NCERT Solutions Class 12 Maths (Continuity And Differentiability) Miscellaneous Exercise

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-5 (Continuity And Differentiability) Miscellaneous Exercise This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

Solutions Class 12 Maths (Relation And Functions) Miscellaneous Exercise

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

$\frac{dy}{dx}=\frac{d }{dx}(3x^2 -9x +5)^9$

Using chain rule, we get

= 9(3x2 – 9 x + 5)8 $\frac{d}{dx}(3x^2 -9x +5)$

= 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. sin³x + cos⁶x

Solution:

Let us assume y = sin3 x + cos6 x

Now, differentiate w.r.t x

$\frac{dy}{dx}= \frac{d}{dx}(sin^3x + cos^6x)$

Using chain rule, we get

= $\frac{d}{dx}sin^3x +\frac{d}{dx} cos^6x$

= $3sin^2x.\frac{d}{dx}(sinx)+6sin^2x.\frac{d}{dx}(cosx)$

= 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 = 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

$\frac{1}{y}\frac{dy}{dx}=3[log5x.\frac{d}{dx}(cos2x)+cos2x.\frac{d}{dx}(log5x)]$

$\frac{dy}{dx}=3y[log 5x(-sin2x).\frac{d}{dx}(2x)+cos2x.\frac{1}{5x}.\frac{d}{dx}(5x)]$

$\frac{dy}{dx}=3y[-2sin2xlog5x+\frac{cos2x}{x}]$

$\frac{dy}{dx}=3y[\frac{3cos2x}{x}-6sin2xlog5x]$

$\frac{dy}{dx}=5x^{3cos2x}[\frac{3cos2x}{x}-6sin2xlog5x]$

Question 4. sin^-1(x√x), 0 ≤ x ≤ 1

Solution:

Let us assume y = sin-1(x√x)

Now, differentiate w.r.t x

$\frac{dy}{dx}=\frac{d}{dx}sin^{-1}(x\sqrt{x})$

Using chain rule, we get

= $\frac{1}{\sqrt{1-(x\sqrt{x})^2}}.\frac{d}{dx}(x\sqrt{x})$

= $\frac{1}{\sqrt{1-x^3}}.\frac{d}{dx}[x^{\frac{3}{2}}]$

= $\frac{1}{\sqrt{1-x^3}}.\frac{3}{2}.x^{\frac{1}{2}}$

= $\frac{3\sqrt{x}}{2\sqrt{1-x^3}}$

= $\frac{3}{2}\sqrt{\frac{x}{1-x^3}}$

Question 5. $\frac{cos^{-1}\frac{x}{2}}{\sqrt{2x+7}}$ ,-2 < x < 2

Solution:

Let us assume y = $\frac{cos^{-1}\frac{x}{2}}{\sqrt{2x+7}}$

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

$\frac{dy}{dx}=\frac{\sqrt{2x+7}\frac{d}{dx}(cos^{-1}\frac{x}{2})-(cos^{-1}\frac{x}{2})\frac{d}{dx}(\sqrt{2x+7})}{(\sqrt{2x+7})^2}$

= $\frac{\sqrt{2x+7}[\frac{-1}{\sqrt{1-(\frac{x}{2})^2}}.\frac{d}{dx}(\frac{x}{2})]-(cos^{-1}\frac{x}{2})\frac{1}{2\sqrt{2x+7}}.\frac{d}{dx}(2x+7)}{2x+\frac{7}{2}}$

= $\frac{\sqrt{2x+7}\frac{-1}{\sqrt{4-x^2}}-[cos^{-1}\frac{x}{2}]\frac{2}{2\sqrt{2x+7}}}{2x+7}$

= $\frac{-\sqrt{2x+7}}{\sqrt{4-x^2}(2x+7)}-\frac{cos^{-1}\frac{x}{2}}{(\sqrt{2x+7})(2x+7)}$

= $-[\frac{1}{\sqrt{4-x^2}\sqrt{2x+7}}+\frac{cos^{-1}\frac{x}{2}}{(2x+7)^\frac{1}{2}}]$

Question 6. $cot^{-1}\frac{\sqrt{1+sinx}+\sqrt{1-sinx}}{\sqrt{1+sinx}-\sqrt{1-sinx}}$ , 0 < x < π/2

Solution:

Let us assume y = $cot^{-1}\frac{\sqrt{1+sinx}+\sqrt{1-sinx}}{\sqrt{1+sinx}-\sqrt{1-sinx}}$ ……(1)

Now solve $\frac{\sqrt{1+sinx}+\sqrt{1-sinx}}{\sqrt{1+sinx}-\sqrt{1-sinx}}$

= $\frac{(\sqrt{1+sinx}+\sqrt{1-sinx})^2}{(\sqrt{1+sinx}-\sqrt{1-sinx})(\sqrt{1+sinx}+\sqrt{1-sinx})}$

= $\frac{(1+sinx)+(1-sinx)+2\sqrt{(1-sinx)(1+sinx)}}{(1+sinx)-(1-sinx)}$

= $\frac{2+2\sqrt{1-sin^2x}}{2sinx}$

= $\frac{1+cosx}{sinx}$

= $\frac{2cos^2\frac{x}{2}}{2sin\frac{x}{2}cos\frac{x}{2}}$

= 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

$\frac{dy}{dx}=\frac{1}{2}\frac{d}{dx}(x)$

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

$\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}[logx.log(logx)]$

$\frac{1}{y}\frac{dy}{dx}=log(logx).\frac{d}{dx}(log x)+logx.\frac{d}{dx}[log(logx)]$

$\frac{dy}{dx}=y[log(logx).\frac{1}{x}+logx.\frac{1}{logx}.\frac{d}{dx}(logx)]$

$\frac{dy}{dx}=y[\frac{1}{x}log(logx)+\frac{1}{x}]$

$\frac{dy}{dx}=(logx)^{logx}[\frac{1}{x}+\frac{log(logx)}{x}]$

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

$\frac{dy}{dx}=\frac{d}{dx}cos(acosx+bsinx)$

By using chain rule, we get

$\frac{dy}{dx}=-sin(acosx+bsinx).\frac{d}{dx}(acosx+bsinx)$

= -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

$\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}[(sinx-cosx)log(sinx-cosx)]$

Using chain rule, we get

$\frac{1}{y}\frac{dy}{dx}=log(sinx-cosx).\frac{d}{dx}(sinx-cosx)+(sinx-cosx).\frac{d}{dx}log(sinx-cosx)$

$\frac{1}{y}\frac{dy}{dx}=log(sinx-cosx).(cosx+sinx)+(sinx-cosx).\frac{1}{(sinx-cosx)}.\frac{d}{dx}(sinx-cosx)$

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 = 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

$\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}+\frac{dw}{dx}+\frac{ds}{dx}$ ……….(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

$\frac{1}{u}\frac{du}{dx}=logx.\frac{d}{dx}(x)+x.\frac{d}{dx}(logx)$

$\frac{du}{dx}=u[logx.1+x.\frac{1}{x}]$

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

Now we solve: v = xa

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

$\frac{dv}{dx}=\frac{d}{dx}(x^a)$

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

$\frac{1}{w}.\frac{dw}{dx}=loga.\frac{d}{dx}(x)$

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, $x^{x^2-3}+(x-3)^{x^2}$ , for x > 3

Solution:

Let us assume y = $x^{x^2-3}+(x-3)^{x^2}$

Also let us considered u = $x^{x^2-3}$ and v = $(v-3)^{x^2}$

so, y = u + v

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

$\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$ …….(1)

So, now we solve, u = $x^{x^2-3}$

Now we are taking logarithm on both sides,

log u = log $(x^{x^2-3})$

log u = (x 2 – 3) log x

On differentiating w.r.t x, we get

$\frac{1}{u}.\frac{du}{dx} = logx.\frac{d}{dx}(x^2-3) + (x^2-3).\frac{d}{dx}(log x)$

$\frac{1}{u}.\frac{dy}{dx}= logx.2x+(x^2-3).\frac{1}{x}$

= $\frac{du}{dx}= x^{x^2-3}.[\frac{x^2-3}{x}+ 2xlogx]$ …….(2)

Now we solve: v = $(x-3)^{x^2}$

Now we are taking logarithm on both sides,

log v = $log(x-3)^{x^2}$

log v = x2 log(x – 3)

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

$\frac{1}{v}.\frac{dv}{dx}= log (x-3).\frac{d}{dx}(x^2)+x^2\frac{d}{dx}[log(x-3)]$

$\frac{1}{v}.\frac{dv}{dx}=log (x-3).2x+x^2.\frac{1}{x-3}.\frac{d}{dx}(x-3)$

$\frac{dv}{dx}=v[2xlog(x-3)+\frac{x^2}{x-3}.1]$

$\frac{dv}{dx}=(x-3)^{x^2}[\frac{x^2}{x-3}+2xlog(x-3)]$ …..(3)

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

$\frac{dy}{dx}= x^{x^2-3}[\frac{x^2-3}{x}+2xlogx]+(x-3)^{x^2}[\frac{x^2}{x-3}+2xlog(x-3)]$

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)

……(3)

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

$\frac{dy}{dt} = \frac{d}{dt} [12 (1 - cost)]$

= $12\frac{d}{dt}(1- cost )$

= 12.[0 – (- sin t)]

= 12 sin t

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

$\frac{dx}{dt}=\frac{d}{dt}[10. (t - sin t)]$

= $10\frac{d}{dt} (t - sin t)$

= 10(1 – cos t)

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

$\frac{dy}{dx}=\frac{12 sin t}{10(1-cost)}$

= $\frac{12.2sin\frac{t}{2}cos \frac{t}{2}}{10.2sin^2\frac{t}{2}}$

= 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

$\frac{dy}{dx}=\frac{d}{dx}[sin^{-1}x+sin ^{-1}\sqrt{1-x^2}]$

Using chain rule, we get

$\frac{dy}{dx}=\frac{d}{dx}(sin^{-1}x)+\frac{d}{dx}(sin ^{-1}\sqrt{1-x^2})$

= $\frac{1}{\sqrt{1-x^2}}+\frac{1}{\sqrt{1-(\sqrt{1-x^2}})^2}.\frac{d}{dx}(\sqrt{1-x^2})$

= $\frac{1}{\sqrt{1-x^2}}+\frac{1}{x}.\frac{1}{2\sqrt{1-x^2}}.\frac{d}{dx}(1-x^2)$

= $\frac{1}{\sqrt{1-x^2}}+\frac{1}{2x\sqrt{1-x^2}}(-2x)$

= $\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}}$

dy/dx = 0

Question 14. If x√1 + y + y√1 + x = 0, for, -1 < x < 1, prove that $\frac{dy}{dx} = -\frac{1}{(1+x)^2}$

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

$\frac{dy}{dx}= -\frac{(1+x)\frac{d}{dx}(x)-x\frac{d}{dx}(1+x)}{(1+x)^2}$

= $\frac{(1+x)-x}{(1+x)^2}$

= $-\frac{1}{(1+x)^2}$

Hence proved.

Question 15. If (x – a)²+ (y – b)² = c ², for some c > 0, prove that $\frac{[1+(\frac{dy}{dx})^2]^\frac{3}{2}} {\frac{d^2y}{dx^2}}$

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

$\frac{d}{dx}[(x-a)^2]+\frac{d}{dx}[(y-b)^2]=\frac{d}{dx}(c^2)$

⇒ 2(x – a). $\frac{d}{dx}(x-a)$ + 2(y – b) $\frac{d}{dx}(y-b)$ = 0

⇒ 2(x – a).1 + 2(y – b). $\frac{dy}{dx}$ = 0

⇒ $\frac{dy}{dx}=\frac{-(x-a)}{y-b}$ …….(1)

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

$\frac{d^2y}{dx^2} = \frac{d}{dx}[\frac{-(x-a)}{y-b}]$

$= -[\frac{(y-b).\frac{d}{dx}(x-a)-(x-a).\frac{d}{dx}(y-b)}{(y-b)^2}]$

$=-[\frac{(y-b)-(x-a).\frac{dy}{dx}}{(y-b)^2}]$

$= -[\frac{(y-b)-(x-a).\{-\frac{(x-a)}{y-b}\}}{(y-b)^2}]$ …….[From equation (1)]

$=-[\frac{(y-b)^2+(x-a)^2}{(y-b)^3}]$

$[\frac{1+[\frac{dy}{dx}]^2}{\frac{d^2y}{dx^2}}]^{\frac{3}{2}}= \frac{[1+\frac{(x-a)^2}{(y-b)^2}]^{\frac{3}{2}}}{-[\frac{(y-b)^2 + (x-a)^2}{(y-b)^3}]}$

= $\frac{[\frac{[(y-b)^2+(x-a)^2]}{(y-b)^2}]^{\frac{3}{2}}}{-[\frac{(y-b)^2+(x-a)^2}{(y-b)^3}]}$

= $\frac{[\frac{c^2}{(y-b)^2}]\frac{3}{2}}{-\frac{c^2}{(y-b)^3}}$

= $\frac{\frac{c^3}{(y-b)^3}}{-\frac{c^2}{(y-b)^3}}$

= – 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 $\frac{dy}{dx}= \frac{cos^2(a+y)}{sina}$

Solution:

According to the question

cos y = x cos (a + y)

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

$\frac{d}{dx}[cos y]$ = $\frac{d}{dx}[x cos(a + y)]$

⇒ – sin y dy/dx = cos (a + y). $\frac{d}{dx}(x)$ + x $\frac{d}{dx}[cos (x+y)]$

⇒ – 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 = $\frac{cosy }{cos (a+y)}$

Now we can reduce eq(1)

$[\frac{cosy}{cos(a+y)}.sin(a+y)-siny]\frac{dy}{dx}$ = 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)

⇒ $\frac{dy}{dx}= \frac{cos^2(a+b)}{sin a}$

Hence proved.

Question 17. If x = a (cos t + t sin t) and y = a (sin t – t cos t), find $\frac{d^2y}{dx^2}$

Solution:

According to the question

x = a (cos t + t sin t) …..(1)

y = a (sin t – t cos t) …..(2)

So,

…..(3)

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

dx/dt = a. $\frac{d}{dt}(cost + t sin t)$

Using chain rule, we get

= a[-sin t +sin t. $\frac{d}{dt}(t)$ + t. $\frac{d}{dt}(sin t)$ ]

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

= at cos t

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

dy/dt = a. $\frac{d}{dt}(sin t - t cos t)$

Using chain rule, we get

= a [cos t – [cost. $\frac{d}{dt}(t)$ + t. $\frac{d}{dt}(cos t)$ ]]

= 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

$\frac{d^2y}{dx^2}= \frac{d}{dx}[\frac{dy}{dx}]$

= $\frac{d}{dx}(tant )$

= sec 2 t. $\frac{dt}{dx}$

= sec2 t. $\frac{1}{at cos t}$ ……..[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| = $\begin{cases} x, \hspace{0.2cm}x\geq0\\ -x,\hspace{0.2cm}x<0 \end{cases}$

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) = $\begin{cases} 6x, \hspace{0.2cm}x\geq0\\ -6x,\hspace{0.2cm}x<0 \end{cases}$

Question 19. Using mathematical induction prove that $\frac{d}{dx}(x^n)$ = (nx)n – 1 for all positive integers n.

Solution:

So, P(n) = $\frac{d}{dx}(x^n)$ = (nx)n – 1

For n = 1:

P(1) : $\frac{d}{dx}(x^1)$ = (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): $\frac{d}{dx}(x^k)$ = (kx)k – 1

For P(k + 1): $\frac{d}{dx}(x^{k+1})$ = ((k + 1)x)(k + 1) – 1

x k $\frac{d}{dx} (x)$ + x. $\frac{d}{dx} (x^k)$ ….(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

$\frac{d}{dx} [sin(A+B)]$ = $\frac{d}{dx}(sin A cos B)$ + $\frac{d}{dx}(cos A sin B)$

⇒ cos (A + B). $\frac{d}{dx}(A+B)$ = cos B. $\frac{d}{dx}(sin A)$ + sin A. $\frac{d}{dx}(cos B)$ + sin B. $\frac{d}{dx}(cos A)$ + cos A. $\frac{d}{dx}(sin B)$

⇒ cos (A+B). $\frac{d}{dx}(A+B)$ = cos B.cos A $\frac{dA}{dx}$ + sin A (-sin B) $\frac{dB}{dx}$ + sin B (-sin A). $\frac{dA}{dx}$ + cos A cos B $\frac{dB}{dx}$

⇒ cos (A + B). $[ \frac{dA}{dx}+ \frac{dB}{dx}]$ =(cos A cos B – sin A sin B). $[ \frac{dA}{dx}+ \frac{dB}{dx}]$

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– $\frac{f(x) - f(1)}{(x-1)}$

L.H.D = limh⇢0 $\frac{f(1-h) - f(1)}{-h}$

= limh⇢0 $\frac{|1-h-1| + |1-h-2| - |1-1|-|1-2|}{-h}$

= limh⇢0 $\frac{|1-h-1| + |1-h-2| - |0|-|-1|}{-h}$

= limh⇢0 $\frac{|-h| + |-h-1| - 1}{-h}$

= limh⇢0 $\frac{h - (-h-1) - 1}{-h}$

= limh⇢0 $\frac {h + h + 1 - 1}{-h}$

= limh⇢0 $\frac{2h}{-h}$

= -2

R.H.D = limx⇢1+ $\frac{f(x) - f(1)}{(x-1)}$

R.H.D = limh⇢0 $\frac{f(1+h) - f(1)}h$

= limh⇢0 $\frac{|1+h-1| + |1+h-2| - |1-1|-|1-2|}{h}$

= limh⇢0 $\frac{|1+h-1| + |1+h-2| - |0|-|-1|}{h}$

= limh⇢0 $\frac{|h| + |h-1| - 1}{h}$

= limh⇢0 $\frac{h - (h-1) - 1}{h}$

= limh⇢0 $\frac{h - h + 1 - 1}{h}$

= limh⇢0 $\frac 0{h}$

= 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 $Miscellaneous Exercise on Chapter-5$ ,prove that $Miscellaneous Exercise on Chapter-5$

Solution:

Given that $y=\begin{bmatrix} f(x) & g(x) & h(x)\\ l & m & n\\ a & b & c \end{bmatrix}$

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

$\frac{dy}{dx}=\frac{d}{dx}$ [(mc -nb) f(x)] – $\frac{d}{dx}$ [(lc – na) g(x)] + $\frac{d}{dx}$ [(lb – ma) h(x)]

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

$= \begin{bmatrix} f'(x) & g'(x) & h'(x)\\ l & m & n\\ a & b & c \end{bmatrix}$

So,

$\frac{dy}{dx}= \begin{bmatrix} f'(x) & g'(x) & h'(x)\\ l & m & n\\ a & b & c \end{bmatrix}$

Hence proved.

Question 23. If y = $e^{a cos^{-1}x}$ ,-1 ≤ x ≤ 1, show that $Miscellaneous Exercise on Chapter-5$

Solution:

According to the question

y = $e^{a cos^{-1}x}$

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

$\frac{1}{y}\frac{dy}{dx}=a.\frac{-1}{\sqrt{1-x^2}}$

⇒ $\frac{dy}{dx}=\frac{-ay}{\sqrt{1-x^2}}$

On squaring both sides,we get

$[\frac{dy}{dx}]^2=\frac{a^2y^2}{1-x^2}$

⇒(1-x 2) $[\frac{dy}{dx}]^2$ =a 2 y 2

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

$[\frac{dy}{dx}]^2 \frac{d}{dx}(1-x^2)+(1-x^2)\frac{d}{dx} [[\frac{dy}{dx}]^2]=a^2 \frac{d}{dx}(y^2)$

⇒ $[\frac{dy}{dx}]^2(-2x)+(1-x^2).2 \frac{dy}{dx}. \frac{d^2y}{dx^2}=a^2.2y \frac{dy}{dx}$

⇒ $-x \frac{dy}{dx}+(1-x^2) \frac{d^2y}{dx^2}= a^2.y$

$(1-x^2)\frac{d^2y}{dx^2}- x \frac{dy}{dx}-a^2y=0$

Hence proved

NCERT Solutions Class 12 Maths (Continuity And Differentiability) Miscellaneous Exercise

NCERT Solutions Class 12 Maths (Continuity And Differentiability) Miscellaneous Exercise

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NCERT Solutions Class 12 Maths (Continuity And Differentiability) Miscellaneous Exercise

NCERT Solutions Class 12 Maths (Continuity And Differentiability) Miscellaneous Exercise

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