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

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  • Last Updated : 14 Feb, 2022

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)\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. sin3 x + cos6 x

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. 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

\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. 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

\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)

So, \frac{dy}{dx}= \frac{\frac{dy}{dt}}{\frac{dx}{dt}}    ……(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, \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}   …..(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 = -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) = \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  y=\begin{bmatrix} f(x) & g(x) & h(x)\\ l & m & n\\ a & b & c \end{bmatrix}    ,prove that  \frac{dy}{dx}= \begin{bmatrix} f'(x) & g'(x) & h'(x)\\ l & m & n\\ a & b & c \end{bmatrix}

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

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

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


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