sample paper (cbse)cbse-12-mathematics ©educomp solutions ltd. 2017 23 a bag contains (2n +1)...

Sample Paper (CBSE) Series SC/SP/017

Code No. SP/017

CBSE-12- Mathematics ©Educomp Solutions Ltd. 2017

Mathematics

Time Allowed: 3 hours Maximum : 70

General Instructions:

(i) There are 26 questions in all.

(ii) All questions are compulsory.

(iii) Marks for each question are indicated against it.

(iv) Question numbers 01 to 04 are very short answer questions carrying 01 mark

each.

(v) Question numbers 05 to 12 are short answer questions carrying 02 marks each.

Out of which one question is a value based question.

(vi) Question numbers 13 to 23 are short answer questions carrying 04 marks each.

Out of which one question is a value based question.

(vii) Question numbers 24 to 29 are long answer questions.

1 What is the principal value of 푡푎푛 푡푎푛 ?

2 A and B are square matrices of order 3 each, |A | = 2 and |B | = 3. Find |3AB |.

3 What is the distance of the point (p, q, r) from the x-axis?

4 Let f : R → R be defined by 푓(푥) = 3푥 − 5 and g : R → R be defined by

푔(푥)= find g of .

CBSE-12-Mathematics ©Educomp Solutions Ltd. 2017

5 How many equivalence relations on the set {1, 2, 3} containing

(1, 2) and (2, 1) are there in all ? Justify your answer.

6 Let li, mi, ni ; i = 1, 2, 3 be the direction cosines of three mutually perpendicular vectors in space. Show that AA’ = I3 , where A = 푙 푚 푛푙 푚 푛푙 푚 푛

.

7 If 푒 (푥 + 1) = 1, show that = −푒 .

8 Find the sum of the order and the degree of the following differential equations:

+ + (1 + 푥) =0

9 Find the Cartesian and Vector equations of the line which passes through the point (2, 4,−5) and parallel to the line given by = =

.

10 Solve the following Linear Programming Problem graphically: Maximize 푍 = 3푥 + 4푦 subject to

푥 + 푦 ≤ 4, 푥 ≥ 0 and 푦 ≥ 0

11 A couple has 2 children. Find the probability that both are boys, if it is known that (i) one of them is a boy (ii) the older child is a boy.

12 The sides of an equilateral triangle are increasing at the rate of 2 cm/sec. Find the rate at which its area increases, when side is 10 cm long.

13 If A + B + C =휋 then find the value of

푆푖푛(퐴 + 퐵 + 퐶) 푆푖푛 퐵 cos C−푆푖푛 퐵 0 tan퐴

cos (퐴 + 퐵) − tan퐴 0

OR

Using properties of determinant, prove that


푏 + 푐 푎 − 푏 a푐 + 푎 푏 − 푐 푏a + b 푐 − 푎 푐

= 3푎푏푐 − 푎 − 푏 − 푐

14 It is given that for the function 푓(푥)=푥 − 6푥 + 푎푥 + 푏 Rolle’s theorem

holds in [1, 3] with c=2+√

Find the value of ‘a’ and ‘b’.

15 Determine for what values of x, the function 푓(푥) = 푥 +

( 푥 ≠ 0) is strictly increasing or strictly decreasing.

OR

Find the point on the curve y =푥 − 11푥 + 5 at which the tangent is 푦 = 푥 − 11

16 Evaluate ∫ (푥 + 3) 푑푥 as limit of sums.

17 Find the area of the region bounded by the y-axis, y = cos 푥 and y = sin 푥, 0 ≤ 푥 ≤

18 Can y = 푎푥 + be a solution of the following differential equation?

y = x + ...............(*)

If no, find the solution of the D.E.(*)

OR

Check whether the following differential equation is homogeneous or not

푥 − 푥푦 = 1 + 푐표푠 , 푥 ≠ 0

Find the general solution of the differential equation using substitution 푦 = 푣푥.


19 If the vectors 푝 ⃗= 푎 횤̂ + 횥̂ + 푘, 푞 ⃗= 횤̂ + 푏횥̂ + 푘, 푟 ⃗= 횤̂ + 횥̂ + 퐶푘 are coplanar, then for a, b, c ≠ 1 show that

11 − 푎 +

11 − 푏 +

11 − 푐 = 1

20 A plane meets the coordinate axes in A B and C such that the centroid of ∆ ABC is the point ( 훼,훽, 훾) . Show that the equation of the plane is

푥훼 +

1훽 +

1훾 = 3

21 If a 20 year old girl drives her car at 25 km/h, she has to spend Rs 4/km on petrol. If she drives her car at 40 km/h, the petrol cost increases to Rs 5/km. She has Rs 200 to spend on petrol and wishes to find the maximum distance she can travel within one hour. Express the above problem as a Linear Programming Problem. Write any one value reflected in the problem.

22 The random variable X has a probability distribution P(X) of the following form, where k is some number:

P(X) =k , if x = 0

2k , if x = 13k , if x = 20 , otherwise

(i) Find the value of k (ii) Find P(X < 2) (iii) Find P(X ≤ 2) (iv) P(X ≥ 2)


23 A bag contains (2n +1) coins. It is known that ‘n’ of these coins have a head on both its sides whereas the rest of the coins are fair. A coin is picked up at random from the bag and is tossed. If the probability that

the toss results in a head is find the a ue of ‘n’.

24 Using properties of integral, evaluate ∫ 푑푥.

OR

Find: ∫ 푑푥

25 Does the following trigonometric equation have any solutions? If Yes, obtain the solution(s):

푡푎푛푥 + 1 푥 − 1

+ 푡푎푛푥 − 1 푥

= 푡푎푛 7

OR

Determine whether the operation * define below on ℚ is binary operation or not. a * b = ab+1

If yes, check the commutative and the associative properties. Also check the existence of identity element and the inverse of all elements in ℚ .

26 Find the value of x, y and z, if A =

0 2푦 푧푥 푦 −푧푥 −푦 푧

satisfies A’ = 퐴

OR

Verify: A(adj A) = (adj A)A=|퐴| matrix A= 1 −1 23 0 −21 0 3


27 Find , if y =푒 2푡푎푛

28 Find the shortest distance between the line 푥 − 푦 + 1 = 0 and the curve 푦 = 푥

29 Define skew lines. Using only vector approach, find the shortest distance between the following two skew lines:

푟 ⃗ = (8 + 3휆)횤̂ − (9 + 16휆)횥̂+ (10 + 7휆)푘

푟 ⃗ = 15횤̂ + 29횥̂ + 5푘 + 휇 3횤̂ + 8횥̂ − 5푘


1 푡푎푛 푡푎푛 = 푡푎푛 −푡푎푛 = −

2 |3AB |=3 |A||B |=27 × 2 × 3 = 162

3 Distance of the point (p, q, r) from the x-axis

= Distance of the point (p, q, r) from the point (p,0,0)

= 푞 + 푟

4 go f(x) = g{ f(x)} = g (3푥 − 5 )=

( ) =

5 Equivalence relations could be the following:

{ (1,1), (2,2), (3,3), (1,2), (2,1)} and { (1,1), (2,2), (3,3), (1,2), (1,3), (2,1), (2,3), (3,1), (3,2)} So, only two equivalence relations.(Ans.)

6 AA’=

푙1 푚1 푛1푙2 푚2 푛2푙3 푚3 푛3

푙1 푙2 푙3푚1 푚2 푚3푛1 푛2 푛3

=1 0 00 1 00 0 1

=푙

because 푙 +푚 + 푛 = 1 for each 푖 = 1, 2, 3 푙 푙 + 푚 푚 + 푛 푛 = 0 (푖 ≠ 푗) for each 푖 푗 = 1, 2, 3

7 푒 (푥 + 1) = 1, w.r.t. x, we get

푒 + (푥 + 1) = 0

푒 + = 0

= −푒

ANSWERS


8 Here, + (1 + 푥) = −

Thus, order is 2 and degree is 3. So, the sum is 5

9 Here, = = is same as = = Cartesian equation of the line is = = Vector equation of the line is 푟 ⃗ = −2횤̂ + 4횥̂ − 5푘 +⋌ 3횤̂ + 5횥̂ + 6푘

10 The feasible region is a triangle with vertices O(0,0), A(4,0) and B(0,4) Zo = 3×0 + 4×0 = 0 ZA = 3×4 + 4×0 = 12 ZB = 3×0 + 4×4 = 16 Thus, maximum of Z is at B(0,4) and the maximum value is 16

11 Sample space = {B1B2, B1G2, G1B2, G1G2} , B1 and G1 are the older boy and girl respectively. Let E1 = both the children are boys; E2 = one of the children is a boy ; E3 = the older child is a boy

Then (i) P(E1/E2)=P ∩ = //

=

(ii) P(E1/E3)=P ∩ = //

=

12 Here, Area (A)=√ 푥 where ‘x’ is the side of the equilateral triangle


So = √ ×

=√ (10)(2) = 10√3 cm /sec

13 As A + B + C = 휋

푆푖푛(퐴 + 퐵 + 퐶) 푆푖푛 퐵 cos C−푆푖푛 퐵 0 tan퐴

cos(퐴 + 퐵) − tan퐴 0=

0 푆푖푛 퐵 cos C−푆푖푛 퐵 0 tan퐴− cos퐶 − tan퐴 0

=0 × 0 tan퐴

−tan퐴 0 − 푆푖푛 퐵 × −푆푖푛 퐵 tan퐴− cos퐶 0 + cos퐶 × −푆푖푛 퐵 0

− cos퐶 −tan퐴 =0 sin B tan A cos C + cos C sin B tan A = 0 (Ans.)

OR

Let ∆=푏 + 푐 푎 − 푏 a푐 + 푎 푏 − 푐 푏a + b 푐 − 푎 푐

Applying C1 → C1 + C3 we get ∆= (푎 + 푏 + 푐)1 푎 − 푏 a1 푏 − 푐 푏1 푐 − 푎 푐

Applying R2 → R2 R1, and R3 → R3 R1 , we get

∆= (푎 + 푏 + 푐)1 푎 − 푏 a0 2푏 − 푎 − 푐 푏 − 푎0 2푎 + 푏 + 푐 푐 − 푎

Expanding Δ along first column we have the result

14 Since Rolle’s theorem holds true, f(1) = f(3)

i.e., (1)3 6(1)2 + a(1) + b = (3)3 6(3)2 + a(3) + b i.e., a + b + 22 = 3a + b ⇒ a = 11 Also, f ’ (x) = 3푥 − 12푥 + 푎 표푟 3푥 − 12푥 + 11 As f ’ (c) = 0 , we have

3 1 +√

− 12 1 +√

+ 11 = 0

As it is independent of b, b is arbitrary


15 Here f ’ (x) = 3푥 − 3푥 = ( )

= (푥 + 1)(푥 − 1)

Critical points are −1 and 1 f ‘(x) > 0 if x >1 or x < −1; and f ‘(x) < 0 if 1< x < 1

3(푥 + 푥 + 1)푥 푎푙푤푎푦푠 + 푖푣푒

Hence, f(x) is strictly increasing for x > 1 Or x < −1 and strictly decreasing for (-1,0)u(0,1) [1]

OR

Here, = 3푥 − 11

So, slope of the tangent is 3푥 − 11 Slope of the given tangent line is 1. Thus, 3푥 − 11 = 1

that gives x =±2

When 푥 = − 2, 푦 = 2 – 11 = −9

When 푥 = − 2, 푦 = −2 – 11 = −13 out of the two points (2,−9 ) and ( −2,−13) only the point (2,−9 ) lies on the curve thus the required point is (2,−9 )

16 Here, f(x) =(푥 + 3), 푎 = 0, 푏 = 2 and 푛ℎ = 푏 − 푎= 2

(푥 + 1)dx = lim→ℎ[푓(푎) + 푓(푎 + ℎ) + 푓(푎 + 2ℎ) + … … … .푓(푎 + 푛 − 1 ℎ)]

= lim

→ℎ[3 + 1 ℎ + 3 + 2 ℎ + 3 + … … … . +(푛 − 1) ℎ + 3]

= lim

→ℎ[3푛 + ℎ {1 + 2 + 3 + … … … . (푛 − 1) }]

= lim→ℎ 3푛ℎ + ℎ

(푛 − 1)푛(2푛 − 1)6


= lim→ℎ 3푛ℎ +

(푛ℎ − ℎ)푛ℎ(2푛ℎ − ℎ)6

= lim→ℎ 3 × 2 +

(2− ℎ)2(4− ℎ)6

=6+

=

17 The rough sketch of the bounded region is shown on the right.

Required area =∫ cos푥⁄ dx − ∫ sin 푥⁄ dx =(cos푥 + sin 푥) ⁄ =sin + cos − sin 0 − cos 0 =√− 1 √2− 1 sq units

18 푦 = 푎푥 + ………(1) Gives = 푎 Substituting this a ue of ‘a’ in (1) we get

푦 = 푥푑푦푑푥 +

푏푑푦푑푥

Thus, 푦 = 푎푥 + is a solution of the following differential equation

푦 = 푥푑푦푑푥 +

푏푑푦푑푥

OR

Given differential equation can be written as

= = + … … … . . (1)


Let f (푥, 푦)= +

Then F(λx, λy) = λλ

+λλ

λ

= y +λ2 ≠ f (푥, 푦)

Hence, the given D.E. is not a homogeneous equation.

Putting y = vx and = 푣 + 푥 in (1), we get

푣 + 푥푑푣푑푥 = 푣 +

1 + 푐표푠 푣푥

= 푑푣1+푐표푠 푣 = 1

푥3 dx

푠푒푐 푑푣= dx

Integrating both sides, we get

2 tan = − 1푥2 + 퐶

or 2 tan = − 1푥2 + 퐶

19 Since the vector 푝 ⃗, 푞 ⃗ and, 푟 ⃗ are coplanar

∴ [푝 ⃗,푞 ⃗, 푟 ⃗]=0 [푝 ⃗, 푞 ⃗, 푟 ⃗] =0

i.e., 푎 1 11 푏 11 1 푐

=0

R2 → R2− R1 R3 → R3− R1

or 푎 1 1

1 − 푎 푏 − 1 01 − a 0 푐 − 1

=0

⇒ 푎( 푏 − 1)(푐 − 1) − 1 (1− 푎)( 푐 − 1) − 1(1 − 푎) 푏 – 1 = 0 i.e., 푎( 1 − 푏)(1 − 푐) + (1 − 푎)( 1 − 푐) + (1− 푎)( 1 – 푏) = 0 Dividing both the sides by (1 − 푎)( 1 − 푏)(1 − 푐), we get 푎

1 − 푎 +1

1 − 푏 +1

1 − 푐 = 0


i. e. − 1−1

1 − 푎 +1

1 − 푏 +1

1 − 푐 i. e.

11− 푎+

11− 푏+

11− 푐 = 1

20 We know that the equation of the plane having intercepts a, b and c on the three coordinate axes is

푥푎 +

푦푏 +

푧푐 = 1

Here, the coordinates of A, B and C are (푎, 0,0), (0,푏, 0) 푎푛푑 (0,0, 푐) respectively. the centroid of Δ ABC is + + . Equating + + to (훼 ,훽, 훾) we get a= 3 훼, b= 3 훽 and c=3 훾 Thus, the equation of the plane is

+

+

=1

Or +

+

= 3

21 Let the distance covered with speed of 25 km/h = x km

and the distance covered with speed of 40 km/h = y km Total distance covered = z km The L.P.P. of the above problem, therefore, is Maximize z = x + y subject to constraints

4푥 + 5푦 ≤ 200 푥

25 +푦

40 ≤ 1 푥 ≥ 0, 푦 ≥ 0 any one value

22 Here, X 0 1 2 P(x) k 2k 3k (i) Since P(0) + P(1) + P(2)= 1, we have

k + 2k + 3k = 1 i.e., 6 k = 1, or k =

(ii) P(X <2)= P(0) + P(1)= k + 2k = 3k= (iii) P(X≤ 2 ) = P(0) + P(1) + P(2)= k + 2k + 3k = 6k = 1 (iv) P(X≥ 2 ) = P(2) = 3k =


23 Let the events be described as follows: E1 : a coin having head on both sides is selected. E2 : a fair coin is selected. A : head comes up in tossing a selected coin P(E1)= ; P(E2)= ; P(A/E1)=1; P(A/E2)= It is given that P(A) =

. So

P(E1 ) P(A/E1 ) + P(E2 ) P(A/E2 ) =

⇒ × 1 + × =

⇒ 푛 + =

⇒ 42(3푛 + 1)= 62(2푛 + 1) 2푛 = 20, or 푛 = 20

24 l =∫ 푑푥 = ∫ ( )푑푥

=휋 ∫ 푑푥 − ∫ 푑푥 ⇒ 2 l= 휋 ∫ 푑푥 ⇒ ∫ 푑푥

⇒ ∫

푑푥

⇒ ∫ 푠푒푐 − 푑푥

⇒ l= −2푡푎푛 −

⇒ l= [2 − (−2)] = 휋


OR

25

OR Given * on 푄 , defined by a*b = ab+1

Let, a 휖 푄, b 휖 푄 then


ab 휖 푄 and (ab+1) 푄 a*b=ab+1 is defined on 푄

∴ * is a binary operation on 푄

Commutative: a*b = ab+1 b*a = ba+1 =ab+1 (∵ ba = ab in 푄) a*b =b*a So * is commutative on 푄 Identity Element : Let e 휖 푄 be the identity element, then for every a 휖 푄 a * e=a and e*a=a ae+1=a and ea+1=a ⇒ e = and e =

e is not unique as it depend on `a’ hence identity e ement does not exist for * Inverse: since there is no identity element hence, there is no inverse.

26. The relation A’ = 퐴 , gives A’A =퐴 A= I

Thus 0 푥 푥

2푦 푦 −푦z −푧 푧

0 2푦 푧푥 푦 −푧푥 −푦 푧

= 1 0 00 1 00 0 1

⇒ 0 + 푥 + 푥 0 + 푥푦 − 푥푦 0− 푥푧 + 푥푧0 + 푥푦 − 푥푦 4푦 + 푦 + 푦 2푦푧 − 푦푧 − 푦푧0 − 푧푥 + 푧푥 −푧 푧 + 푧 + 푧

=1 0 00 1 00 0 1

⇒ 2푥 0 0

0 6푦 00 0 3푧

=1 0 00 1 00 0 1

⇒ 2푥 = 1; 6푦 = 1 and 3푧 = 1 ⇒ 푥 = ±

√ ; 푦 = ±

√ and 푧 = ±

√

OR

Here, |퐴| =1 −1 23 0 −21 0 3

= 1(0+0) +1( +2) +2(0 0) =11

⇒ |퐴|I=11 0 00 11 00 0 11


adj A= 0 3 2

−11 1 80 −1 3

Now A(adj A)= 1 −1 23 0 −21 0 3

0 3 2−11 1 8

0 −1 3=

11 0 00 11 00 0 11

And (adj A) A= 0 3 2

−11 1 80 −1 3

1 −1 23 0 −21 0 3

=11 0 00 11 00 0 11

Thus, it is verified that A(adj A)= (adj A) A=|퐴|I

27 Putting x = cos 2휃 in 2푡푎푛 ,푤푒 푔푒푡

2푡푎푛1 − cos 2휃1 + cos 2휃

=

2푡푎푛 = 2푡푎푛 (tan휃) = 2휃 = 푐표푠 푥

Hence, y = 푒 푐표푠 푥 퐿표푔 y= 푠푖푛 푥 + 푙표푔 (푐표푠 푥)

1푦 ×

푑푦푑푥 = 2푠푖푛푥 cos푥 +

1푐표푠 푥 ×

−1√1 − 푥

= 푠푖푛2푥 −1

푐표푠 푥√1 − 푥

= =푒 푐표푠 푥 푠푖푛2푥 −√

28 Let (t2, t) be any point on the curve y2 = 푥. Its distance (S) from the line 푥 − 푦 + 1 = 0 is given by S=

√

=√

(∵ t − 푡 + 1 = (t − ) + > 0 )

= =√

(2푡 − 1)

And = √2 > 0

Now = 0

⇒ √

(2푡 − 1)=0 푖. 푒. t= Thus, S is minimum at t =

So, the required shortest distance is √

=√

,= √


29 1) the line which are neither intersecting nor parallel.

2) The given equations are 푟 ⃗ = 8횤̂ − 9횥̂ + 10푘 + 휇 3횤̂ − 16횥̂ + 7푘 … … … … … . (1)

푟 ⃗ = 15횤̂ + 29횥̂ + 5푘 + 휇 3횤̂ + 8횥̂ − 5푘 … … … … … . (2)

Here 푎 ⃗ = 8횤̂ − 9횥̂ + 10푘 ; 푎 ⃗ = 15횤̂ + 29횥̂ + 5푘

푏 ⃗ =3횤̂ − 16횥̂ + 7푘; 푏 ⃗ = 3횤̂ + 8횥̂ − 5푘

Now 푎 ⃗ − 푎 ⃗ = (15 − 8)횤̂ + (29 + 9)횥̂ + (5 − 10)푘 = 7횤̂ + 38횥̂ − 5푘

And

푏 ⃗ × 푏 ⃗ = 횤̂ 횥̂ 푘3 −16 73 8 −5

= 24횤̂ + 36횥̂ + 72푘

=( 푎 ⃗ − 푎 ⃗ ).( 푏 ⃗ × 푏 ⃗ ) =( 7횤̂ + 38횥̂ − 5푘).( 24횤̂+ 36횥̂ + 72푘)=1176

Shortest distance = ( ⃗ ⃗ ).( ⃗ × ⃗ )( ⃗ × ⃗ )

=√

=√

= =

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