cbse mathematics – 2017 · 2020. 6. 8. · cbse mathematics – 2017 comptt. ... unable to get...
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mathematics – 2017 comptt. (delhi) ■ 197
set i
section A
Question numbers 1 to 4 carry one mark each.
Q.1. Let A and B be the matrices of order 3 × 2 and 2 × 4 respectively. Write the order of matrix
(AB).
Q.2. Write the equation of tangent drawn to the curve y = sin x at the point (0, 0).
Q.3. Find : 1
1x x( log )+∫ dx
Q.4. Write the angle between the vectors a b→ →× and b a
→ →× .
section B
Question numbers 5 to 12 carry 2 marks each. Q.5. In the following matrix equation use elementary operation R2 → R2 + R1 and write the
equation thus obtained.
2 31 4
1 02 1
8 39 4
−
=
−−
.
Q.6. Find the value of k for which the function f (x) = x x
xx
k x
2 3 102
2
2
+ −−
≠
=
,
, is continuous at x = 2.
Q.7. The radius r of a right circular cone is decreasing at the rate of 3 cm/minute and the height h is increasing at the rate of 2 cm/minute. When r = 9 cm and h = 6 cm, find the rate of change of its volume.
Q.8. Find : x x2 2−∫ dx.
Time allowed : 3 hours Maximum marks : 100
General InstructIons Same as in Delhi Board 2015.
cbse
mathematics – 2017Comptt. (DELHI)
198 ■ shiv das senior secondary series (Xii)
Q.9. Find the differential equation of the family of curves y2 = 4ax.
Q.10. Find the general solution of the differential equation dydx
+ 2y = e3x.
Q.11. If the points with position vectors 10 3 12 5 11ˆ ˆ, ˆ ˆ ˆ ˆi j i j i j+ − +and λ are collinear, find the value of λ.
Q.12. A firm has to transport atleast 1200 packages daily using large vans which carry 200 packages each and small vans which can take 80 packages each. The cost for engaging each large van is ` 400 and each small van is ` 200. Not more than ` 3,000 is to be spent daily on the job and the number of large vans cannot exceed the number of small vans. Formulate this problem as a LPP given that the objective is to minimize cost.
section c
Question numbers 13 to 23 carry 4 marks each.
Q.13. Prove that : tan–1 1 11 1 4
12
12
1 1+ − −+ + −
= − − ≤ ≤−x x
x xx xπ cos , .
Q.14. If a b y c z
a x b c z
a x b y c
− −− −− −
= 0, then using properties of determinants, find the value of
ax
by
cz
+ + , where x, y , z ≠ 0.
Or
Using elementary operations, find the inverse of the following matrix A
A = 1 22 1−
.
Q.15. If x = a(cos θ + θ sin θ) and y = a(sin θ – θ cos θ), then find d y
dx
2
2 .
Q.16. Find the equation of tangent to the curve y = cos(x + y), – 2π ≤ x ≤ 0, that is parallel to the line x + 2y = 0.
Q.17. Find : x
x x
++ −∫ 5
3 13 102 dx.
Or
Evaluate : 1
42 20
4
cos sin
/
x x+∫π
dx.
Q.18. Find : x dx
x x
2
21 1( )( )− +∫ Q.19. Find the general solution of the following differential equation :
x cos yx
dydx
= y cos yx
+ x
mathematics – 2017 comptt. (delhi) ■ 199
Q.20. If four points A, B, C and D with position vectors 4 3 3 5 7 5 3ˆ ˆ ˆ, ˆ ˆ ˆ, ˆ ˆi j k i x j k i j+ + + + + and 7 6ˆ ˆ ˆi j k+ + respectively are coplanar, then find the value of x.
Q.21. Find the value of p so that the lines
13
7 142
31
− − −= =x yp
z and 7 7
35
111
7− − −= =x
py z
are at right angles.
Or Find the equation of the plane through the line of intersection of the planes x + y + z = 1
and 2x + 3y + 4z = 5 and twice of its y-intercept is equal to three times its z-intercept. Q.22. Solve the following Linear Programming problem graphically : Minimize : z = 6x + 3y
Subject to the constraints :
4 805 115
3 2 1500 0
x y
x y
x y
x y
+ ≥+ ≥+ ≤≥ ≥
,
Q.23. There are three categories of students in a class of 60 students : A : Very hard working students B : Regular but not so hard working C : Careless and irregular 10 students are in category A, 30 in category B and rest in category C. It is found that
probability of students of category A, unable to get good marks in the final year examination is, 0.002, of category B it is 0.02 and of category C, this probability is 0.20. A student selected at random was found to be the one who could not get good marks in the examination. Find the probability that this student is of category C.
section D
Question numbers 24 to 29 carry 6 marks each.
Q.24. Let f : IR – − →
43
IR be a function defined as f (x) = 43 4
xx +
. Show that, in
f : IR – − →
43
Range of f, f is one-one and onto. Hence find f –1. Range f → − −
IR 43
.
Or
Let A = IR × IR and be the binary operation on A defined by (a, b) (c, d) = (a + c, b + d). Show that is commutative and associative. Find the identity element for on A, if any. (Not in syllabus)
Q.25. Find matrix A, if 2 11 03 4
1 8 101 2 59 22 15
−
−
=− − −
− −
A
Q.26. Find the intervals in which the function f given by f (x) = sin x + cos x, 0 ≤ x ≤ 2π is strictly increasing or strictly decreasing.
Or
200 ■ shiv das senior secondary series (Xii)
Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height h and semi-vertical angle α, is one-third that of the cone. Hence find the greatest volume of the cylinder.
Q.27. Using integration find the area of the region {(x, y) : y2 ≤ 4x, 4x2 + 4y2 ≤ 9}.
Q.28. Find the equation of plane containing the lines x y z− +
−−= =8
319
1610
7 and
x y z− + −−
= =383
298
55 .
Q.29. A fair coin is tossed 8 times, find the probability of (Not in syllabus) (i) exactly 5 heads (ii) at least six heads (iii) at most six heads
Or Three cards are drawn successively with replacement from a well shuffled pack of 52 cards.
Find the mean and variance of the number of red cards.
set ii
Note: Except for the following questions, all the remaining questions have been asked in Set I. Q.10. The position vectors of points A, B and C are λ µˆ ˆ, ˆ ˆi j i j+ +3 12 and 11 3ˆ ˆi j− respectively.
If C divides the line segment joining A and B in the ratio 3 : 1, find the value of λ and μ.
Q.11. Find : 2 2x x−∫ dx.
Q.12. Find the value of p for which the function f (x) = 1 4
2 0
0
−
≠
=
cos ,
,
x
xx
p x is continuous at x = 0.
Q.20. Solve the following Linear Programming problem graphically :
Maximise : z = 8000x + 12000y
Subject to the constraints : 3 4 60
3 300 0
x y
x y
x y
+ ≤+ ≤≥ ≥
,
Q.21. Find : x
x x
++ +∫ 7
3 25 282 dx
Q.22. If x = 3 cos t – 2 cos3 t and y = 3 sin t – 2 sin3 t then find d y
dx
2
2 .
Q.28. Find matrix X if : X1 2 34 5 6
7 8 92 4 6
11 10 9
=
− − −
.
Q.29. Show that the lines x y z+−
− −= =33
11
55
and x y z+−
− −= =11
22
55
are coplanar. Hence find
the equation of the plane containing these lines.
mathematics – 2017 comptt. (delhi) ■ 201
set iii
Note: Except for the following questions, all the remaining questions have been asked in Set I and Set II.
Q.10. Find the value of k for which the function f (x) =
sin cos ,
,
x xx
x
k x
−−
≠
=
4 4
4
ππ
π .
Q.11. Find : 1
42x x−∫ dx
Q.12. Find the general solution of the differential equation dydx x
+ 2 y = x.
Q.20. Maximize z = 105x + 90y graphically under the following constraints :
x + y ≤ 50, 2x + y ≤ 80, x ≥ 0, y ≥ 0
Q.21. Find : x
x x
+
− +∫ 3
5 4 2 dx
Q.22. Prove that : cot–1 1 11 1 2 4
0+ + −+ − −
∈
=sin sin
sin sin, ,x x
x xx
xπ
.
Q.28. Using Integration, find the area of the following region : {(x, y) : y2 ≥ ax, x2 + y2 ≤ 2ax, a > 0} Q.29. Find the equation of the plane passing through the point (–1, 3, 2) and perpendicular to
each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.
SolutionS
1. The order of matrix (AB) is 3 × 4. 2. y = sin x
Differentiating both sides w.r.t. x,
dydx
= cos x
dydx
at (0, 0) = cos 0 = 1
∴ Slope of tangent, m = 1Equation of the tangent at point (0, 0) and
slope 1 is (y – 0) = 1(x – 0) ⇒ y – x = 0
3. 1
1x x( log )+∫ dx [Let p = 1 + log x, dp = 1x
dx
= dpp∫ = log |p| + c
= log|1 + log x| + c
4. b a a b→ → → →× = − ×( )
∴ The angle between the vector a b→ →× and
b a→ →× = π.5. I A = A
2 31 4
1 02 1−
=
8 39 4
−−
Applying R2 → R2 + R1
2 3
3 7
1 0
2 1−
=
8 3
17 7
−−
6. L.H.L = lim ( )x
f x→ −2
= limx
x xx→ −
+ −−2
2 3 102
= lim( )( )
( )x
x xx→ +
+ −−2
5 22
= lim ( )x
x→ −
+2
5 = 2 + 5 = 7
x x
x x x
x x x
x x
2
2
3 10
5 2 10
5 2 5
5 2
+ −
= + − −= + − += + −
( ) ( )
( )( )
202 ■ shiv das senior secondary series (Xii)
R.H.L = lim ( )x
f x→ +2
= limx
x xx→ +
+ −−2
2 3 102
= lim( )( )
( )x
x xx→ +
+ −−2
5 22
= lim ( )x
x→ +
+2
5 = 2 + 5 = 7
At x = 2, f (x) = k
f (2) = kf (x) is continuous at x = 2. (Given)
∴ LHL = RHL = f (2) 7 = 7 = k ∴ k = 7
7. Given : drdt
= –3 cm/min; and
dhdt
= 2 cm/min; r = 9 cm; h = 6 cm
Volume of the Cone, V = 13
πr2h
Differentiating both sides w.r.t. time,
ddtV =
π3
2 2r h rdhdt
drdt
+
.
= π3
[(81(2) + 6(18) (– 3)]
[ r = 9 cm, h = 6 cm)
= π3
[162 – 324] = π3
(–162)
∴ Change in volume of cone = – 54π cm3/min.
∴ Volume is decreasing at the rate of 54π cm3/minute.
8. x x2 2−∫ dx
= x x2 2 22 1 1− + − dx = ( )x − −∫ 1 12 2 dx
y a dy y y a2 2 2 21
2− =
−∫− + − +a
y y a c2
2 22
log
= 1
21 1 12 2( ) ( )x x− − − −
12 log|( ) ( ) |x x c− + − −
+1 1 12 2
= 1
21 22( )x x x− − −
log|( )x x x c− + −
+1 22
9. y2 = 4ax …[Given equation of family of curves
⇒ yx
2 = 4a
Differentiating both sides w.r.t. x
x y
dydx
y
x
. .2 12
2
− = 0
y 2x ydydx
−
= 0
2x dydx
– y = 0
2x dydx
= y
dydx
yx
=2
or 2x dydx
– y = 0
10. Given differential equation is
dydx
+ 2y = e3x
Here ‘P’ = 2, Q = e3x
I.F = e edx dxP∫ ∫= 2 = e2x
Hence, the solution is y(I.F) = Q I.F.( ) dx∫ y(e2x) = e e dxx x3 2.∫ [am . an = am + n
y(e2x) = e dxx5∫ y(e2x) = e x5
5 + c
y = e
e
c
e
x
x x
5
2 25+
y = e x3
5 + ce–2x
11. Let A( 10 3ˆ ˆi j+ ), B( 12 5ˆ ˆi j− ) and C( λ ˆ ˆi j+ 11 ).
Points A, B, C are collinear.
AB→
= m BC→
(12 – 10) i +(–5 – 3) j = m[(λ – 12) i + (11 + 5) j ] 2 8ˆ ˆi j− = m(λ – 12) i + 16m j
2 = m(λ – 12) – 8 = 16m
2 = − 12
(λ – 12) − 12
= m …(i) [From (i)]
⇒ – λ + 12 = 4 ⇒ – λ = – 8 ⇒ λ = 8
12. Let the firm has x number of large vans and y number of small vans.
From the given information, we have the following corresponding constraint table.
mathematics – 2017 comptt. (delhi) ■ 203
Packages Cost Large van, x 200 ` 400 Small van, y 80 ` 200 1200 ` 3000 (at least) (maximum) Minimum Cost, Z = 400x + 200ySubject to the constraints : 200x + 80y ≥ 1200 ⇒ 5x + 2y ≥ 30 400x + 200y ≤ 3000 ⇒ 2x + y ≤ 15and x ≤ y, x, y ≥ 0Thus, required LPP to minimize cost is Z = 400x + 200ySubject to constraints : 5x + 2y ≥ 30 2x + y ≤ 15 x ≤ y; x, y ≥ 0
13. L.H.S. = tan–1 1 11 1+ − −+ + −
x xx x
= tan–1 1 2 1 21 2 1 2+ − −+ + −
cos coscos cos
θ θθ θ
Let x
x
x
=
⇒ =
⇒ =
−
−
cos
cos
cos
2
2
12
1
1
θ
θ
θ
= tan–1 2 2
2 2
2 2
2 2
cos sin
cos sin
θ θ
θ θ
−
+
1 2 2
1 2 2
2
2
+ =
− =
cos cos
cos sin
θ θ
θ θ;
= tan–1 22
(cos sin )(cos sin )
θ θθ θ−+
Dividing numerator and denominator by cos θ, we have
= tan–1 11−+
tantan
θθ
= tan–1 tan π θ4−
11 4−+
= −
tantan
tanθθ
π θ
= π4
– θ = π4
12
− cos–1 x = R.H.S. [ tan–1 (tan A) = A(Hence proved)
14. a b y c z
a x b c z
a x b y c
− −− −− −
= 0
[Applying R2→ R2 – R1; R3 → R3 – R1
a b y c z
x y
x z
− −−−
00
= 0
Expanding along R1, we have
⇒ a(yz) – (b – y) (–xz) + (c – z) (xy) = 0⇒ ayz + bxz – xyz + cxy – xyz = 0⇒ ayz + bxz + cxy = 2xyz
Dividing both sides by xyz, we have
ayzxyz
bxzxyz
cxyxyz
xyzxyz
+ + = 2
ax
by
cz
+ + = 2Or
Let A = 1 22 1−
A = I A [By Elementary Row Transformation
1 22 1−
= 1 00 1
A
1 20 5−
= 1 02 1−
A [R R R2 2 12→ −
1 20 1
= 1 02
5
1
5−
R R2 21
5→ −
1 0
10
=
1
5
2
52
5
1
5−
A [R R R1 1 22→ −
Now, I = A–1 A
∴ A–1 =
1
5
2
52
5
1
5−
15. x = a(cos t + t sin t)
Differentiating both sides w.r.t. t, dxdt
= a(– sin t + t cos t + sin t . 1) = a(t.cos t) …(i)
Again differentiating both sides w.r.t. t, 2
2d x
dt
204 ■ shiv das senior secondary series (Xii)
= a[cos t . 1 + t (– sin t)] = a[cos t – t sin t] y = a(sin t – t cos t)
Differentiating both sides w.r.t. t, dydt
= a[cos t – (cos t . 1 + t(– sin t))] = a(t . sin t) …(ii)
Differentiating both sides w.r.t. t, 2
2d y
dt
= a[t . cos t + sin t]
Now, dydx
= dy dtdt dx
×
= at sin t × 1cosat t
[From (i) and (ii)]
= tan t
Differentiating w.r.t. x, 2
2d y
dx
= sec2 t . dtdx
= sec2 t × 1cosat t
= cos31
at t
16. y = cos(x + y) …(i)
Differentiating both sides w.r.t. xdydx
= – sin(x + y) . 1 +
dydx
dydx
= – sin(x + y) – sin(x + y) dydx
dydx
+ sin(x + y) dydx
= – sin(x + y)
[1 + sin(x + y)] dydx
= – sin(x + y)
dydx
= − ++ +sin( )sin( )
x yx y1 …(ii)
Since tangent is parallel to the line x + 2y = 0,
∴ Slope of tangent = − 12
…[From (i) and (ii)
∴ − ++ +sin( )sin( )
x yx y1 = − 1
2 2 sin(x + y) = 1 + sin(x + y) sin(x + y) = 1 …(iii)Since sin2 θ + cos2 θ = 1
sin2(x + y) + cos2(x + y) = 1 (1)2 + (y2) = 1 [From (i) and (iii)
y2 = 0 y = 0From (i), y = cos(x + y) 0 = cos(x + 0) (y = 0)
cos x = 0
∴ x = − 32π , [–2π ≤ x ≤ 0 (given)
∴ Point is −
32
0π ,
∴ Equation of tangent at −
32
0π , is
(y – 0) = −
+1
232
x π
2y = –x – 32π
4y = –2x – 3π∴ 2x + 4y + 3π = 0 is the required equation
of tangent.
17. x
x x
++ −∫ 5
3 13 102 dx
= 16
6 53 13 102
( )x
x x
++ −∫ dx
= 16
6 30 13 133 13 102
( )x
x x
+ + −+ −∫ dx
= 16
6 13
3 13 10176 3 13 102 2
( )x dx
x x
dx
x x
++ − + −∫ ∫+
Let p x x
dp x dx
= + −= +
3 13 10
6 13
2
( )
= 16
176 3 13
3103
2
dpp
dx
x x∫ ∫+
× + −
= 16
1718 13
3136
136
103
22 2log| |p dx
x x
++ +
−
−
∫
= 16
1718 13
6176
2 2log| |p dx
x
++
−
∫
mathematics – 2017 comptt. (delhi) ■ 205
= 16
1718
12 17
6
log| |p + × × logx
x
+ −
+ +
136
176
136
176
+ c
dxx a a
x ax a
c2 21
2−
= −+
+∫ log c
= 16
16
6 13 176 13 17
log| | logp cxx
+ ++ −+ +
= 16
16
6 46 30
3 13 102log| | logx x cxx
+ − + +−+
= 16
16
2 3 26 5
3 13 102log| | log ( )( )
x x cxx
+ − + +−+
= 1
6
1
6
3 2
3 53 2 13 10log| | ( )
( )x x c
x
x+ − ++ −
+
= 16
16
3 25
3 13 102log| | logx x xx
+ − + −+
− +16
3log c
= 16
16
3 25
3 13 102log| | logx x xx
+ − + +−+
K
… Where K = c – 16
log 3
Or
142
0
4
cos sin
/
x x+ 2∫π
dx
Dividing numerator and denominator by cos2 x, we have
= sec
tan
/ 2
20
4
1 4
x dx
x+∫π
= 12 1 2
0
2dp
p+∫
Let
When
When
p x
dp x dx
x p
x p
=
=
= =
= =
2
12
2
42
0 0
2
tan
sec
,
,
π
= 12
10
2tan ( )−
p
dxx
x c1 2
1+
= +
−∫ tan
= 1
2[tan–1(2) – tan–1 (0)]
= 1
2tan–1 2 tan− =
1 0 0
18. Let x
x x xx
x
2
2 21 1 1 1( )( )− + −++
= +A B C …(i)
x
x x
x x x
x x
2
2
2
21 11 1
1 1( )( )) ( )( )
( )( )− ++ + + −
− += A( B C
x2 = A(x2 + 1) + (Bx + C) (x – 1)
x2 = A(x2 + 1) + Bx(x – 1) + C(x – 1)Comparing the coefficients of x2, x and
constant term respectively,1 = A + B; 0 = –B + C; 0 = A – C1 = 2C ⇒ B = C ⇒ A = C 12
= C …(ii) …(iii)
[From (ii & (iii)]
∴ A = B = C = 12
Putting the value of A, B, C in (i), we get
x
x x x
x
x
2
2 21 1
12
1
12
121( )( )− + −
+
+= +
x
x x
2
21 1( )( )− +∫ dx
= 12 1
12
112
dxx
x
x−++
+∫ ∫ dx
= 12 1
12 1
12 12 2
dxx
x
x
dx
xdx
− + ++ +∫ ∫ ∫
Let p x
dpx dx
= +
=
2 1
2
= 12 1
14
12 1 2
dxx
dpp
dx
x− ++ +∫ ∫ ∫
= 12
14
12
1 1log| | log| | tanx p x c− + + +−
dxx
x1 2
1+
=
−∫ tan
= 12
14
12
1 12 1log| | log( ) tanx x x c− + + + +−
19. x cos yx
dydx
= y cos
yx
+ x
dydx
yyx
x
xyx
=+cos
cos
206 ■ shiv das senior secondary series (Xii)
⇒ dydx
yx y
x
= +
1
cos Let y vx
yx
v
dydx
v xdvdx
= ⇒ =
∴ = +
. .1
⇒ v + x dvdx v
v= + 1cos
⇒ xdvdx v
= 1cos
⇒ cos v dv = dxx
⇒ cos v dvdx
x=∫ ∫
⇒ sin v = log |x| + log |C|
⇒ sin yx
= log |cx|
20. Let A, B, C, D be the given points.
AB→
= ( 5 7ˆ ˆ ˆi x j k+ + ) – ( 4 3 3ˆ ˆ ˆi j k+ + ) = ˆ ( ) ˆ ˆi x j k+ − +3 4
AC→
= ( 5 3ˆ ˆi j+ ) – ( 4 3 3ˆ ˆ ˆi j k+ + ) = ˆ ˆi k− 3
AD→
= ( 7 6ˆ ˆ ˆi j k+ + ) ( 4 3 3ˆ ˆ ˆi j k+ + ) = 3 3 2ˆ ˆ ˆi j k+ −
The given points are coplanar if AB AC AD→ → →
, , are coplanar
∴ ( , , )AB AC AD→ → →
= 0
1 3 41 0 33 3 2
x −−−
= 0
Expanding along R1⇒ 1(9) – (x – 3) (–2 + 9) + 4(3) = 0⇒ 9 – 7x + 21 + 12 = 0⇒ –7x = –42⇒ x = 6
21. 1
37 14
23
1− − −= =x y
pz
= 0 …(i)
⇒ x y
pz−
−− −= =1
32
27
31
Direction ratios of line (i) are –3, 27p , 1.
7 73
51
117
− − −= =xp
y z …(ii)
⇒ xp
y z−−
−−
−−
= =137
51
117
Direction ratios of line (ii) are − 37
p , –1, –7. Lines are at right anglesUsing a1a2 + b1b2 + c1c2 = 0
− + − + −−
3 1 1 737
27
p p ( ) ( ) = 0
9 2 497
p p− − = 0
⇒ 7p – 49 = 0⇒ 7p = 49⇒ p = 7
OrThe required equation of plane is
(x + y + z – 1) + p(2x + 3y + 4z – 5) = 0⇒ (1 + 2p)x + (1 + 3p)y
+ (1 + 4p)z – 1 – 5p = 0⇒ (1 + 2p)x + (1 + 3p)y + (1 + 4p)z = 1 + 5p …(i)Dividing both sides by (1 + 5p), we have
⇒ ( ) ( ) ( )1 2
1 51 31 5
1 41 5
++
++
++
+ +p xp
p yp
p zp
= 1
⇒ x
pp
ypp
zpp
1 51 2
1 51 3
1 51 4
++
++
++
+ + = 1
2(y-intercept) = 3(z-intercept) …(Given)
2 31 51 3
1 51 4
++
++
=pp
pp p ≠ −
15
3(1 + 3p) = 2(1 + 4p)⇒ 3 + 9p = 2 + 8p⇒ 9p – 8p = 2– 3⇒ p = –1Putting the value of p in (i), we have⇒ (1 – 2)x + (1 – 3)y + (1 – 4)z = 1 – 5⇒ –x – 2y – 3z = –4or x + 2y + 3z = 422. 4x + y ≥ 80; x + 5y ≥ 0, 3x + 2y ≤ 150Let 4x + y = 80; x + 5y = 115; 3x + 2y = 150
x 0 20 15 2y 80 0 20 72
x 0 115 40y 23 0 15
x 0 50 40y 75 0 15
mathematics – 2017 comptt. (delhi) ■ 207
C(40, 1
5)
A
(15, 20)
80
90100
70
60
50
40
30
20
10
0 XX′Y′
Y
B(2, 72)
3x + 2y = 150
x + 5y = 115
Solution set
4x + y = 80
10 20 30 40 50 60 70 80 90 100110 120
Corner points z = 6x + 3y
A(15, 20)B(2, 72)C(40, 15)
90 + 60 = 150 (Minimum)12 + 216 = 228240 + 45 = 285
Minimum value of Z is 150 at A (15, 20)i.e., x = 15, y = 20.23. Let E1 : Very hard working studentsE2 : Regular but not so hard working studentsE3 : Careless and irregular studentsA : Student who could not get good marks
in the examination
P(E1) = 1060
16
= , P(E2) = 3060
36
= ,
P(E3) = 20
6026
= ,
P(A/E1) = 0.002, P(A/E2) = 0.02
P(A/E3) = 0.20
By Bayes’ theorem,
P(E3/A) = P E P A E
P E P A E P E P A EP E P A E
( ) ( / )[ ( ) ( / )] [ ( ) ( / )]
[ ( ) ( /
3 3
1 1 2 2
3 3
×× + ×
+ × ))]
=
26
0 20
16
0 00236
0 0226
0 20
×
×
+ ×
+ ×
.
. . .
= 0 40
0 002 0 06 0 40.
. . .+ + = 0 40
0 462..
= 400462
= 200231
24. See Q. 25 (Or), 2017 Comptt. (I Outside Delhi). [Page 226
OrFor commutativeLet (a, b), (c, d) ∈ A(a, b) (c, d) = (a + c, b + d)(c, d) (a, b) = (c + a, d + b) = (a + c, b + d)∴ (a, b) (c, d) = (c, d) (a, b)∴ is commutative.For associative :Let (a, b), (c, d), (e, f) ∈ A((a, b) (c, d)) (e, f) = (a + c, b + d) (e, f) = (a + c + e, b + d + f)(a, b) ((c, d) (e, f) = (a, b) (c + e, d + f) = (a + c + e, b + d + f)∴ {(a, b) (c, d)} (e, f) = (a, b) {(c, d) (e, f)}∴ is associative.For (a, b) ∈ R × R, if (c, d) ∈ R × R is identity
element, then(a, b) (c, d) = (c, d) (a, b) = (a, b)(a + c, b + d) = (c + a, d + b) = (a, b)∴ a + c = a and b + d = b c = 0 and d = 0⇒ (0, 0) R × R is the identity element.
25.
2 11 03 4 3 2
−
−
×
A =
− − −− −
×
1 8 101 2 59 22 15 3 3
From the above equation, order of A should be 2 × 3.
Let A = a b c
d e f
2 11 03 4
−
−
a b c
d e f
=
− − −− −
1 8 101 2 59 22 15
2 2 2
3 4 3 4 3 4
a d b e c f
a b c
a d b e c f
− − −
− + − + − +
= − − −
− −
1 8 101 2 59 22 15
By equality of matrices, we have
208 ■ shiv das senior secondary series (Xii)
A
B C
D
x
x
h
α
A
B
D
C
FEx
x
y
(h – y)h
r
a = 1, b = –2, c = –52a – d = – 1 2b – e = – 8 2c – f = – 102(1) + 1 = d 2(– 2) + 8 = c 2(– 5) + 10 = f3 = d 4 = e 0 = f
∴ A = 1 2 5
3 4 0
− −
26. f (x) = sin x + cos x, 0 ≤ x ≤ 2π
∴ f ′(x) = cos x – sin x f (x) = 0, ∴ cos x – sin x = 0⇒ – sin x = – cos x ⇒ tan x = 1
∴ x = π π4
54
, as 0 ≤ x ≤ 2π
0 π4
π2
π 2π54π 3
2π
The points x = π4
and x = 54π , divide the
interval [0, 2π] into 3 disjoint intervals —
04
, π
,
π π4
54
,
,
54
2π π,
Intervals Testvalue
Sign of f ′(x) Nature of function
04
, π
x = π6
cos π6
– sin π6
= 32
12
− > 0
f is strictly increasing
π π4
54
,
x = π2
cos π2
– sin π2
= 0 – 1 < 0
f is strictly decreasing
54
2π π,
x = 3
2π cos 3
2π – sin 3
2π
= 0 + 1 > 0
f is strictly increasing
Hence f is strictly increasing in the intervals
04
,π
∪
54
2π π,
and f is strictly decreasing
in the interval π π4
54
,
.
Or1st part :Let cylinder of radius x
and height y is inscribed in a cone of radius r.
ΔABC ~ ΔADF(AA rule)
∴ ABAD
BCDF
= (sides are proportional)
h yh
xr
− =
r h yh
( )− = x …(i)
Vol. of cylinder, V = πx2y
V = π r
h
2
2 .(h – y)2y [From (i)]
ddyV
= πr
h
2
2 [(h – y)2 . 1 + y.2(h – y).(–1)]
= πr
h
2
2 (h – y) [h – y – 2y]
= πr
h
2
2 (h – y) (h – 3y)
For maximum, ddyV
= 0
πr
h
2
2 (h – y) = 0 h – 3y = 0
h – y = 0 ⇒ –3y = –h
⇒ h = y, which ⇒ y = h3
is not possible.
d
dy
r
h
2
2
2
2V = π [(h – y) (–3) + (h – 3y) (–1)]
d
dy at yh
2
2
3
V
=
= πr
h
h hh h2
2 333
3 1− − + − −
( ) ( )
= πr
h
h2
223
3
−( ) < 0 (–ve)
Volume is maximum at y = 13
h.
∴ Height of cylinder = 13
(Height of cone)2nd part :Let x be the radius of the
cylinder which is inscribed in the cone of given height ‘h’ and semi-verticle angle ‘α’.
In rt. ΔABC,
tan α = BCAB
tan α = xAB
mathematics – 2017 comptt. (delhi) ■ 209
OD B
A12
2, −
C12
2, −
4x2 + 4y2 =
9 y2 = 4x
x = 0
x = 12
x = 32
X′
Y′
X
Y
•
• ••
•
⇒ 4x2 + 16x – 9 = 0⇒ 4x2 + 18x – 2x – 9 = 0⇒ 2x(2x + 9) – 1(2x + 9) = 0
⇒ x = − 92
(rejected) or x = 12
Putting x = 12
in (i), we get
y2 = 412
= 2 ∴ y = ± 2
∴ Points of intersection are
12
12
2 2, ,
−and .
From equation (i), y = 4x = 2x1/2
From equation (ii), y = 94
2− x
= 32
22
− x
Area of the bounded region OADCO,
A1 = 2 2 2 1 2
0
1 2
0
1 2
y dx x dx( ) ///
parabola = ∫∫
= 4 4 0
32
0
12
3 2
32
23
12
x
= −./
= 83
12
12
43
12
. .
=
= 4
3 222
4 26
2 23
. = = sq. units
AB = xtanα
= x cot α
Height of cylinder, BD = AD – AB = (h – x cot α) …(i)Volume of cylinder = π(BC)2 (BD) V = πx2(h – x cot α) [From (i)
V = πhx2 – πx3 cot α …(ii)Differentiating both sides w.r.t. x, we have
ddxV = 2πhx – 3πx2 cot α
For maxima or minima, ddxV = 0
πx(2h – 3x cot α) = 0 πx = 0 2h – 3x cot α = 0⇒ x = 0 –3x cot α = –2h
(rejected) x = 23
23
h hcot α
= tan α
d
dx
2
2V = 2πh – 6πx cot α
d
dx xh
2
2 23
V
at
= tanα = 2πh – 6π
23h tanα
cot α
= 2πh – 4πh = 2πh < 0 (–ve)
∴ Volume is maximum at x = 23h tan α.
Now from (ii),Volume of greatest cylinder,
V = π α π αh h h49
827
22
33tan tan
− cot α
= 49
827
32
32π πα αh htan tan−
= 12 827
3 2 3 2π α π αh htan tan−
= 427
πh3 tan2 α cu. units
27. Given curves are y2 = 4x …(i) 4x2 + 4y2 = 9
⇒ x2 + y2 = 94
…(ii)
Equation (ii) is a circle with centre (0, 0) and
radius = 32
.
Solving equations (i) and (ii), we get 4x2 + 4(4x) = 0
210 ■ shiv das senior secondary series (Xii)
Area of the bounded region ABCDA,
A2 = 2 21 2
3 2 22
1 2
3 232
y dx x dx(circle) = −∫ ∫
/
/
/
/
= 22
32
322 3
2
22
2
1
1 2
3 2
x xx
− + −( ) sin
/
/
= 22
94
98
23
2 1
1 2
3 2x xx− + −
sin
/
/
= 2 0 98
23
32
1+ −
sin .
− − +
−122
94
14
98
23
12
1sin .
= 2 198
14
84
98
13
1 1sin ( ) sin− −− −
= 2 298 2
14
98
13
1. sinπ − − −
= 98
22
94
13
1π − − −sin
∴ Total area of the bounded region OABCO = A1 + A2
= 2 23
98
22
94
13
1+ − − −π sin
= 26
98
94
13
1+ − −π sin Sq. units
28. Given lines : x y z− +
−−= =8
319
1610
7, and
x y z− + −−
= =383
298
55
Here x1 = 8, y1 = –19, z1 = 10 a1 = 3 b1 = –16 c1 = 7 a2 = 3 b2 = 8 c2 = –5The equation of plane containing given line is
x x y y z z
a b c
a b c
− − −1 1 1
1 1 1
2 2 2
= 0
x y z− + −
−−
8 19 103 16 73 8 5
= 0
Expanding along R1, we have (x – 8) (80 – 56) – (y + 19) (–15 – 21)
+ (z – 10) (24 + 48) = 0 24(x – 8) + 36(y + 19) + 72(z – 10) = 0Dividing both sides by 12 2(x – 8) + 3(y + 19) + 6(z – 10) = 0 2x – 16 + 3y + 57 + 6z – 60 = 0 2x + 3y + 6z – 19 = 0or 2x + 3y + 6z = 19 is the required
equation of the plane.
29. Here n = 8, p = P(a head) = 12
,
q = 1 12
12
− =
P(r) = nCr.qn – r pr
(i) P(exactly 5 heads) = P(5)
= 8C5q3p5
= 56 12
12
3 5
85
83
8 7 63 2 1
56C C= = =. .. .
= 56 18
132
× × = 732
(ii) P(at least 6 heads) = P(6) + P(7) + P(8) = 8C6q
2p6 + 8C7q1p7 + 8C8q
0p8
= 82
2 68
1
78
0
812
12
12
12
12
C C C
+
+
= 12
8
(8C2 + 8C1 + 8C0)
= 1256
8 72 1
81
1..+ +
= 37256
(iii) P(at most 6 heads) = P(0) + P(1) + … + P(6) = 1 – [P(7) + P(8)]
= 1 – [8C7 q1p7 + 8C8 q0p8]
= 1 – 81
1 78
0
812
12
12
C C
+
= 1 8 1 112
92568− + = − =( ) 247
256Or
p = P(red card) = 2652
12
= ; q = 1 12
12
− =
Also, n = 3
Here random variable X can take values 0,
1, 2, 3.
mathematics – 2017 comptt. (delhi) ■ 211
P(r) = nCr qn – r pr
P(X = 0) = 3C0 q3p0 = 12
3
= 1
8
P(X = 1) = 3C1 q2p1 = 3 12
12
2
= 38
P(X = 2) = 3C2 q1p2 = 3 12
12
2
= 3
8
P(X = 3) = 3C3 q0p3 = 12
3
= 18
Xi Pi PiXi PiXi2
0 18
0 0
1 38
38
38
2 38
68
128
3 18
38
98
ΣPiXi = 12
8 ΣPiXi
2 = 248
∴ Mean = ΣPiXi = 128= 3
2
∴ Variance = ΣPiXi2 – (Mean)2
= 248
32
2−
= 3 94
12 94
− = =− 34
or 0.75
set ii10.
A
3 1
C B
12 ˆ ˆi j+ µ11 3ˆ ˆi j−λ ˆ ˆi j+ 3
Using Section Formula,Position vector of point C = Position vector
of point C
3( 12 ˆ ˆi j+ µ ) + 1(λ ˆ ˆi j+ 3 )
3 + 1 = 11 3ˆ ˆi j−
36 3 34
ˆ ˆ ˆ ˆi j i j+ + +µ λ = 11 3ˆ ˆi j−
(36 + λ) i (3μ +3) j
4 4+ = 11 3ˆ ˆi j−
Equating the corresponding components
364+ λ = 11 3 3
4µ + = – 3
36 + λ = 44 3μ + 3 = –12 ⇒ 3μ = –15
λ = 8 μ = –5
11. 2 2x x−∫ dx
= − −∫ ( )x x2 2 dx
= − − + −∫ [( ) ]x x2 2 22 1 1 dx
= − − −∫ [( ) ]x 1 12 2 dx
= 1 12 2− −∫ ( )x dx
= 1
2
1
11 1 1 12 2 2 1( ) ( ) sinx x c
x− − − +
+− −
… a x dx x a x a c2 2 2 2 212
− = − +( ) +
∫
= ( )sin ( )
xxx x c− −− + +−1
212
12 2 1
12. lim ( )x
f x→0
= lcosim
x
x
x→
−0 2
1 4
= lsinim
x
x
x→0
2
22 2
1 22
2− =
cos sinAA
= lsinim
x
x
x→
×0
2
24 2 2
4
= 8 lsinim
xx
xx→
→
0
2 0
222
= 8 × 1 limsin
A
AA→
=0
1
= 8
At x = 0, f (x) = p ⇒ f (0) = p
Since f (x) is continous at x = 0, (Given)
∴ lim ( )x
f x→0
= f (0)
∴ 8 = p
212 ■ shiv das senior secondary series (Xii)
20. 3x + 4y ≤ 60, x + 3y ≤ 30Let 3x + 4y = 60 Let x + 3y = 30
x 0 20 12 y 15 0 6
x 0 30 12 y 10 0 6
Corner Points
Z = 8000x + 12000y
A(0,0)B(0, 10)C(12, 6)D(20, 0)
01,20,00096,000 + 72,000 = 1,68,000 (Maximum)1,60,000
X′ X
Y
Y′
60
50
40
20
30
10
(0, 0) A 10 20 30 40 50 60
B
D(20, 0)
C(12, 6)(0, 10)
3x + 4y = 60
x + 3y = 30 →
Maximum value of Z is 1,68,000 at C(12, 6)
i.e., x = 12, y = 6.
21. x
x x
++ +∫ 7
3 25 282 dx
= 16
6 73 25 282
( )x
x x
++ +∫ dx
= 16
6 42 25 253 25 282
( )x
x x
+ + −+ +∫ dx
= 16
6 25
3 25 28176 3 25 282 2
( )x dx
x x
dx
x x
++ + + +∫ ∫+
Let p = 3x2 + 25x + 28 dp = (6x + 25) dx
= 16
176 3 25
3283
2
dpp
dx
x x∫ ∫+
× + +
= 16
1718 25
3256
283
256
22 2log| |p dx
x x
++ +
+ −
∫
= 16
1718 25
6176
2 2log| |p dx
x
++
−
∫
= 16
1718
1
2176
256
176
256
176
log| | logp cx
x+ +×
×
+ −
+ +
dxx a a
x ax a
c2 21
2−= −
++
∫ log
= 16
16
6 25 176 25 17
log| | logp cxx
+ ++ −+ +
= 16
16
2 3 46 7
log| | log ( )( )
p cxx
+ +++
= 16
16
3 47
3 25 282log| | logx x xx
+ + + ++
− +16
3log c
= 16
16
3 47
3 25 282log| | logx x xx
k+ + + ++
+ ,
where k = c – 16
log 3
22. x = 3 cos t – 2 cos3 t
Differentiating w.r.t. t, we have
dxdt
= –3 sin t + 6 cos2 t sin t
= 3 sin t(2 cos2 t – 1) = 3 sin t . cos 2t [ cos cos 2 1 22 t t− =
…(i) y = 3 sin t – 2 sin3 t
Differentiating w.r.t. t, we have
dydt
= 3 cos t – 6 sin2 t . cos t
= 3 cos t(1 – 2 sin2 t)
= 3 cos t . cos 2t [1 – 2 sin2 t = cos 2t]
…(ii)
From (i) and (ii), dydx
= dydt
dtdx
×
= 3 cos t cos 2t × 13 2sin . cost t
∴ dydx
= cot t
Differentiating both sides w.r.t. x, we have
d y
dx
2
2 = – cosec2 t . dtdx
mathematics – 2017 comptt. (delhi) ■ 213
= −×
1 13 22sin sin cost t t
[From (i)
= − 1
3 23sin . cost t
28. X = 1 2 34 5 6 2 3
×
= − − −
×
7 8 92 4 6
11 10 9 3 3
From the above equation, order of X should be 3 × 2.
Let X =
a b
c d
e f
a b
c d
e f
1 2 34 5 6
=
− − −
7 8 92 4 6
11 10 9
a b a b a b
c d c d c d
e f e f e f
+ + ++ + ++ + +
4 2 5 3 64 2 5 3 64 2 5 3 6
= − − −
7 8 92 4 6
11 10 9
By equating the matrices, we havea + 4b = –7 c + 4 d = 2 e + 4f = 11 …(i) …(iii) …(v)2a + 5b = –8 2c + 5d = 4 2e + 5f = 10 …(ii) (iv) …(vi)From (i) and (ii), From (iii) and (iv), From (v) and (vi),a = 1 and b = –2 c = 2 and d = 0 e = –5 and f = 4
∴ X =
1 2
2 0
5 4
−
−
29. Here x1 = – 3, y1 = 1, z1 = 5 x2 = – 1, y2 = 2, z2 = 5a1 = – 3, b1 = 1, c1 = 5 a2 = – 1, b2 = 2, c2 = 5
Now, 2 1 2 1 2 1
1 1 1
2 2 2
x x y y z z
a b c
a b c
− − −
= 2 1 03 1 51 2 5
−−
= 2(5 – 10) – 1(– 15 + 5) (Expanding along R1)
= –10 + 10 = 0
∴ Given lines are coplanar.Equation of plane containing given lines is
( 3) 1 53 1 51 2 5
x y x− − − −−−
= 0
⇒ (x + 3) (5 – 10) – (y – 1) (– 15 + 5) + (z – 5) (– 6 + 1) = 0
( Expanding along R1)
⇒ – 5(x + 3) + 10(y – 1) – 5(z – 5) = 0⇒ – 5x – 15 + 10y – 10 – 5z + 25 = 0⇒ – 5x + 10y – 5z = 0or x – 2y + z = 0 is the required equation of
the plane.
set iii
10. lim ( )x
f x→π
4
= lim sin cos
x
x xx→
−−π π
44
= limsin cos
p
p p
p→
−
− −
−
−
0
4 4
44
π π
π π
Let
when
π
π
π
4
4
40
− =
− =
→
→
x p
p x
x
p
= limsin cos cos sin cos cos sin sin
p
p p p p
→
−
− +
0
4 4 4 4π π π π
π −− −4p π
= limcos sin cos sin
p
p p p p
p→
− − −
−0
12
12
12
12
4
= limsin
p
p
p→
−
−0
22
4 =
24 0
lim sin
p
pp→
= 24
limsin
p
pp→
=
01
At x = π4
, f (x) = k ⇒ f π4
= k
Since f (x) is continuous at x = π4
, …(Given)
lim ( )x
f x f→
=
π
π
44
214 ■ shiv das senior secondary series (Xii)
∴ 24
= k
11. dx
x x2 4−∫
= dx
x x2 2 24 2 2− + −∫
= dx
x( )− −∫2 22 2
= log |( ) ( ) |x x c− + − − +2 2 22 2
dx
x ax x a c
2 22 2
−= + − +
∫ log | |
= log | |x x x c− + − +2 42
12. dydx x
y+ 2 = x
Here ‘P’ = 2x
, ‘Q’ = x
I.F. = e e e e xdx xdx x xP∫ ∫
= = = =2
2 22log| | log
Here the solution is y(I.F) = Q I.F.( ) dx∫ y(x2) = x x dx. 2∫ y(x2) = x dx3∫ y(x2) = x c
4
4+
y = x
x
c
x
4
2 24+
y = xcx
22
4+ −
20. x + y ≤ 50, 2x + y ≤ 80Let x + y = 50; Let 2x + y = 80
x 0 50 30y 50 0 20
x 0 40 30y 80 0 20
Corner points Z = 105x + 90y
A(0, 50)B(30, 20)C(40, 0)D(0, 0)
0 + 4500 = 45003150 + 1800 = 4950 (Maximum) 4200 + 0 = 42000 + 0 = 0
80
•
70
60
50
40
30
20
10
0 10 20 30 40 50 60 70 80X′ X
Y′
Y
A(0, 50)
B(30,
0)C(4
0, 20
)
D(0, 0)
2x + y = 80
x + y = 50
Maximum value of Z is 4950 at B (30, 20)i.e., x = 30, y = 20.
21. x
x x
+
− +∫ 3
5 4 2 dx
= 12
2 3
4 52
( )x
x x
+
− +∫ dx
= 12
2 6 4 4
4 52
( )x
x x
+ − +
− +∫ dx
= 12
2 4
4 5
102 4 52 2
( )x dx
x x
dx
x x
−
− + − +∫ ∫+
Let p x x
dp x dx
= − += −
2 4 5
2 4( )
= 12 4 2 5 2
52 2 2
dpp
dx
x x+∫ ∫ − + + −
= 12 2 1
1 22 2
5p dp dx
x− +∫ ∫ − +
/
( )
= 12
21
1 2 25 2 2 1× + − + − + +p x x c/ log ( ) ( )
dz
x ax x a c
2 22 2
+= + + +
∫ log
= x x2 4 5 5− + + log |( ) |x x x c− + − + +2 4 52
22. Ist method
L.H.S. = cot–1 1 sin 1 sin1 sin 1 sin
x xx x
+ + − + − −
mathematics – 2017 comptt. (delhi) ■ 215
cot–1 1 sin 1 sin 1 sin 1 sin1 sin 1 sin 1 sin 1 sin
x x x xx x x x
+ + − + + −× + − − + − −
= cot–1 1 sin 1 sin 2 1 sin 1 sin1 sin (1 sin )
x x x xx x
+ + − + + − + − −
= cot–1 2 2 cos2 sin
xx
+
= cot–1 2(1 cos )2 sin
xx
+
21 cos 2 cos2
sin 2 sin cos2 2
xx
x xx
+ = =
= cot–1
22 cos2
2 sin cos2 2
x
x x
= cot–1 2 2
cot x x = = R.H.S.
2nd Method
L.H.S. = cot–1 2 2
2 2
cos sin 2 sin cos2 2 2 2
cos sin 2 sin cos2 2 2 2
x x x x
x x x x
+ +
+ +
2 2
2 2
cos sin 2 sin cos2 2 2 2
cos sin 2 sin cos2 2 2 2
x x x x
x x x x
+ + −
− + −
= cot–1 cos sin cos sin
2 2 2 2
cos sin cos sin2 2 2 2
x x x x
x x x x
+ + − + − −
= cot–1 2 cos
2
2 sin2
x
x
= cot–1 2
cot x
= 2x = R.H.S.
28. See Q. 23, 2016 (I Delhi). [Page 141
Ans. a2π4
23
−
29. Let a, b, c be the direction ratios of the normal to the required plane.
Equation of plane through the point (–1, 3, 2) is a(x + 1) + b(y – 3) + c(z – 2) = 0 …(i) Plane (i) is ⊥ to plane x + 2y + 3z = 5∴ a + 2b + 3c = 0 …(ii) Plane (i) is ⊥ to plane 3x + 3y + z = 0 ∴ 3a + 3b + c = 0 …(iii)
[ a1a2 + b1b2 + c1c2 = 0
Solving (ii) and (iii), we get
a b c2 33 1
1 33 1
1 23 3
= =−
a b c2 9 1 9 3 6−
−− −
= =
⇒ a b c−
−− −
= =7 8 3
or a b c7 8 3= =−
= λ (Say)
∴ a = 7λ, b = –8λ, c = 3λ
Putting the values of a, b and c in (i), we get
7λ(x + 1) + (–8λ) (y – 3) + 3λ(z – 2) = 0Dividing both sides by λ, we have 7(x + 1) – 8(y – 3) + 3(z – 2) = 0 7x + 7 – 8y + 24 + 3z – 6 = 0⇒ 7x – 8y + 3z + 25 = 0 is the required
equation of the plane.
216 ■ shiv das senior secondary series (Xii)
set i
section A
Question numbers 1 to 4 carry one mark each.
Q.1. If the following function f (x) is continuous at x = 0, then write the value of k.
f (x) = sin
,
,
32 0
0
x
xx
k x
≠
=
Q.2. If A is an invertible matrix of order 2 and det (A) = 4, then write the value of det(A–1).
Q.3. If ( ) ( . )a b a b→ → → →× + =2 2 225 and | |a
→= 5 then write the value of | |b
→.
Q.4. Find 3
3 1x
x −∫ dx.
section B
Question numbers 5 to 12 carry 2 marks each. Q.5. Find the values of x and y from the following matrix equation :
25
7 33 41 2
7 615 14
x
y −
+
−
=
Q.6. If f (x) = sin 2x – cos 2x, find ′ f π
6.
Q.7. A particle moves along the curve 6y = x3 + 2. Find the points on the curve at which y-coordinate is changing 2 time as fast as x-cordinate.
Q.8. Find : x
x32 2−∫ dx.
Time allowed : 3 hours Maximum marks : 100
General InstructIons Same as in Delhi Board 2015.
cbse
mathematics – 2017Comptt. (outside deLHi)
mathematics – 2017 comptt. (outside delhi) ■ 217
Q.9. Find the general solution of the differential equation log dydx
= 3x + 4y.
Q.10. Find the integrating factor of the differential equation dydx
yx
y+ = +1 .
Q.11. If the vectors a i j k b i j k→ →
= + + = − −ˆ ˆ ˆ, ˆ ˆ ˆ3 2 and c i j k→
= + +λ ˆ ˆ ˆ7 3 are coplanar, then find the value of λ.
Q.12. An aeroplane can carry a maximum of 250 passengers. A profit of ` 1,500 is made on each executive class ticket and a profit of ` 1,000 is made on each economy class ticket. The airline reserves at least 25 seats for executive class. However, at least 3 times as many passengers prefer to travel by economy class than by executive class. Frame the Linear Programing Problem to determine how many tickets of each type must be sold in order to maximize the profit for the airline.
section c
Question numbers 13 to 23 carry 4 marks each. Q.13. A school wants to award its students for regularity and hardwork with a a total cash award
of ` 6,000. If three times the award money for hardwork added to that given for regularity amount to ` 11,000, represent the above situation algebraically and find the award money for each value, using matrix method.
Q.14. Find the real solution of the equation tan–1 11
12
−+
=
xx
tan–1 x, (x > 0).
Q.15. If y = (cos x)x + sin–1 3x , find dydx
.
Or
If y = (sec–1 x)2, then show that x2(x2 – 1) d y
dx
2
2 + (2x3 – x) dydx
= 2.
Q.16. Find the intervals in which the function given by f (x) = 2x3 – 3x2 – 36x + 7 is (a) Strictly increasing (b) Strictly decreasing
Q.17. Evaluate : x x
x
sincos1 2
0+∫
π dx.
Q.18. Find : x x
x x
2
21
1 2+ +
+ +∫ ( ) ( ) dx.
Or
Find : ( )x x x− − −∫ 3 3 2 2
dx.
Q.19. Find the general solution of the differential equation 2xy
dydx
= x2 + y2.
Q.20. If a b c→ → →
, and are mutually perpendicular vectors of equal magnitudes, find the angles
which the vector 2 a b c→ → →+ + 2 makes with the vectors a b c
→ → →, and .
218 ■ shiv das senior secondary series (Xii)
Q.21. Find the co-ordinates of the foot of perpendicular drawn from a point A(1, 8, 4) to the line joining the points B(0, –1, 3) and C(2, –3, –1).
Q.22. Solve the following Linear Programming problem graphically : Minimize : z = 3x + 9y When : x + 3y ≤ 60, x + y ≥ 10, x ≤ y, x ≥ 0, y ≥ 0 Q.23. Bag A contains 3 red and 2 black balls, while bag B contains 2 red and 3 black balls. A ball
drawn at random from bag A is transferred to bag B and then one ball is drawn at random from bag B. If this ball was found to be a red ball, find the probability that the ball drawn from bag A was red.
Or If A and B are two independent events, then prove that the probability of occurrence of
atleast one of A and B is given by 1 – P(A′).P(B ′).
section D
Question numbers 24 to 29 carry 6 marks each.
Q.24. If a + b + c ≠ 0 and
a b c
b c a
c a b = 0, then using properties of determinants, prove that
a = b = c.
Q.25. Given a non-empty set X, consider the binary operation : P(X) × P(X) → P(X) given by A B = A ∩ B, ∀ A, B ∈ P(X), where P(X) is the power set of X. Show that is commutative and associative and X is the identity element for this operation and X is the only invertible in P(X) with respect to the operation . (Not in syllabus)
Or
Let f : IR – −
43
→ IR be a function defined as f(x) = 43 4
xx +
. Show that f is a one-
one function. Also check whether f is an onto function or not. Hence f –1 in
(Range of f) → IR – −
43
.
Q.26. Show that the right circular cone of least curved surface and given volume has an altitude equal to 2 times the radius of the base.
Or (i) Find the equation of tangent to the curve x = a cos θ + a θ sin θ, y = a sin θ – a θ cos θ
at any point θ of the curve. (ii) Also show that at any point θ of the curve the normal is at a constant distance from
origin.
Q.27. Find the distance between the line x y z− − −= =53
43
81
and the plane determined by the
points A(2, –2, 1), B(4, 1, 3) and C(–2, –2, 5). Q.28. A coin is biased so that the head is 4 times as likely to occur as tail. If the coin is tossed
thrice, find the probability distribution of number of tails. Hence find the mean and variance of the distribution.
mathematics – 2017 comptt. (outside delhi) ■ 219
Q.29. Using method of integration find the area of the triangle ABC, co-ordinates of whose vertices are A(1, –2), B(3, 5) and C(5, 2).
Or Evaluate the following definite integral as limit of sums :
( )3 2 12
0
4
x x+ +∫ dx
set ii
Note: Except for the following questions, all the remaining questions have been asked in Set I.
Q.10. Find the general solution of the differential equation xy dydx
= (x + 2) (y + 2).
Q.11. Find : cos
sin
x dx
x8 2−∫ Q.12. The radius r of a right circular cylinder is increasing at the rate of 5 cm/min and its height
h, is decreasing at the rate of 4 cm/min. When r = 8 cm and h = 6 cm, find the rate of change of the volume of cylinder.
Q.21. Evaluate : x xx xtan
sec tan+∫0
π dx
Q.22. Solve the differential equation y e dx xe y dy y
xy
xy. . ( ) , ( )= + ≠2 0 .
Q.23. Find the co-ordinates of the foot of perpendicular drawn from the point (2, 3, –8) to the line 4
2 61
3− −= =x y z
.
Q.28. From a lot of 30 bulbs, which includes 6 defectives, 3 bulbs are drawn one by one at random, with replacement. Find the probability distribution of number of defective bulbs. Hence find the mean and variance of the distribution.
Q.29. Find the distance between the line r i j k i j k→
= + − + + +3 5 2 3 3ˆ ˆ ˆ ( ˆ ˆ ˆ)λ and the plane
determined by the points A(1, 1, 0), B(1, 2, 1) and C(–2, 2, –1).
set iii
Note: Except for the following questions, all the remaining questions have been asked in Set I and Set II.
Q.10. If the vectors ˆ ˆ ˆ, ˆ ˆ ˆ ˆ ˆ ˆi j k i j k i j k− + + + + −3 2 3and λ are coplanar, then find the value of λ.
Q.11. Find the particular solution of the differential equation dydx
y
x= +
+1
1
2
2, given that y(0) = 3 .
Q.12. A balloon, which always remains spherical, has a variable diameter 23
(3x + 1). Find the
rate of change of its volume with respect to x.
220 ■ shiv das senior secondary series (Xii)
Q.21. Solve the following linear programming problem graphically : Find the maximum value of z = 105x + 90y when x + y ≤ 50, 2x + y ≤ 80, x ≥ 20, x ≥ 0, y ≥ 0
Q.22. Find the particular solution of the differential equation 2xy + y2 – 2x2 dydx
= 0; y = 2
when x = 1.
Q.23. Evaluate : x dxx1
0+∫ sin
π.
Q.28. Find the co-ordinates of the point of intersection P of the line r i j k→
= − +2 2ˆ ˆ ˆ +
λ( 3 4 2ˆ ˆ ˆi j k+ + ) and the plane determined by points A(1, –2, 2), B(4, 2, 3) and C(3, 0, 2).
Q.29. A biased die is such that P(4) = 1
10 and other scores are equally likely. The die is tossed
twice. If X is the ‘number of fours obtained’, find the variance of X.
SolutionS
1. 0
lim ( )x
f x→
= 0
3sin
2limx
x
x→
= 0
30
2
33 2lim 32
2x
x
x
x→
→
[at x = 0, f(x) = k, f (0) = k
= 32
(1) = 32
θ→
θ = θ
0
sinlim 1
f (x) is continuous at x = 0. [Given]
⇒ 0
( ) (0)limx
f x f→
= .
∴ 32
= k
2. det (A–1) = 1det (A)
= 14
3. 2 2( ) ( . )a b a b→ → → →× + = 225
But 2 2( ) ( . )a b a b→ → → →× + = 2 2| | | |a b
→ →
∴ 2 2| | | |a b→ →
= 225
(5)2 2| |b→
= 225
2| |b→
= 22525
= 9 ∴ →
| |b = 3
4. 33 1
xx −∫ dx
= [(3 1) 1]3 1x
x− +−∫ dx
= 13 1
1x
−
+∫ dx
= x + 13
log|3x – 1| + c
5. 25 3 4 7 6
7 3 1 2 15 14x
y
− + = −
2 10 3 4 7 614 2 6 1 2 15 14
x
y
− + = −
2 3 6 7 615 2 4 15 14
x
y
+ = −
2x + 3 = 7 2y – 4 = 14 2x = 4 2y = 18∴ x = 2 ∴ y = 96. f (x) = sin 2x – cos 2xDifferentiating both sides w.r.t. x, we get f´ (x) = 2 cos 2x + 2 sin 2x
6
f π ′ = 2 cos
3π + 2 sin
3π
= 1 32 2
2 2
+ = +(1 3)
mathematics – 2017 comptt. (outside delhi) ■ 221
7. y-coordinate is changing 2 times as fast as x-coordinate.
∴ dydt
= 2 dxdt
…(i)
6y = x3 + 2
y = 3 2
6 6x + …(ii)
Differentiating both sides w.r.t. t
dydt
= 216
. 3 dxdt
x
2 dxdt
= 212
dxdt
x [From (i)]
x2 = 4 ⇒ x = ± 2
From (ii), when x = 2, y = 53
From (ii), when x = –2, y = –1
∴ Required points are
53
2, and (–2, –1).
8. 232
x
x−∫ dx
= – 12
dpp∫
= −
= − = −
2Let 32
2
2
p x
dp x dx
dpx dx
= 121
2p dp−
− ∫ =
121 2
2 1p c− × +
= 232 x− − + c
9. log dydx
= 3x + 4y
dydx
= e3x + 4y ⇒ dy
dx = e3x . e4y
⇒ dy
e y4 = e3x dx
e dy e dxy x−∫ ∫=4 3
e ey x−
−=
4 3
4 3 + c
10. 1dy ydx x
y ++ =
1dy ydx x x
y+ = +
1dy ydx x x
y+ − =
1 11dydx x x
y + − =
Comparing with dydx
+ Py = Q
‘P’ = 1 – 1x
and ‘Q’ = 1x
Integrating Factor (I.F)
= −∫ ∫ =
11 dx
Pdx xe e = −log| |orx
x x ee
x11. Vector , anda b c
→ → →are coplanar
1 3 12 1 1
7 3− −
λ= 0
Expanding along C, we have⇒ 1(–3 + 7) – 2(9 – 7) + λ (–3 + 1) = 0⇒ 4 – 4 – 2λ = 0 ⇒ 2λ = 0 ⇒ λ = 012. Passengers Profit per seat (`)Executive class x 1500 Economy class y 1000 250 (maximum) Let, no. of executive class tickets be x and
economy class tickets be y. Maximise, Z = 1500x + 1000ySubject to the constraints, x + y ≤ 250 x ≥ 25, y ≥ 3x, x, y ≥ 0 13. Let award money for regularity = ` xLet award money for hardwork = ` y x + y = 6000 x + 3y = 11000
1 11 3
x
y
= 6000
11000
A X = B A–1A X = A–1 B X = A–1 B …(i)
222 ■ shiv das senior secondary series (Xii)
|A| = 3 – 1 = 2 ≠ 0
adj A = 3 11 1
− −
As we know, A–1 = 1|A|
. adj A
= 12
3 11 1
− −
From (i), X = A–1 B
x
y
= 1
2
3 1 60001 1 11000
− −
= 1
2
18000 110006000 11000
− − +
= 1
2
70005000
= 35002500
∴ Award money for regularity, x = ` 3500Award money for hardwork, y = ` 2500
14. tan–1 11 1
12
−+
=
xx
, x > 0
⇒ tan–1 (1) – tan–1 (x) = 12
tan–1 x
tan tan tan− − −−+
= −1 1 11A B
ABA B
⇒ π4
12
= tan–1 x + tan–1 x
⇒ π4
32
= tan–1 x ⇒ tan–1 x = π4
23
×
⇒ x = tan π6
⇒ x =
13
15. Let A = (cos x)x …(i)Taking log on both sides log A = x.log (cos x)Differentiating both sides w.r.t. x
1 AA
ddx
= x . 1cos x
(– sin x) + log cos x
Addx
= A(log cos x – x tan x)
Addx
= (cos x)x (log cos x – x tan x)…(ii) [From (i)]
y = (cos x)x + sin–1 3x
Differentiating both sides w.r.t. x
dydx
= ddx x
ddx
xA + ×−
1
1 3 23
( )( ) …[From (ii)
= (cos x)x (log cos x – x tan x) +
11 3
12 3
3−
× ×x x
∴ dydx
= (cos x)x (log cos x – x tan x)
+ 32 (1 3 )x x−
Or y = (sec–1 x)2
Differentiating both sides w.r.t. x
dydx
= 2(sec–1 x) . 2
1
1x x −
x 2 1x − dydx
= 2 sec–1 x
Differentiating both sides w.r.t. x
x 2 1x −2
2 2
1.2 1
2d y dydxdx x
x x−
+ ×
+ 2 1x − dydx
.1 = 2.2
1
1x x −
Multiplying both sides by x 2 1x − , we have
x2 (x2 – 1) 2
32
d y dydxdx
x+ + x(x2 – 1) dydx
= 2
x2 (x2 – 1) 2
3 32 ( )d y dy
dxdxx x x+ + − = 2
x2 (x2 – 1) + −2
32 (2 )d y dy
dxdxx x = 2
(Hence proved)16. f (x) = 2x3 – 3x2 – 36x + 7Differentiating both sides w.r.t. x f´(x) = 6x2 – 6x – 36 = 6(x2 – x – 6) = 6(x2 – 3x + 2x – 6) = 6[x(x – 3) + 2(x – 3)] = 6(x + 2) (x – 3)When f´(x) = 0, x = –2, 3
mathematics – 2017 comptt. (outside delhi) ■ 223
Intervals Check-ing
points
Value of Sign of
f’(x)
Nature of f (x)
6 (x+2) (x–3)
(–∞, –2) –3 + – – > 0 st. increasing
(–2, 3) 1 + + – < 0 st. decreasing
(3, ∞) 4 + + + > 0 st. increasing
∴ f (x) is strictly increasing in (– ∞, –2) ∪ (3, ∞) f (x) is strictly decreasing in (–2, 3)
17. See Q. 17, 2017 (I Delhi). [Page 165
Ans. I = 2
4π
18. Let x x
x x x x x
2
2 21
1 2 1 1 2+ +
+ + + + += + +
( ) ( ) ( )A B C
…(i)
x x
x x
x x x x
x x
2
2
2
21
1 21 2 2 1
1 2+ +
+ ++ + + + + +
+ +=
( ) ( )( )( ) ( ) ( )
( ) ( )A B C
x2 + x + 1 = A (x2 + 3x + 2) + B (x + 2) + C (x2 + 2x + 1)Comparing the coefficients of x2, x and
constant term1 = A + C, 1 = 3A + B + 2C, 1 = 2A + 2B + C …(ii) …(iii) …(iv)Multiplying (iii) by 2 and (iv) by 1, we get 6A + 2B + 4C = 2 2A + 2B + C = 1 – – – – 4A + 3C = 1 …(v) Multiplying (ii) by 4 and (v) by 1, we get 4A + 4C = 4 4A + 3C = 1 – – – C = 3 …(v)
Putting the value of C in (ii) A + 3 = 1 ⇒ A = –2Putting the value of A and C in (iii) 3 (– 2) + B + 2 (3) = 1 ∴ B = 1Putting the value of A, B and C in (i),
x x
x x x x xdxdx
2
2 21
1 22
11
13
2+ +
+ +−+ +
++
∫ ∫= +( ) ( ) ( ) (
= – 2 log |x + 1| – 11x +
+ 3 log |x + 2| + C
Or
2( 3) 3 2x x x− − −∫ dx
= 21
22( 3) 3 2x x x
−− − − −∫ dx
= 21
2( 2 6 2 2) 3 2x x x
−− + − + − −∫ dx
= 21
2( 2 2) 3 2x x x
−− − − −∫ dx
21
2. 8 3 2x x− − −∫ dx
= − −
= − −
2Let 3 2
( 2 2 )
p x x
dp x dx
= 21
24 ( 2 3)p dp x x
−− − + −∫ dx
= 1
2 2 221
24 [ 2 1 3 1 ]p dp x x
−− − + + − −∫ ∫
dx
= 3/2 21 2
2 3. 4 [( 1) 4]p x
−− − + −∫ dx
= 3/2 2 21
34 2 ( 1)p x
−− − +∫ dx
= 3/2 2 11 1 1
3 2 24 ( 1) 4 ( 1) 4 sin
xp x x −− + − × + − + +
+ c
− − = − + + ∫ 2 2 2 2 2 11sin
2x
a x dx x a x a ca
= −
− − − + − −2 3/2 21
3(3 2 ) 2( 1) 3 2x x x x x
– 8 sin–1 + 12
x + c
19. 2xy dydx
= x2 + y2
dydx
= 22
2 2yx
xy xy+
dydx
= 1 12 2
yxy x+
= ⇒ = = +
Let
. 1 .
yy vx v
xdy dv
v xdx dx
v + x dvdx
= 1 12 2v
v+
x dvdx
= 12 2
vv
v+ −
224 ■ shiv das senior secondary series (Xii)
x dvdx
= 12 2
vv−
x dvdx
= 21
2vv
−
22
1
v
vdv
− = dx
x
22
1
v
vdv
−∫ = dxx
−∫ = −
=
2Let 1
2
p v
dp v dv
dpp∫ = dx
x−∫
log |p | = – log |x| + log |c|
log |p | = log cx
⇒ p = cx
±
x(v2 – 1) = c± ⇒ x2
2 1y
x
− = k
x2 2
2y x
x
−
= k ⇒ y2 – x2 = kx
20. Let | | | | | |a b c→ → →
= = = m (Let) …(i)
Since , ,a b c→ → →
are mutually perpendicular vectors
∴ .a b→ →
= 0, .b c→ →
= 0, .c a→ →
= 0
2|2 2 |a b c→ → →+ +
= (2 2 ) . (2 2 )a b c a b c→ → → → → →+ + + +
→ → →=
2| | .x x x
2|2 2 |a b c→ → →+ + = 4 . 2 . 4 .a a a b a c
→ → → → → →+ +
+ 2 . . 2 .a b b b b c→ → → → → →
+ +
+ 4 . 2 . 4 .c a c b c c→ → → → → →
+ +→ → →
→ → → →
=
= =
2. | |
Given: . . 0
a a a
a b b a
2|2 2 |a b c→ → →+ + = 2 2 24| | | | 4| |a b c
→ → →+ +
2|2 2 |a b c→ → →+ + = 4m2 + m2 + 4m2 [From (i)
2|2 2 |a b c→ → →+ + = 9m2
|2 2 |a b c→ → →+ + = 3m …(ii)
Let 2 2a b c→ → →+ + makes angles α, β, γ with
, anda b c→ → →
respectively, then
cos α = (2 2 )
| ||2 2 |
a a b c
a a b c
→ → → →
→ → → →+ +
+ +
→ →
→ →
θ =
.cos
| || |
a b
a b
= 2 . . 2 .
| |(3 )
a a a b a c
a m
→ → → → → →
→+ + [From (i) and (ii)
cos α = 22| | 0 0
| |(3 )
a
a m
→
→+ +
→ → →
→ → → →
=
= =
2. | |
. . 0
a a a
a b a c
= 22 2
(3 ) 3m
m m=
∴ α = cos–1
23
cos β = .(2 2 )
| ||2 2 |
b a b c
b a b c
→ → → →
→ → → →+ +
+ +
= 2 . . 2 .( ) (3 )
b a b b b cm m
→ → → → → →+ +
= 2
20 | | 0
3
b
m
→+ + =
2
2133
m
m=
∴ β = cos–1
13
cos γ = .(2 2 )
| ||2 2 |
c a b c
c a b c
→ → → →
→ → → →+ +
+ +
= 2 . . 2 .
( ) (3 )c a c b c c
m m
→ → → → → →+ +
= 2
20 0 2| |
3
c
m
→+ + =
2
22 2
33m
m=
∴ γ = cos–1 23
mathematics – 2017 comptt. (outside delhi) ■ 225
21.
•
•
•
A(1, 8, 4)
B D C
(0, –1, 3) (2, –3, –1)
Direction ratios of BC are 2 – 0, –3 + 1, – 1 – 3 2, –2, –4 or 1, –1, –2Equation of line BC is
11 1y yx x z za b c
−− −= =
10 31 1 2
yx z+− −− −
= = = p (let)
Let D(p, –p – 1, –2p + 3) …(i) be the general point of line BC.
Direction ratios of AD are p – 1, –p – 1 – 8, –2p + 3 – 4 p – 1, –p – 9, –2p – 1Given that BC ⊥ AD,Using a1a2 + b1b2 + c1c2 = 0
∴ (1(p – 1) – 1(–p – 9) – 2(–2p – 1) = 0 p – 1 + p + 9 + 4p + 2 = 0 6p + 10 = 0
p = 10 56 3
− −=
From (i), D = 5 5 103 3 3
, 1, 3−
− +
= 5 2 193 3 3
, ,−
22. x + 3y ≤ 60, x + y ≥ 10, x ≤ y Let x + 3y = 60; Let x + y = 10; Let x = y
x 0 60 15y 20 0 15
x 0 10 5y 10 0 5
x 0 20 5y 0 20 5
Corner points Z = 3x + 9y
A(0, 10)B(0, 20)C(15, 15)D(5, 5)
0 + 90 = 900 + 180 = 18045 + 135 = 18015 + 45 = 60 (Minimum)
Y
XX′
60
50
40
30
20
10
10 20 30 40 50 600
D(5, 5)
B(0, 2
0)
C(15, 1
5)
x + 3y = 60
x + y = 10
x = y
Y′
∴ Minimum value of Z is 60 at (5, 5) i.e. x = 5, y = 5 of the feasible region.
23. Let E1 : Red ball is transferred from bag A to bag B.
E2 = Black ball is transferred from bag A to bag B.
A = A red ball is drawn from bag B Bag A Bag B
Red Black 3 2
Red Black 2 3
P(E1) = 35
; P(A/E1) = 36
P(E2) = 25
; P(A/E2) = 26
By Bayes’ theorem,
P(E1/A) = ×× + ×
1 1
1 1 2 2
P(E ) P(A / E )[P(E ) P(A)E ] [P(E ) P(A)E ]
=
3 35 6
3 3 2 25 6 5 6
×
× + ×
= 99 4+
= 913
OrA and B are Independent events (Given)
∴ P(A ∩ B) = P(A) × P(B) …(i)P (at least one of A and B) = P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = P(A) + P(B) – P(A) × P(B) [From (i)
226 ■ shiv das senior secondary series (Xii)
= P(A) + P(B)[1 – P(A)] = 1 – P(A′) + P(B).P(A′)
+ =′⇒ = −′
P(A) P(A ) 1
P(A ) 1 P(A)
= 1 – P(A′) [1 – P(B)] = 1 – P(A′)P(B′) [Hence proved]
[ 1 P(B) P(B )− = ′
24.
a b c
b c a
c a b
= 0 [C1 → C1 + C2 + C3
a b c b c
a b c c a
a b c a b
+ ++ ++ +
= 0
Taking common (a + b + c) from C1
(a + b + c) 111
b c
c a
a b
= 0
(a + b + c) 100
b c
c b a c
a b b c
− −− −
= 0
2 2 1
3 3 1
R R R
R R R
→ − → −
(Expanding along C1)
⇒ (a + b + c) [(c – b) (b – c) – (a – b) (a – c)] = 0
⇒ (a + b + c) [bc – c2 – b2 + bc – a2 + ac + ba – bc) = 0
⇒ (a + b + c) (–a2 – b2 – c2 + ab + bc + ca) = 0
⇒ – (a + b + c) (a2 + b2 + c2 – ab – bc – ca) = 0
⇒ – 12
(a + b + c) (2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca) = 0
⇒ – 12
(a + b + c) (a2 + a2 + b2 + b2 + c2
+ c2 – 2ab – 2bc – 2ca) = 0⇒ (a + b + c) [(a2 + b2 – 2ab) + (b2 + c2
– 2bc) + (c2 + a2 – 2ca)] = 0⇒ (a + b + c) [(a – b)2 + (b – c)2
+ (c – a)2] = 0Either a + b + c = 0
or (a – b)2 + (b – c)2 + (c – a)2 = 0But a + b + c ≠ 0 …(Given)
(a – b)2 = 0 (b – c)2 = 0 (c – a)2 = 0 a – b = 0 b – c = 0 c – a = 0 a = b b = c c = a ∴ a = b = c Hence proved25. (i) CommutativeFor any two elements A, B ∈ P(X), we have A ∩ B = B ∩ A⇒ A B = B A for A, B ∈ P(X)Thus, is commutative on P(X).(ii) AssociativeFor any three elements A, B, C ∈ P(X), we
have (A B) C = (A ∩ B) C = (A ∩ B) ∩ Cand A (B C) = A (B ∩ C) = A ∩ (B ∩ C)Since (A ∩ B) ∩ C = A(B ∩ C)Thus, is associative on P(X).(iii) Identity elementLet E be the identity element in P(X) w.r.t.
. Then, A E = A = E A for all A ∈ P(X) A ∩ E = A = E ∩ A for all A ⊂ X E = XThus, X is the identity element with respect
to on P(X).(iv) Invertible elementLet A be an invertible element in P(X) andLet S be its inverse. Then A S = X = S A A ∩ S = X = S ∩ A A = S = XThus, X is the only invertible element in
P(X) with respect to and it is the inverse of itself.
Or
f (x) = 43 4
xx +
For f is one-one—Let x1, x2 ∈ R F(x1) = F(x2)
+ +
=1 2
1 2
4 43 4 3 4
x xx x
⇒ x1(3x2 + 4) = x2(3x1 + 4)⇒ 3x1x2 + 4x1 = 3x1x2 + 4x2⇒ 4x1 = 4x2⇒ x1 = x2∴ f is one-one.
mathematics – 2017 comptt. (outside delhi) ■ 227
For f is onto —1st methodLet y be any element of R.
Let y = f (x)
y = 43 4
xx +
⇒ 3xy + 4y = 4x⇒ 3xy – 4x = – 4y⇒ x(3y – 4) = – 4y
⇒ x = 43 4
yy−−
or 44 3
yy−
We have, f (x) = 43 4
xx +
f 44 3
yy
−
=
44
4 3
43 4
4 3
yy
yy
−
+ −
= 1612 16 12
yy y+ −
= 1616
y
f (x) = y ∴ f is onto
f –1 y = 44 3
yy−
26. Let radius of the base of cone = xLet height of the cone = yLet volume of the cone
= 13
π k (a constant)
13
πx2y = 13
πk π
1 2Volume of cone = 3
r
x2y = k …(i)
y = 2k
x …(ii)
Curved Surface Area of cone, S = πrl
S = πx 2 2x y+
S = πx 2
24
k
xx + [From (ii)
S2 = π2x2 2
24
k
xx
+ [Squaring both sides
Let S2 = z ∴ z = π2x4 + π2k2x–2
Differentiating both sides w.r.t. x, we get
dzdx
= 4π2x3 – 2π2k2x–3 …(iii)
For stationary points, dzdx
= 0
4π2x3 – 2 2
32 k
x
π = 0
2 6 2 2
34 2x k
x
π − π = 0
4π2x6 – 2π2k2 = 0 4π2x6 = 2π2k2
2x6 = k2
⇒ 2x6 = (x2y)2 [From (i) 2x6 = x4y2
2x2 = y2
2 x y=Differentiating (iii) both sides w.r.t. x, we
get
2
2d z
dx = 12π2x2 + 6π2k2x–4
22 2 22
2 4at
2
62
4
12 vey
x
yd z k
dx y=
π
= π + = +
∴ z is minimum at x = y/ 2
and S2 is minimum at x = y/ 2
∴ S is minimum at x = 2
y or y = 2 x
∴ Curved Surface Area of cone is least
when y = 2 x.
or Altitude = 2 (radius of the base)Or
Part I : x = a cos θ + aθ sin θDifferentiating both sides w.r.t. θ, we getdxdθ
= – a sin θ + aθ cos θ + a sin θ = aθ cos θ
y = a sin θ – aθ cos θDifferentiating both sides w.r.t. θ, we get
θdyd
= a cos θ – (aθ sin θ + a cos θ) = aθ sin θ
y
x
2nd method
For y = 43
There is no X
such that
f (x) = 43
∴ f is not
invertible
But f : R – 4
4 3y
y−Range of f is onto
so invertible and
f–1(y) = 4
4 3y
y−
228 ■ shiv das senior secondary series (Xii)
Slope of tangent, dydx
= dyd
ddxθθ×
= aa
θ θθ θ
θθ
sincos
sincos
× =1
Equation of tangent is (y – y1) = slope of tangent (x – x1) y – (a sin θ – aθ cos θ)
= sincos
θθ
[x – (a cos θ + aθ sin θ)]
⇒ sin θ[x – a cos θ – a θ sin θ] = cos θ[y – a sin θ + aθ cos θ]⇒ x sin θ – a sin θ cos θ – aθ sin2 θ = y cos θ – a sin θ cos θ + a cos2 θ⇒ x sin θ – y cos θ = aθ(sin2 θ + cos2 θ)⇒ x sin θ – y cos θ = aθ [ sin2 θ + cos2 θ = 1]
Part II :
Slope of normal = − 1tan θ
= –cot θ = – cossin
θθ
Equation of normal is y – y1 = Slope of normal (x – x1) y – (a sin θ – aθ cos θ)
= – cossin
θθ
[x – (a cos θ + aθ sin θ)]
y – sin θ – a sin2 θ + aθ cos θ sin θ = –x cos θ + a cos2 θ + aθ cos θ sin θ x cos θ + y sin θ – a(sin2 θ + cos2 θ) = 0 x cos θ + y sin θ – a = 0Perpendicular distance from the origin, (0, 0)
= 0 02 2
(cos ) (sin )
cos sin
θ θ
θ θ
+ −
+
a
⊥ = + +
+
distanceax by c
a b1 1
2 2
= − a
1 = a, which is a constant
27. Given : Points A(2, –2, 1), B(4, 1, 3) and C(–2, –2, 5)
Using equation of plane
1 1 1
2 1 2 1 2 1
3 1 3 1 3 1
x x y y z z
x x y y z z
x x y y z z
− − −− − −− − −
= 0
2 2 14 2 1 2 3 12 2 2 2 5 1
x y z− + −− + −
− − − + − = 0
2 2 1
2 3 24 0 4
x y z− + −
− = 0
(expanding along R1)
⇒ (x – 2) (12 – 0) – (y + 2) (8 + 8) + (z – 1) (0 + 12) = 0
⇒ 12x – 24 – 16y – 32 + 12z – 12 = 0⇒ 12x –16y + 12z – 68 = 0⇒ 3x – 4y + 3z – 17 = 0 …(i)
(Dividing both sides by 4)
D.r’s of normal to the plane (i) are 3, –4, 3.D.r’s of the given line are 3, 3, 1.Now, a1a2 + b1b2 + c1c2
= 3(3) – 4(3) + 3(1) = 9 – 12 + 3 = 0∴ Normal to the plane ⊥ to the given line So given line is parallel to the plane (i).Required distance = Distance between point
(5, 4, 8) and plane (i) Given line (5, 4, 8)
Plane (i)
= 1 1 12 2 2
ax by cz d
a b c
+ + +
+ +
= 3(5) 4(4) 3(8) 179 16 9
− + −+ +
= = = = = − = = −
1 1 1Here 5, 4, 8
3, 4, 3, 17
x y z
a b c d
= 15 16 24 1734
− + −
= 6 34 6 343434 34
× = = 3 3417
units
28. P(a head) = 4P(a tail) P(H) = 4P(T) …(i) P(H) + P(T) = 1 4P(T) + P(T) = 1 [From (i)
5P(T) = 1 P(T) = 15
Let p = 15
, q = 1 – 15
= 45
, n = 3
mathematics – 2017 comptt. (outside delhi) ■ 229
Here random variable X (no. of tails) can take values 0, 1, 2, 3.
Using P(r) = nCr.qn– rpr
P(X = 0) = 3C0q3p0 = q3 =
34 645 125
=
P(X = 1) = 3C1q2p1 = 3q2p = 3
24 1 485 5 125
=
P(X = 2) = 3C2qp2 = 3qp2 = 3 4 1 125 5 125
=
P(X = 3) = 3C3q0p3 = p3 =
31 15 125
=
Probability distribution is
Xi 0 1 2 3P(Xi) 64
12548
12512125
1125
Xi Pi PiXi PiXi2
0 64125
0 0
1 48
125
48125
48125
2 12125
24125
48125
3 1125
3125
9125
ΣPiXi = 75
125 ΣPiXi
2 = 105125
∴ Mean = ΣPiXi = 35
=75125
Variance = ΣPiXi2 – (ΣPiXi)2
= 2105 3
125 5 −
= 21 925 25
− = 1225
29. Equation of AB is 1
1 2 13 5 1
x y
− = 0
⇒ x(– 2 – 5) – y(1 – 3) + 1(5 + 6) = 0
⇒ – 7x + 2y + 11 = 0 ⇒ 2y + 11 = 7x
⇒ x = 2 117
y + …(i)
Y
X
Y′
X′
5
5
3
2
1
0 1 2 3 4 5–1
–2
C(5, 2)
B(3, 5)
B(3, 5)
A(1, –2)
Equation of BC is 1
3 5 15 2 1
x y
= 0
⇒ x(5 – 2) – y(3 – 5) + 1(6 – 25) = 0
⇒ 3x + 2y – 19 = 0
⇒ x = 19 23
y− … (ii)
Equation of AC is 1
1 2 15 2 1
x y
− = 0
⇒ x(– 2 – 2) – y(1 – 5) + 1(2 + 10) = 0⇒ 4y + 12 = 4x⇒ x = y + 3 …(iii)From (i), (ii) and (iii),Area of the shaded region
=
2 5
2 2
19 23
( 3) yy dy dy−
−
+ +∫ ∫
– 5
2
2 117
y dy−
+ ∫
230 ■ shiv das senior secondary series (Xii)
= 2 5 52 2 2
2 2 2
2 21 12 3 2 7 2
3 19 11y y yy y y− −
+ + − − +
= 4 4 12 2 3
6 6 − + − + [(95 – 25) – (38 – 4)]
– 17
[(25 + 55) – (4 – 22)]
= 8 + 4 + 13
(70 – 34) – 17
(80 + 18)
= 12 + 12 – 14 = 10 sq. unitsOr
f (x) = 3x2 + 2x + 1, b = 4, a = 0, nh = b – a = 4 – 0 = 4
f (a) = f (0) = 0 + 0 + 1 = 1f (a + h) = f (0 + h) = 3h2 + 2h + 1f (a + 2h) = f (0 + 2h) = 3(2h)2 + 2(2h) + 1
= 12h2 + 4h + 1: : :: : :: : :f (a + (n – 1)h) = f (0 + (n – 1)h)
= 3[(n – 1)h]2 + 2(n – 1)h + 1
( )b
a
f x dx∫ = 0
limh
h→
[f (a) + f (a + h) + f (a + 2h) +
… + f (a + (n – 1)h]4
2
0
(3 2 1)x x dx+ +∫ = 0
limh
h→
[1 + 3h2
+ 2h + 1) + (12h2 + 4h + 1) + …
+ 3(n – 1)2h2 + 2(n – 1)h + 1]
= 0
limh
h→
[n + 3h2(12 + 22 + … + (n – 1)2)
+ 2h(1 + 2 + … + (n – 1)]
= 0
limh
h→
2 2 . ( 1)( 1)(2 1)6 2
3 h n nn n nn h −− −
+ +
= 0
limh
h→ 2
( )(2 ) ( )hnnh nh h nh h nh nh h
+ − − + −
= 0
limh
h→
42
4 (4 )(8 ) 4(4 )h h h
+ − − + −
(nh = 4)
= 4 + 2(4 – 0) (8 – 0) + 4(4 – 0)= 4 + 64 + 16 = 84
set ii
10. xy dydx
= (x + 2) (y + 2)
2y
y + dy = ( 2)xx+ dx
( 2) 2
2y
y
+ − + ∫ dy =
2xx x
+∫ dx
2
21
y
+
−∫ dy = 21x
+∫ dx
∴ y – 2 log |y + 2| = x + 2 log |x| + c
11. 2
cos
8 sin
x
x−∫ dx
= 2 2(2 2)
dp
p−∫ =
=
Let sin
cos
p x
dp x dx
= sin–12 2
p + c −
= + −∫ 1
2 2sin
dx xc
aa x
= sin–1
sin2 2
x + c
12. drdt
= 5 cm/min, dhdt
= – 4 cm/min
r = 8 cm, h = 6 cm
Volume of cylinder, v = πr2h Differentiating w.r.t. t, we get
dvdt
= 2. . 2dh drdt dt
r h r
π +
= π [64 (–4) + 6(16) (5)]
= π [–256 + 480]
= 224π cm3/min
∴ Volume is increasing at the rate of
224π cm3/min.
21. Let I = 0
tansec tan
x xx x
π
+∫ dx …(i)
I = 0
( ) tan( )sec( ) tan( )
x xx x
ππ − π −π − + π −∫ dx
= −∫ ∫
0 0
( ) ( )a a
f x dx f a x dx
I = 0
( )( tan )( sec ) ( tan )
x xx x
ππ − −
− + −∫ dx
mathematics – 2017 comptt. (outside delhi) ■ 231
I = 0
( ) tansec tan
x xx x
ππ −
+∫ dx …(ii)
Adding (i) and (ii), we have
2I = 0 0
tan ( ) tansec tan sec tan
x x x xx x x x
dx dxπ π
π −+ +
+∫ ∫
2I = 0
tansec tan
x dxx x
ππ
+∫
∴ I = 0
tan1 sec tan2 sec tan sec tan
x x xx x x x
π−×
+ −π∫ dx
= 2
2 20
(sec tan tan )2 sec tan
x x x
x x
ππ −
−∫ dx
…(sec2 x – tan2 x = 1)
= 2
02
[sec tan (sec 1)]x x xπ
π − −∫ dx
= π π
2 0sec tanx x−[ ]
= 2
[(sec tan ) (sec 0 tan 0 0)]π π − π + π − − +
= 2π [(–1 – 0 + π) – (1 – 0)] = π
2( π – 2)
22. y . .xye dx = 2( )
xyxe y dy+
dxdy
= 2
.
xy
xyxe y
y e
+
v + y dvdy
= 2v
vv ye y
ye
+
= ⇒ = = +
Let
.1 .
xx vy v
y
dx dvv y
dy dy
y dvdy
= ( )v
vy ve y
ye
+ – v
y dvdy
= v v
vve y ve
e
+ −
ev dv = dy
vdy e dv=∫ ∫
y = ev + c
y = e
xy + c
23.
A(2, 3, – 8)
Given line
B CD
4 12 6 3
yx z− −= =
4 12 6 3
yx z− −− −
= = = p (Let) …(i)
Let D(–2p + 4, 6p, –3p + 1) …(ii) be the general point of given line BC.
Direction ratios of AD are –2p + 4 – 2, 6p – 3, –3p + 1 + 8 –2p + 2, 6p – 3, –3p + 9Direction ratios of given line BC are –2, 6, –3.AD ⊥ BC, using a1a2 + b1b2 + c1c2 = 0⇒ –2(–2p + 2) + 6(6p – 3) – 3(–3p + 9) = 0⇒ 4p – 4 + 36p – 18 + 9p – 27 = 0⇒ 49p = 49 ⇒ p = 1
From (ii), Required foot of ⊥ D
= (–2 + 4, 6, –3 + 1) or D(2, 6, –2)
28. p = P(defective bulb) = 6 130 5
=
q = 1 – p = 1 – 1 45 5= , n = 3
Here random variable X can take values 0, 1, 2, 3.
Using P(r) = nCr qh – r pr
P(X = 0) = 3C0q3p0 = q3 =
34 645 125
=
P(X = 1) = 3C1q2p1 = 3
24 1 485 5 125
=
P(X = 2) = 3C2q1p2 = 3
24 1 125 5 125
=
232 ■ shiv das senior secondary series (Xii)
⇒ (x – 1) (–1 – 1) – (y – 1) (3) + z(3) = 0 (Expanding along R1)
⇒ –2x + 2 – 3y + 3 + 3z = 0⇒ –2x – 3y + 3z + 5 = 0or 2x + 3y – 3z – 5 = 0 …(i)Direction ratios of normal to the plane (i)
are 2, 3, –3.Direction ratios of given line are 3, 1, 3Now, a1a2 + b1b2 + c1c2 = 2(3) + 3(1) + (–3) (3) = 6 + 3 – 9 = 0∴ Given line is parallel to plane (i)Required distance = Distance between point
(3, 5, –2) and plane (i)
= 1 1 12 2 2
ax by cz d
a b c
+ + +
+ +
= = = − = = = − = −
1 1 13, 5, 2
2, 3, 3, 5
x y z
a b c d
= 2(3) 3(5) 3( 2) 54 9 9
+ − − −+ +
= + + −6 15 6 522
= 2222
= 22 units
set iii10. Given vectors are coplanar
1 1 13 1 21 3
−
λ − = 0
⇒ 1(–3 – 2λ) + 1(–9 – 2) + 1(3λ – 1) = 0(Expanding along R1)
⇒ –3 – 2λ – 11 + 3λ – 1 = 0
⇒ λ = 15
11. 2
21
1
dy ydx x
++
=
2 21 1
dy dx
y x+ +=
2 21 1
dy dx
y x+ +=∫ ∫ …(By integrating)
⇒ tan–1 y = tan–1 x + c …(i)
⇒ tan–1 3 = tan–1 0 + c =∴ = =
(0) 3
0, 3
y
x y
P(X = 3) = 3C3q0p3 =
31 15 125
=
Probability distribution is
Xi 0 1 2 3P(Xi) 64
12548
12512125
1125
Xi Pi PiXi PiXi2
0 64125
0 0
1 48
125
48125
48125
2 12125
24125
48125
3 1125
3125
9125
ΣPiXi = 75
125 ΣPiXi
2 = 105125
Mean = ΣPiXi = 35
=75125
Variance = ΣPiXi2 – (ΣPiXi)2
= 2105 3
125 5 −
= 21 925 25
− = 1225
29. Equation of plane through points A, B, C is
1 1 1
2 1 2 1 2 1
3 1 3 1 3 1
x x y y z z
x x y y z z
x x y y z z
− − −− − −− − −
= 0
1 1 0
1 1 2 1 1 02 1 2 1 1 0
x y z− − −− − −
− − − − − = 0
1 1
0 1 13 1 1
x y z− −
− −= 0
mathematics – 2017 comptt. (outside delhi) ■ 233
3π = 0 + c
Putting the value of c in (i), we get
tan–1 y = tan–1 x + 3π
12. radius, r = 13
(3x + 1)
Volume of Sphere, V = 43
πr3
V = 43
π × 127
(3x + 1)3
Differentiating both the sides w.r.t. x,Vd
dx = 24
813(3 1) 3xπ × + × = π4
9(3x + 1)2
21. x + y ≤ 50, 2x + y ≤ 80, x ≥ 20 Let x + y = 50, Let 2x + y = 80, Let x = 20
x 0 50 20y 50 0 30
x 0 40 30y 80 0 20
X
2x + y = 80
X′
Y′
80
70
60
50
40
30
20
10
0 10 20 30 40 50 60
x =
20
x + y = 50
C(30, 20)
B(20, 30)
A(20, 0) D(40, 0)
Y
Corner points
Z = 105x + 90y
A(20, 0)B(20, 30)C(30, 20)D(40, 0)
2100 + 0 = 21002100 + 2700 = 48003150 + 1800 = 4950 (Maximum)4200 + 0 = 4200
Maximum value of Z is 4950 at C(30, 20) i.e., x = 30, y = 20.
22. 2xy + y2 – 2x2 dydx
= 0
–2x2 dydx
= – (2xy + y2)
⇒ 2
22
2
dy xy ydx x
+=
⇒ 2 2
22 ( )
2dv x vx v xdx x
v x ++ = = ⇒ = = +
Let
.1 .
yy vx v
xdy dv
v xdx dx
⇒ 2 2
2(2 )
2dv x v vdx x
x v+= −
22 22
dv v v vdx
x + −=
22dv dx
xv=
− =∫ ∫ 122
xv dv dx
⇒ 12
1v−
− = log|x| + c
⇒ 2v− = log|x| + c
⇒ 2xy
− = log|x| + c
∴ –2x = y log|x| + cy …(i)
⇒ –2(1) = 2 log 1 + c(2) [Putting y = 2, x = 1]
c = –1 [ log 1 = 0]
Putting the value of c in (i), we get –2x = y log|x| – y –2x = –y(1 – log|x|)
y = 2
1 log | |x
x− , x ≠ 0, x ≠ e
23. Let I = 0
1 sinx
xdx
π
+∫ …(i)
I = 0
( )1 sin( )
xx
dxπ
π −+ π −∫ ( ) ( )
0 0
a af x dx f a x dx
= −∫ ∫
I = 0
1 sinx
xdx
ππ −+∫
…(ii)
234 ■ shiv das senior secondary series (Xii)
Adding (i) and (ii), we get
2I = 0
1 sin xdx
ππ
+∫
I = 0
1 sin2 1 sin 1 sin
xx x
dxπ
π π −+ −
×∫
= 20
1 sin2 cos
x
xdx
ππ −∫
= 2
0
sec tan )2
(sec x x xπ
π −∫
= [ ]02tan secx x ππ −
= [ ]2
(tan sec ) (tan 0 sec 0)π π − π − −
= 2
(0 ( 1)) (0 1)π − − − −
= 2
(2)π = π
28. Equation of plane through points A, B, C is
1 1 1
2 1 2 1 2 1
3 1 3 1 3 1
x x y y z z
x x y y z z
x x y y z z
− − −− − −− − −
= 0
⇒ 1 2 2
4 1 2 2 3 23 1 0 2 2 2
x y z− + −− + −− + −
= 0
1 2 2
3 4 12 2 0
x y z− + −= 0
⇒ (x – 1) (– 2) – (y + 2) (– 2) + (z – 2) (6 – 8) = 0
⇒ –2(x – 1) + 2(y + 2) – 2(z – 2) = 0
Dividing both sides by –2
⇒ (x – 1) – (y + 2) + (z – 2) = 0⇒ x – 1 – y – 2 + z – 2 = 0⇒ x – y + z = 5 …(i)
Point (2, –1, 2) lies on the given line, andDirection ratio of given line are 3, 4, 2.Cartesian equation of given line is
+− −= =12 23 4 2
yx z = q (Let)
Let P(3q + 2, 4q – 1, 2q + 2) …(ii) be the required point
Then point P lies on the plane (i) 3q + 2 – (4q – 1) + (2q + 2) = 5⇒ 3q + 2 – 4q + 1 + 2q + 2 = 5⇒ q = 0Putting the value of q in (ii), we have∴ Point P(0 + 2, 0 – 1, 0 + 2)∴ Point (2, –1, 2) is the required point.
29. p = P(4) = 110
, q = 1 – p = 1 – 1 910 10
=
Here random variable X can take values 0, 1, 2.
Using P(r) = nCr qn – r pr
P(X = 0) = 2C0q2p0 = q2 =
29 8110 100
=
P(X = 1) = 2C1q1p1 = 2qp = 2
9 1 1810 10 100
=
P(X = 2) = 2C2q0p2 = p2 =
21 110 100
=
Xi Pi PiXi PiXi2
0 81100
0 0
1 18100
18100
18100
2 1
100 2
100 4
100
ΣPiXi = 20
100 ΣPiXi
2 = 22100
Variance = ΣPiXi2 – (ΣPiXi)2
= 222 20
100 100 − = 22 4
100 100− = 18
100 or 0.18
250 ■ SHIV DAS SENIOR SECONDARY SERIES (XII)
CBSE
MATHEMATICS – 2018 (COMPTT.)
Time allowed : 3 hours Maximum marks : 100
GENERAL INSTRUCTIONS
(i) All questions are compulsory.
(ii) The question paper consists of 29 questions divided into four sections A, B, C and D. Section Acomprises of 4 questions of one mark each, Section B comprises of 8 questions of two marks each,Section C comprises of 11 questions of four marks each and Section D comprises of 6 questions ofsix marks each.
(iii) All questions in Sections A are to be answered in one word, one sentence or as per the exactrequirement of the question.
(iv) There is no overall choice. However, internal choice has been provided in 3 questions of four markseach and 3 questions of six marks each. You have to attempt only one of the alternatives in all suchquestions.
(v) Use of calculators is not permitted. You may ask for logarithmic tables, if required.
SECTION A
Question numbers 1 to 4 carry one mark each.
Q.1. Find the value of tan–1 3 – sec–1(– 2).
Q.2. If A =
1 2 22 12 2 1
x
− −
⎛
⎝
⎜⎜
⎞
⎠
⎟⎟ is a matrix satisfying AA′ = 9I, find x.
Q.3. Find the value of [ˆ, ˆ, ˆ]i k j .
Q.4. Find the identity element in the set Q+ of all positive rational numbers for the operation ✻
defined by a ✻ b =
32ab for all a, b ∈ Q+. (Not in syllabus)
SECTION B
Question numbers 5 to 12 carry 2 marks each.
Q.5. Prove that : 3 cos–1 x = cos–1 (4x3 – 3x), x ∈
12
1,⎡⎣⎢
⎤⎦⎥.
(Common for Delhi & Outside Delhi)
MATHEMATICS – 2018 (COMPTT.) ■ 251
Q.6. If A =
2 35 2−
⎡
⎣⎢
⎤
⎦⎥ be such that A–1 = kA, then find the value of k.
Q.7. Differentiate tan–1
cos sincos sin
x xx x
−+
⎛⎝⎜
⎞⎠⎟ with respect to x.
Q.8. The total revenue received from the sale of x units of a product is given by R(x) = 3x2 + 36x + 5in rupees. Find the marginal revenue when x = 5, where by marginal revenue we mean therate of change of total revenue with respect to the number of items sold at an instant.
Q.9. Find :
3 52
−∫ sincos
xx
dx.
Q.10. Solve the differential equation cos
dydx
⎛⎝⎜
⎞⎠⎟
= a, (a ∈ IR).
Q.11. If a b c→ → → →
+ + = 0 and | |a→
= 5, | |b→
= 6 and | |c→
= 9, then find the angle between a→
and b→
.
Q.12. Evaluate P(A ∪ B), if 2P(A) = P(B) =
513
and P(A/B) =
25
.
SECTION C
Question numbers 13 to 23 carry 4 marks each.Q.13. Using properties of determinants, prove that
5 2 22 5 22 2 5
a a b a c
b a b b c
c a c b c
− + − +− + − +− + − +
= 12(a + b + c) (ab + bc + ca)
Q.14. If sin y = x cos(a + y), then show that
dydx
a ya
= +cos ( )cos
2.
Also, show that
dydx
= cos a, when x = 0.
Q.15. If x = a sec3 θ and y = a tan3 θ, find
d y
dx
2
2 at θ =
π3
.
Or
If y = extan−1, prove that
( ) ( )1 2 12
2
2+ + −x xd ydx
dydx
= 0.
Q.16. Find the angle of intersection of the curves x2 + y2 = 4 and (x – 2)2 + y2 = 4, at the point inthe first quadrant.
OrFind the intervals in which the function f (x) = – 2x3 – 9x2 – 12x + 1 is
(i) Strictly increasing (ii) Strictly decreasingQ.17. A window is in the form of a rectangle surmounted by a semicircular opening. The total
perimeter of the window is 10 metres. Find the dimensions of the window to admitmaximum light through the whole opening.
252 ■ SHIV DAS SENIOR SECONDARY SERIES (XII)
Q.18. Find :
42 42( )( )x x− +∫ dx
Q.19. Solve the differential equation (x2 – y2) dx + 2xydy = 0.Or
Find the particular solution of the differential equation (1 + x2)
dydx
+ 2xy =
11 2+ x
, given that
y = 0 when x = 1.Q.20. Find x such that the four points A(4, 4, 4), B(5, x, 8), C(5, 4, 1) and D(7, 7, 2) are coplanar.
Q.21. Find the shortest distance between the lines
x y z− − −= =12
23
34
and
x y z− − −= =23
44
55
.
Q.22. Two groups are competing for the positions of the Board of Directors of a corporation. Theprobabilities that the first and second groups will win are 0.6 and 0.4 respectively. Further,if the first group wins, the probability of introducing a new product is 0.7 and thecorresponding probability is 0.3 if the second group wins. Find the probability that the newproduct introduced was by the second group.
Q.23. From a lot of 20 bulbs which include 5 defectives, a sample of 3 bulbs is drawn by random,one by one with replacement. Find the probability distribution of the number of defectivebulbs. Also, find the mean of the distribution.
SECTION D
Question numbers 24 to 29 carry 6 marks each.
Q.24. Show that the relation R on the set Z of all integers defined by (x, y) ∈ R ⇔ (x - y) is divisibleby 3 is an equivalence relation.
Or
A binary operation ✻ on the set A = {0, 1, 2, 3, 4, 5} is defined as a ✻ b =
a b a b
a b a b
+ + <+ − + ≥
⎧⎨⎩
,,
ifif
66 6
.
Write the operation table for a ✻ b in A.Show that zero is the identity for this operation ✻ and each element ‘a’ ≠ 0 of the set isinvertible with 6 – a, being the inverse of ‘a’. (Not in syllabus)
Q.25. Given A =
5 0 42 3 21 2 1
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
, B–1 =
1 3 31 4 31 3 4
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
, compute (AB)–1.
Or
Find the inverse of the matrix A =
1 2 21 3 00 2 1
−−
−
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
by using elementary row transformations.
MATHEMATICS – 2018 (COMPTT.) ■ 253
Q.26. Using integration, find the area of the region : {(x, y) : 0 ≤ 2y ≤ x2, 0 ≤ y ≤ x, 0 ≤ x ≤ 3}
Q.27. Evaluate:
x x x
x x
sin cossin cos
/
4 40
2
+∫π
dx
Or
Evaluate:
( )3 2 12
1
3
x x dx+ +∫ as the limit of a sum.
Q.28. Find the vector equation of the line passing through (1, 2, 3) and parallel to each of the planes
r i j k→
− +.(ˆ ˆ ˆ)2 = 5 and r i j k→
+ +. ( ˆ ˆ ˆ)3 = 6. Also find the point of intersection of the line thus
obtained with the plane r i j k→
+ +.( ˆ ˆ ˆ)2 = 4.
Q.29. A company produces two types of goods, A and B, that require gold and silver. Each unitof type A requires 3 g of silver and 1 g of gold while that of B requires 1 g of silver and2 g of gold. The company can use atmost 9 g of silver and 8 g of gold.If each unit of typeA brings a profit of � 40 and that of type B � 50, find the number of units of each type thatthe company should produce to maximize the profit. Formulate and solve graphically theLPP and find the maximum profit.
SOLUTIONS
3. 1st method
[ˆ ˆ ˆ]i k j =
ˆ ( ˆ ˆ) ˆ ( ˆ)i k j i i• •× = − = – 12nd method
[ˆ ˆ ˆ]i k j =1 0 00 0 10 1 0
Expanding along C1, 1(0 – 1) = – 1
4. a ✻ b =
32ab …[Given]
a ✻ e =
32ae …(i)
Identity element, a ✻ e = a
⇒
32ae = a …[From (i)
⇒
32e = 1
⇒ e =
23
5. R.H.S. = cos–1(4x3 – 3x)= cos–1(4 cos3 θ – 3 cos θ) Let x = cos θθθθθ= cos–1 (cos 3θ) ∴ cos–1 x = θθθθθ= 3θ= 3 cos–1 x = L.H.S. (Hence proved)
1. tan–1 ( )3 – sec–1 (–2)
=
π3
– [π – sec–1 (2)]
Q sec ( ) sec ,| |−− −−−− == −− ≥≥[[ ]]1 1 1x x xππ
=
π3
– π +
π3
=
−− ππ3
2. AA′ = 9I
1 2 22 12 2 1
1 2 22 1 22 1
91 0 00 1 00 0 1
x
x− −
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
−
−
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
=⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
1 4 4 2 2 2 2 4 22 2 2 4 1 4 2
2 4 2 4 2 4 4 1
2+ + + + − + −
+ + + + − + −− + − − + − + +
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
x
x x x
x
=
9 0 00 9 00 0 9
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
Equating the corresponding elements4 + 2x = 0 5 + x2 = 92x = – 4 ⇒ x = – 2 x2 = 4 ⇒ x = ± 2.
Therefore, x = – 2 is the only common solution.
⎡
⎣⎢
254 ■ SHIV DAS SENIOR SECONDARY SERIES (XII)
6. A =
2 35 2−
⎡
⎣⎢
⎤
⎦⎥
|A| = – 4 – 15 = – 19 ≠ 0, ∴ A–1 exists.
adj A =
− −−
⎡
⎣⎢
⎤
⎦⎥
2 35 2
As we know, A–1 =
1| |A
. adj A
=
− − −−
⎡
⎣⎢
⎤
⎦⎥
1
19
2 35 2
=
219
319
519
219−
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
Now,A–1 = kA …(Given)
219
319
519
219−
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥ = k
2 35 2−
⎡
⎣⎢
⎤
⎦⎥
219
319
519
219−
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥ =
2 35 2
k k
k k−⎡
⎣⎢
⎤
⎦⎥
⇒ 2k =
2
19(Equating the corresponding elements)
∴ k =
119
7. Let y = tan–1
cos sincos sin
x xx x
−+
⎛⎝⎜
⎞⎠⎟
Dividing Numerator and Denominator by cos x,we have
y = tan–1
11
−+
⎛⎝⎜
⎞⎠⎟
tantan
xx
y = tan–1
tan π4
−⎛⎝⎜
⎞⎠⎟
⎛⎝⎜
⎞⎠⎟
x
y =
π4
– x
Differentiating both sides w.r.t. x, we have
dydx
= – 1
8. R(x) = 3x2 + 36x + 5Differentiating both sides w.r.t. x, we have
Marginal Revenue, R′(x) = 6x + 36When x = 5, R′(5) = 6(5) + 36
= 30 + 36 = � 66∴ Marginal Revenue is � 66.
9.
3 52
−∫ sincos
x
x dx
=
3 52 2cos
sincosx
x
x−⎛
⎝⎜⎞⎠⎟∫ dx
=
( sec sec tan )3 52 x x x−∫ dx
= 3 tan x – 5 sec x + C
10. cos
dydx
⎛⎝⎜
⎞⎠⎟
= a
dydx
= cos–1 a
dy = cos–1 a . dx
dy∫ = cos–1 a
dx∫ (Integrating both sides)
y = x . cos–1 a + Cy – C = x cos–1 a
yx− C = cos–1 a
cos
yx−−⎛⎛
⎝⎝⎜⎜⎞⎞⎠⎠⎟⎟
C = a is the solution of the
differential equation.
11. a b c→ → →
+ + = 0 …(Given)
⇒ a b c→ → →
+ = −Squaring both sides, we get
( ) ( ) ( ) ( )a b a b c c→ → → → → →
+ + = − −• •
[ . | |Q a a a a→→ →→ →→ →→
== ==2
2
a a a b b a b b c c→ → → → → → → → → →
• • • • •+ + + =
| | | | | |a a b b c→ → → → →
+ + =•2 2 22
| | | || |cos | | | |a a b b c→ → → → →
+ + =2 2 22 θ52 + 2(5) (6) cos θ + 62 = 92
… [ | | , | | , | |Q a b c→→ →→ →→
== == ==5 6 9
60 cos θ = 81 – 25 – 36
⇒ cos θ =
2060
13
=
∴ Required angle, θθθθθ = cos–1
13
⎛⎝⎜
⎞⎠⎟
12. Given: 2P(A) = P(B) =
513
⇒ P(A) =
526
and P(A/B) =
25
MATHEMATICS – 2018 (COMPTT.) ■ 255
As we know, P(A/B) =
P A BP B
( )( )∩
25 5
13
= ∩P A B( )
P(A ∩ B) =
25
513
213
× =
∴ P(A ∪∪∪∪∪ B) = P(A) + P(B) – P(A ∩ B)
=
526
513
213
+ −
=
5 10 426
+ − = 1126
13. L.H.S. =
5 2 22 5 22 2 5
a a b a c
b a b b c
c a c b c
− + − +− + − +− + − +
Applying C1 → C1 + C2 + C3
=
5 2 2 2 22 5 2 5 22 2 5 2 5
a a b a c a b a c
b a b b c b b c
c a c b c c b c
− + − + − + − +− + + − + − +− + − + + − +
=
a b c a b a c
a b c b b c
a b c c b c
+ + − + − ++ + − ++ + − +
2 25 2
2 5
Taking common (a + b + c) from C1, we get
= (a + b + c)
1 2 21 5 21 2 5
− + − +− +
− +
a b a c
b b c
c b c
Applying R2 → R2 – R1 and R3 → R3 – R1
= (a + b + c)
1 2 20 2 4 2 20 2 2 2 4
− + − ++ −− +
a b a c
a b a b
a c a c
Expanding along C1, we get= (a + b + c) [(2a + 4b) (2a + 4c) –
(2a – 2b) (2a – 2c)]= 4(a + b + c) [(a + 2b) (a + 2c) –
(a – b) (a – c)]= 4(a + b + c) [a2 + 2ac + 2ab + 4bc –
a2 + ac + ab – bc]= 4(a + b + c) (3ab + 3bc + 3ca)= 12 (a + b + c) (ab + bc + ca) = R.H.S.
(Hence proved)
14. sin y = x cos(a + y) …(i)
sincos( )
ya y+
= x
Differentiating both sides w.r.t. x, we have
cos( ). cos sin ( sin( ))
cos ( )
a y ydydx
y a ydydx
a y
+ − − +
+2 = 1
cos(a + y – y)
dydx
= cos2(a + y)
…[ Q cos A cos B + sin A sin B = cos(A – B)]
dydx
=
cos ( )cos
2 a ya++ (Hence proved)
Putting the value of x = 0 in (i), we getsin y = 0.cos(a + y)sin y = 0 ⇒ y = 0
∴ Point A(x, y) = A(0, 0)
Now,
dydx
⎤⎤⎦⎦⎥⎥atA( , )0 0
=
cos ( )cos
2 0aa+ =
coscos
2 aa
= cos a (Hence proved)15. x = a sec3 θ y = a tan3 θDifferentiating Differentiatingboth sides w.r.t. θ both sides w.r.t. θ
dxdθ
= a.3 sec2 θ.sec θ tan θ
dydθ
= 3a tan2 θ . sec2 θ
dydx
=
dyddxd
θ
θ
=
33
2 2
2a
a
tan secsec sec tan
θ θθ θ θ
dydx
=
tansec
sincos
cosθθ
θθ
θ= ×1
dydx
= sin θ
Again differentiating both sides w.r.t. x,
d y
dx
2
2 = cos θ .
ddxθ
= cos θ .
13 3a sec . tanθ θ
=
13
4
acostan
θθ
∴
d y
dx
2
2
3
⎤⎤
⎦⎦⎥⎥⎥⎥ ==at θθ ππ
=
13
3
3
4
a
cos
tan
π
π
⎛⎝⎜
⎞⎠⎟
⎛⎝⎜
⎞⎠⎟
=
13
12
3
4
a
⎛⎝⎜
⎞⎠⎟
( )=
13
116
13
33a
× × × =
3144 a
256 ■ SHIV DAS SENIOR SECONDARY SERIES (XII)
x m x m
2x m
y m
Or
y = extan−1
…(i)Differentiating both sides w.r.t. x, we have
dydx
= extan−1.
11 2+ x
(1 + x2).
dydx
= y [From (i)
Differentiating both sides w.r.t. x, we have
(1 + x2).
d y
dx
dydx
dydx
x2
2 2+ =.
(1 + x2).
d y
dx
dydx
dydx
x2
2 2+ −. = 0
(1 + x2).
d y
dx
dydx
x2
2 2 1++ −−( ) = 0 (Hence Proved)
16. Point of intersectionx2 + y2 = 4 (x – 2)2 + y2 = 4y2 = 4 – x2 …(i) (x – 2)2 + 4 – x2 = 4
[From (i)x2 – 4x + 4 – x2 = 0– 4x = – 4
⇒ x = 1
Putting the value of x in (i), y = ± 3
∴ y = 3 , but y ≠ – 3 ( Q Pt . lies in Ist Quadrant)
∴ Point of intersection is A(1, 3 )x2 + y2 = 4 (x – 2)2 + y2 = 4
Differentiating both Differentiating bothsides w.r.t. x, sides w.r.t. x,
2x + 2y
dydx
= 0 2(x – 2) + 2y
dydx
= 0
2y
dydx
= – 2x 2y
dydx
= – 2(x – 2)
dydx
xy
= − = m1
dydx
xy
= − −( )2 = m2
m1 at (1, 3 ) =
− 13
m2 at (1, 3 ) =
13
Let θ be the required angle.
tan θ =
m mm m
2 1
2 11−
+ .
tan θ =
13
13
113
13
− −⎛⎝⎜
⎞⎠⎟
+⎛⎝⎜
⎞⎠⎟
−⎛⎝⎜
⎞⎠⎟
=
23
3 13−
tan θ =
23
32
3× =
θ = 60° or
π3
∴ Required angle of intersection of the curves
is
ππ3
.
Orf (x) = – 2x3 – 9x2 – 12x + 1
Differentiating both sides w.r.t. x, we havef ′(x) = – 6x2 – 18x – 12
= – 6(x2 + 3x + 2)= – 6(x2 + 2x + x + 2)= – 6[x(x + 2) + 1(x + 2)]= – 6(x + 2) (x + 1)
When f ′(x) = 0, x = – 2, – 1
Intervals Checking Value of Sign of Nature ofPoint – 6 (x + 2) (x + 1) f ′′′′′(x) f(x)
(– ∞, – 2) – 3 – – – < 0 strictly decreasing(– 2, – 1) – 1.5 – + – > 0 strictly increasing(– 1, ∞) 1 – + + > 0 strictly decreasing
(i) f is strictly increasing in (– 2, – 1).(ii) f is strictly decreasing in (– ∞∞∞∞∞, – 2) ∪∪∪∪∪ (– 1, ∞∞∞∞∞).
17.
Let length of rectangular window = 2x mLet breadth of rectangular window = y m∴ Radius of rectangle window = x mPerimeter of window = 10 m∴ y + 2x + y + πx = 10 [ Q Perimeter of semi-circle = πππππr
2y = 10 – 2x – πx
y =
10 22
− −x xπ …(i)
y =
52
− −⎛⎝⎜
⎞⎠⎟
x xπ
Area of window = Area of rectangle+ Area of semi-circle
A = (2x)y +
12
πx2
A = 2x
10 22
12
− −⎛⎝⎜
⎞⎠⎟
+x xπ πx2 [From (i)]
A = 10x – 2x2 – πx2 +
12
πx2
A = 10x – 2x2 –
12
πx2
MATHEMATICS – 2018 (COMPTT.) ■ 257
ddxA = 10 – 4x – πx
For Stationary points, 10 – 4x – πx = 0
⇒ – (4 + π)x = – 10 ⇒ x =
104 + π
m
ddx
2
2A = – 4 – π
ddx
x
2
2 104
A
at
⎤
⎦⎥⎥ =
+π
= – (4 + π) < 0 (–ve)
∴∴∴∴∴ Area is maximum at x =
104 ++ ππ
m.
∴ Maximum light will enter the window
when radius =
104 ++ ππ
m.
Length of rectangle window = 2x =
204 ++ ππ
m,
Breadth of rectangle window, y
= 5 –
104 2
104+ +
⎛⎝⎜
⎞⎠⎟
−π
ππ
=
20 5 10 54
+ − −+
π ππ
=
104 ++ ππ
m
18. Let
42 42( )( )x x− +
=
A B Cx
x
x−+ +
+2 42 …(i)
42 42( )( )x x− +
=
A B C( ) ( )( )( )( )
x x x
x x
2
24 2
2 4+ + + −
− +4 = A(x2 + 4) + Bx(x – 2) + C(x – 2)
Comparing the coefficient of x2, coefficent of xand constant terms respectively,
Coefficient of x2 Coeficient of x Constant term 0 = A + B 0 = – 2B + C 4 = 4A – 2C – B = A ...(ii) 2B = C ...(iii) 4 = 4(– B) – 2(2B)Putting the Putting the (From (ii) and (iii))
value of B in value of B in 4 = – 4B – 4B (ii), (iii), 4 = – 8B
–
−⎛⎝⎜
⎞⎠⎟
12
= A 2
−⎛⎝⎜
⎞⎠⎟
12
= C ⇒ B =
− 12
⇒
12
= A ⇒ C = – 1
Putting the value of A, B, and C in (i),
42 42( ) ( )x x
dx− +∫ =
12
2
12
1
42x
x
xdx
−+
− −
+
⎛
⎝
⎜⎜⎜
⎞
⎠
⎟⎟⎟
∫
=
12
12 2
22 2 2log| |x dxx
x
dx
x− −
× +−
+∫ ∫4 22
=
12
14
12 2
2 42 1log| | log( ) tanx x x−− −− ++ −− ++−− C
Q
Q
f xf x
dx f x
dx
x a axa
’( )( )
log| ( )|
tan
== ++
++== ++
⎡⎡
⎣⎣
⎢⎢⎢⎢⎢⎢⎢⎢
∫∫
∫∫ −−
C
C2 211
19. (x2 – y2)dx + 2xy dy = 0 [Given]2xy dy = – (x2 – y2)dx
⇒
dydx
x yxy
= − +2 2
2 ⇒
dydx
xy
yx
= +−2 2
v + x
dvdx
x v xx vx
= − +×
2 2 2
2
let y vxyx
v
dydx
v xdvdx
== ⇒⇒ ==
== ++
⎡⎡
⎣⎣
⎢⎢⎢⎢⎢⎢⎢⎢ . .1
⇒ x v xdv
dxx v
x v
dvdx
vv
= − ⇒ =− −2 2
2
212
12
( ) – v
⇒ x
dvdx
v vv
dvdx
vv
x= ⇒ =− − − −2 2 21 22
12
⇒
21 2
vdv
v
dxx( )+
−=
21 2
v
v
dxx
dv+∫ ∫= − …[Integrating both sides]
dpp
dxx
=∫ ∫− …
Let p v
dp vdv
== ++==
⎡⎡
⎣⎣⎢⎢⎢⎢
1
2
2
log|p| = –log|x|+ log c
⇒ log|p| = log
cx
⇒ p =
cx
⇒ 1 + v2 =
cx
⇒ 1 +
yx
cx
2
2 = ⇒
x yx
cx
2 2
2+ =
⇒
x yx
2 2+ = c ⇒ x2 + y2 = cx …(i)
∴ 12 + 12 = c(1) [ Q x = y = 1 (given)]⇒ 2 = cPutting the value of c in (i),
x2 + y2 = 2xOr
Given differential equation is
(1 + x2)
dydx
+ 2xy =
11 2( )+ x
Dividing both sides by (1 + x2), we get
dydx
xyx x
+ =+ +2
11
12 2 2( )
On comparing with
dydx
+ Py = Q,
Here ‘P’ =
21 2
xx+
, ‘Q’ =
11 2 2( )+ x
I.F. = e e edxxx
dx xP∫ ∫= =+ +2
1 12 2log| | = 1 + x2
Q
f xf x
dx f x c′′
== ++⎡⎡
⎣⎣⎢⎢⎢⎢
⎤⎤
⎦⎦⎥⎥⎥⎥∫∫ ( )
( )log| ( )|
258 ■ SHIV DAS SENIOR SECONDARY SERIES (XII)
Hence the solution is y(I.F.) =
Q(I.F.)∫ dx
y (1 + x2) =
11 2 2
21( )
( )+
+∫ xx dx
y (1 + x2) =
11 2+∫ x
dx
y (1 + x2) = tan–1 x + c …(i)0.(1 + 1) = tan–1 (1) + c [ Qy = 0, x = 1]
0 =
π4
+ c ⇒ c =
− π4
Putting the value of c in (i), we get
y.(1 + x2) = tan–1 x –
ππ4
is the particular
solution.
or y =
tan( )
–1
2 21 4 1x
x x++ ++−− ππ .
20. AB→
= ( ) ˆ ( ) ˆ ( ) ˆ5 4 4 8 4− + − + −i x j k
= ˆ ( ) ˆ ˆi x j k+ − +4 4
AC→
= ( ) ˆ ( ) ˆ ( ) ˆ5 4 4 4 1 4− + − + −i j k
= ˆ ˆ ˆi j k+ −0 3
AD→
= ( ) ˆ ( ) ˆ ( ) ˆ7 4 7 4 2 4− + − + −i j k= 3 3 2ˆ ˆ ˆi j k+ −
Points A, B, C and D are coplanar …(Given)
[ ]AB AC AD→ → →
= 0
1 4 41 0 33 3 2
x −−−
= 0
Expanding along R1⇒ 1(9) – (x – 4) (– 2 + 9) + 4(3) =0⇒ 9 – (x – 4)7 + 12 = 0⇒ 9 – 7x + 28 + 12 = 0⇒ – 7x = – 49 ⇒ x = 7
21. Herex1 = 1, y1 = 2, z1 = 3; x2 = 2, y2 = 3, z2 = 5,a1 = 2, b1 = 3, c1 = 4; a2 = 3, b2 = 4, c2 = 5Shortest distance between the lines, d
x x y y z z
a b c
a b c
b c b c c a c a a b a b
2 1 2 1 2 1
1 1 1
2 2 2
1 2 2 12
1 2 2 12
1 2 2 12
− − −
− + − + −( ) ( ) ( )
d =− + − + −
1 1 22 3 43 4 5
3 5 4 4 4 3 5 2 2 4 3 32 2 2[ ( ) ( )( )] [( ( ) ( )] [ ( ) ( )]
d =
− − + −
+ +
1 15 16 1 10 12 2 8 9
1 4 1
( – ) ( ) ( )
d =
− +1 2 2
6
– =
− 1
6
∴ d == 1
6 or
d == 6
6 units.
22. Let E1, E2, A be the following events.E1 : Ist group will winE2 : 2nd group will winA : Introducing a new productP(E1) = 0.6, P(A/E1) = 0.7P(E2) = 0.4, P(A/E2) = 0.3
Using Bayes’ theorem,
P(E2/A) =
P E P A EP(E )P(A / E ) + P(E )P(A / E
2
1 1 2 2
( ) ( / ))
2
=
0 4 0 30 6 0 7 0 4 0 3
. .( . . ) ( . . )
×× + ×
=
0 120 42 0 12
.. .+
=
0 120 54
1254
.
.= =
29
23. p = P(defective bulb) =
520
14
=
q = 1 – p = 1 –
14
34
= , n = 3
Here random variable X can take values 0, 1, 2, 3.Using P(r) = nCr.qn – r (p)r
P(X = 0) = 3C0.q3p0 = q3 =
34
2764
3⎛⎝⎜
⎞⎠⎟
=
P(X = 1) = 3C1.q2p1 = 3q2p = 3
34
14
2764
2⎛⎝⎜
⎞⎠⎟
⎛⎝⎜
⎞⎠⎟
=
P(X = 2) = 3C2.q1p2 = 3qp2 = 3
34
14
964
2⎛⎝⎜
⎞⎠⎟
⎛⎝⎜
⎞⎠⎟
=
P(X = 3) = 3C3.q0p3 = p3 =
14
164
3⎛⎝⎜
⎞⎠⎟
=
Probability distribution isXi 0 1 2 3
P(Xi)
2764
2764
964
164
∴ Mean = ΣPiXi
=
0 1 2 32764
2764
964
164
× + × + × + ×⎛⎝⎜
⎞⎠⎟
⎛⎝⎜
⎞⎠⎟
⎛⎝⎜
⎞⎠⎟
⎛⎝⎜
⎞⎠⎟
=
0 27 18 364
+ + + =
4864
= 34
MATHEMATICS – 2018 (COMPTT.) ■ 259
1 ✻ 5 = 1 + 5 – 6 = 02 ✻ 5 = 2 + 5 – 6 = 1 …5 ✻ 5 = 5 + 5 – 6 = 44 ✻ 5 = 4 + 5 – 6 = 3
Identity :Let a be an arbitrary element ofset A = {0, 1, 2, 3, 4, 5}
a ✻ 0 = a + 0 = a …(i)0 ✻ a = 0 + a = a …(ii)
From (i) and (ii),a ✻ 0 = 0 ✻ a = a ∀ a ∈{0, 1, 2, 3, 4, 5}
Hence, 0 is the identity for binary operation ✻.Inverse :
1 ✻ 5 = 1 + 5 – 6 = 02 ✻ 4 = 2 + 4 – 6 = 03 ✻ 3 = 3 + 3 – 6 = 04 ✻ 2 = 4 + 2 – 6 = 05 ✻ 1 = 5 + 1 – 6 = 0
Therefore, Inverse of 1 is 5 or (6 – 1)Inverse of 2 is 4 or (6 – 2)Inverse of 3 is 3 or (6 – 3)Inverse of 4 is 2 or (6 – 2)Inverse of 5 is 4 or (6 – 1)Inverse of a is (6 – a)
Here, for each element a ≠≠≠≠≠ 0, (6 – a) is its inverse.
25. A =
5 0 42 3 21 2 1
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
|A| = 5(3 – 4) + 4(4 – 3) (Expanding along R1) = – 5 + 4 = –1 ≠ 0
∴ A–1 does exist.a11 = 3 – 4 a12 = – (2 – 2) a13 = 4 – 3
= – 1 = 0 = 1a21 = – (0 – 8) a22 = 5 – 4 a23 = – (10 – 0)
= 8 = 1 = – 10a31 = 0 – 12 a32 = – (10 – 8) a33 = 15 – 0
= – 12 = – 2 = 15
adj A =
− −−
−
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
1 8 120 1 21 10 15
Now, A–1 =
1| |A
. adj A
=
1
1
1 8 120 1 21 10 15
−
− −−
−
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
=
1 8 120 1 21 10 15
−−
− −
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
✻ 0 1 2 3 4 5
0 0 1 2 3 4 5
1 1 2 3 4 5 0
2 2 3 4 5 0 1
3 3 4 5 0 1 2
4 4 5 0 1 2 3
5 5 0 1 2 3 4
a b+ <⎧
⎨⎪
⎩⎪
0
24. R = {(x, y) : x, y ∈ Z and (x – y) is divisibleby 3 }
Reflexive : R is reflexive, for x ∈ Zx – x = 0, which is divisible by 3(x, x) ∈ R ∴ R is reflexive.
Symmetric :R is symmetric, (x, y) ∈ R where x, y ∈ Zx – y is divisible by 3.
Let (x – y) = 3m, where m ∈ Z– (y – x) = 3m(y – x) = – 3m
∴ (y – x) is divisible by 3.(y, x) ∈ R ∴ R is symmetric.
Transitive :R is transitive, (x, y) ∈ R where x, y ∈ Zx – y is divisible by 3x – y = 3m, where m ∈ Z …(i)
For (y, z) ∈ Z, where y, z ∈ Zy – z is divisible by 3.y – z = 3n, where n ∈ Z …(ii)
Adding (i) and (ii),(x – y) + (y – z) = 3m + 3nx – z = 3(m + n)
⇒ (x – z) is divisible by 3.(x, y) ∈ R, (y, z) ∈ R
∴ (x, z) ∈ R where x, y, z ∈ Z∴ R is transitive.Since R is reflexive, symmetric and transitive.∴ R is an equivalence relation.
OrOperation table for ✻ is :
0 ✻ 0 = 0 + 0 = 00 ✻ 1 = 0 + 1 = 10 ✻ 2 = 0 + 2 = 2
…
0 ✻ 5 = 0 + 5 = 5
a b+ ≥⎧
⎨⎪
⎩⎪
6
260 ■ SHIV DAS SENIOR SECONDARY SERIES (XII)
Applying R3 → 5R3
=
1 0 1
0 1
0 0 1
1 0 1
0
2 2 5
2
5
1
5
1
5
−
−
⎡
⎣
⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥
=
⎡
⎣
⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥
A
Applying R1 → R1 + R3
=
1 0 0
0 1
0 0 1
3 2 6
0
2 2 5
2
5
1
5
1
5−
⎡
⎣
⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥
=
⎡
⎣
⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥
A
Applying R2 → R2 +
25
R3
=
1 0 00 1 00 0 1
3 2 61 1 22 2 5
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
=⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥A
I = A–1A
∴ A–1 =
3 2 6
1 1 2
2 2 5
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
26. 0 ≤ 2y ≤ x2; 0 ≤ y ≤ x; 0 ≤ x ≤ 3Let 2y = 0, 2y = x2 Let y = x Let x = 0⇒ y = 0 x = 3For Point of intersection,We have x2 = 2y y = x …(i)∴ x2 = 2x …[from (i)
x2 – 2x = 0x(x – 2) = 0x = 0 or x = 2
Now, 2y = x2
y =
12
x2
As we know, (AB)–1 = B–1A–1
(AB)–1 =
1 3 31 4 31 3 4
1 8 120 1 21 10 15
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
−−
− −
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
=
1 0 3 8 3 30 12 6 451 0 3 8 4 30 12 8 451 0 4 8 3 40 12 6 60
+ − − − + + −+ − − − + + −+ − − − + + −
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
=
− −− −− −
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
2 19 272 18 253 29 42
Or
Let A =
1 2 21 3 0
0 2 1
−−
−
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
We know that A = IA
=
1 2 21 3 0
0 2 1
1 0 00 1 00 0 1
−−
−
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
=⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥ A
Applying R2 → R2 + R1
=
1 2 20 5 20 2 1
1 0 01 1 00 0 1
−−
−
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
=⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
A
Applying R2 →
15
R2
=
1 2 2
0 1
0 2 1
1 0 0
0
0 0 1
2
5
1
5
1
5
−
−
−
⎡
⎣
⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥
=
⎡
⎣
⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥
A
Applying R1 → R1 + R3
=
1 0 1
0 1
0 2 1
1 0 1
0
0 0 1
2
5
1
5
1
5
−
−
−
⎡
⎣
⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥
=
⎡
⎣
⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥
A
Applying R3 → R3 + 2R2
=
1 0 1
0 1
0 0
1 0 1
0
1
2
51
5
1
5
1
52
5
2
5
−
−
⎡
⎣
⎢⎢⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥⎥⎥
=
⎡
⎣
⎢⎢⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥⎥⎥
A X′ X
Y
8
7
6
5
4
3
2
1
–4 –3 –2 –1 0 1 2 3 4
Y′ x = 3
x 0 3 2y 0 3 2
2y =
x2
y = x
x 0 ±2 ±4y 0 2 8
MATHEMATICS – 2018 (COMPTT.) ■ 261
The area of the shaded region
=
12
2
0
2
2
3
x dx x∫ ∫+
=
12
13
12
30
2 22
3× +[ ] [ ]x x dx
=
16
12
2 0 3 23 2 2( ) ) ( )− + −
=
86
12
+ (9 – 4) =
8 156
+ = 236
sq. units
27. Let I =
x x xx x
sin cossin cos
/
4 40
2
+∫π
dx …(i)
=
π π π
π π
π2 2 2
2 24 4
0
2 −⎛⎝⎜
⎞⎠⎟
−⎛⎝⎜
⎞⎠⎟
−⎛⎝⎜
⎞⎠⎟
−⎛⎝⎜
⎞⎠⎟
+ −⎛⎝⎜
⎞⎠⎟
∫x x x
x x
sin . cos
sin cos
/
dx
Q f x dx f a x dxa a
( ) ( )0 0∫∫ ∫∫
⎡⎡
⎣⎣
⎢⎢⎢⎢
== −−
I =
ππ
24 4
0
2 −⎛⎝⎜
⎞⎠⎟
+∫x x x
x x
cos sin
sin cos
/
dx …(ii)
Adding (i) and (ii), we get
2I =
x x x x x x
x x
sin cos cos sin
sin cos
/ + −⎛⎝⎜
⎞⎠⎟
+∫π
π2
4 40
2
dx
2I =
ππ2
4 40
2 sin cos
sin cos
/ x x
x x+∫ dx
Dividing num. and deno. by cos4 x, we get
2I =
ππ
2 1
2
40
2tan . sectan
/x x
x +∫ dx
=
π2
12 12
0
×+
∞
∫dp
p
=
π4
10
tan ( )− ∞[ ]p
2I =
π4
[tan–1 (∞) – tan–1 (0)]
∴ I =
π π8 2
0−⎡⎣⎢
⎤⎦⎥ =
ππ2
16
OrLet f (x) = 3x2 + 2x + 1, b = 3, a = 1,
nh = b – a = 3 – 1 = 2
f (a) = 3(1)2 + 2(1) + 1 = 6f (a + h) = 3(1 + h)2 + 2(1 + h) + 1
= 3 + 3h2 + 6h + 2 + 2h + 1= 3h2 + 8h + 6
f (a + 2h) = 3(1 + 2h)2 + 2(1 + 2h) + 1= 3 + 12h2 + 12h + 2 + 4h + 1= 12h2 + 16h + 6
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .f (a + (n – 1)h) = 3[1 + (n – 1)h]2
+ 2[1 + (n – 1) h] + 1= 3 + 3(n – 1)2h2 + 6(n – 1)h + 2 +
2(n – 1)h + 1= 3(n – 1)2h2 + 8(n – 1)h + 6
f x dx
a
b
( )∫ = limh
h→0
[f (a) + f (a + h) + f (a + 2h)
+ ... + f(a + (n –1)h)]
( )3 2 12
1
3
x x dx+ +∫=
limh
h→0
[6 + (3h2 + 8h + 6) + (12h2 + 16h +
6) + … + 3(n – 1)2h2 + 8(n – 1)h + 6]=
limh
h→0
[6n + 3h2(12 + 22 + … + (n – 1)2]
+ 8h[1 + 2 + … + (n – 1)]
= limh→0
h
6 3 2 1 2 1
6
8 1
2n h
n n n hn n+ +⎡⎣⎢
⎤⎦⎥
− − −( )( ) ( )
= limh→0
6 21
2nh nh nh h nh h+ − − +⎡
⎣⎢( )( )
4nh nh h+ − ⎤⎦⎥
( )
=
6 21
22 2 4 4 2 2( ) ( )( )( ) ( )( )+⎡
⎣⎢⎤⎦⎥
− − + −h h h
( Q nh = 2)= 12 + 1(2 – 0) (4 – 0) + 8(2 – 0)= 12 + 8 + 16= 36 sq. units
28. Part I : Pt. (1, 2, 3) ∴ A→
= ˆ ˆ ˆi j k+ +2 3
Let D.rs of the required line are a, b, cRequired line is || to the given two planes∴ Required line is ⊥ to the normal to the
given two planes
Let
when
when
p xdp x x dxdp
x xdx
x p
x p
====
==
== == ∞∞
== ==
⎡⎡
⎣⎣
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
tantan sec
tan sec
,
,
2
2
2
2
2
20 0
ππ
262 ■ SHIV DAS SENIOR SECONDARY SERIES (XII)
P
Y
X′ X
Y′
9
8
7
6
5
4
3
2
1
0 1 2 3 4 5 6 7 8
A(0, 4)
D(0, 0)
B(2, 3)
3x + y = 9
x + 2y = 8
�
C(3, 0)•
∴ a – b + 2c = 0and 3a + b + c = 0
a b c− −
−− +
= =1 2 1 6 1 3
= p (Let)
a = – 3p, b = 5p, c = 4p⇒ a = – 3, b = 5, c = 4
∴ B→
= − + +3 5 4ˆ ˆ ˆi j kVector equation of a line is
r→ =
→+
→A Bλ
∴ r i j k i j k→→ == ++ ++ ++ −− ++ ++( ˆ ˆ ˆ) ( ˆ ˆ ˆ)2 3 3 5 4λλPart II :
Now we have,
r i j k i j k→ = + + + − + +(ˆ ˆ ˆ) ( ˆ ˆ ˆ)2 3 3 5 4λ
r i j k→ = − + + + +( ) ˆ ( ) ˆ ( ) ˆ1 3 2 5 3 4λ λ λ
Let P(1 – 3λ, 2 + 5λ, 3 + 4λ) …(i) be the requiredpoint of intersection, must satisfy the givenequation of plane i.e.,
r i j k→ + +.( ˆ ˆ ˆ)2 = 4
⇒ [( ) ˆ ( ) ˆ ( ) ˆ]1 3 2 5 3 4− + + + +λ λ λi j k .
[ˆ ˆ ˆ]2 i j k+ + = 4
⇒ 2(1 – 3λ) + (2 + 5λ) + (3 + 4λ) = 4⇒ 2 – 6λ + 2 + 5λ + 3 + 4λ = 4⇒ 3λ = 4 – 7⇒ 3λ = – 3 ⇒ λλλλλ = – 1Putting the value of λ in (i), we get
P(1 + 3, 2 – 5, 3 – 4)Therefore, P(4, – 3, – 1) is the required point
of intersection.
29. Silver Gold ProfitType A, x 3g 1g � 40Type B, y 1g 2g � 50
9g 8g(at most) (at most)
Maximise, Z = � (40x + 50y)Subject to the constraints, 3x + 1y ≤ 9
1x + 2y ≤ 8x, y ≥ 0
Let 3x + y = 9 Let x + 2y = 8
x 0 3 x 0 8
y 9 0 y 4 0
Corner Points Z = �(40x + 50y)
A (0, 4) 0 + 200 = � 200
B (2, 3) 80 + 150 = � 230 →Maximum
C (3, 0) 120 + 0 = � 120
D (0, 0) 0 + 0 = � 0
Maximum value of Z is ����� 230 at B(2, 3) i.e.
The company should produce 2 units ofType A and 3 units of Type B goods tomaxize the Profit.