Solution

PRACTICE SHEET

Class 12 - Mathematics
1. We observe that
(LHL at x = 2) = lim f (x) = lim(2x − 1) = 2 × 2 − 1 = 3
− x→2
x→2

(RHL at x = 2) = lim f (x) = lim(x + 1) = 2 + 1 = 3

+ x→2
x→2

and, f(2) = a
Since f(x) is continuous at x = 2.
Therefore,
lim f (x) = lim f (x) = f (2)
− +
x→2 x→2

⇒ 3 = 3 = a⇒ a= 3

Thus, f(x) is continuous at x = 2, if a = 3

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

′ 1 1
⇒ f (x) = ( )
loge x x

′ 1 1
⇒ f (e) = ( ) [∵ x = e]
log e e
e

′ 1
⇒ f (e) = [∵ loge e = 1]
e
2
x −3x+2
, if x ≠ 1
3. Given function is, f (x) = { x−1

k , if x = 1
If f(x) is continuous at x = 1, then,
lim f (x) = f (1)
x→1
2
x −3x+2
lim = k
x−1
x→1

(x−2)(x−1)
lim = k
x−1
x→1

lim(x − 2) = k
x→1

k = -1
4. Given:
x = f(t) and y = g(t)
dx dy
′ ′
= f (t); = g (t)
dt dt
dy
′
dy dt g (t)
= = ′
dx dx
f (t)
dt

d dy
( )
2 dt dx
d y
=
dx2 dx

dt

1 1 ′′ ′ ′′ ′
= { (g (t)f (t) − f (t)g (t))}
′ ′ 2
f (t) f (t )

′′ ′ ′′ ′
(g (t) f (t)− f (t) g (t))

=
′ 3
(f (t))

5. We have, y = sin-1x + cos-1x

π −1 −1 π
⇒ y = [∵ sin x + cos x = ]
2 2
dy
⇒
dx
= 0 [ d

dx
(constant=0)]
6. Let, y = cos-1 2x and u = 2x
Therefore, y = cos-1 u
Differentiating above equation w.r.t. x,
dy dy du
∴
dx
=
du
⋅
dx
(By chain rule)
dy d d
∴
dx
= du
(cos
−1
u) ⋅
dx
(2x)

1/8
 −1
= ⋅ 2
√1−u2

−2
=
2
√1−(2x)

−2
=
√1−4x2

dy −2
∴ =
dx √1−4x2
−−−−−
7. Given, y = log{sin (√x + 1)} .....(i) 2

On differentiating both sides of (i) with respect to x,

dy 1
−−−−− 1

dx
= 2
2
⋅ cos(√x + 1) ⋅
2
⋅ 2x
sin √x +1 2√x +1

dy −−−−− x
⇒ = cot 2
(√x + 1) ⋅
dx √x2 +1

2
dy x cot( √x +1)
∴ =
dx √x2 +1

8. Here we have, x = a sec3θ and y = a tan3θ

dy d
Now, dθ
= 3a tan
2
θ
dθ
(tan θ)

dy
2 2
= 3a tan θ sec θ
dθ
dx

dθ
= 3a sec
2
θ
d

dθ
(sec θ)
dx

dθ
= 3a sec 2
θ sec θ tan θ = 3a sec
3
θ tan θ

dy

dy 2 2
dθ 3a tan θ sec θ
= =
dx dx 3
3a sec θ tan θ
dθ

tan θ
= = sin θ
sec θ
dy
= sin θ
dx

d dy
( )
2 dθ dx
d y
Now, 2
=
dx
dx
dθ
d
(sin θ)
dθ
=
3
3a sec θ tan θ
5
cos θ
=
3a sin θ
π
At θ = 4
5
1
( )
2
d y √2
1
= =
2 12a
dx 1
3a( )
√2

9. Given: f (x) = 2x − 1 2

2
Continuity at x = 3 lim f (x) = lim(2x 2
− 1) = 2(3) − 1 = 18 − 1 = 17
x→3 x→3

And f (3) = 2(3) 2

− 1 = 18 − 1 = 17

Since lim f (x) = f (x) , therefore, f(x) is continuous at x = 3.

x→3
−1

10. y = e tan
x

dy −1 y −1

dx
=e tan x
(
1

2
) = 2
[y = e tan
x ]
1+x 1+x

dy
(1 + x )
2

dx
=y
differentiating both sides w.r.t. x, we get
2
d y dy dy
⇒ (1 + x )
2

2
+ 2x
dx
= dx
dx
2
d y dy
⇒ (1 + x )
2
2
+ (2x − 1)
dx
=0
dx

11. We have, y = ex + 10x + xx

x
x x log x
⇒ y = e + 10 + e

⇒ y = ex + 10x + ex log x
Differentiating with respect to x,
dy d d d
x x x log x
= (e ) + (10 ) + (e )
dx dx dx dx

x x x log x d
= e + 10 log 10 + e (x log x)
dx

x x x log x d d
= e + 10 log 10 + e [x (log x) + log x (x)]
dx dx
x
x x log x 1
= e + 10 log 10 + e [x ( ) + log x]
x

= ex + 10x log 10 + xx[1 + log x]

2/8
 = ex + 10xlog10 + xx [log e + log x] ... [∵ loge e = 1]

= ex + 10x log 10 + xx (log ex) ...[∵ log A + log B = log AB]

The differentiation of the given function y is as above.
12. Given equation,
y = sin-1 (cos x) + cos-1 (sin x)
Let s = sin-1 (cos x) and t = cos-1 (sin x)
Therefore, y = s + t ....(i)
For sin-1(cos x)
Let u = cos x
Therefore, s = sin-1 u
Differentiating above equation w.r.t. x,
∴
ds

dx
= ⋅ (By chain rule)
ds

du
du

dx

∴
ds

dx
= d

du
(sin
−1
u) ⋅
d

dx
(cos x)

= 1
⋅ (− sin x)
√1−u2

= 1
⋅ (− sin x)
2
√1−(cos x)

= 1
⋅ (− sin x) (∵ sin2 x + cos2 x = 1)
√sin2 x

= 1

sin x
⋅ (− sin x)

= -1
∴
ds

dx
= -1 .....(i)
For cos-1 (sin x)
Let u = sin x
Therefore, t = cos-1 u
Differentiating above equation w.r.t.x,
∴
dt

dx
= ⋅ (By chain rule)
dt

du
du

dx

∴
dt

dx
= d

du
(cos
−1
u) ⋅
dx
d
(sin x)

−1
= ⋅ (cos x)
√1−u2

−1
= ⋅ (cos x)
2
√1−(sin x)

(∵ sin2 x + cos2 x = 1)
−1
= ⋅ (cos x)
√cos 2 x

−1
= cos x
⋅ (cos x)

= -1
∴
dt

dx
= -1 .....(i)
Differentiating eq(i) w.r.t. x,
dy d
∴ = (s + t)
dx dx

= ds

dx
+
dt

dx

= -1 - 1 ......[From (ii) and (iii)

= -2
dy
∴
dx
= -2
Hence proved.
2
6x−4√1−4x
13. Given y = sin −1
[
5
]

Put, x = 1

2
sin θ

2
sin θ sin θ
⎡ 6× −4√1−4×( ) ⎤
2 2

Therefore, y = sin −1
⎢
5
⎥
⎣ ⎦

2
3 sin θ−4√1− sin θ
−1
= sin ( )
5

−1 3 sin θ−4 cos θ

= sin ( )
5

3/8
 3
= sin
−1
(
5
sin θ −
4

5
cos θ) ...........(i)
−−−−−− −−
2
−−−−−−− −
Let cos ϕ = 3

5
, therefore,sinϕ = √1 − cos 2 3
ϕ = √1 − ( )
5

−−−−− −−
9 16 4
= √1 − = √ =
25 25 5

Now, Eq(i) becomes,

y = sin (cos ϕ sin θ − sin ϕ cos θ)
−1

−1
= sin [sin(θ − ϕ)] = θ − ϕ

−1 −1 3
⇒ y = sin (2x) − cos ( )
5

Therefore,on differentiating both sides w.r.t x, we get,

dy 1 d
= (2x) − 0
dx 2 dx
√1−(2x)

2
=
√1−4x2
−−−−− −−−−−
14. According to the question, x = √a and y = √a
−1 −1
sin t cos t

1/2
−1

Consider, x = (a sin t
)

Differentiating both sides w.r.t x,

−1/2
−1 −1
⇒
dx

dt
=
1

2
(a
sin t
)
d

dt
(a
sin t
) [ Using chain rule of derivative]
−1/2
1 −1 −1 d
sin t sin t −1
= (a ) a log a (sin t)
2 dt

−1/2
−1 −1
1 sin t sin t 1
= (a ) a log a ⋅
2 √1−t2

1/2
1 −1 1
sin t
= (a ) log a ⋅
2 √1−t2

1 −1 t
√asin ⋅log a
dx
⇒ =
2
..........(i)
dt √1−t2

1/2
−1
Consider , y = (a cos t
)

Differentiating both sides w.r.t x,

−1/2
dy −1 d −1

dt
=
1

2
(a
cos t
)
dt
(a
cos t
) [Using chain rule of derivative]
−1/2
−1 −1
1 cos t cos t d −1
= (a ) a log a (cos t)
2 dt

1/2
1 −1 (−1)
cos t
= (a ) log a ⋅
2 √1−t2

1 −1 t
− √ac os ⋅log a
dy
⇒
dt
=
2

2
..........(ii)
√1−t

Dividing Eq.(ii) by Eq.(i),

1 √ −1
⎛ − ac os t log a ⎞
2
⎜ ⎟
dy ⎜ ⎟
( )
⎝ √1−t2 ⎠
dy dt

⇒ = =
dx dx 1 −1 t
( ) ⎛ √asin log a ⎞
dt 2
⎜ ⎟

⎝ √1−t2 ⎠

√ c os−1 t
a y
= − = −
x
−1 t
√asin

Hence proved
15. Let us differentiate the whole equation w.r.t. x
According to product rule of differentiation
d(xy) xd(y) yd(x) dy

dx
= dx
+
dx
=x× dx
,+ y
Therefore,
d(xy×log x+y) d(1)

dx
= dx
d(xy) d(log x+y) d(1)
⇒ log x + y × dx
+ xy × dx
= dx
dy dy
⇒ log x + y [x dx
+ y] + xy [ x+y
1
× (1 +
dx
)] =0
dy xy dy
⇒
dx
[x × log x + y] + y × log (x + y) + x+y
(1 +
dx
) =0

4/8
 dy xy xy
⇒
dx
(x log(x + y) +
x+y
) = -(y log (x + y) + x+y
)

[(x2 + xy)log (x + y) + xy] = -[(y2 + xy) log(x + y) + xy]

dy
⇒
dx
2
dy −y log(x+y)−xy log(x+y)−xy
⇒
dx
= ×
x

x
(Multiply and divide by x)
x[(x+y) log(x+y)+y]

2
dy −yxy log(x+y)−xxy log(x+y)− x y
⇒
dx
= 2
x [(x+y) log(x+y)+y]

2
dy −y(1)−x(1)− x y
⇒ =
dx x2 [(x+y) log(x+y)+y]

2
dy −(x+y+ x y)
⇒
dx
= 2
x {y+(x+y) log(x+y)}

16. y = xn {a cos (log x) + b sin (log x)}

y= axn cos (log x) + bxn sin (log x)
= an xn-1 cos (log x) - axn-1 sin (log x) + bnxn-1 sin (log x) + bxn-1 cos (log x)
dy

dx

= xn-1 cos (log x) (na + b) + xn-1 sin (log x) (bn - a)

dy

dx
2

(xn-1 cos (log x) (na + b) + xn-1 sin (log x) (bn - a))

d y d
−
2 dx
dx
2

= (na + b) [(n - 1) xn-2 cos (log x) - xn-2 sin (log x)] + (bn - a) (n - 1) xn-2 sin (log x) + xn-2 cos (log x)]
d y

2
dx
2

= (na + b) xn-2 [(n - 1) cos (log x) - sin (log x)] + (bn - a) xn-2 [(n - 1) sin (log x) + cos (log x)]
d y

2
dx
dy dy
2 2
x + (1 − 2n) + (1 + n ) y
2 dx
dx

= (na + b) xn [(n - 1) cos (log x) - sin (log x)] + (bn - a) xn [(n - 1) sin (log x) + cos (log x)]
+ (1 - 2n) xn-1 cos (log x) (na + b) + (1 - 2n) xn-1 sin (log x) (bn - a)
+ a(1 + n2) xn cos (log x) + b (1 + n2) xn sin (log x)
=0
2
d y dy
Hence x 2
2
+ (1 − 2n)
dx
+ (1 + n ) y = 0
2

dx

17. When x < 1, we have f(x) = 1 - x.

We know that a polynomial function is everywhere continuous and differentiable. So, f (x) is
continuous and differentiable for all x < 1
Similarly, f (x) is continuous and differentiable for all x ∈ (1, 2) and x > 2.
Thus, the possible points where we have to check the continuity and differentiability of f (x) are
x = 1 and x = 2.
Continuity at x = 1:
We observe that:
lim f (x) = lim(1 − x) [∵ f (x) = 1 − x for x < 1]
− x→1
x→1

=1-1=0
lim f (x) = lim(1 − x)(2 − x) =0 [∵ f (x) = (1 − x)(2 − x), for 1 ≤ x ≤ 2]
+ x→1
x→1

and, f(1) = (1 - 1)(2 - 1) = 0

∴ lim f (x) = lim f (x) = f (1)
− +
x→1 x→1

So, f(x) is continuous at x = 1.

Continuity at x - 2: We observe that:
lim f (x) = lim(1 − x)(2 − x) [∵ f (x) = (1 − x)(2 − x) for 1 ≤ x ≤ 2]
− x→2
x→2

and, lim f (x) = lim(3 − x) = 3 − 2 = 1 [f (x) = 3 − x for x > 2]

+ x→2
x→2

∴ lim f (x) ≠ lim f (x)

− x→2
x→2

So, f (x) is not continuous at x = 2.

Differentiability at x = 1:
we observe that:
f (x)−f (1)
(LHD at x = 1) = lim
x−1
−
x→1

(1−x)−0
⇒ (LHD at x = 1) = lim x−1
[Using definition of f(x)]
x→1

x−1
⇒ (LHD at x = 1) = − lim x−1
= −1
x→1

5/8
 f (x)−f (1)
and (RHD at x = 1) = lim
x−1
+
x→1

(1−x)(2−x)−0
⇒ (RHD at x = 1) = lim x−1
[Using definition of f(x)[
x→1

(x−1)(x−2)
⇒ (RHD at x = 1) = lim x−1
x→1

⇒ (RHD at x = 1) = lim x − 2 = 1 − 2 = −1
x→1

Clearly, (LHD at x = 1) = (RHD at x = 1). So, f(x) is differentiable at x = 1.

Differentiability at x = 2:
Since, f (x) is not continuous at x = 2. So, it is not differentiable at x = 2.
18. Given: xy + yx = 1
Let y = xy + yx = 1
Let u = xy and v = yx
Then, u + v = 1
du dv
⇒ + = 0
dx dx

For, u = xy
Taking log on both sides, we get
log u = log xy
⇒ log u = y ⋅ log(x)

Now, differentiating both sides with respect to x

d d
(log u) = [y ⋅ log(x)]
dx dx

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

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

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

For v = yx
Taking log on both sides, we get
log v = log yx
⇒ log v = x ⋅ log(y)

Now, differentiate both sides with respect to x

d d
(log v) = [x ⋅ log(y)]
dx dx

1 dv d d
⇒ = {x ⋅ (log y) + log y ⋅ x}
v dx dx dx

dv 1 dy dx
⇒ = v [x ⋅ ⋅ + log y ⋅ ( )]
dx y dx dx

dv x dy
x
⇒ = y [ ⋅ + log y]
dx y dx

du dv
because, dx
+
dx
= 0

y dy dy
So, x y
[
x
+ log x ⋅ (
dx
)] + y
x
[
x

y
⋅
dx
+ log y] = 0

dy
y x−1 y−1 x
⇒ (x log x + xy )⋅ + (yx + y log y) = 0
dx
dy
y x−1 y−1 x
⇒ (x log x + xy )⋅ = − (yx + y log y)
dx
y−1 x
dy (y x +y log y)
= −
dx y x −1
(x log x+x y )

−−−−− −−−−−
19. 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√x2 +1 dx √x2 +1−x 2√x2 +1

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

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

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

6/8
 f (1+h)−f (1)
20. i. R.H.D. of f(x) at x = 1 = lim h
h→0

|1+h−3|−|−2| 2−h−2
= lim
h
= lim
h
= -1
h →0 h →0

f (1−h)−f (1)
ii. L.H.D. of f(x) at x = 1 = lim −h
h→0
2
(1−h) 3(1−h) 13
[ − + −2]
4 2 4

= lim −h
h→0
2
h −2h+1−6+6h+13−8
= lim [ −4h
]
h→0
2
h +4h
= lim [ −4h
] = -1.
h→0

iii. Since L.H.D. of f(x) at x = 1

is same as R.H.D. of f(x) at x = 1,
f(x) is differentiable at x = 1.
OR
⎧ x − 3, x ≥ 3
⎪ ⎪
⎪

f(x) = ⎨ 3 − x, 1 ≤ x < 3
⎪ 2
⎪
⎩
⎪ x 3 x 13
− + ,x< 1
4 2 4

[f'(x)]x=2 = 0 - 1 = -1
2(−1)
[f'(x)]x=–1 = 4
−
3

2
= -2
21. i. f'(x) = x

2
−
3

2
,x<1
−1
∴ f'(-1) = 2
−
3

2
= -2
ii. f'(x) = -1, 1 ≤ x ≤ 3
∴ f'(2) = - 1

⎧ x − 3, x ≥ 3
⎪
⎪

iii. We have, f (x) = ⎨ 3 − x, 1 ≤ x < 3

⎪
⎩ 2
⎪ x 3x 13
− + , x < 1
4 2 4
f (1−h)−f (1)
LHD at x = 1 = lim −h
h→0

2
−1 (1−h) 3(1−h) 13
= lim [ − + − 2]
h 4 2 4
h→0

2
1+ h −2h−6+6h+13−8
= lim ( )
−4h
h→0
2
h +4h
= limh→0 (
−4h
) =1
f (1+h)−f (1)
RHD at x = 1 = lim h
h→0

3−(1+h)−2
= lim h
h→0

3−(1+h)−2
= lim h
h→0

h
= lim − h
= -1
h→0

Since, LHD = RHD

Thus, f(x) is differentiable at x = 1.
OR
We have
3
x − 1, 1 < x < ∞
f(x) = {
x − 1, −∞ < x ≤ 1

LHD at x = 1
f (1−h)−f (1)
f'(1) = lim −h
h→0

(1−h)−1−0
= lim −h
h→0

RHD at x = 1
f (1+h)−f (1)
f'(1) = lim h
h→0
3
(1+h ) −1−0
= lim h
h→0

7/8
 = lim h2 + 3 + 3h = 3
h→0

Clearly, LHD ≠ RHD

Thus, f(x) is not differentiable at x = 1.
⎧x − 3
⎪
,x ≥ 3

22. i. We have, f (x) = ⎨ 3 − x ,1 ≤ x < 3

2
⎩
⎪ x 3x 13
− + ,x < 1
4 2 4
f (1+h)−f (1)
Rf'(1) = lim h
h→0

3−(1+h)−2
= lim h
= lim − h

h
= -1
h→0 h→0

f (1−h)−f (1)
ii. Lf'(1) = lim −h
h→0

2
−1 (1−h) 3(1−h)
= lim h
[
4
−
2
+
13

4
− 2]
h→0

2
1+ h −2h−6+6h+13−8
= lim( −4h
)
h→0
2
h +4h
= lim( −4h
) =1
h→0

iii. Since, R.H.D. at x = 3 is 1 and L.H.D. at x = 3 is -1

∴ f(x) is non-differentiable at x = 3

OR
we have
f'(x) = − , x < 1
x

2
3

2
−1
′
∴ f (−1) =
2
−
3

2
= -2

8/8