MHT-CET ✦ TEST PAPERS MATHEMATICS ✦ Test Paper Answers

Answers,
11
Test
Paper
Hints and Solutions
1. (c) 2. (b) 3. (b) 4. (b) 5. (d) 26. (a) 27. (c) 28. (b) 29. (a) 30. (d)
6. (b) 7. (c) 8. (b) 9. (b) 10. (c) 31. (b) 32. (a) 33. (a) 34. (b) 35. (c)
11. (c) 12. (b) 13. (d) 14. (a) 15. (c) 36. (b) 37. (b) 38. (b) 39. (a) 40. (c)
16. (a) 17. (d) 18. (c) 19. (d) 20. (d) 41. (a) 42. (b) 43. (b) 44. (c) 45. (b)
21. (c) 22. (b) 23. (c) 24. (a) 25. (a) 46. (b) 47. (c) 48. (b) 49. (a) 50. (b)

1. Equation of curve is y = 4x2. y" 2

y= x + (y' – y" x) x
2
Y
y = 4x.2 ∴ y" x2 – 2xy' + 2y = 0
y=4
which is the required differential equation.
5. Put x + y = u.
y=1
dy du dy du
∴ 1+ = ∴ = –1
dx dx dx dx
O X du
∴ – 1 = sin u + cos u
∴ Required area = 2 z 1
4 1
2
y dy
∴
dx
du
dx
= sin u + cos u + 1

=
LM 2 y OP
3/ 2
4
=
2
[ 8 – 1] =
14
. 1
N3 Q 1 3 3 ∴
1 + sin u + cos u
du = dx

3. Integrating both sides, we get

Y
log 1 + tan
FG x + y IJ = x+c
3 H 2 K
2 If x = y = π, then c = – π.
1 ∴ Particular solution is
O 1 2 3 4 5 6
X
log 1 + tan
FG x + y IJ = x – π.
4. Given equation is
H 2 K
y = c1 x2 + c2 x ........ (1) 6. Given equation : dy = x 100 – x 2 dx
∴ y' = 2c1 x + c2 ........ (2) Integrating both sides, we get
1
∴ y " = 2c1 ........ (3) y = – (100 – x 2 )3/ 2 + c '
3
∴ Equation (2) becomes
∴ 3y = c – (100 – x2)3/2 .... [3c' = c, say]
y' = y" x + c2
which is the general solution.
∴ c2 = y' – y" x ........ (4)
∴ From equation (3) and (4), equation (1)
becomes

( M-1 )
MHT-CET ✦ TEST PAPERS MATHEMATICS ✦ Test Paper Answers

8. Dividing by x2 to given equation, we get Y C(3,5)

LM y cos FG y IJ + FG y IJ sin FG y IJ OP
2

MN x H x K H x K H x K PQ (1,2)
L F y I y F y I O dy = 0
+ Mcos G J – sin G J P A
N H x K x H x K Q dx B(5,1)
y dy dv
Put = v ∴ = v+ x X
x dx dx O
∴ (v cos v + v2 sin v) + (cos v – v sin v) x=1 x=3 x=5

FG v + x dv IJ = 0 ∴ Required area
H dx K = z 3
(line AC) dx + z 5
(line BC) dx
2 dx cos v – v sin v 1 3
∴
x
+
v cos v
dv = 0
– z
1
5
(line AB) dx
Integrating both sides, we get
2 log |x| + log |v| + log |cos v| = log c
∴ A=
1
2 z
1
3
(3 x + 1) dx + z 5
3
(11 – 2 x) dx

∴ x2 v cos v = c ∴ xy cos
FG y IJ = c
H xK –
1
4 z1
5
(9 – x) dx

∴ f ( x) = cos
y
.
=
LM
1 3x 2
+x
OP 3

+ 11x – x 2
5
–
1 LM
9x –
x2 OP 5

x
9. Given equation :
dy ax + 3
=
2 2 N Q 1
3 4 N 2 Q 1

dx 2 y + f 1 1
= (14) + (6) – (24) = 7
∴ (2y + f) dy = (ax + 3) dx 2 4
Alternative method :
Integrating both sides, we get
1 2 1
2 y2 ax 2 1
+ fy = + 3x + c A (∆ ABC) = 5 1 1 = 7 sq. units.
2 2 2
3 5 1
ax 2
∴ – + y 2 – 3x + fy – c = 0 12. Here, P = 2 tan x, Q = sin x
2
The equation represent a circle if –
a
=1
∴ z P dx = 2 log sec x
2
(i.e., coefficient of x2 = coefficient of y2). ∴ I.F. = e
z P dx
= e 2 log sec x = sec 2 x
∴ General solution is
∴ a = – 2.

10. Let I.F. = tan–1 x ∴ e z P dx

= tan – 1 x
z
y • sec 2 x = sin x • sec 2 x dx + c

z
2
∴ y sec x = sec x + c.
∴ P dx = log tan –1 x
π
d 1 1 If x = , y = 0 ∴ c = – 2.
∴ P= (log tan –1 x) = • 3
dx –1
tan x 1 + x 2 ∴ Required particular solution is
11. The vertices of ∆ ABC are A(1, 2), B(5, 1) and
y sec2 x = sec x – 2
C(3, 5).
i.e., y = cos x – 2 cos2 x.
∴ Equation of AB is x + 4y = 9
13. Given general solution is
Equation of BC is 2x + y = 11, and
y = c1 e2x + c2 + c3 ex + c4 sin (x + c5)
Equation of AC is 3x – 2y = –1.
= c1 e2x · ec2 + c3 ex + c4 sin (x + c5)

( M-2 )
MHT-CET ✦ TEST PAPERS MATHEMATICS ✦ Test Paper Answers

Taking c1 ec2 = a, c3 = b, c4 = c and c5 = d. dP

2x x
16. From given : = kP ∴ dP = k dt
∴ y = ae + b e + c sin (x + d) dt
As general solution contains four arbitrary Integrating, we get log P = kt + c1
constants a, b, c and d. Hence, the order of ∴ P = ekt + c1 = ekt · ec1
its differential equation is 4. ∴ P = c ekt (ec1 = c, say)
14. Given equation is dy y + x 2 + x 2
1 dy 1 17. Given equation : =
• + cos x • n – 1 = sin 2 x dx x
n
y dx y dy dv
Put y = vx ∴ = v+ x
1 1 dy 1 dt dx dx
Put n – 1 = t ∴ • = •
y y n
dx 1 – n dx dv
∴ v+ x = v + v2 + 1
1 dt dx
∴ • + cos x t = sin 2 x 1 1
1 – n dx ∴ dv = dx
dt 2
v +1 x
∴ + (1 – n) cos xt = (1 – n) sin 2x
dx Integrating both sides, we get
I. F. = e z
(1 – n) cos x dx (1 – n) sin x
∴ =e
log v + v2 + 1 – log x = log c
∴ General solution is

z
t • e(1 – n) sin x = e(1 – n) sin x • (1 – n) sin 2 x dx ∴
v + v2 + 1
=c ∴ y+ y 2 + x 2 = x 2 c.
∴
1
yn – 1
• z
e (1– n) sin x = (1 – n) e (1 – n) v • 2v dv + c
x
18. Given equation is
R 2v U y = a + be5x + ce– 7x ........ (1)
= (1 – n) S
T1 – n
• e ( 1 – n) v
– z e (1 – n) v
1– n
• 2 dvV + c
W Differentiate w.r.t. x, we get
2 y' = 5b e5x – 7ce– 7x ........ (2)
= 2 v e ( 1 – n) v – •e
( 1 – n) v
+c 5x – 7x
1– n y" = 25b e + 49c e ........ (3)
1 2 and y''' = 125b e 5x
– 343c e – 7x
........ (4)
∴ = 2v – + c e( n – 1) v
yn – 1 1– n
Equation (2), (3) and (4) are consistent.
2
= 2 sin x – + c • e ( n – 1) sin x y' 5 –7
1– n
∴ P (x) = sin x. ∴ y'' 25 49 =0
15. y ' ' ' 125 – 243
Y
y' 1 –1
∴ y'' 5 7 =0
y=e .x y ' ' ' 25 – 49
∴ – 420 y' + 24 y" + 12y''' = 0
∴ y''' + 2y" – 35y' = 0
O X ∴ A = 1, B = 2, C = – 35
x = –1 x=1 ∴ A + B + C = – 32.
Required area = z
–1
1
e x dx = e –
1
e 19. Given equation of curves x2 + y2 = 25 and
16
e2 – 1 y2 = – (x – 7) and the lines x = ± 4.
= sq. units 3
e
∴ Required area is as shown in figure.

( M-3 )
MHT-CET ✦ TEST PAPERS MATHEMATICS ✦ Test Paper Answers

Y 22.
Y
y=4 y = x.2

y=0 2
X
O C
y=–4 1 B

X
O A

∴ Required area = 2 z
0
4
xRC – xLC dy x=1

=2 z 4 LMFG7 – 3 y IJ –
2
25 – y 2 OP dy Required area = A ( OABC) – z 1 2
x dx
NH 16 K
0
0
Q LM x OP
3 1
1
= 1– = 1–
LM 3
= 2 7y – •
y
–
3
y
25 – y 2 –
25
sin –1
y OP 4
N3Q 0
3
N 16 3 2 2 5 Q 0 2
=sq. units
L 1
= 2 M28 – • 64 –
4
9 –
25
sin –1
4 OP 3
N 16 2 2 5 Q 23. From given condition :
4 dA 1
= 36 – 25 sin –1 = –k A ∴ dA = – k dt
5 dt A
20. From given condition Integrating, we get 2 A = – k t + c
dh If t = 0, A = 25 lakh ∴ c = 2 25 = 10
= – 0.5 h (height is decreasing)
dt If t = 2, A = 6.25 lakh,
1
∴ dh = – 0.5 dt 5
h ∴ 2 6 . 25 = – k (2) + 10 ∴ k=
2
Integrating, we get 5
If A = 0 ∴ 2 0 = – (t) + 10
2 h = – 0.5 t + c ........ (1) 2
Initially, t = 0, h = 25 ∴ c = 10 5
∴ t = 10 ∴ t = 4 years.
2
∴ 2 h = – 0.5 t + 10 ........ (2)
24. The circle passing through the points A(2, 0)
If t = 10, then 2 h = – 0.5 (10) + 10 and B(–2, 0). ∴ Its centre lies on Y-axis.
25
∴ 2 h =5 ∴ h= = 6.25 cm. Consider C(0, a) be the centre of the circle.
4
dy ∴ Radius of circle = 4 + a 2 = AC
21. We have, y 2 = (1 + x 2 )
dx ∴ The equation of family of circles passing
1 1 through the points A and B is
∴ dx = 2 dy which is S.V. form.
1 + x2 y
x2 + (y + a)2 = a2 + 4
Integrating both sides, we get
where, a is arbitrary constant.
1 1
tan–1 x = – +c ∴ tan–1 x + =c Y
y y
which is the general solution.

C(0,a)
X
B O A
(–2, 0) (2, 0)

( M-4 )
MHT-CET ✦ TEST PAPERS MATHEMATICS ✦ Test Paper Answers

Differentiate w.r.t. x, we get dy 4 xy

27. Given differential equation is = 2
2x + 2 (y + a) y1 = 0 dx x + 3
1 4x
∴ x + (y + a) y1 = 0 ∴ dy = 2 dx
y x +3
x x F I 2

∴ y+a=–
y1
and a 2 = y + GH JK Integrating both sides, we get

∴ Equation (1) becomes,

y1 1
yzdy = 2 2
2x
x +3
z
dx

F x I = Fy+ x I + 4
+ G–
2 2 ∴ log y = 2 log (x2 + 3) + log c

H y JK GH y JK
2
x y
1 1 ∴ log = log ( x 2 + 3) 2
c
x 2
=G
F yy + x I + 4 2
∴ y = c (x2 + 3)2 ........ (1)
H y JK
1
∴ x2 + 2
y 1 1 This curve passes through the point (0, 9).
2 2 2
∴ x y1 + x = (yy1 + x) + 4y1 2 2 ∴ 9 = c (32) ∴ c=1
2 2
∴ (x – y – 4) y1 = 2xy ∴ Equation (1) becomes, y = (x2 + 3)2
which is the required differential equation. which is the equation of required curve.
25. Equation of lines are ± 4x + y = 8 and y = 2. 29. From given condition,

∴ A –
FG 3
,2 , B
IJ FG
3
,2
IJ and C (0, 8).
dP
=
rP
∴
1
dP =
r
dt
H 2 2K H K dt 100 P 100
r
Y Integrating, we get log P = t+c
100
4x + y = 8 – 4x + y = 8 If t = 0, P = 1000 ∴ c = log 1000
P r
∴ log = t
1000 100
C(0,8)
– 3 ,2 A B
3 ,2 P 5 × 10
2 2 If r = 5, t = 10 ∴ log =
1000 100
y=2 P 1 P
∴ log = ∴ = e1 / 2
X 1000 2 1000
(–2,0) O (2,0)
∴ P = 1000 × e = 1000 × 1.648
∴ P = 1648.
∴ Required area = A (∆ ABC) 30. Order Degree Yes/Not
1
= × base × height (a) 3 1 Not same 2
2
1 (b) 4 2 Not same 2
= × 3 × 6 = 9 sq. units (c) 2 1 Not same 2
2
dy dv (d) 1 1 Same 3
26. Put x + y = v. ∴ = –1
dx dx 31.
dv Y
∴ = 1 + tan 2 v = sec 2 v
dx
∴ cos2 v dv = dx. –1 O 1
X
Integrating both sides, we get
2 (y – x) + sin 2 (x + y) = c
x

y=
–e–

which is the required general solution.

–e
y=

.x

( M-5 )
MHT-CET ✦ TEST PAPERS MATHEMATICS ✦ Test Paper Answers

Required area = 2 z 1
0
– e x dx = 2 – e x
1

0
Y

= 2 |– (e – 1)|
= 2 (e – 1) sq. units
32. Let m be the amount of radium at any time t. O X

∴ From given condition

dm 1
= – km ∴ dm = – k dt ∴ Equation (1) becomes,
dt m
∴ log m = – kt + c. xy1 = 2y ∴ xy1 – 2y = 0
Initial amount of radius is m0. ∴ A + B = 1 – 2 = –1.
∴ c = log m0 dy
34. Given equation : x + yx – y = 1
∴ log m = – kt + log m0 ........ (1) dx
Pm0 ∴
dy
+
x–1 FG y=
1 IJ 1
At t = 1, m = m0 –
100
dx x H x
∴ P = 1–
K x
FG
∴ log m0 –
Pm0 IJ
= k (1) + log m0 z P dx z FGH 1–
1
dx
IJ
K
H 100 K ∴ I.F. = e
x – log x
=e
x
x
= e x – log x
log (1 / x )
FG
∴ – k = log 1 –
P IJ =e e =e • e
H 100 K 1 e
= ex= •
x

∴ Equation (1) becomes, x x

log m = t • log 1 –
FG P IJ
+ log m0
35. Equation of given curve is y = (x –1) (2 – x).
H 100 K ∴ It cuts the X-axis at A(1, 0), B(2, 0).
If t = 2, then m = ? Y

∴ log m = 2 log 1 –
FG P IJ
+ log m0 (1,0) (2,0)
H 100 K O A B
X

LM F P IJ 2 OP
MN GH
∴ log m = log m0 1 –
100 K PQ
∴ Required area
FG 1 – P IJ 2
∴ m = m0
H 100 K × 100 %
= z
1
2
(– x 2 + 3 x – 2) dx

∴ m = m0
FG 10 – P IJ 2
% LM x + 3x – 2xOP 3 2 2

H 10 K = –
N 3 2 Q 1
F P IJ
After 2 years radium left G 10 –
2
% F – 8 + 12 – 4IJ – FG – 1 + 3 – 2IJ
=G
∴
H 10 K H3 2 K H 3 2 K
of the original amount.
1
33. The vertex of the parabola is at origin and =
6
axis along Y-axis, then equation of parabola 1 dx 1
is 36. Given equation • + + 1= 0
x2 = ± 4ay ........ (1) x2 dy xy
1 dx 1
where, a is arbitrary constant. ∴ 2
• + = –1
x dy xy
Differentiate equation (1), w.r.t. x,
1 1 dx dt
2x Put = t ∴ – 2 • =
2x = ± 4ay1, i.e., ± 4a = x x dy dy
y1
( M-6 )
MHT-CET ✦ TEST PAPERS MATHEMATICS ✦ Test Paper Answers

dt t dt 1 8
∴ – + = –1 ∴ – t=1 ∴ A2 = 2 3 –
dy y dy y 3

∴ I. F. = e z
P dy
=e
– z 1
y
dy
=e – log y
=
1 ∴ Required area =
1
3
+2 3 –
8
3
y 7
z
∴ t • (I.F. ) = Q • (I. F. ) dy
= 2 3 – sq. units
3

∴ t•
1
y
1
z
= 1 • dy + log c 1
y
39. Let r, S, V be the radius, surface area and
volume of a spherical rain drop at time t.
t ∴ From given condition
∴ = log y c 1
y dV dV
∝S ∴ = kS
1 dt dt
∴ = log y + log c 1 dV
xy ∴ = k (4 π r 2 )
1 dt
∴ log y – =c 4 dV dr
xy V = πr3 ∴ = 4π r2 •
3 dt dt
dy y e– 2 x 2 dr
37. Given equation + = ∴ 4πr 2
= k (4 π r )
dx x x dt
is linear differential equation where dr
∴ =k ∴ dr = k dt
1 e– 2 x dt
P= and Q = . Integrating, r = kt + c ........ (1)
x x
If r = 3 mm, t = 0 ∴ c = 3.
z = ez
1
P dx dx
∴ I.F. = e x
= e2 x
. ∴ r = kt + 3 ........ (2)
38. At t = 1, r = 2 mm
Y
y.2 = 1 + x ∴ k=–1 ∴ r=–t+3
1 x ∴ r = 3 – t, 0 ≤ t ≤ 3.
y=
A2 x 40. From given condition
X dy
–1 O 1 2 y=x ∴ y dy = x dx
A1 dx
x –1 Integrating both sides, we get
y=–
x y2 x2 c
x = –1 x=0 x=2
= + ∴ y2 = x2 + c
2 2 2
Required area = A1 + A2 This curve passes through the point (5, 3).
A1 = 1 – z–1
0
1 + x dx ∴ c = – 16.
∴ The required curve is x2 – y2 = 16 which
2 0
= 1 – (1 + x) 3 / 2 is a hyperbola.
3 –1
2 1 42. The given equation is
∴ A1 = 1 – [ 1 – 0 ] =
3 3 x2 + y2 – 2ay = 0 ........ (1)
A2 =
2
z 2
0
1 + x dx – 2
2
Differentiate w.r.t. x, we get
2x + 2yy1 – 2ay1 = 0
= (1 + x) 3 / 2 – 2 2x + 2 y y 1
3 0
2 ∴ 2a =
= 3 3–1 –2 y1
3 ∴ Equation (1) becomes

( M-7 )
MHT-CET ✦ TEST PAPERS MATHEMATICS ✦ Test Paper Answers

F 2x + 2 y y I y = 0 46. Given equation of curve is y = 16 – x 2 , i.e.,

x2 + y2 – GH y JK 1
1
x2 + y2 = 16, i.e., the standard circle.
2 xy Y
∴ x 2 + y2 = + 2 y2
y1
∴ (x2 – y2) y1 = 2xy.
This is the required differential equation. O
X
∴ Order = Degree = 1.
∴ Sum of order and degree = 2. x=4
dy x 2 + y 2 x=0
43. Let = π a 2 π × 16
dx x 2 + xy ∴ Required area = =
dy dv 4 4
Put y = vx ∴
= v+ x = 4π sq. units
dx dx
dv 1 + v2 dv 1 – v 47. Given equation is (by + k) dy = (ax + h) dx
∴ v+ x = ∴ x =
dx 1+ v dx 1 + v Integrating both sides, we get
Integrating both sides, we get by 2 ax 2
+ ky = + hx + c
∴ z1+ v
1– v
dv =
1
x
dx z 2 2
i.e., by2 – ax2 – 2hy + 2kx – 2c = 0
∴ – v – 2 log (1 – v) = log x + log c 2 (a) If a = b = 0, then equation is
y
∴ – – 2 log
x– y FG
= log x + c
IJ kx – hy – c = 0 which is not parabola.
x x H K 2 (b) If a = 1, b = 2, then
F x I= y+c
∴ log GH (x – y) JK x
2
2y2 – x2 – 2hy + 2kx – 2c = 0
Compare with
44. Given differential equation is
Ax2 + 2Hxy + By2 + 2Gx + 2Fy + c = 0
sin x dx + sin x dy = cos x dx – cos x dy
∴ H 2 – AB ≠ 0 which is not parabola.
∴ e sin x – j
cos x dx = – e j
sin x + cos y dy 3 (c) If a = 0, b ≠ 0,
dy cos x – sin x ∴ ∆ ≠ 0 and H 2 – AB = 0
∴ =
dx sin x + cos y This equation represent parabola.
∴ Order = 1 and degree = 1. 2 (d) If a = 2, b = 1, then equation is
dy y2 – 2x2 – 2hy + 2kx – 2c = 0
45. Given equation : sec2 y + x tan y = x3
dx ∴ H 2 – AB ≠ 0
dy dv
Put tan y = v ∴ sec 2 y = ∴ Equation does not represent parabola.
dx dx
∴
dv
+ xv = x 3 ∴ I. F. = e
x dx 2
= ex / 2 z Note ....✍
dx
∴ General solution is The general equation of conic is
ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents
2

z 2
v • e x / 2 = x 3 • e x / 2 dx + c . Put
x2
2
=t (i) a parabola, if ∆ ≠ 0 and h2 – ab = 0

z
∴ v e t = 2 t e t dt = 2 e t (t – 1) + c (ii) an ellipse, if ∆ ≠ 0 and h2 – ab < 0
(iii) a hyperbola, if ∆ ≠ 0 and h2 – ab > 0
∴ v = 2 (t – 1) + c · e– t
2
i.e., tan y = x 2 – 2 + c e – x /2
.
( M-8 )
MHT-CET ✦ TEST PAPERS MATHEMATICS ✦ Test Paper Answers

(iv) a rectangular hyperbola,

2
if ∆ ≠ 0 and h – ab > 0 and a + b = 0
= z
1
t (1 + t)
dt =
1
–
t 1+ t
1
z FGH
dt
IJ
K
(v) a pair of lines, if ∆ = 0, h2 – ab ≥ 0. = log
t FG = log
IJ cos 2 x

are nesessary conditions.

1+ t H K
1 + cos 2 x

48. Given equation is

∴ z P dx = log
cos 2 x
2 cos 2 x
y 2 = 2c ( x + c ) ........ (1) ∴ I.F. = e
z
P dx
=
cos 2 x
2 cos 2 x
Differentiate w.r.t. x, we get
∴ General solution is
dy dy
2y
dx
= 2c ∴ c = y

∴ Equation (1) becomes,

dx y•
cos 2 x
2
2 cos x
= cos 2 x •z cos 2 x
2 cos 2 x
dx

1
dy F dy I = sin 2 x + c 1
4
y2 = 2 y
dx
x+ GH y
dx JK ∴ 2y •
cos 2 x
= sin 2 x + c .... [4c1 = c]
2 dy cos 2 x
y – 2 xy ∴ P(x) = sin 2x.
dx = dy
∴ y
dy dx 50. Given function y = tan x
2y
dx
Y
∴ 4
y – 4 xy
dy3
+ 4 x2 y2
dy FG IJ 2 y = tan x

dx dx H K
F dy I dy
= 4y G J y 2
2

H dx K dx
∴ y 4 – 4 xy 3
dy F dy I
+ 4x y G J = 4 y G J
2 2 F dy I
2
3
3
O p
dx H dx K H dx K 4
X

dy
∴ The higher order derivative is , its
dx
power is 3. p
∴ Order = 1 and degree = 3. x=0 x=
4

49. Given equation

dy tan 2 x
–
dx cos 2 x
y = cos 2 x ∴ Required area = z0
π/4
π/4
tan x dx
tan 2 x = log|sec x|
Here, P = – . 0
cos 2 x
π
∴ z
P dx = – z sin 2 x
cos 2 x • cos 2 x
dx
= log sec
4
– log|sec 0|

=– z2 sin 2 x
cos 2 x (1 + cos 2 x)
dx = log 2 – log 1 =
1
2
log 2 .

❏❏❏

( M-9 )