Solution

MATH

Class 12 - Mathematics

1. Let u = cos-1(4x3 - 3x)

Put x = cosθ
⇒ θ = cos-1x
Now, u = cos-1(4 cos3θ - 3cosθ)
⇒ u = cos-1(cos 3θ) ...(i)
√1−x2

Let, v = tan −1
(
x
)

√1− cos 2 θ
−1
⇒ v = tan ( )
cos θ

−1 sin θ
⇒ v = tan ( )
cos θ

⇒ v = tan-1(tan θ) ...(ii)
Here,
1
< x < 1
2
1
⇒ < cos θ < 1
2
π
⇒ 0 < θ<
3

So, from equation (i),

u = 3θ .. [Since cos-1(cos θ) = θ, if θ ∈ [0, π ]]
⇒ u = 3 cos-1x
Differentiating it with respect to x,
−3
du
= ...(iii)
dx √1−x2

From equation (ii),

v = θ … . [ since , tan if θ ∈ (−
−1 π π
(tan θ) = θ, , )]
2 2

⇒ v = 3 cos-1x
Differentiating it with respect to x,
dv −1
= ...(iv)
dx √1−x2

Dividing equation (iii) by (iv),

du
√1−x2
dx −3
= ( ) (− )
dv 2 1
√1−x
dx

du
∴
dv
= 3 .
2. Given,
x = a (cosθ + θ sinθ) ...(1)
y = a (sinθ - θ cosθ) ...(2)
To prove:
2
d x
(i) 2
= a(cos θ − θ sin θ),
dθ
2
d y
(ii) 2
= a(sin θ + θ cos θ)
dθ
2 3
d y
(iii) 2
=
sec

aθ
θ

dx

We notice a second order derivative in the expression to be proved so first take the step to find the second order derivative.
2
d y
Let's find 2
dx
2
d y dy
As 2
=
dx
d
(
dx
)
dx

dx d
= a(cos θ + θ sin θ)
dθ dθ

= a (-sinθ + θ cosθ + sinθ)

[differentiated using product rule for θ sinθ]
= aθ cosθ ...(3)

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 Again differentiating w.r.t. θ using product rule:
2
d x
= a(−θ sin θ + cos θ)
2
dθ
2

∴
d x

2
= a(cos θ − θ sin θ) ...proved (i)
dθ

Similarly,
dy d d d
= a(sin θ − θ cos θ) = a sin θ − a (θ cos θ)
dθ dθ dθ dθ

= a cosθ + a θ sinθ - a cosθ

dy
∴
dθ
= aθ sin θ ...(5)
Again differentiating w.r.t θ using product rule:
2
d x
= a(θ cos θ + sin θ)
2
dθ
2

∴
d x

2
= a(sin θ + θ cos θ) ...proved (ii)
dθ
dy

dy dθ
∵ =
dx dx

dθ

Using equations (4) and (5)

dy aθ sin θ
= = tan θ
dx aθ cos θ
2
d y dy
As 2
=
d

dx
(
dx
)
dx

∴ Again differentiating w.r.t x:

2
d y d
= tan θ
dx2 dx

dθ
= sec
2
θ
dx
[using chain rule]
dx dθ 1
∵ = aθ cos θ => =
dθ dx aθ cos θ

Putting a value in the above equation

We have:
2
d y 1
2
= sec θ×
2 aθ cos θ
dx
2
d y 3

2
=
sec

aθ
θ
...proved (iii)
dx

3. We have, y = {logcos x sin x}{logsin x cos x}-1 + sin-1( 2x

2
)
1+x

y = {logcos x sin x}{logcos x sin x} + sin-1( [∵ logab = (log ba)-1]

2x
⇒ )
2
1+x

2
log sin x 2x log b
−1
⟹ y = [ ] + sin ( ) … [∵ log b = ]
log cos x 2 a log a
1+x

Differentiating with respect to x,

2
dy d log sin x d 2x
−1
= [ ] + {sin ( )}
dx dx log cos x dx 1+x2

dy log sin x d log sin x 1 d 2x

⇒ = 2[ ] ( ) + × [ ]
dx log cos x dx log cos x 2 dx 2
1+x
2x
√1−( )
1+x2

d d
(log cos x) (log sin x)−log sin x (log cos x) 2 2
dy log sin x (1+ x ) (1+ x )(2)−(2x)(2x)
dx dx
⇒ −2 [ ][ ] +[ ][ ]
2 2
dx log cos x (log cos x) √1+ x4 −2x2 2
(1+ x )

1 d 1 d
log cos x× (sin x)−log sin x× (cos x) 2 2
dy log sin x c os x (1+ x ) (1+ x )(2)−(2x)(2x)
sin x dx dx
⇒ = 2[ ][ ] +[ ][ ]
dx log cos x 2 2
(log cos x) √1+ x4 −2x2 2
(1+ x )

c os x sin x
log cos x×( )+log sin x×( ) 2 2
dy log sin x sin x c os x (1+ x ) (1+ x )(2)−(2x)(2x)
⇒ 2[ ][ ] +[ ][ ]
dx log cos x 2 4 2 2 2
(log cos x) √1+ x −2x (1+ x )

dy log sin x
⇒
dx
= 2
3
(cot x log cos x + tan x log sin x) + 2

2
(log cos x) 1+x

π
put x = 4
π
log sin
dy 4 π π π π 1
⇒ = 2{ } (cot log cos + tan log sin ) + 2{ }
dx 3 4 4 4 4 2
π π
(log cos ) 1+( )
4 4

⎧ ⎫
⎪ ⎪
dy 1 1 1 16
⇒ = 2⎨ ⎬ (1 × log + 1 × log ) + 2( )
dx 2 √2 √2 2
16+π
⎩
⎪ (log
1
)
⎭
⎪
√2

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 1
2 log( )
dy √2 32
⇒ = 2 × +
dx 2 2
16+π
1
{log( )}
√2

dy 1 32
⇒ = 4 +
dx 2
1 16+π
log( )
√2

dy 1 32
⇒ = 4 +
dx 1 2
16+π
− log 2
2

dy 8 32
⇒ = − +
dx log 2 16+π 2

dy
So, ( )
π
= 8[
4
−
1
] .
dx 16+π 2 log 2
x=
4

4. Here, x = a sin 2t(1 + cos 2t)

and y = b cos 2t(1 - cos 2t)
dx

dt
= a[2cos2t (1 + cos2t) + sin2t(-2sin2t)]
= 2a[cos 2t + (cos2 2t - sin2 2t)]
= 2a(cos 2t + cos 4t)
dy
and di
= b[-2sin2t(1 - cos 2t) + cos2t(2sin 2t)
= 2b[-sin 2t + 2 sin 2t cos 2t]
= 2b(- sin 2t + sin 4t)
dy dy
∴ =
dx dt

dx

dy 2b(− sin 2t+sin 4t)

⇒ =
dx 2a(cos 2t+cos 4t)

At t = π

4
,
π
b(− sin +sin π)
dy 2
=
dx π
a(cos +cos π)
2

b(−1+0)
= = b

a
a(0−1)

1
x (1+ )
5. Let y = (x + 1

x
) + x
x

1
x (1+ )
Also, Let u = (x + 1

x
) and v = x x

∴ y=u+v
dy
⇒
dx
=
du

dx
+
dv

dx
...(i)
1 x
Then, u = (x + x
)
1 x
⇒ log u = log (x + )
x
1
⇒ log u = x log(x + )
x

Differentiating both sides with respect to x,

1 du 1 d d 1
= log(x + ) (x) + x [log(x + )]
u dx x dx dx x

1 du 1 1 d 1
⇒ = log(x + )+ x (x + )
u dx x 1 dx x
(x+ )
x

du 1 x 1
⇒ = u [log(x + )+ × (1 − )]
dx x 1 2
x
(x+ )
x

1
(x− )
x x
du 1 1
⇒ = (x + ) [log(x + )+ ]
dx x x 1
(x+ )
x

x 2
du 1 1 x −1
⇒ = (x + ) [log(x + )+ ]
dx x x 2
x +1

x 2
du 1 x −1 1
⇒ = (x + ) [ + log(x + )]
dx x 2 x
x +1

1
(1+ )
Again, v = x x

1
(1+ )
x
⇒ log v = log[x ]

1
⇒ log v = (1 + ) log x
x

Differentiating both sides with respect to x,

1 dv d 1 1 d
= log x (1 + ) + (1 + ) log x
v dx dx x x dx

1 dv 1 1 1
⇒ = (− ) log x + (1 + )( )
v dx 2 x x
x

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 1 dv log x 1 1
⇒ = − + +
v dx 2 x 2
x x

dv − log x+x+1
⇒ = v[ ]
dx 2
x
1
(1+ ) x+1−log x
⇒
dv

dx
= x
x
(
2
) ....(iii)
x

From (i), (ii) and (iii), we obtain

1
dy x 2 x+1−log x
1 x −1 1 (1+ )
x
= (x + ) [ + log(x + )] + x ( )
dx x 2 x 2
x +1 x

−−−−− −−−−−
6. 2 2
y √x + 1 − log(√x + 1 − x) = 0

differentiating both sides w.r.t x

1
−−−−− dy 1 1(2x)
2
y. (2x) + √x + 1. − [ − 1] = 0
2 dx √x2 +1−x 2
2√x +1 2√x +1

xy −−−−− dy 1 x− √x +1
2
2
+ √x + 1. − [ ]= 0
√x2 +1 dx √x2 +1−x √x2 +1

2
2 −( √x +1−x)
xy+( x +1) dy
=
√x2 +1 dx
( √x2 +1−x) √x2 +1

dy
2
xy + (x + 1) = −1
dx
dy
2
(x + 1) + xy + 1 = 0
dx

7. Given: (log x)x + xlog x

Let y = (log x)x + xlog x
Let y = u + v
⇒ u = (log x)x and v = xlog x
For, u = (log x)x
Taking log on both sides, we get
log u = log (log x)x
⇒ log u = x.log (log (x))

Now, differentiate both sides with respect to x

d d
(log u) = [x ⋅ log(log x)]
dx dx

1 du d d
⇒ = x⋅ log(log x)) + log(log x) ⋅ (x)
u dx dx dx

du 1 d
⇒ = u [x ⋅ (log x) + log(log x) ⋅ (1)]
dx log x dx

du x x 1
⇒ = (log x) [ ⋅ + log(log x) ⋅ (1)]
dx log x x

1+log(log x)⋅(log x)
du x
⇒ = (log x) [ ]
dx log x

du x−1
⇒ = (log x) [1 + log x ⋅ log(log x)]
dx

For, v = xlog x
Taking log on both sides, we get
log v = log (xlog x)
⇒ log v = log x. log x
Now, differentiate both sides with respect to x
d d 2
(log v) = [(log x) ]
dx dx
1 dv d
⇒ = 2 ⋅ log x (log x)
v dx dx

dv log x
⇒ = v [2 ⋅ ]
dx x

dv log x
log x
⇒ = x [2 ⋅ ]
dx x

dv log x−1
⇒ = 2 ⋅ x ⋅ log x
dx

Because, y = u + v
dy du dv
⇒ = +
dx dx dx
dy
x−1 log x−1
⇒ = (log x) [1 + log x ⋅ log(log x)] + 2 ⋅ x ⋅ log x
dx

8. We have,
ax2 + 2 hxy + by2 + 2gx + 2fy + c = 0 ...(i)
Differentiating both sides of this respect to x, we get

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 d 2 d d 2 d d d d
(ax ) + (2hxy) + (by ) + (2gx) + (2f y) + (c) = (0)
dx dx dx dx dx dx dx
d 2 d d 2 d d
⇒ a (x ) + 2h (xy) + b (y ) + 2g (x) + 2f (y) + 0 = 0
dx dx dx dx dx

dy dy dy
⇒ 2ax + 2h (x + y) + b2y + 2g × 1 + 2f × = 0
dx dx dx

dy
⇒
dx
(2hx + 2by + 2f) + 2a + 2hy + 2g = 0
dy 2(ax+hy+g) ax+hy+g
⇒
dx
= − = −(
hx+by+f
) ...(ii)
2(hx+by+f )

Differentiating both sides of (i) with respect to y, we obtain

d 2 d d 2 d d d d
(ax ) + (2hxy) + (by ) + (2gx) + (2f y) + (c) = (0)
dy dy dy dy dy dy dy

d 2 d d 2 d d d
⇒ a (x ) + 2h (xy) + b (y ) + 2g (x) + 2f (y) + (c) = 0
dy dy dy dy dy dy

dx dx dx
⇒ a (2x ) + 2h (y + x) + b(2y) + 2g + 2f × 1 + 0 = 0
dy dy dy

2(hx+by+f ) hx+by+f
⇒
dx

dy
= −
2(ax+hy+g)
= −(
ax+hy+g
) ...(iii)
From (i) and (iii), we obtain
dy ax+hy+g hx+by+f

dx
⋅
dx

dy
= −(
hx+by+f
) × −(
ax+hy+g
) = 1 .
dy π
9. If x = a (θ - sin θ) and y = a (1 + cos θ), then,we have to find find dx
at θ = 3
.
Now, x = a (θ - sin θ)
and y = a (1 + cos θ)
On differentiating both sides w.r.t θ, we get
dy
dx

dθ
= a(1 − cos θ) and dθ
= −a sin θ

dy dy/dθ −a sin θ
∴ = =
dx dx/dθ a(1−cos θ)

θ θ
−2a sin cos
dy 2 2
⇒ =
dx 2 θ
a×2 sin
2

dy θ
⇒ = − cot
dx 2

On putting θ = π

3
, we get
dy π –
[ ] = − cot = − √3
dx π 6
θ=
3

dy π –
Hence, dx
at θ = 3
is − √3 .
−−
2
−−
dy x +1
10. We have, dx
, if y = x tan x
+ √
2
−−
2
−−
x +1
Taking, u = x tan x
and v = √
2

log u = tan x log x ...(ii)

2
x +1
and v 2
=
2
...(iii)
On, differentiating Eq. (ii) w.r.t. x, we get
1 du 1 2
. = tan x. + log xsec x
u dx x

du tan x 2
⇒ = u[ + log x. sec x]
dx x

= x
tan x
[
tan x

x
+ log x. sec x]
2
...(iv)
also, differentiating Eq. (iii) w.r.t. x, we get
dv 1 dv 1
2v. = (2x) ⇒ = . (2x)
dx 2 dx 4v

dv 1 x. √2
⇒ = .2x =
dx 2
x2 +1 2√x +1
4. √
2

⇒
dv

dx
=
x
...(v)
2
√2(x +1)

Now, y = u + v
dy du dv
∴ = +
dx dx dx

tan x tan x 2 x
= x [ + log x. sec x] +
x 2
√2(x +1)

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