏ½ is a unit vector such that ΞΌββ β οΏ½ββοΏ½ = 0 and οΏ½βοΏ½ β οΏ½ββοΏ½ = 0,
then prove that |οΏ½βββοΏ½ β οΏ½ββοΏ½| = 3.
3. If 4πΜ + πΜ + 3οΏ½ΜοΏ½, πΜ β 2πΜ + 4οΏ½ΜοΏ½ and 3πΜ + 4πΜ are the position vectors of the vertices, A, B and C of the β ABC
respectively, find the angle between the median AD and the side BC of the βABC.
4. Three vectors οΏ½βοΏ½, οΏ½ββοΏ½ and π satisfy the condition οΏ½βοΏ½ + οΏ½ββοΏ½ + π = 0ββ . Evaluate οΏ½βοΏ½ β οΏ½ββοΏ½ + οΏ½ββοΏ½ β π + π β οΏ½βοΏ½ , if
|οΏ½βοΏ½| = 1, |οΏ½ββοΏ½| = 4 and |π| = 2.
5. Show that the point A, B, C with position vectors 2πΜ β πΜ + οΏ½ΜοΏ½ , πΜ β 3πΜ β 5οΏ½ΜοΏ½ and 3πΜ β 4πΜ β 4οΏ½ΜοΏ½
respectively are the vertices of a right angled triangle.
6. Find the unit vector in the direction of the sum of the vectors 2πΜ β πΜ + 2οΏ½ΜοΏ½ and βπΜ + πΜ β οΏ½ΜοΏ½ .
7. If οΏ½βοΏ½ = 2πΜ + 2πΜ β 3οΏ½ΜοΏ½ , οΏ½ββοΏ½ = β1πΜ + 2πΜ + οΏ½ΜοΏ½ and π = 3πΜ + πΜ such that οΏ½βοΏ½ + ποΏ½ββοΏ½ is perpendicular to π then
find the value of Κ»πΚΌ.
8. Show that |οΏ½βοΏ½|οΏ½ββοΏ½ + |οΏ½ββοΏ½|οΏ½βοΏ½ is perpendicular to |οΏ½βοΏ½|οΏ½ββοΏ½ β |οΏ½ββοΏ½|οΏ½βοΏ½ for any two non-zero vectors οΏ½βοΏ½ and οΏ½ββοΏ½ .
9. If οΏ½βοΏ½ . οΏ½βοΏ½ = 0 and οΏ½βοΏ½ . οΏ½ββοΏ½ = 0 , then what can be concluded about the vector οΏ½ββοΏ½ ?
10. If οΏ½βοΏ½, οΏ½ββοΏ½, π are unit vectors such that οΏ½βοΏ½ + οΏ½ββοΏ½ + π = 0ββ , then find the value of οΏ½βοΏ½ . οΏ½ββοΏ½ + οΏ½ββοΏ½ . π + π . οΏ½βοΏ½ .
11. If the vertices A, B, C of a triangle ABC have coordinates (1, 2, 3), (-1, 0, 0) and (0, 1, 2)
respectively, then find β ABC.
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12. Show that the points A (1, 2, 7), B (2, 6, 3) and C (3, 10, -1) are collinear.
13. Find a unit vector perpendicular to each of the vectors οΏ½βοΏ½ + οΏ½ββοΏ½ and οΏ½βοΏ½ β οΏ½ββοΏ½ where
οΏ½βοΏ½ = 3πΜ + 2πΜ + 2οΏ½ΜοΏ½ and οΏ½ββοΏ½ = πΜ + 2πΜ β 2οΏ½ΜοΏ½ .
14. Find the area of the triangle with vertices A(1, 1, 1), B (1, 2, 3) and C(2,3,1).
15. Find the area of the parallelogram whose adjacent sides are determined by the vectors
οΏ½βοΏ½ = πΜ β πΜ + 3οΏ½ΜοΏ½ and οΏ½ββοΏ½ = 2πΜ β 7πΜ + οΏ½ΜοΏ½.
16. Prove that (οΏ½βοΏ½ + οΏ½ββοΏ½) . (οΏ½βοΏ½ + οΏ½ββοΏ½) = |οΏ½βοΏ½|2 + |οΏ½ββοΏ½|2
if and only if οΏ½βοΏ½ πππ οΏ½ββοΏ½ are perpendicular to each other
given οΏ½βοΏ½ β 0ββ , οΏ½ββοΏ½ β 0ββ.
17. If βΞΈβ is the angle between any two vectors οΏ½βοΏ½ and οΏ½ββοΏ½ such that |οΏ½βοΏ½ . οΏ½ββοΏ½| = |οΏ½βοΏ½ Γ οΏ½ββοΏ½| . Evaluate " ΞΈ ".
18. Find the area of a parallelogram with diagonals π1βββββ = πΜ + 2πΜ + 3οΏ½ΜοΏ½ and π2βββββ = 3πΜ β 2πΜ + οΏ½ΜοΏ½.
19. Find |οΏ½βοΏ½ β οΏ½ββοΏ½| , if two vectors οΏ½βοΏ½ πππ οΏ½ββοΏ½ are such that |οΏ½βοΏ½| = 2, |οΏ½ββοΏ½| = 3 and οΏ½βοΏ½ β οΏ½ββοΏ½ = 4.
20. If |οΏ½βοΏ½ β οΏ½ββοΏ½| = 40, |οΏ½βοΏ½ + οΏ½ββοΏ½|= 60 and |οΏ½βοΏ½| = 22, then find |οΏ½ββοΏ½|.
21. Let οΏ½βοΏ½, οΏ½ββοΏ½ and π be three vectors such that |οΏ½βοΏ½| = 3, |οΏ½ββοΏ½| = 4 and |π| = 5 and each one of them
being perpendicular to the sum of other two, find |οΏ½βοΏ½ + οΏ½ββοΏ½ + π |.
22. Find the value of βxβ if the points A(3, 2,1), B(4, x, 5), C(4,2,β2) and D (6, 5,-1) are coplanar. ,
23. Find βπΎβ if the vectors πΜ β πΜ + οΏ½ΜοΏ½ , 3πΜ + πΜ + 2οΏ½ΜοΏ½ πππ πΜ + πΎπΜ β 3οΏ½ΜοΏ½ are coplanar.
24. The volume of a parallelepiped whose edges are β12πΜ + πΎοΏ½ΜοΏ½, 3πΜ β οΏ½ΜοΏ½ πππ 2πΜ + πΜ β 15οΏ½ΜοΏ½ is 546
cubic units. Find the value βπΎβ
25. Find a vector whose magnitude is 3 units and which is perpendicular to each of 3πΜ + πΜ β 4οΏ½ΜοΏ½
and 6πΜ + 5πΜ β 2οΏ½ΜοΏ½.
ANSWERS
1Mark Questions
1. x =2 & y = -5 2. 1
2 (3πΜ + οΏ½ΜοΏ½) 3.
8
β30(5πΜ β πΜ + 2οΏ½ΜοΏ½) 4. <β
1
3, β
2
3,2
3>
5. (i) β1
3π Μ +
2
3οΏ½ΜοΏ½ +
1
3οΏ½ΜοΏ½ (ii) β3πΜ + οΏ½ΜοΏ½ 6. π=
π
4 7. 0
8. 1 9. 19β2 10. π= 3 & ΞΌ = 27
2 11.
(οΏ½ΜοΏ½+οΏ½ΜοΏ½+οΏ½ΜοΏ½)
β3
12.1 13. Both are Zero (Collinear) 14. βππ 15. p =2
3
16. 1 17. p=7
2 19. 2β
2
63 & 16β
2
63 20. π=
π
4
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2 Marks Questions
3. π
2 4. β
21
2 6.
1
β2 (πΜ + οΏ½ΜοΏ½) 7. π= 8
9. οΏ½ββοΏ½ Can be any vector 10. β3
2 11. cosβ1(β
10
β102) 13.
2
3π Μ β
2
3οΏ½ΜοΏ½ β
1
3οΏ½ΜοΏ½
14. 1
2β21 Sq units 15. 15β2 Sq units 17. π=
π
4 18. 4β3 Sq units
19. β5 20. 46 21. β50 22. x = 5
23. 15 24. πΎ = -3 &179 25. 2πΜ β 2πΜ + οΏ½ΜοΏ½
Hints to some selected problems ( 2 Marks Question):
2. οΏ½ββοΏ½ is perpendicular to both ΞΌββ πππ οΏ½βοΏ½. => οΏ½ββοΏ½ = ΞΌββ Γ οΏ½βοΏ½. Then use concept of unit vector.
8. Take (|οΏ½βοΏ½|οΏ½ββοΏ½ + |οΏ½ββοΏ½|οΏ½βοΏ½ ). (|οΏ½βοΏ½|οΏ½ββοΏ½ β |οΏ½ββοΏ½|οΏ½βοΏ½ )
16. οΏ½βοΏ½. οΏ½ββοΏ½ = 0
17. sinπ = cos π
18. Area of Parallelogram = π
π | π1βββββ Γ π2βββββ|
21. Use |οΏ½βοΏ½ + οΏ½ββοΏ½ + π |2
= ( οΏ½βοΏ½ + οΏ½ββοΏ½ + π). ( οΏ½βοΏ½ + οΏ½ββοΏ½ + π)
22. Volume of parallelepiped = [οΏ½βοΏ½ , οΏ½ββοΏ½, π ]
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TOPIC-: 3 β DIMENSIONAL GEOMETRY CLASS 12
Q. No Problems of 1 Mark
1. Find the equation of a straight line passing through the point (3, 1, 2) and
parallel to the vector 2πΜ + 3πΜ β 4οΏ½ΜοΏ½ in vector form.
2. Find the equation of the line passing through the point (2, β1, 6), and
parallel to the line π₯+1
2=
π¦β2
3=
π§β4
β2.
3. Find the equation of the line passing through the points (-3, 1, 2) and (1, -2, -2).
4. If a line makes angles 900, 1350, 450 with the x, y and z-axes respectively, find its
direction cosines.
5. Find the direction cosines of the line whose direction ratios are 2,β1,β2.
6. Find the direction cosines of x, y and z axes.
7. The Cartesian equation of a line is π₯+3
2=
π¦β5
4=
π§+6
2. Find the vector equation
for the line.
8. The vector equation of a line is π = β2πΜ + 3πΜ β 3οΏ½ΜοΏ½ + π(2πΜ + πΜ + 2οΏ½ΜοΏ½). Find the
Cartesian equation of the line.
9. Find the Cartesian equation of the line which passes through the point (-2, 4, -5)
and parallel to the line given by π₯+3
3=
π¦β4
5=
π§+8
6.
10. Find the equation of the line passing through the origin and (5, -2, 3).
11. Find the angle between the lines π₯+3
3=
π¦β1
5=
π§+3
4 πππ
π₯+1
1=
π¦β4
1=
π§β5
2.
12. Find the value of p so that the lines π₯+1
β6=
3βπ¦
3=
π§β4
β2 πππ
π₯β2
π=
π¦β1
1=
π§+1
3 are
at right angles.
13. Find the direction cosines of the normal to the plane 2x + 3y β z = 5
14. Find the direction cosines of the normal to the plane z = 5
15. Find the distance of the plane x + y + 3z = 5 from the origin.
16. Find the distance of the plane π₯ + 2π¦ β 2π§ = 9, ππππ π‘βπ πππππ‘ (2, 3, β5).
17. Find the vector equation of the plane which is at a distance of 7 units from the
origin and normal to the vector 3πΜ + 5πΜ β 6οΏ½ΜοΏ½.
18. Find the Cartesian equation of the plane π. (2πΜ + 3πΜ β 4οΏ½ΜοΏ½) = 1
19. Find the intercepts cut off by the plane 2x + y β z = 5 on the coordinate axis.
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20. Find the equation of the line passing through the point (1, 2, 3) and
perpendicular to the plane π₯ + 2π¦ β 5π§ + 9 = 0.
21. Find the direction ratios of the line which is perpendicular to the lines π₯β2
3=
π¦β1
β1=
π§+3
7 and
π₯β15
3=
π¦β29
8=
π§β5
β5
22. Find the direction ratios of the line which is parallel to the planes 3π₯ β π¦ +
7π§ = 1 πππ 3π₯ + 8π¦ β 5π§ + 5 = 0.
23. Find the direction ratios of the line which is parallel to the plane 3π₯ β π¦ + 7π§ =
1 and perpendicular to the line π₯β15
3=
π¦β29
8=
π§β5
β5.
24. Find the intersection point of the line π₯
3=
π¦
2=
π§
5 with the plane π₯ + π¦ + π§ = 20
25. Write the equation of the plane that makes intercepts 2, 3, and 5 on the
coordinate axes respectively.
3 β DIMENSIONAL GEOMETRY
Q. No Problems of 2 Marks
1. Find the equation of a straight line passing through the point (2, 1, 3) and
parallel to the vector 2πΜ + 3πΜ β 4οΏ½ΜοΏ½ in vector and Cartesian form.
2. Find the equation of a straight line passing through the point (-1, 2, 5) and
parallel to the line through the points (3, 1, 2) and (-2, -1, 4).
3. Find the equation of the line passing through the point whose position vector is
β2πΜ + πΜ β 2οΏ½ΜοΏ½ and parallel to the line through the points whose position vectors
are 3πΜ + πΜ + 2οΏ½ΜοΏ½ andβ2πΜ β πΜ + 4οΏ½ΜοΏ½ in vector and Cartesian form.
4. Find the equation of the line passing through the point (-1, 2, 6) and parallel to
the line π₯+1
2=
3βπ¦
3=
π§β4
β2.
5. Find the equation of the line passing through the points whose position vectors
are β3πΜ + 2πΜ + 3οΏ½ΜοΏ½ πππ 3πΜ + πΜ β οΏ½ΜοΏ½
6. Find the direction cosines of a line which makes equal angles with the
coordinate axes.
7. Show that the points (2, 3, 4), (-1, 1, 2) and (5, 5, 6) are collinear.
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8. Find the direction cosines of the line joining the points (2, -1, 4) and (-1, 3, 3).
9. Show that the line through the points (4, 7, 8) and (2, 3, 4) is parallel to the line
through the points (-1, -2, 1) and (1, 2, 5).
10. Find the angle between the lines π = 3πΜ + πΜ β 2οΏ½ΜοΏ½ + π(πΜ β πΜ β 2οΏ½ΜοΏ½) πππ π =
2πΜ β πΜ β 56οΏ½ΜοΏ½ + π(3πΜ β 5πΜ β 4οΏ½ΜοΏ½)
11. Find the angle between the lines π₯β2
2=
π¦β1
5=
π§+3
β3 πππ
π₯+2
β1=
π¦β4
8=
π§β5
4
12. Find the intersection point of the lines π₯
2=
π¦
2=
π§
1 πππ
π₯β5
4=
π¦β9
1=
π§β3
8
13. Find the Cartesian equations of the plane that passes through the point
(1, 4, 6) and normal vector to the plane is πΜ β 2πΜ + οΏ½ΜοΏ½.
14. Find the angle between the planes π. (2πΜ + 2πΜ β 3οΏ½ΜοΏ½) = 5 πππ π. (3πΜ β 3πΜ +
5οΏ½ΜοΏ½) = 3
15. Find the angle between the lines 2π₯ β π¦ + 3π§ β 1 = 0 πππ 3π₯ β π¦ + π§ + 3 =
0.
16. Find the distance between the planes 2π₯ + 3π¦ + 4π§ = 4 πππ 4π₯ + 6π¦ + 8π§ =
12
17. Find the angle between the lines π₯β2
2=
π¦β1
5=
π§+3
β3 πππ π‘βπ πππππ 2π₯ + 3π¦ +
π§ = 1.
18. Find the angle between the lines whose direction ratios are a, b, c and
π β π, π β π, π β π.
19. Find the coordinates of the point where the line π₯β2
2=
π¦β1
5=
π§+3
β3 meets the YZ
plane.
20. Find the equation of the line passing through the point (1, 2, 3) and parallel to
the plane is π. (πΜ + πΜ + οΏ½ΜοΏ½) = 2
21. Find the equation of the plane passing through the point (-1, 3, 2) and
perpendicular to each of the planes π₯ + 2π¦ + 3π§ = 5 πππ 3π₯ + 3π¦ + π§ = 0.
22. Find the equation of the plane passing through the point (-1, 3, 2) and parallel
to the line π₯β2
1=
π¦β1
2=
π§+3
3 and perpendicular to the plane 3π₯ + 3π¦ + π§ = 0
23. Find the equation of the plane passing through the point (-1, 3, 2) and parallel
to the lines π₯β2
1=
π¦β1
2=
π§+3
3 and
π₯β3
3=
π¦+1
3=
π§β5
1
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63 | P a g e
24. Find the equation of the plane passing through the line π₯+1
1=
π¦β3
2=
π§β2
3 and
parallel to the line π₯β3
3=
π¦+1
3=
π§β5
1.
25. Find the perpendicular distance from the line π₯β2
1=
π¦β1
2=
π§+3
2 to the plane
2π₯ + 3π¦ β 4π§ = 3
3-DIMENSIONAL GEOMETRY
CLASS 12
Answers for 1 mark Problems.
Q. No Answer
1. π = 3πΜ + πΜ + 2οΏ½ΜοΏ½ + π(2πΜ + 3πΜ β 4οΏ½ΜοΏ½)
2. π₯β2
2=
π¦+1
3=
π§β6
β2.
3. π₯+3
4=
π¦β1
β3=
π§β2
β4.
4. 0, β1
β2,1
β2.
5. 2
3, β
1
3, β
2
3.
6. d.c.βs of X axis: 1, 0, 0; d.c.βs of Y axis: 0, 1, 0; d.c.βs of z axis: 0, 0, 1;
7. π = β3πΜ + 5πΜ β 6οΏ½ΜοΏ½ + π(2πΜ + 4οΏ½ΜοΏ½ + 2οΏ½ΜοΏ½).
8. π₯+2
2=
π¦β3
1=
π§β3
2.
9. π₯+2
3=
π¦β4
5=
π§β5
6.
10. π₯β5
5=
π¦+2
β2=
π§β3
3.
11. cosβ1 (8β3
15).
12. β1
2.
13. 2
β14,3
β14, β
1
β14.
14. 0, 0, 1
15. 5
β11.
16. 3
17. π. (3πΜ + 5πΜ β 6οΏ½ΜοΏ½) = 7.
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18. 2π₯ + 3π¦ β 4π§ = 1.
19. x-intercept = 5/2, y-intercept = 5, z-intercept = -5
20. π₯β1
1=
π¦β2
2=
π§β3
β5.
21. -51, 36, 27
22. -51, 36, 27
23. -51, 36, 27
24. (6, 4, 10)
25. π₯
2+π¦
3+
π§
5= 1.
3-DIMENSIONAL GEOMETRY
CLASS 12
Answers for 2 marks Problems.
Q. No Solution
1. π = 2πΜ + πΜ + 3οΏ½ΜοΏ½ + π(2οΏ½ΜοΏ½ + 3πΜ β 4οΏ½ΜοΏ½).
π₯β2
2=
π¦β1
3=
π§β3
β4.
2. π = βπΜ + 2οΏ½ΜοΏ½ + 5οΏ½ΜοΏ½ + π(5πΜ + 2πΜ β 2οΏ½ΜοΏ½).
π₯+1
5=
π¦β2
2=
π§β5
β2.
3. π = β2πΜ + 1οΏ½ΜοΏ½ β 2οΏ½ΜοΏ½ + π(5πΜ + 2πΜ β 2οΏ½ΜοΏ½).
π₯+2
5=
π¦β1
2=
π§+2
β2.
4. π₯βπ₯1
2=
π¦βπ¦1
β3=
π§βπ§1
β2.
π₯+1
2=
2βπ¦
3=
π§β6
β2.
5. π = β3πΜ + 2πΜ + 3οΏ½ΜοΏ½ + π( 6πΜ β πΜ β 4οΏ½ΜοΏ½).
π₯+3
6=
π¦β2
β1=
π§β3
β4.
6. If l, m, n are the d.c.βs of the required line then l = m = n and
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65 | P a g e
π2 +π2 + π2 = 1.
π2 + π2 + π2 = 1.
π = Β±1
β3
Therefore required d.c.βs are Β±1
β3, Β±
1
β3, Β±
1
β3
7. Let A(2, 3, 4), B(-1, 1, 2), C(5, 5, 6) be the given points
d.rs of AB line: 3, 2, 2
d.rs of BC line: 6, 4, 4
d.rs of AB line are proportional to d.rs of BC line
Therefore AB line is parallel to BC line and B is a common point
Hence given points are collinear.
8. (2, -1, 4) and (-1, 3, 3).
d.rs of given line: 3, -4, 1
d.c.βs of given line: 3
β26,β4
β26,1
β26
9. Let A (4, 7, 8), B (2, 3, 4), C (-1, -2, 1), D (1, 2, 5).
d.rs of AB line: 2, 4, 4
d.rs of BC line: 2, 4, 4
d.rs of AB line are proportional to d.rs of BC line
Therefore AB line is parallel to BC line.
10.
οΏ½βοΏ½1 = 3πΜ + πΜ β 2οΏ½ΜοΏ½, οΏ½ββοΏ½1 = πΜ β πΜ β 2οΏ½ΜοΏ½.
οΏ½βοΏ½2 = 2πΜ β πΜ β 56οΏ½ΜοΏ½, οΏ½ββοΏ½2 = 3πΜ β 5πΜ β 4οΏ½ΜοΏ½.
Let be the required angle.
cos π =οΏ½ββοΏ½1.οΏ½ββοΏ½2
|οΏ½ββοΏ½1|.|οΏ½ββοΏ½2|.
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66 | P a g e
cos π =(πΜ β πΜ β 2οΏ½ΜοΏ½). (3πΜ β 5πΜ β 4οΏ½ΜοΏ½)
|πΜ β πΜ β 2οΏ½ΜοΏ½||3πΜ β 5πΜ β 4οΏ½ΜοΏ½| =
3 + 5 + 8
β5β50=
16
5β10
11. π1 = 2, π1 = 5, π1 = β3, π2 = β1, π2 = 8, π2 = 4,
Let be the required angle.
cos π = |π1π2 + π1π2 + π1π2
βπ12 + π1
2 + π12βπ2
2 + π22 + π2
2| = |
(2)(β1) + (5)(8) + (β3)(4)
β4 + 25 + 9β1 + 64 + 16|
=26
β38β101
π = cosβ1 (26
β38β101)
12. Find the intersection point of the lines π₯
2=
π¦
2=
π§
1 πππ
π₯β5
4=
π¦β9
1=
π§β3
8
πΏππ‘ πΏ1 β‘π₯
2=π¦
2=π§
1= π‘(π ππ¦)
πππ πΏ2 β‘π₯ β 5
4=π¦ β 9
1=π§ β 3
8
π΄ππ¦ πππππ‘ ππ πΏ1 = π(2π‘, 2π‘, π‘)
πΌπ π ππ πππ‘πππ πππ‘πππ πππππ‘ π‘βππ π ππππ ππ πΏ2
πβπππππππ π‘ β 5
4=π‘ β 9
1=π‘ β 3
8βπ‘ β 5
4=π‘ β 9
1πππ
π‘ β 9
1=π‘ β 3
8
β π‘ =31
3πππ π‘ =
69
7
Therefore there is no intersection point between the given lines.
13. Let A (1, 4, 6) and normal vector to the plane is πΜ β 2πΜ + οΏ½ΜοΏ½.
Equation of the plane is
π΄(π₯ β π₯1) + π΅(π¦ β π¦1) + πΆ(π§ β π§1) = 0
βΉ 1(π₯ β 1) β 2(π¦ β 4) + 1(π§ β 6) = 0 .
β π₯ β 2π¦ + π§ + 3 = 0.
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67 | P a g e
14. π. (2πΜ + 2πΜ β 3οΏ½ΜοΏ½) = 5 πππ π. (3πΜ β 3πΜ + 5οΏ½ΜοΏ½) = 3
οΏ½ββοΏ½1 = 2πΜ + 2πΜ β 3οΏ½ΜοΏ½, οΏ½ββοΏ½2 = 3πΜ β 3πΜ + 5οΏ½ΜοΏ½.
Let be the required angle.
cos π =οΏ½ββοΏ½1.οΏ½ββοΏ½2
|οΏ½ββοΏ½1|.|οΏ½ββοΏ½2|.
cos π = |(2πΜ + 2πΜ β 3οΏ½ΜοΏ½). (3πΜ β 3πΜ + 5οΏ½ΜοΏ½)
|2πΜ + 2πΜ β 3οΏ½ΜοΏ½||3πΜ β 3πΜ + 5οΏ½ΜοΏ½|| = |
6 β 6 β 15
β17β43| =
15
β17β43
π = cosβ1 (15
β17β43)
15. 2π₯ β π¦ + 3π§ β 1 = 0 πππ 3π₯ β π¦ + π§ + 3 = 0.
π1 = 2, π1 = β1, π1 = 3, π2 = 3, π2 = β1, π2 = 1,
Let be the required angle.
cos π = |π1π2 + π1π2 + π1π2
βπ12 + π1
2 + π12βπ2
2 + π22 + π2
2| = |
(2)(3) + (β1)(β1) + (3)(1)
β4 + 1 + 9β9 + 1 + 1|
=10
β14β11
π = cosβ1 (10
β14β11)
16. 2π₯ + 3π¦ + 4π§ = 4 πππ 4π₯ + 6π¦ + 8π§ = 12
4π₯ + 6π¦ + 8π§ = 8 πππ 4π₯ + 6π¦ + 8π§ = 12
π = 4, π = 6, π = 8, π1 = 8 π2 = 12
πππ π‘ππππ =|π2 β π1|
βπ2 + π2 + π2=
|12 β 8|
β16 + 36 + 64=
4
β116=
2
β29π’πππ‘π
17. πΏππ‘ πΏ β‘
π₯ β 2
2=π¦ β 1
5=π§ + 3
β3 πππ Ξ β‘ 2π₯ + 3π¦ + π§ = 1
π1 = 2, π1 = 5, π1 = β3, π2 = 2, π2 = 3, π2 = 1,
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68 | P a g e
Let be the required angle.
sin π = |π1π2 + π1π2 + π1π2
βπ12 + π1
2 + π12βπ2
2 + π22 + π2
2| = |
(2)(2) + (5)(3) + (β3)(1)
β4 + 25 + 9β4 + 9 + 1|
=16
β38β14
π = sinβ1 (16
β38β14)
18. π1 = π, π1 = π, π1 = π, π2 = π β π, π2 = π β π, π2 = π β π,
Let be the required angle.
cos π = |π1π2 + π1π2 + π1π2
βπ12 + π1
2 + π12βπ2
2 + π22 + π2
2|
= |(π)(π β π) + (π)(π β π) + (π)(π β π)
βπ2 + π2 + π2β(π β π)2 + (π β π)2 + (π β π)2|
=ππ β ππ + ππ β ππ + ππ β ππ
βπ2 + π2 + π2β(π β π)2 + (π β π)2 + (π β π)2= 0
π =π
2
19. πΏππ‘ πΏ β‘
π₯ β 2
2=π¦ β 1
5=π§ + 3
β3= π‘
Any point on the line L is π = (2π‘ + 2, 5π‘ + 1, β3π‘ β 3)
Let the intersection point is (0, π¦, π§)
Therefore 2t+2 = 0 t = -1
Required point = π = (2π‘ + 2, 5π‘ + 1, β3π‘ β 3) =
(0, 5(β1) + 1, β3(β1) β 3) = (0, β4 0)
20. Equation of the plane is
π΄(π₯ β π₯1) + π΅(π¦ β π¦1) + πΆ(π§ β π§1) = 0 β 1(π₯ β 1) + 1(π¦ β 1) +
1(π§ β 1) = 0 β π₯ + π¦ + π§ = 3
21. Find the equation of the plane passing through the point (-1, 3, 2) and
perpendicular to each of the planes π₯ + 2π¦ + 3π§ = 5 πππ 3π₯ + 3π¦ + π§ = 0.
Equation of plane is
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69 | P a g e
|
π₯ β π₯1 π¦ β π¦1 π§ β π§1π1 π1 π1π2 π2 π2
| = 0 β |π₯ + 1 π¦ β 3 π§ β 11 2 33 3 1
| = 0
β (π₯ + 1)(2 β 9) β (π¦ β 3)(1 β 9) + (π§ β 1)(3 β 6) = 0
β β7π₯ + 8π¦ β 3π§ β 25 = 0 β 7π₯ β 8π¦ + 3π§ + 25 = 0
22. Find the equation of the plane passing through the point (-1, 3, 2) and parallel
to the line π₯β2
1=
π¦β1
2=
π§+3
3 and perpendicular to the plane 3π₯ + 3π¦ + π§ = 0
Equation of plane is
|
π₯ β π₯1 π¦ β π¦1 π§ β π§1π1 π1 π1π2 π2 π2
| = 0 β |π₯ + 1 π¦ β 3 π§ β 11 2 33 3 1
| = 0
β (π₯ + 1)(2 β 9) β (π¦ β 3)(1 β 9) + (π§ β 1)(3 β 6) = 0
β β7π₯ + 8π¦ β 3π§ β 25 = 0 β 7π₯ β 8π¦ + 3π§ + 25 = 0
23. Find the equation of the plane passing through the point (-1, 3, 2) and parallel
to the lines π₯β2
1=
π¦β1
2=
π§+3
3 and
π₯β3
3=
π¦+1
3=
π§β5
1
Equation of plane is
|
π₯ β π₯1 π¦ β π¦1 π§ β π§1π1 π1 π1π2 π2 π2
| = 0 β |π₯ + 1 π¦ β 3 π§ β 11 2 33 3 1
| = 0
β (π₯ + 1)(2 β 9) β (π¦ β 3)(1 β 9) + (π§ β 1)(3 β 6) = 0
β β7π₯ + 8π¦ β 3π§ β 25 = 0 β 7π₯ β 8π¦ + 3π§ + 25 = 0
24. Find the equation of the plane passing through the line π₯+1
1=
π¦β3
2=
π§β2
3 and
parallel to the line π₯β3
3=
π¦+1
3=
π§β5
1.
Equation of plane is
|
π₯ β π₯1 π¦ β π¦1 π§ β π§1π1 π1 π1π2 π2 π2
| = 0 β |π₯ + 1 π¦ β 3 π§ β 11 2 33 3 1
| = 0
β (π₯ + 1)(2 β 9) β (π¦ β 3)(1 β 9) + (π§ β 1)(3 β 6) = 0
β β7π₯ + 8π¦ β 3π§ β 25 = 0 β 7π₯ β 8π¦ + 3π§ + 25 = 0
25. Find the perpendicular distance from the line π₯β2
1=
π¦β1
2=
π§+3
2 to the plane
2π₯ + 3π¦ β 4π§ = 3
π1 = 1, π1 = 2, π1 = 2, π2 = 2, π2 = 3, π2 = β4,
π1π2 + π1π2 + π1π2 = (1)(2) + (2)(3) + (2)(β4) = 0
Therefore the given line is parallel to the plane.
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70 | P a g e
A point on the line (2, 1, β3)
Required distance = |ππ₯1+ππ¦1+ππ§1+π
βπ2+π2+π2| = |
2(2)+3(1)β4(β3)β3
β4+9+16| =
8
β29π’πππ‘π
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71 | P a g e
TOPIC-: LINEAR PROGRAMMING Very short answer type questions carrying 1 mark
1. What is linear programming?
2. What is an objective function in linear programming?
3. What is feasible solution?
4. What is an optimum solution in an LPP?
5. What are the methods to solve the linear programming problems?
6. What are bounded and unbounded region in the linear programming problems?
7. What are linear inequalities? Give examples.
8. The region represented by the inequation system π₯, π¦ β₯ 0 , π₯ + π¦ β€ 3 is bounded or
unbounded?
9. What are the corner points in the linear programming problems?
10. The point (π₯, 3)satisfies the ineqality; β5π₯ β 2π¦ β€ 13 . Find the smallest possible value of
?
11. Find , if possible , the minimum value of the objective function 3π₯ β 4π¦ , subject to
constraints 2π₯ + π¦ β€ 12 , π₯ β π¦ β€ 2 , π₯ β₯ 0 , π¦ β₯ 0.
12. Draw the graph of inequality 2π₯ + 3π¦ β€ 6 , π₯, π¦ β₯ 0.
13. Find the feasible region of π¦ β€ 2π₯ , π₯, π¦ β₯ 0.
14. What is feasible region in linear programming?
15. Find the point at which the maximum value of π₯ + π¦, subject to constraints π₯ + 2π¦ β€ 70 ,
2π₯ + π¦ β€ 95 , π₯, π¦ β₯ 0 is obtained ?
16. Find the point at which the minimum value of π₯ + π¦, subject to constraints π₯ + π¦ β€ 2 ,
2π₯ + 5π¦ β€ 7 , π₯, π¦ β₯ 0 is obtained ?
17. What are the constraints in linear programming problems? Give examples.
18. Find the corner points of the region bounded by the inequalities 5π₯ + π¦ β€ 100 , π₯ + π¦ β€ 60
π₯, π¦ β₯ 0 .
19. Maximize π§ = 3π₯ + 4π¦ , subject to constraints π₯ + π¦ β€ 4 , π₯, π¦ β₯ 0.
20. Solve graphically the inequalities π₯ + 2π¦ β€ 4 , π₯, π¦ β₯ 0.
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72 | P a g e
Short Answer type questions carrying 2 marks
1. A man has `1500 to purchase rice and wheat. A bag of rice and a bag of wheat cost `180
and `120 resp. He has storage capacity of 200 bags only. He earns a profit of `11 and
`9 per bag of rice and wheat respectively. Formulate an LPP to maximize the profit.
2. Two tailors A and B are paid `225 and `300 per day respectively. A can stitch 9 shirts and 6
pants while B can stitch 15 shirts and 6 pants per day. Form an LPP to minimize the labour
cost to produce 90 shirts and 48 paints.
3. How many corner points are possible for the region bounded by the inequalities,
5π₯ + π¦ β€ 100 , π₯ + π¦ β€ 60 , π₯, π¦ β₯ 0.
4. Show graphically the feasible region bounded by 2π₯ + π¦ β₯ 3 , π₯ + 2π¦ β€ 6 , π₯, π¦ β₯ 0.
5. Check whether the region bounded by the inequality π₯ β₯ 3 , π₯ + π¦ β₯ 5 π¦ β₯ 0 is bounded or
unbounded.
6. Minimize π = β3π₯ + 4π¦, subject to constraints π₯ + 2π¦ β€ 8 ,3π₯ + 2π¦ β€ 12 , π₯, π¦ β₯ 0.
7. Maximize π = 4π₯ + π¦ , subject to constraints π₯ + π¦ β€ 50 , 3π₯ + π¦ β€ 90 , π₯ β₯ 0 , π¦ β₯ 0.
8. Maximize π = 3π₯ + 2π¦ , subject to constraints π₯ + π¦ β₯ 8 , 3π₯ + 5π¦ β€ 15 , π₯ β₯ 0 , π¦ β₯ 0
9. Solve the linear inequalities graphically π₯ + 2π¦ β€ 10 , 3π₯ + π¦ β€ 30 , π₯ β₯ 0 , π¦ β₯ 0.
10. Maximize and minimize π = 2π₯ β π¦, subject to constraints π₯ + 2π¦ β€ 2000 , π₯ + π¦ β€ 1500,
π₯ β₯ 0 , π¦ β₯ 0.
11. Corner points of feasible region determined by 2π₯ + π¦ β€ 10 , π₯ + 3π¦ β€ 15 , π₯, π¦ β₯ 0
are (0,0), (5,0), (3,4)πππ (0,5). Let π = ππ₯ + ππ¦ , where p,q> 0. Find the condition on p
and q so that the maximum value of Z occurs at (3,4) and (0,5).
12. Maximize π = π₯ + π¦ , subject to constraints π₯ β π¦ β€ β1 , βπ₯ + π¦ β€ 0 , π₯, π¦ β₯ 0.
13. Check whether maximum value of π = 2π₯ β π¦, subject to constraints 3π₯ β π¦ β₯ 0 , π₯ β₯
π¦, π₯, π¦ β₯ 0 exists or not.
14. Maximize π = π₯ + 2π¦ , subject to constraints π₯ β π¦ β₯ 0 , 2π¦ β€ π₯ + 2 , π₯, π¦ β₯ 0.
15. A housewife wishes to mix two types food F1 and F2 in such a way that the mixture
contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C.
The vitamin content of one kg of food F1 and F2 are as follows :
Food Vitamin A Vitamin B Vitamin C
Food F1 1 2 3
Food F2 2 2 1
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73 | P a g e
One kg of food F1 cost `16 and One kg of Food F2 costs `20. Formulate the above problem
as a linear programming problem.
16. To maintain oneβs health, a person must fulfill certain minimum daily requirements for the
following three nutrients β calcium, protein and calories. His diet consist of only food items
I and II whose prices and nutrient contents are as shown below :
Food I cost
` 6 per unit
Food II cost
` 10 per unit
Minimum daily
requirement
Calcium 10 4 20
Protein 5 5 20
calories 2 6 12
Formulate the above problem as an LPP.
17. Maximize π = 2π₯ + 3π¦ , subject to constraints π₯ + π¦ β₯ 2 , π₯ + 2π¦ β₯ 3 , π₯ β₯ 0 , π¦ β₯ 0.
18. Show that π = 2π₯ + 5π¦ has neither maximum nor minimum value subject to constraints
π₯ + π¦ β€ 2 , 3π₯ + 3π¦ β₯ 18 , π₯ β₯ 0 , π¦ β₯ 0
19. If a young man drives his vehicle at 25 km/hr, he has to spend ` 2 per km on petrol. If he
drives it at a faster speed of 40 km/hr, the petrol cost increases to `5 per km. He has `100
To spend on petrol and travel within one hour. Express this is an LPP.
20. An aeroplane can carry a maximum of 200 passengers. A profit of `1000 is made on each
executive class ticket and a profit of ` 600 is made on each economy class ticket. The airline
reserves at least 20 seats for the executive class. However at least 4 times as many
passengers prefer to travel by economy class. Determine how many tickets of each type
must be sold in order to maximize the profit. Make an LPP.
Answer key of 1 mark questions β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦
1. Linear programming is a mathematical technique to maximize or minimize a linear function of
several variables under the certain conditions and that conditions are linear equalities or
inequalities.
2. A linear function whose value is to be maximized or minimized in a linear programming problem
is known as an objective function.
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74 | P a g e
3. A general solution which satisfies all the constraints of a linear programming problem is known
as feasible solution of that linear programming problem.
4. A feasible solution of an LPP at which the minimum or maximum value of an objective function
exists is called an optimum solution.
5. There are two methods to solve a linear programming problem which are as follow :
1. Corner point method. 2. Iso-profit Or Iso -cost method.
6. If the feasible region is closed, then it is bounded and if the feasible region is not closed , then it
is unbounded.
7. A linear inequality is an inequality which involves a linear function and contain one of the
symbol : < less than, > greater than, β€ less than equal to or β₯ greater than equal to.
8. Bounded region
9.
The coordinates from which the feasible region is formed are known as corner points.
10. Smallest possible value of π₯ isβ18
5.
11. Minimum value of Z is βππ at the point (0, 12).
12. It is in the first quadrant bounded by the points (0, 0), (0, 2) and (3, 0).
13. The feasible region of given inequality is unbounded region in first quadrant passing through
(0, 0) and towards right to the line.
14. A feasible region is an area defined by a set of coordinates that satisfy a system of inequalities.
The region satisfies all restrictions imposed by linear programming problems. The concept is an
optimization technique
15. (40,15).
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75 | P a g e
16. Minimum value of Z is 0 at (0,0).
17. The restrictions, in the form of linear inequalities, imposed on the optimization of an objective
function in a linear programming problem are known as the constraints.
18. The corner points are (0, 0), (20,0),(0,60) and (10,50).
19. Maximum value of Z is 16 at (0, 4).
20. It is the region bounded by the points (0,2), (0,0) and (4,0).
Answer key of two marks questions
1. Let he buys π₯ and π¦ be the no. of bags of rice and wheat resp. Then objective function ,
which is to maximize is π = 11π₯ + 9π¦ ,
subject to constraints 18π₯ + 15π¦ β€ 150 ,
π₯ + π¦ β€ 200,
π₯, π¦ β₯ 0.
2. Let cost of stitching of 1 shirt is `π₯ and that of 1 pant is `π¦. Then the objective function ,
which is to minimize is π = 90π₯ + 48π¦ ,
subject to constraints , 9π₯ + 6π¦ β€ 225 ,
15π₯ + 6π¦ β€ 300 ,
π₯, π¦ β₯ 0 .
3. Four corner points.
4. The feasible region is unbounded formed by the points (0,3)and (6,0) away from
the origin .
5. The region formed by the given inequalities is unbounded.
6. The minimum value of π = β3π₯ + 4π¦ is β12 at the point (4,0).
7. The corner points of the feasible region are (0, 50), (0, 0), (30, 0) and (20, 30). The
maximum value of Z is 120 at the point (30, 0).
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76 | P a g e
8. There is no feasible region corresponding to the given inequalities. Hence there is
no maximum value of Z subject to the given constraints.
9. The solution of the given inequalities is the region bounded by the points (0, 0),
(10, 0) and (0, 5).
10. The minimum value of Z is β1000 at the point (0, 1000) and maximum value is
3000 at the point (1500,0).
11. q = 3p.
12. There is no common region formed by given inequalities and hence no maximum
value exists subjected to these constraints.
13. Exist.
14. There is no region which is common to given inequalities and hence no maximum or
minimum value exists.
15. Let π₯ πππ π¦ be the number of units of food F1 and food F2 resp. to mix together to
form the mixture. Then the following LPP is formed :
The objective function is π = 16π₯ + 20π¦ ,
subject to constraints , π₯ + 2π¦ β₯ 10 ,
2π₯ + 2π¦ β₯ 12,
3π₯ + π¦ β₯ 8,
and π₯ β₯ 0 , π¦ β₯ 0.
16. Let the personβs diet consists of π₯ unit of food I and π¦ units of food II. Then the
following LPP is formed :
The objective function is π = 16π₯ + 20π¦ ,
subject to constraints , 10π₯ + 4π¦ β₯ 20 ,
5π₯ + 5π¦ β₯ 20,
2π₯ + 6π¦ β₯ 12 ,
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77 | P a g e
and π₯, π¦ β₯ 0 .
17. The feasible region which is common to the given constraints are unbounded and
the corner points are (0, 2) , (1, 1) and (3, 0). Since the region is unbounded
therefore Z = 6 may or may not be the maximum value. To check whether Z = 6 is
maximum value, draw the graph of 2π₯ + 5π¦ > 6 and which has so many common
points with the given feasible region. Hence maximum value of Z is 6 at the line
joining of (0, 2) and (3, 0).
18. Since the given inequalities have no common feasible region, therefore Z has neither
maximum value nor minimum value under these constraints.
19. Let the young man covers π₯ km at the speed of 25 km/hr and π¦ km at the speed of
40 km/hr.
The total distance travelled by the man is π₯ + π¦ , which we have to maximize
Under the certain constraints.
Here , the objective function is π = π₯ + π¦ ,
Subject to constraints , 2π₯ + 5π¦ β€ 100 ,
π₯
25+
π¦
40 β€ 1 , or 8π₯ + 5π¦ β€ 200 ,
π₯ β₯ 0 , π¦ β₯ 0.
20. Let π₯ passengers travel by executive class ticket and π¦ passengers travel by
economy class ticket. Then following LPP is formed :
The objective function is , π = 1000π₯ + 600π¦ ,
subject to constraints , π₯ + π¦ β€ 200 ,
π₯ β₯ 20 ,
π¦ β 4π₯ β₯ 0 or π¦ β₯ 4π₯
π₯ β₯ 0 , π¦ β₯ 0.
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Answer key of 1 mark questions
β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦
1. Linear programming is a mathematical technique to maximize or minimize a linear function of
several variables under the certain conditions and that conditions are linear equalities or
inequalities.
2. A linear function whose value is to be maximized or minimized in a linear programming problem
is known as an objective function.
3. A general solution which satisfies all the constraints of a linear programming problem is known
as feasible solution of that linear programming problem.
4. A feasible solution of an LPP at which the minimum or maximum value of an objective function
exists is called an optimum solution.
5. There are two methods to solve a linear programming problem which are as follow :
3. Corner point method. 4. Iso-profit Or Iso -cost method.
6. If the feasible region is closed, then it is bounded and if the feasible region is not closed , then it
is unbounded.
7. A linear inequality is an inequality which involves a linear function and contain one of the
symbol : < less than, > greater than, β€ less than equal to or β₯ greater than equal to.
8. Bounded region
9.
The coordinates from which the feasible region is formed are known as corner points.
10. Smallest possible value of π₯ isβ18
5.
11. Minimum value of Z is βππ at the point (0, 12).
12. It is in the first quadrant bounded by the points (0, 0), (0, 2) and (3, 0).
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79 | P a g e
13. The feasible region of given inequality is unbounded region in first quadrant passing through
(0, 0) and towards right to the line.
14. A feasible region is an area defined by a set of coordinates that satisfy a system of inequalities.
The region satisfies all restrictions imposed by linear programming problems. The concept is an
optimization technique
15.
(40,15).
16. Minimum value of Z is 0 at (0,0).
17. The restrictions, in the form of linear inequalities, imposed on the optimization of an objective
function in a linear programming problem are known as the constraints.
18. The corner points are (0, 0), (20,0),(0,60) and (10,50).
19. Maximum value of Z is 16 at (0, 4).
20. It is the region bounded by the points (0,2), (0,0) and (4,0).
Answer key of two marks questions
1. Let he buys π₯ and π¦ be the no. of bags of rice and wheat resp. Then objective function ,
which is to maximize is π = 11π₯ + 9π¦ ,
subject to constraints 18π₯ + 15π¦ β€ 150 ,
π₯ + π¦ β€ 200,
π₯, π¦ β₯ 0.
2. Let cost of stiching of 1 shirt is Rs. π₯ and that of 1 pant is Rs. π¦. Then the objective function ,
which is to minimize is π = 90π₯ + 48π¦ ,
subject to constraints , 9π₯ + 6π¦ β€ 225 ,
15π₯ + 6π¦ β€ 300 ,
π₯, π¦ β₯ 0 .
3. Four corner points.
4. The feasible region is unbounded formed by the points (0,3)and (6,0) away from
the origin .
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80 | P a g e
5. The region formed by the given inequalities are unbounded.
6. The minimum value of π = β3π₯ + 4π¦ is β12 at the point (4,0).
7. The corner points of the feasible region are (0,50) , (0,0) ,(30,0) and (20 ,30). The
maximum value of Z is 120 at the point (30 ,0).
8. There is no feasible region corresponding to the given inequalities. Hence Ther is no
maximum value of Z subject to the given constraints.
9. The solution of the given inequalities is the region bounded by the points (0,0) ,
(10,0) and (0,5).
10. The minimum value of Z is β1000 at the point (0,1000) and maximum value is 3000
at the point (1500,0).
11. q = 3p.
12. There is no common region formed by given inequalities and hence no maximum
value exists subjected to these constraints.
13. Exist.
14. There is no region which is common to given inequalities and hence no maximum or
minimum value exists.
15. Let π₯ πππ π¦ be the number of units of food F1 and food F2 resp. to mix together to
form the mixture. Then the following LPP is formed :
The objective function is π = 16π₯ + 20π¦ ,
subject to constraints , π₯ + 2π¦ β₯ 10 ,
2π₯ + 2π¦ β₯ 12,
3π₯ + π¦ β₯ 8,
and π₯ β₯ 0 , π¦ β₯ 0.
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81 | P a g e
16. Let the personβs diet consists of π₯ unit of food I and π¦ units of food II. Then the
following LPP is formed :
The objective function is π = 16π₯ + 20π¦ ,
subject to constraints , 10π₯ + 4π¦ β₯ 20 ,
5π₯ + 5π¦ β₯ 20,
2π₯ + 6π¦ β₯ 12 ,
and π₯, π¦ β₯ 0 .
17. The feasible region which is common to the given constraints are unbounded and
the corner points are (0,2) , (1,1) and (3,0). Since the region is unbounded therefore
Z = 6 may or may not be the maximum value. To check whether Z = 6 is maximum
value, draw the graph of 2π₯ + 5π¦ > 6 and which has so many common points with
the given feasible region. Hence maximum value of Z is 6 at the line joining of (0,2)
and (3,0).
18. Since the given inequalities have no common feasible region, therefore Z has neither
maximum value nor minimum value under these constraints.
19. Let the young man covers π₯ km at the speed of 25 km/hr and π¦ km at the speed of
40 km/hr.
The total distance travelled by the man is π₯ + π¦ , which we have to maximize
Under the certain constraints.
Here , the objective function is π = π₯ + π¦ ,
Subject to constraints , 2π₯ + 5π¦ β€ 100 ,
π₯
25+
π¦
40 β€ 1 , or 8π₯ + 5π¦ β€ 200 ,
π₯ β₯ 0 , π¦ β₯ 0.
20. Let π₯ passengers travel by executive class ticket and π¦ passengers travel by
economy class ticket. Then following LPP is formed :
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82 | P a g e
The objective function is , π = 1000π₯ + 600π¦ ,
subject to constraints , π₯ + π¦ β€ 200 ,
π₯ β₯ 20 ,
π¦ β 4π₯ β₯ 0 or π¦ β₯ 4π₯
π₯ β₯ 0 , π¦ β₯ 0.
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83 | P a g e
TOPIC-: PROBABILITY S.No. QUESTIONS
1. Given two events E and F such that π(πΈ) = 0.6, π(πΉ) = 0.3, π(πΈ β© πΉ) = 0.2, then find
π(πΈ|πΉ) and π(πΉ|πΈ).
2. If π(π΄) =7
13 , π(π΅) =
9
13, π(π΄ β© π΅) =
4
13, then evaluate π(π΄|π΅).
3. If π(π΄) = 0.8, π(π΅ π΄β ) = 0.4, then find π(π΄ β© π΅).
4. Two coins are tossed once. E is the event that no tail appears and F is the event that no head
appears. Find π(πΈ|πΉ).
5. Mother, father and son line up at random for a family picture. Find π(πΈ|πΉ) if
E: son on one end and F: father in middle.
6. An instructor has a question bank consisting of 300 easy True/False questions, 200 difficult
True/False questions, 500 easy multiple choice questions and 400 difficult multiple choice
questions. If a question is selected at random from the question bank, what is the
probability that it is a difficult question given that it is a True/False question?
7. Find π(π΄|π΅) if P(A) = 0.5 and P(B) = 0.
8. One card is drawn at random from a well-shuffled deck of 52 cards. Events E and F are
defined below β¦.
E: the card drawn is a spade
F: the card drawn is an ace
Check whether the events E and F are dependent or independent.
9. An urn contains 6 red and 3 black balls. Two balls are randomly drawn. Let X represents the
number of black balls. What are the possible values of X?
10. State whether the following table is a probability distribution of a random variable or not:
X 3 2 1 0 -1
P(X) 0.3 0.2 0.4 0.1 0.05
11. State whether the following table is a probability distribution of a random variable or not:
X -1 0 1 2
P(X) 0.6 0.1 0.2 -0.1
12. Find the value of k for the probability distribution of a discrete random variable X given
below:
X 2 3 4 5
P(X) 5/k 7/k 9/k 11/k
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13. If a leap year is selected at random, then what is the probability that it will contain 53
Tuesdays?
14. Two cards are drawn at random and without replacement from a pack of 52 playing cards.
Find the probability that both the cards are black.
15. An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the
other without replacement. What is the probability that both the drawn balls are black?
16. The probability of solving a specific problem independently by A and B are 1
2 and
1
3
respectively. If both try to solve the problem independently, find the probability that both
solve the problem.
17. What is the probability of obtaining an even prime number on each die when a pair of dice is
rolled?
18. Six balls are drawn successively from an urn containing 7 red and 9 black balls. If, after each
draw, the ball drawn is not replaced in the urn, tell whether the trials of drawing balls are
Bernoulli Trials or not.
19. Find the variance of the Binomial Distribution π΅ (4,1
3).
20. If A and B are any two events such that P(A) + P(B) β P(A and B) = P(A), then find π(π΄|π΅).
21. If P(A) = 0.4, P(B) = x and P(AUB) = 0.7, then find the value of x if A and B are independent
events.
S.No. SOLUTIONS
1. P(πΈ|πΉ) =2
3 and P(πΉ|πΈ) =
2
6=
1
3
2. π(π΄|π΅) =4
9
3. π(π΄ β© π΅) = 0.32
4. π(πΈ|πΉ) =014β= 0
5. π(πΈ|πΉ) = 1
6. Required Probability = 200
500=
2
5
7. Not defined
8. π(πΈ β© πΉ) = π(πΈ) Γ π(πΉ) and hence, the events are independent.
9. X = 0,1,2
10. No because sum of Probabilities is 1.05
11. No because P( X=2) = -0.1
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12. 32
π= 1
Implies k = 32
13. Required Probability = 2
7
14. Required Probability = 1
2Γ25
51=
25
102
15. Required Probability = 10
15Γ
9
14=
3
7
16. Required Probability = 1
2Γ1
3=
1
6
17. 1
36
18. no
19. n = 4, p = 1
3 and q = 1 β
1
3=
2
3
Therefore, variance = npq = 4 Γ1
3Γ2
3=
8
9
20. π(π΄|π΅) = 1
21. P(AUB) = P(A) + P(B) β P(A).P(B)
Which implies that x = 0.5
S.No. QUESTIONS
1. The probability that at least one of the two events A and B occurs is 0.6. If A and B occur
simultaneously with probability 0.3, the evaluate π(οΏ½Μ οΏ½) + π(οΏ½Μ οΏ½).
2. Evaluateπ(π΄ βͺ π΅), if 2π(π΄) = π(π΅) =5
13 and π(π΄|π΅) =
2
5.
3. A and B are two events such that π(π΄) β 0. Find π(π΅|π΄) if
(i) A is a subset of B. (ii) Aβ©B = β
4. Three events A, B and C have probabilities 2
5,1
3 and
1
2 respectively. If P (Aβ© πΆ) =
1
5 and
P(π΅ β© πΆ) =1
4, then find the values of P(C/B) and P(οΏ½Μ οΏ½ β© πΆΜ ).
5. If A and B are two events such that P( A/B ) = P ( B/A ) , then prove that P(A) = P(B)
6. If P(A) = 0.8, P(B) = 0.5 and P( B/A ) = 0.4, then find the value of P(AUB).
7. 10% of the bulbs produced in a factory are red and 2% are red and defective. If one bulb is
picked up at random, then determine the probability of its being defective, given that it is
red.
8. A couple has two children. Find the probability that both the children are males, if it is
known that at least one of the children is male.
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86 | P a g e
9. A die is thrown three times. Events A and B are defined as belowβ¦.
A: 4 on the third throw
B: 6 on first and 5 on second throw.
Find the probability of A given that B has already occurred.
10. A bag contains 19 tickets, numbered from 1 to 19. A ticket is drawn and then another
ticket is drawn without replacement. Find the probability that both tickets will show even
numbers.
11. Three cards are drawn successively without replacement from a pack of 52 cards. What is
the probability that the first two cards are king and the third card drawn is an ace?
12. A bag contains 5 white, 7 red and 8 black balls. If four balls are drawn one by one without
replacement, then find the probability of getting all white balls?
13. Events A and B are such that P (A) = 1
2 , P(B) =
7
12 and P( not A or not B ) =
1
4 . State whether
the events A and B are independent or not.
14. A fair coin and an unbiased die are tossed. Let A be the event βhead appears on the coinβ
and B be the event β3 on the dieβ. Check whether A and B are independent events or not.
15. A die is tossed thrice. Find the probability of getting an even number at least once.
16. A can hit a target 4 times out of 5, B can hit the target 3 times out of 4 and C can hit the
target 2 times out of 3. They fire simultaneously. Find the probability that none of them
will hit the target.
17. Probability of solving specific problem independently by A and B are 1
2 πππ
1
3 respectively.
If both try to solve the problem independently, then find the probability that the problem
is solved.
18. If A and B are two events such that P(A) = 1
4, P(B) =
1
2 and P(Aβ©B) =
1
8, then find P( not A and
not B).
19. Two balls are drawn at random with replacement from a box containing 10 black and 8 red
balls. Find the probability that first ball is black and second ball is red.
20. There are 25 tickets bearing numbers from 1 to 25. One ticket is drawn at random. Find the
probability that the number on it is a multiple of 5 or 6.
21. If P(π΄|π΅) > π(π΄), then prove that P(π΅|π΄) > π(π΅).
22. A die is thrown thrice. Getting an even number is considered success. What is the mean
and variance of the Binomial Distribution?
23. If X follows a Binomial Distribution with mean 4 and variance 2, find P(Xβ₯ 7).
24. A die is thrown 6 times. If getting an odd number is a success, then what is the probability
of 5 success?
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87 | P a g e
25. If the probability that a person is not a swimmer is 0.3, then find the probability that out of
5 persons, 4 are swimmers.
S.No. SOLUTIONS
1. P(A) + P(B) = P(AUB) + P ( Aβ©B)
= 0.6 + 0.3
= 0.9
P( οΏ½Μ οΏ½) + π(οΏ½Μ οΏ½) = 1 β P(A) + 1 β P(B)
= 2 β 0.9 = 1.1
2. P(A) = 5
26 , P(B) =
5
13
P( Aβ©B) = P(B).P(A/B) = 5
13Γ2
5 =
2
13
P( AUB) = 5
26+
5
13β
2
13 = 11
26
3. .(i) P(B/A) = π(π΄β©π΅)
π(π΄) = π(π΄)
π(π΄)= 1
(ii) P(B/A) = π(π΄β©π΅)
π(π΄) =
0
π(π΄)= 0
4. P(C/B) =
π(π΅β©πΆ)
π(π΅)=
1
41
3
=3
4
P(οΏ½Μ οΏ½ β© πΆΜ ) = 1 β P ( AU C )
= 1 β { 2
5+1
2β1
5 }
= 1 - 7
10 =
3
10
5. P ( A/B ) = P ( B/A )
π(π΄β©π΅)
π(π΅)=
π(π΅β©π΄)
π(π΄)
Which implies that P(A) = P(B)
6. P( A β© π΅) = P(A). P(B/A) = 0.8 Γ 0.4 = 0.32
Therefore, P(AUB) = 0.8 + 0.5 β 0.4
= 0.9
7. P ( D/R ) =
π(π·β©π )
π(π )=
2100β
110β=
1
5
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88 | P a g e
8. P(A/B) =
14β
34β=
1
3
9. P(A/B) =
π(π΄β©π΅)
π(π΅)=
1216β
6216β
= 1
6
10. Let A is the event of drawing an even numbered ticket in the first draw and B is the event of
drawing an even numbered ticket in the second draw. Then
P(A) = 9
19 and P(B/A) =
8
18=
4
9
Therefore, P ( π΄ β© π΅) = P(A).P(B/A) = 9
19Γ4
9=
4
19
11. Required Probability = 4
52Γ
3
51Γ
4
50=
2
5525
12. Total number of balls = 20
Required Probability = 5
20Γ
4
19Γ
3
18Γ
2
17=
1
969
13. P( οΏ½Μ οΏ½ποΏ½Μ οΏ½) = π(π΄ β© π΅Μ Μ Μ Μ Μ Μ Μ )
14. A = {(π», 1), (π», 2), (π», 3), (π», 4), (π», 5), (π», 6)} , B = {(π», 3), (π, 3)} , Aβ©B = {(π», 3)}
P(A) = 6
12=
1
2 , P(B) =
2
12=
1
6 and P(Aβ©B ) =
1
12
P(A). P(B) = 1
2Γ1
6=
1
12= π(π΄ β© π΅)
Therefore, A and B are independent events.
15. P( an even number ) = 3
6=
1
2
Required Probability = 1 β Probability of getting an even number in none of the throws
= 1 - 1
2Γ1
2Γ1
2
= 7
8
16. P(A) = 4
5 , P(B) =
3
4 , P(C) =
2
3
Therefore, P( οΏ½Μ οΏ½) = 1 β4
5=
1
5
P( οΏ½Μ οΏ½) = 1 β3
4=
1
4
P( πΆΜ ) = 1 β2
3=
1
3
Therefore, P( οΏ½Μ οΏ½ β© οΏ½Μ οΏ½ β© πΆΜ ) = π(οΏ½Μ οΏ½) Γ π(οΏ½Μ οΏ½) Γ π(πΆΜ )
= 1
5Γ1
4Γ1
3=
1
60
17. P ( problem is solved )
= P ( at least one of them solves the problem )
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89 | P a g e
= 1 β P ( none of them solves the problem )
= 1 β P(οΏ½Μ οΏ½). π(οΏ½Μ οΏ½)
= 1 - 1
2Γ2
3
= 2
3
18. P( not A and not B ) = P( οΏ½Μ οΏ½ β© οΏ½Μ οΏ½)
= π(οΏ½Μ οΏ½) Γ π(οΏ½Μ οΏ½) { not A and not B are independent events }
= (1 β1
4) (1 β
1
2)
= 3
8
19. P ( Bβ©R ) = P(B) Γ P(R)
= 10
18Γ
8
18=
20
81
20. P( EUF ) = P(E) + P(F) β P( Eβ©F)
= 5
25+
4
25β 0 =
9
25
21. P(π΄|π΅) > π(π΄)
Implies π(π΄β©π΅)
π(π΅)> π(π΄)
Implies π(π΄β©π΅)
π(π΄)> π(π΅)
Implies P(π΅|π΄) > π(π΅)
22. n = 3
p = 3
6=
1
2 which implies that q = 1- p = 1 -
1
2=
1
2
Therefore, Mean = np = 3 Γ1
2=
3
2
Variance = npq = 3 Γ1
2Γ1
2=
3
4
23. πππ
ππ=
2
4 implies q =
1
2 which implies p =
1
2.
Also, np = 4 which implies n Γ1
2= 4 which implies n = 8
Therefore, P(X = r) = 8Cr(1
2)π(1
2)8βπ
= 8Cr (1
2)8
Therefore, P(Xβ₯ 7) = P(X=7) + P(X=8)
= 8C7(1
2)8+ 8C8(
1
2)8
= 1
28(8 + 1) =
9
28
24. n = 6
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90 | P a g e
p = 3
6=
1
2 which implies that q =
1
2
Therefore, P(X=5) = 6C5 p5q1 = 1 Γ (1
2)5Γ (
1
2)1= (
1
2)6=
1
64
25. q = 0.3, p = 1 β 0.3 = 0.7 and n = 5
Therefore, P(X=4) = 5C4(0.7)4(0.3) = 1.5(0.7)4