Solution

TEST SERIES-1 DIFFERENTIATION

Class 12 - Mathematics
Section A
1.
(b) ex cot ex
Explanation:
ex cot ex
y = log (sin ex)
log (sin ex)
dy d
=
dx dx

= 1

sine
x
dx
d
sin ex

my
= 1
cos e
x d ex
x
sine dx

= cot e x
(e )
x

= ex cot ex

e
2. (a) f ‘ (x) = g ‘ (x)

ad
Explanation:
1+x 1 (1−x).1−(1+x).(−1) 1
−1 ′
g(x) = tan ( ) ⇒ g (x) = =
1−x 1+x 2 2 2
(1−x) (1+ x )
1+(
1−x
)
Ac
3. (a) 2x (log 2)
Explanation:
Given that y = 2x
ine

Taking log both sides, we get

loge y = x loge 2 (Since loga bc = c loga b)
Differentiating with respect to x, we get
dy dy
1
or
ash

= log 2 = log 2 × y
y dx e dx e

dy
Hence dx
= 2
x
loge 2

4.
(b) (tan x)cot x . cosec2 x (1 - log tan x)
dy

Explanation:
Given that y = (tan x)cot x
Taking log both sides, we obtain
Vi

loge y = cot x × loge tan x (Since loga bc = cloga b)

Differentiating with respect to x, we obtain
tanx × cosec2x = cosec2x (1 - loge tanx)
1 dy 1 2
= cot x × × sec x − log
y dx tan x e

cosec2x (1 - loge tan × y) = cosec2x (1 - loge tanx) (tanx)cotx

dy
Therefore, dx
=

5.
(b) 1

Explanation:
Given that, x = at2, y = 2at
dy
dx

dt
= 2at and dt
= 2a
dy

dy
Therefore, dx
= dt

dx
= 2a

2at
= 1

dt

1/5
Contact - +91-9999906710
 6.
(c) tan θ
Explanation:
x = a(cos θ + θ sin θ) ,we get
dx
∴ = a(− sin θ + sin θ + θ cos θ)
dθ
dθ 1
⇒ =
dx aθ cos θ

y = a(sin θ − θ cos θ) ,we get

dy
∴ = a(cos θ − (cos θ + θ(− sin θ))
dθ
dy
⇒ = acos θ − acos θ + θ asin θ
dθ
dy
⇒ = aθ sin θ
dθ
dy dy dθ
⇒ = ×
dx dθ dx
dy 1
⇒ = aθ sin θ ×
dx aθ cos θ
dy
⇒ = tan θ
dx

my
7.
(b) 3 (xy2 + y1) y2
Explanation:

e
ax+b
y =
2

ad
x +c

⇒ y(x2 + c) = ax + b
Differentiating both sides w.r.t. x we get
y1 (x2 + c) + 2xy = a
Ac
Again differentiating w.r.t to x we get
y2 (x2 +c) + 2xy1 + 2y + 2xy1 = 0

⇒ y2 (x2 + c) = -(4xy1 + 2y) ...(i)

ine

Again differentiating w.r.t to x we get

y3 (x2 + c) + 2xy2 + 4xy2 + 4y1 + 2y1 = 0

⇒ y3 (x2 + c) = -(6xy2 + 6y1) ...(ii)

ash

y2 2x y1 +y
=
y 3x y +3y
3 2 1

⇒ y3 (2xy1 + y) = 3y2 (xy2 + y1)

dy

8.
(c) a function of y only
Explanation:
Vi

y = ax2 + bx + c
dy
= 2ax + b
dx
2
d y
= 2a
dx2
2
d y
y
3
2
= 2ay
3
= A function of y only
dx

9. (a) neither continuous nor differentiable at x = 1

Explanation:
2

We know that domain of sin-1 x is |x| < 1 and so sin-1 (

1+x

2x
) defined only for |x| < 1.
Hence, f(x) is neither continuous nor differentiable at x = 1.
10.
√3+1
(b) 2

Explanation:

2/5
Contact - +91-9999906710
 Given, f(x) = |cos x - sin x|,
We know that, < x < , sin x > cos x π

4
π

∴ cos x - sin x < 0

i.e. f(x) = -(cos x - sin x)
f'(x) = -[-sin x - cos x]
π − √3 1 √3+1
∴ f'( 3
) = -( 2
−
2
) =( 2
)

11. Let y = log(tan-1x)

differentiating both sides with respect x, we get,
dy d −1
= log(tan x)
dx dx
d
=
1

−1
×
dx
(tan
−1
x) [Using chain rule]
tan x
1
=
2 −1
(1+ x ) tan x

So, dx
d
{log(tan
−1
x)} =
2
1

−1
(1+ x ) tan x

my
12. We have, f(x) = loge(logex)
differentiating both sides with respect x, we get,
1 d
′
f (x) =
loge x dx
(loge x) [using chain rule]

e
′ 1 1
⇒ f (x) = ( )
loge x x

1 1

ad
′
⇒ f (e) = ( ) [∵ x = e]
loge e e

′ 1
⇒ f (e) = [∵ loge e = 1]
e

13. Let y = sin3 x + cos6 x

Ac
Differentiating both sides with respect to x
dy d d
3 6
= (sin x) + (cos x)
dx dx dx

= 3 sin 2
x ×
d

dx
(sin x) + 6 cos
5
x ×
dx
d
(cos x)

= 3 sin 2
x × cos x + 6 cos
5
x × (− sin x) [∵
d
(sin x) = cos x &
d
(cos x) = − sin x]
ine

dx dx

= 3 sin x cos x (sin x − 2 cos 4

x)

dy
4
∴ = 3 sin x cos x (sin x − 2 cos x)
dx

Section B
ash

2
14. Let y = e + 3 cos
sec
x
x −1

This is defined at every real number in [-1,1]. Therefore

dy 2 d 1
sec x 2
= e ⋅ (sec x) + 3 (− )
dx dx √1−x2
dy

2 d 1
=e sec x
⋅ 2 sec x (sec x) + 3 (− )
dx √1−x2

2 1
= 2sec x (secxtanx) e sec x
+ 3 (− )
√1−x2
Vi

2 1
= 2sec 2
x tan x e
sec x
+ 3 (− )
√1−x2

Observe that the derivative of the given function is valid only in (– 1, 1) as the derivative of cos–1 x exists only in
(– 1, 1).
15. Let y = log (xx + cosec2x). Then,
dy

dx
=
x
1

2
×
d

dx
(x
x
+ cosec x)
2
[using chain rule]
x +cose c x

dy 1 d d
x 2
⇒ = { (x ) + (cosec x)}
dx x 2 dx dx
x +cose c x

[ ab = ebloga]
dy 1 d d
x log x 2
⇒ = { (e )+ (cosec x)}
dx x 2 dx dx
x +cose c x

dy 1 d d
x log x
⇒ = {e (x log x) + 2cosecx (cosecx)}
dx x 2 dx dx
x +cose c x

dy 1 x 2
⇒ = {x (1 + log x) − 2cosec x cot x}
dx x 2
x +cose c x

16. Let u = log x

Differentiating both sides with respect to x, we obtain
du

dx
= 1

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Contact - +91-9999906710
 Let v = cot x
Differentiating both sides with respect to x, we obtain
dv

dx
= -cosec2 x
Now, du

dv
= du

dx
×
dx

dv
= 1

x
× (−
1

2
)
cose c x
2

⇒
du

dv
=− sin

x
x

17. Given, y = sec x - tan x .....(i)

Differentiating both sides of (i) w.r.t. x,
= sec x tan x - sec2 x ⇒
dy dy

dx dx
= sec x (tan x - sec x)
dy dy
⇒
dx
= cos x
1
{-(sec x - tan x)} ⇒ cos x dx
= -y [From (i)] ....(ii)
Again Differentiating both sides of (ii) w.r.t. x,
2
d y dy dy
cos x ⋅
2
+
dx
(− sin x) =- dx
dx
2
d y dy dy
⇒ cos x ⋅ =− + sin x
dx2 dx dx
2
d y dy

my
⇒ cos x ⋅
2
= dx
(-1 + sin x)
dx
2
d y (sec x−tan x)
⇒ cos x ⋅
2
= {-(1 - sin x)} {− cos x
}
dx
2
d y 1−sin x
⇒ cos x ⋅
2
=( cos x
) (sec x - tan x)
dx

e
2
d y
⇒ cos x ⋅
2
= (sec x - tan x) (sec x - tan x)
dx

ad
2

= y2
d y
∴ cos x ⋅
dx2

Section C
18. According to the question, x y x−y
= e
Ac
Taking log both sides ,
⇒ ylo ge x = (x − y)lo ge e

⇒ ylo ge x = (x − y)

⇒ y(1 + logx) = x
ine

x
⇒ y =
1+log x

Differentiating both sides w.r.t x,

d d
(1+log x) (x)−x (1+log x)
dy
[ Using quotient rule of derivative]
dx dx
⇒ =
2
ash

dx (1+log x)

1
1+log x−x⋅ 1+log x−1
=
2
x
= 2
(1+log x) (1+log x)

dy log x
∴ =
dx 2
(1+log x)

Hence Proved
dy

19. According to the question,we are given that x = 2 cosθ - cos 2θ

and y = 2 sin θ - sin 2θ
dy
then we have to prove that 3θ
.
Vi

= tan( )
dx 2

Therefore, differentiating both sides w.r.t θ, we get,

dx
= −2 sin θ + 2 sin 2θ
dθ
dy
and = 2 cos θ − 2 cos 2θ
dθ
dy dy/dθ 2(cos θ−cos 2θ)
∴ = =
dx dx/dθ 2(− sin θ+sin 2θ)

C+D D−C
θ+2θ 2θ−θ
2 sin( ) sin( ) ⎡ ∵ cos C − cos D = 2 sin( ) sin( ) ⎤
2 2
2 2
= ⎢ ⎥
2θ+θ 2θ−θ
C+D C−D
2[cos(
2
) sin(
2
)]
⎣ and sin C − sin D = 2 cos(
2
) sin(
2
) ⎦

3θ θ
sin( ) sin( )
2 2
=
3θ θ
cos( ) sin( )
2 2

3θ
= tan( )
2

20. We have,
y2 = a2 cos2 x + b2 sin2 x
⇒ 2y2 = a2 (2 cos2 x) + b2 (2 sin2 x)

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Contact - +91-9999906710
⇒ 2y2 = a2 (1 + cos 2x) + b2 (1 - cos 2x)
⇒ 2y2 = (a2 + b2) + (a2 - b2) cos 2x ....(i)
Differentiating with respect to x, we get
= -2 (a2 - b2) sin 2x
dy
4y dx

= -(a2 - b2) sin 2x ....(ii)

dy
⇒ 2y dx

From (i), we obtain

2y2 - (a2 + b2) = (a2 - b2) cos 2x ...(iii)
Squaring (ii) and (iii) and adding, we get
2

= (a2 - b2)2 {sin2 2x + cos2 2x}

dy 2
2 2 2 2
4y ( ) + {2y − (a + b )}
dx

+ 4y4 - 4y2 (a2 + b2) + (a2 + b2)2 = (a2 - b2)2

dy
2
⇒ 4y ( )
dx

= (a2 - b2)2 - (a2 + b2)2

dy
2 2 2 2
⇒ 4y {( ) + y − (a + b )}
dx

= -4a2b2

my
dy
2 2 2 2
⇒ 4y {( ) + y − (a + b )}
dx

2 2 2
dy a b
2 2 2
⇒ ( ) + y − (a + b )= −
dx y2

Differentiating both sides with respect to x, we get

e
2 2 2
dy d y dy 2a b dy
2( ) + 2y =
dx 2 dx 3 dx
dx y

ad
2 2 2
d y dy
⇒
2
+ y =
a b

3
..[Dividing both side by 2 dx
]
dx y
Ac
ine
ash
dy
Vi

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Contact - +91-9999906710