# Class 12 NCERT Solutions- Mathematics Part I – Chapter 5 Continuity And Differentiability – Miscellaneous Exercise on Chapter 5

Last Updated : 14 Feb, 2022

### Question 1. (3 x2 – 9x – 5)9

#### Solution:

Let us assume y = (3x2 – 9x – 5)9

Now, differentiate w.r.t x

Using chain rule, we get

= 9(3x2 – 9 x + 5)

= 9(3x2 – 9x + 5)8.(6x – 9)

= 9(3x 2 – 9x + 5)8.3(2x – 3)

= 27(3x2 – 9x + 5)8 (2x – 3)

### Question 2. sin3 x + cos6 x

#### Solution:

Let us assume y = sin3 x + cos6 x

Now, differentiate w.r.t x

Using chain rule, we get

= 3 sin2 x. cos x + 6 cos5 x.(-sin x)

= 3 sin x cos x(sin x – 2 cos4 x)

### Question 3. 5x3 cos 2 x

#### Solution:

Let us assume y = 5x3 cos 2x

Now we’re taking logarithm on both the sides

logy = 3 cos 2 x log 5 x

Now, differentiate w.r.t x

### Question 4. sin-1(xâˆšx), 0 â‰¤ x â‰¤ 1

#### Solution:

Let us assume y = sin-1(xâˆšx)

Now, differentiate w.r.t x

Using chain rule, we get

=

=

=

=

=

### Question 5. ,-2 < x < 2

#### Solution:

Let us assume y =

Now, differentiate w.r.t x and by quotient rule, we obtain

### Question 6. , 0 < x < Ï€â€‹/2

#### Solution:

Let us assume y =            ……(1)

Now solve

= cotx/2

Now put this value in eq(1), we get

y = cot-1(cotx/2)

y = x/2

Now, differentiate w.r.t x

dy/dx = 1/2

### Question 7. (log x) log x, x > 1

#### Solution:

Let us assume y = (log x)log x

Now we are taking logarithm on both sides,

log y = log x .log(log x)

Now, differentiate w.r.t x on both side, we get

### Question 8. cos(a cos x + b sin x), for some constant a and b.

#### Solution:

Let us assume y = cos(a cos x + b sin x)

Now, differentiate w.r.t x

By using chain rule, we get

= -sin x(a cos x + b sin x).[a (-sin x) + b cos x]

= (a sin x – b cos x).sin (a cos x + b sin x)

### Question 9. (sin x – cos x) (sin x – cos x), Ï€â€‹/4 < x < 3Ï€â€‹/4

#### Solution:

Let us assume y = (sin x – cos x)(sin x – cos x)

Now we are taking logarithm on both sides,

log y = (sin x – cos x).log(sin x – cos x)

Now, differentiate w.r.t x, we get

Using chain rule, we get

dy/dx = (sinx – cosx)(sinx – cosx)[(cosx + sinx).log(sinx – cosx) + (cosx + sinx)]

dy/dx = (sinx – cosx)(sinx – cosx)(cosx + sinx)[1 + log (sinx – cosx)]

### Question 10. xx + x a + a x + aa for some fixed a > 0 and x > 0

#### Solution:

Let us assume y = xx + xa + ax + aa

Also, let us assume xx = u, xa = v, ax = w, aa = s

Therefore, y = u + v + w + s

So, on differentiating w.r.t x, we get

……….(1)

So first we solve: u = xx

Now we are taking logarithm on both sides,

log u = log xx

log u = x log x

On differentiating both sides w.r.t x, we get

du/dx = xx[logx + 1] = xx(1 + logx)    …….(2)

Now we solve: v = xa

On differentiating both sides w.r.t x, we get

dv/dx = ax(a – 1)   ……(3)

Now we solve: w = ax

Now we are taking logarithm on both sides,

log w =log a x

log w = x log a

On differentiating both sides w.r.t x, we get

dw/dx = w loga

dw/dx = axloga ………(4)

Now we solve: s = a a

So, on differentiating w.r.t x, we get

ds/dx = 0 ………(5)

Now put all these values from eq(2), (3), (4), (5) in eq(1), we get

dy/dx = xx(1 + logx) + ax(a – 1) + axloga + 0

= xx (1 + log x) + axa -1 + ax log a

### Question 11. Differentiate w.r.t x, , for x > 3

Solution:

Let us assume y =

Also let us considered u = and v =

so, y = u + v

On differentiating both side w.r.t x, we get

…….(1)

So, now we solve, u =

Now we are taking logarithm on both sides,

log u = log

log u = (x 2 – 3) log x

On differentiating w.r.t x, we get

=      …….(2)

Now we solve: v =

Now we are taking logarithm on both sides,

log v =

log v = x2 log(x – 3)

On differentiating both sides w.r.t x, we get

…..(3)

Now put all these values from eq(2), and (3) in eq(1), we get

### Question 12. Find dy/dx , if y = 12(1 – cos t), x = 10 (t – sin t), -Ï€â€‹/2 < t < Ï€â€‹/2

Solution:

According to the question

y = 12(1 – cos t)   ……(1)

x = 10 (t – sin t)  ……(2)

So, \frac{dy}{dx}= \frac{\frac{dy}{dt}}{\frac{dx}{dt}}    ……(3)

On differentiating eq(1) w.r.t t, we get

= 12.[0 – (- sin t)]

= 12 sin t

On differentiating eq(2) w.r.t t, we get

= 10(1 – cos t)

Now put the value of dy/dt and dx/dt in eq(3), we get

= 6/5 cot t/2

### Question 13. Find dy/dx, if y = sin-1 x + sin-1âˆš1-x2, 0 < x < 1

Solution:

According to the question

y = sin-1 x + sin-1âˆš1 – x2

On differentiating w.r.t x, we get

Using chain rule, we get

dy/dx = 0

### Question 14. If xâˆš1 + y + yâˆš1 + x = 0, for, -1 < x < 1, prove that

Solution:

According to the question

xâˆš1 + y = -yâˆš1 + x

On squaring both sides, we get

x2 (1 + y) = y2 (1 + x)

â‡’ x2 + x2 y = y2 + x y2

â‡’ x2 – y2 = xy 2 – x2 y

â‡’ x2 – y2 = xy (y – x)

â‡’ (x + y)(x – y) = xy (y – x)

â‡’ x + y = -xy

â‡’ (1 + x) y = -x

â‡’ y = -x/(1 + x)

On differentiating both sides w.r.t x, we get

Hence proved.

### Question 15. If (x – a)2 + (y – b)2 = c 2, for some c > 0, prove that  is a constant independent of a and b.

Solution:

According to the question

(x – a)2+ (y – b)2= c2

On differentiating both side w.r.t x, we get

â‡’ 2(x – a). + 2(y – b) = 0

â‡’ 2(x – a).1 + 2(y – b).= 0

â‡’  …….(1)

Again on differentiating both side w.r.t x, we get

…….[From equation (1)]

=

=

=

= – c, which is constant and is independent of a and b.

Hence proved.

### Question 16. If cos y = x cos (a + y), with cos a â‰  Â±1, prove that

Solution:

According to the question

cos y = x cos (a + y)

On differentiating both side w.r.t x, we get

â‡’ – sin y dy/dx = cos (a + y).  + x

â‡’ – sin y dy/dx = cos (a + y) + x [-sin (a + y)]dy/dx

â‡’ [x sin (a + y) – sin y] dy/dx = cos (a + y) ……..(1)

Since cos y = x cos (a + y), x =

Now we can reduce eq(1)

= cos(a + y)

â‡’ [cos y.sin (a + y)- sin y.cos (a + y)].dy/dx = cos2(a + y)

â‡’ sin(a + y – y)dy/dx = cos2(a + b)

â‡’

Hence proved.

### Question 17. If x = a (cos t + t sin t) and y = a (sin t – t cos t), find

Solution:

According to the question

x = a (cos t + t sin t) …..(1)

y = a (sin t – t cos t) …..(2)

So, \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}   …..(3)

On differentiating eq(1) w.r.t t, we get

dx/dt = a.

Using chain rule, we get

= a[-sin t +sin t. + t.]

= a [-sin t + sin t + t cos t]

= at cos t

On differentiating eq(2) w.r.t t, we get

dy/dt = a.

Using chain rule, we get

= a [cos t – [cost. + t.]]

= a[cos t – {cos t – t sin t}]

= at sin t

Now put the values of dx/dt and dy/dt in eq(1), we get

dy/dx = at sin t/at cos t = tan t

Again differentiating both side w.r.t x, we get

= sec 2 t.

= sec2 t.……..[dx/dt = atcost â‡’ dt/dx = 1/atcost]

= sec3t/at

### Question 18. If f(x) = |x|3, show that f”(x) exists for all real x and find it.

Solution:

As we know that |x| =

So, when x â‰¥ 0, f(x) = |x|3 = x3

So, on differentiating both side w.r.t x, we get

f'(x) = 3x2

Again, differentiating both side w.r.t x, we get

f”(x) = 6 x

When x < 0, f(x) = |x|3 = -x

So, on differentiating both side w.r.t x, we get

f'(x) = – 3x

Again, differentiating both side w.r.t x, we get

f”(x) = -6 x

So, for f(x) = |x|3, f”(x) exists for all real x, and is given by

f”(x) =

### Question 19. Using mathematical induction prove that  = (nx)n – 1 for all positive integers n.

Solution:

So, P(n) = = (nx)n – 1

For n = 1:

P(1) : = (1x)1 – 1 =1

Hence, P(n) is true for n = 1

Let us considered P(k) is true for some positive integer k.

So, P(k): = (kx)k – 1

For P(k + 1): = ((k + 1)x)(k + 1) – 1

x k + x. ….(Using applying product rule)

= x k .1 + x . k . x k-1

= x k + k x k

= (k + 1) x k

= (k + 1) x(k + 1) – 1

Hence, P(k+1) is true whenever P(k) is true.

So, according to the principle of mathematical induction, P(n) is true for every positive integer n.

Hence proved.

### Question 20. Using the fact that sin(A + B) = sin A cos B + cos A sin B and the differentiation, obtain the sum formula for cosines.

Solution:

According to the question

sin(A + B) = sin A cos B + cos A sin B

On differentiating both sides w.r.t x, we get

=

â‡’ cos (A + B).= cos B.  + sin A.  + sin B.+ cos A.

â‡’ cos (A+B). = cos B.cos A+ sin A (-sin B) + sin B (-sin A).+ cos A cos B

â‡’ cos (A + B).=(cos A cos B – sin A sin B).

Hence, cos (A + B) = cos A cos B – sin A sin B

### Question 21. Does there exist a function which is continuous everywhere but not differentiable to exactly two points? Justify your answer.

Solution:

Let us consider a function f given as

f(x) = |x – 1| + |x – 2|

As we already know that the modulus functions are continuous at every point

So, there sum is also continuous at every point but not differentiable at every point x = 0

Let x = 1, 2

Now at x = 1

L.H.D = lim xâ‡¢ 1

L.H.D = limhâ‡¢0

= limhâ‡¢0

= limhâ‡¢0

= limhâ‡¢0

= limhâ‡¢0

= limhâ‡¢0

= limhâ‡¢0

= -2

R.H.D = limxâ‡¢1

R.H.D = limhâ‡¢0

= limhâ‡¢0

= limhâ‡¢0

= limhâ‡¢0

= limhâ‡¢0

= limhâ‡¢0

= limhâ‡¢0

= 0

Since L.H.D â‰  R.H.D

So given function f is not differentiable at x = 1.

Similarly, we get that the given function is not differentiable at x = 2.

Hence, there exist a function which is continuous everywhere but not differentiable to exactly two points.

### Question 22. If ,prove that

Solution:

Given that

â‡’ y =(mc – nb) f(x)- (lc – na )g(x) +(lb – ma) h(x)

[(mc -nb) f(x)] –  [(lc – na) g(x)] + [(lb – ma) h(x)]

= (mc – nb) f'(x) – (lc – na) g'(x) + (lb – ma ) h’ (x)

So,

Hence proved.

### Question 23. If y = ,-1 â‰¤ x â‰¤ 1, show that

Solution:

According to the question

y =

Now we are taking logarithm on both sides,

log y = a cos-1 x log e

log y = a cos -1

On differentiating both sides w.r.t x, we get

â‡’

On squaring both sides,we get

â‡’(1-x 2) =a 2 y 2

On differentiating again both the side w.r.t x, we get

â‡’

â‡’

Hence proved

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