# 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 x^{2} – 9x – 5)^{9}

#### Solution:

Let us assume y = (3x

^{2}– 9x – 5)^{9}Now, differentiate w.r.t x

Using chain rule, we get

= 9(3x

^{2}– 9 x + 5)^{8 }= 9(3x

^{2 }– 9x + 5)^{8}.(6x – 9)= 9(3x

^{2 }– 9x + 5)^{8}.3(2x – 3)= 27(3x

^{2 }– 9x + 5)^{8 }(2x – 3)

### Question 2. sin^{3 }x + cos^{6 }x

#### Solution:

Let us assume y = sin

^{3 }x + cos^{6}xNow, differentiate w.r.t x

Using chain rule, we get

=

=

= 3 sin

^{2 }x. cos x + 6 cos^{5 }x.(-sin x)= 3 sin x cos x(sin x – 2 cos

^{4 }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 }= sTherefore, 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

^{x}loga ………(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^{x}loga + 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 xOn differentiating w.r.t x, we get

= …….(2)

Now we solve: v =

Now we are taking logarithm on both sides,

log v =

log v = x

^{2}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-x^{2}, 0 < x < 1

**Solution:**

According to the question

y = sin

^{-1}x + sin^{-1}√1 – x^{2}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

x

^{2 }(1 + y) = y^{2 }(1 + x)⇒ x

^{2}+ x^{2 }y = y^{2}+ x y^{2}⇒ x

^{2}– y^{2}= xy^{2}– x^{2 }y⇒ x

^{2}– y^{2}= 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}= c^{2}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 = cos

^{2}(a + y)⇒ sin(a + y – y)dy/dx = cos

^{2}(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.= sec

^{2 }t.……..[dx/dt = atcost ⇒ dt/dx = 1/atcost]= sec

^{3}t/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}= x^{3}So, on differentiating both side w.r.t x, we get

f'(x) = 3x

^{2 }Again, differentiating both side w.r.t x, we get

f”(x) = 6 x

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

^{3}= -x^{3 }So, on differentiating both side w.r.t x, we get

f'(x) = – 3x

^{2 }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 byf”(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}=1Hence, 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 elog y = a cos

^{-1 }xOn 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