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

Differentiate w.r.t x the function in Exercise 1 to 11.

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)⁸ 

= 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. 5x^{3 cos 2 x}

Solution:

Let us assume y = 5x^{3 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(x√x), 0 ≤ x ≤ 1

Solution:

Let us assume y = sin^-1(x√x)

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 < π​/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)}, π​/4 < x < 3π​/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. x^x + x ^a + a ^x + a^a for some fixed a > 0 and x > 0

Solution:

Let us assume y = x^x + x^a + a^x + a^a

Also, let us assume x^x = u, x^a = v, a^x = w, a^a = s

Therefore, y = u + v + w + s

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

  ……….(1)

So first we solve: u = x^x

Now we are taking logarithm on both sides,

log u = log x^x

log u = x log x

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

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

Now we solve: v = x^a

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

dv/dx = ax^{(a – 1)}   ……(3)

Now we solve: w = a^x

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 = a^xloga ………(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 = x^x(1 + logx) + ax^{(a – 1)} + a^xloga + 0

= x^x (1 + log x) + ax^{a -1} + a^x 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), -π​/2 < t < π​/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^-1√1-x², 0 < x < 1

Solution:

According to the question

y = sin^-1 x + sin^-1√1 – x²

On differentiating w.r.t x, we get 

Using chain rule, we get

= 

= 

= 

= 

dy/dx = 0

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

Solution: 

According to the question

x√1 + y = -y√1 + x 

On squaring both sides, we get

x² (1 + y) = y² (1 + x)

⇒ x² + x² y = y² + x y²

⇒ x² – y² = xy ² – x² y

⇒ x² – y² = xy (y – x)

⇒ (x + y)(x – y) = xy (y – x)

⇒ x + y = -xy

⇒ (1 + x) y = -x

⇒ 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 

⇒ 2(x – a). + 2(y – b) = 0 

⇒ 2(x – a).1 + 2(y – b).= 0

⇒  …….(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 ≠ ±1, prove that 

Solution:

According to the question

cos y = x cos (a + y)

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

= 

⇒ – sin y dy/dx = cos (a + y).  + x

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

⇒ [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)

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

⇒ sin(a + y – y)dy/dx = cos²(a + b)

⇒

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 ⇒ 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 ≥ 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 = -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 

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

 = + 

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

⇒ cos (A+B). = cos B.cos A+ sin A (-sin B) + sin B (-sin A).+ cos A cos B 

⇒ 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 _{x⇢ 1^–}

L.H.D = lim_h⇢0 

= lim_h⇢0

= lim_h⇢0

= lim_h⇢0

= lim_h⇢0

= lim_h⇢0

= lim_h⇢0

= -2

R.H.D = lim_x⇢1⁺ 

R.H.D = lim_h⇢0

= lim_h⇢0

= lim_h⇢0

= lim_h⇢0

= lim_h⇢0

= lim_h⇢0

= lim_h⇢0

= 0

Since L.H.D ≠ 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

⇒ 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 ≤ x ≤ 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 x 

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

⇒

On squaring both sides,we get

⇒(1-x ²) =a ² y ²

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

⇒

⇒

Hence proved