5 Marks Questions - Maths STD 12 Science Questions - Vidyadip

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17 questions · self-marked practice — reveal the answer and mark yourself.

Question 15 Marks

If ${\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {c^2}$ for some c > 0 prove that $\frac{{{{\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]}^{\frac{3}{2}}}}}{{\frac{{{d^2}y}}{{d{x^2}}}}}$ is a constant independent of a and b.

Answer

Given: ${\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {c^2},$……….(i)

$\therefore 2\left( {x - a} \right) + 2\left( {y - b} \right)\frac{{dy}}{{dx}} = 0$

$\Rightarrow 2\left( {x - a} \right) = - 2\left( {y - b} \right)\frac{{dy}}{{dx}}$

$\Rightarrow \frac{{dy}}{{dx}} = - \left( {\frac{{x - a}}{{y - b}}} \right)$ ……….(ii)

Again $\frac{{{d^2}y}}{{d{x^2}}} = \frac{{ - \left[ {\left( {y - b} \right).1 - \left( {x - a} \right)\frac{{dy}}{{dx}}} \right]}}{{{{\left( {y - b} \right)}^2}}}$

$\Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = \frac{{ - \left[ {\left( {y - b} \right).1 - \left( {x - a} \right)\left( {\frac{{ - \left( {x - a} \right)}}{{y - b}}} \right)} \right]}}{{{{\left( {y - b} \right)}^2}}}$ [From eq. (ii)

$\Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = \frac{{ - \left[ {\left( {y - b} \right) + \left( {\frac{{{{\left( {x - a} \right)}^2}}}{{y - b}}} \right)} \right]}}{{{{\left( {y - b} \right)}^2}}}$

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

$= \frac{{ - {c^2}}}{{{{\left( {y - b} \right)}^3}}}$ ……….(iii) [using (i)]

Putting values of $\frac{{dy}}{{dx}}$ and $\frac{{{d^2}y}}{{d{x^2}}}$ in the given expression,

$\frac{{{{\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]}^{\frac{3}{2}}}}}{{\frac{{{d^2}y}}{{d{x^2}}}}}$

$ = \frac{{{{\left[ {1 + {{\frac{{\left( {x - a} \right)}}{{{{\left( {y - b} \right)}^2}}}}^2}} \right]}^{\frac{3}{2}}}}}{{\frac{{ - {c^2}}}{{{{\left( {y - b} \right)}^3}}}}}$

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

which is a constant and is independent of a and b.

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Question 25 Marks

Find $\frac{d y}{d x}$ , if y = 12 (1 - cos t), x = 10 (t - sin t), $-\frac{\pi}{2}<t<\frac{\pi}{2}$

Answer

Clearly, $\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$
Given, y = 12 (1 - cos t) ...(i)
and, x = 10 (t - sin t) ...(ii)
Differentiating (ii), with respect to t.
$\frac{d x}{d t}=\frac{d}{d t}[10(t-\sin t)]$
$\Rightarrow \frac{d x}{d t}=10 \times \frac{d}{d t}(t-\sin t)=10(1-\cos t)$
Differentiating (i), with respect to t.
$\frac{d y}{d t}=\frac{d}{d t}[12(1-\cos t)]$
$\Rightarrow \frac{d y}{d t}=12 \times \frac{d}{d t}(1-\cos t)=12 \times[0-(-\sin t)]=12 \sin t$
$\therefore$ $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}=\frac{12 \sin \mathrm{t}}{10(1-\cos t)}=\frac{12 \times 2 \sin \frac{\mathrm{t}}{2} \cos \frac{\mathrm{t}}{2}}{10 \times 2 \sin ^{2} \frac{\mathrm{t}}{2}}=\frac{6}{5} \cot \frac{\mathrm{t}}{2}$
$or ~ \frac{d y}{d x}=\frac{6}{5} \cot \frac{t}{2}$

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Question 35 Marks

Differentiate the function $x^{x^{2}-3}+(x-3)^{x^{2}}, \text { for } x>3$ w.r.t to x.

Answer

Let $y = x^{x^{2}-3}+(x-3)^{x^{2}}$
And let $u = x^{x^{2}-3}$ and $v = (x-3)^{x^{2}}$
$\therefore y = u + v$
Differentiating both sides w.r.t. x we get
$\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}.....(i)$
Now,
$u = x^{x^{2}-3}$
Taking logarithm both sides
$\log u = \log x^{x^{2}-3}$
$\Rightarrow \log u = (x^2 - 3) \log x$
Differentiating w.r.t. x, we get
$\frac{1}{\mathrm{u}} \frac{\mathrm{du}}{\mathrm{dx}}=\log _{\mathrm{X}} \times \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}-3\right)+\left(\mathrm{x}^{2}-3\right) \times \frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{x})$
$\Rightarrow \frac{d u}{d x}=u\left[\log x \times 2 x+\left(x^{2}-3\right) \times \frac{1}{x}\right]$
$\Rightarrow \frac{d u}{d x}=x^{x^{2}-3}\left[\frac{x^{2}-3}{x}+2 x \log x\right]$.....(ii)
Also,
$v = (x-3)^{x^{2}}$
Taking logarithm both sides
$\log v = \log (x-3)^{x^{2}}$
$\Rightarrow \log v = x^2 \log(x-3)$
Differentiating both sides w.r.t. x
$\frac{1}{\mathrm{v}} \frac{\mathrm{dv}}{\mathrm{dx}}=\log (\mathrm{x}-3) \times \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}\right)+\mathrm{x}^{2} \times \frac{\mathrm{d}}{\mathrm{dx}}[\log (\mathrm{x}-3)]$
$\Rightarrow \frac{d v}{d x}=v\left[\log (x-3) \times 2 x+x^{2} \times \frac{1}{(x-3)} \times \frac{d}{d x}(x-3)\right]$
$\Rightarrow \frac{d v}{d x}=(x-3)^{x^{2}}\left[2 x \log (x-3)+\frac{x^{2}}{(x-3)} \times 1\right]$
$\Rightarrow \frac{d v}{d x}=(x-3)^{x^{2}}\left[\frac{x^{2}}{(x-3)}+2 x \log (x-3)\right]$....(iii)
Substituting (ii) and (iii) in (i)
$\therefore \frac{d y}{d x}=x^{x^{2}-3}\left[\frac{x^{2}-3}{x}+2 x \log x\right]$ + $(x-3)^{x^{2}}\left[\frac{x^{2}}{(x-3)}+2 x \log (x-3)\right]$

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Question 45 Marks

If $y = (\tan^{-1}x)^2$ show that $(x^2 + 1)^2 y_2 + 2x(x^2 + 1)y_1 = 2$

Answer

$y = (\tan^{-1}x)^2 ($given$)$Differentiating both sides w.r.t to $x$
${y_1} = 2{\tan ^{ - 1}}x.\frac{1}{{1 + {x^2}}}$
$(1 + x^2)y_1 = 2 \tan^{-1}x$
Again differentiating both sides w.r.t to $x$
$\left( {1 + {x^2}} \right){y_2} + {y_1}\left( {2x} \right) = 2.\frac{1}{{1 + {x^2}}}$
$(1 + x^2)^2 y_2 + 2x(x^2 + 1)y_1 = 2$

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Question 55 Marks

If $e^y(x + 1) = 1,$ show that $\frac { d ^ { 2 } y } { d x ^ { 2 } } = \left( \frac { d y } { d x } \right) ^ { 2 }$

Answer

According to the question, $e^y (x + 1) = 1$
Taking \log both sides,
$\Rightarrow \log [e^y (x + 1) ]= \log 1$
$\Rightarrow \log e^y + \log(x + 1) = \log 1$
$\Rightarrow y + \log(x + 1) = \log 1$ [$\because \log e^y = y]$
differentiating both sides w.r.t. x,
$\Rightarrow\frac { d y } { d x } + \frac { 1 } { x + 1 } = 0$.........(i)
Differentiating both sides w.r.t. $'x',$
$\Rightarrow \frac { d ^ { 2 } y } { d x ^ { 2 } } - \frac { 1 } { ( x + 1 ) ^ { 2 } } = 0$
$\Rightarrow \quad \frac { d ^ { 2 } y } { d x ^ { 2 } } - \left( - \frac { d y } { d x } \right) ^ { 2 } = 0 [$ From Equation$(i)]$
$\Rightarrow \quad \frac { d ^ { 2 } y } { d x ^ { 2 } } - \left( \frac { d y } { d x } \right) ^ { 2 } = 0$
$\Rightarrow \quad \frac { d ^ { 2 } y } { d x ^ { 2 } } = \left( \frac { d y } { d x } \right) ^ { 2 }$

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Question 65 Marks

If $y = 500e^{7x} + 600e^{-7x}$ show that $\frac{{{d^2}y}}{{d{x^2}}} = 49y$.

Answer

Given: $y = 500e^{7x} + 600e^{-7x} ...(i)$
$\therefore \frac{{dy}}{{dx}} = 500e^{7x}(7) + 600e^{-7x}(-7) = 500(7)e^{7x} - 600(7)e^{-7x}$
$\Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = 500(7)e^{7x}(7) - 600(7)e^{-7x}(-7) = 500(49)e^{7x} + 600(49){e^{-7x}}$
$\Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = 49[500e^{7x} + 600e^{-7x}]$
$= 49y [$From eq. $(i)]$
$\Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = 49y$ Hence proved.

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Question 75 Marks

If $y = Ae^{mx} + Be^{nx}$, show that $\frac{{{d^2}y}}{{d{x^2}}} - \left( {m + n} \right)\frac{{dy}}{{dx}} + mny = 0$

Answer

Given: $y = Ae^{mx} + Be^{nx}$ ....(i)To prove: $\frac{{{d^2}y}}{{d{x^2}}} - \left( {m + n} \right)\frac{{dy}}{{dx}} + mny = 0$
$\therefore \frac{{dy}}{{dx}} =A{e^{mx}}\frac{d}{{dx}}\left( {mx} \right) + B{e^{nx}}\frac{d}{{dx}}\left( {nx} \right)$ $\left[ {\because \frac{d}{{dx}}{e^{f\left( x \right)}} = {e^{f\left( x \right)}}\frac{d}{{dx}}f\left( x \right)} \right]$
$\Rightarrow \frac{{dy}}{{dx}} = Am{e^{mx}} + Bn{e^{nx}}$….(ii)
Again $\frac{{{d^2}y}}{{{d}x^2}} = Am{e^{mx}}.m + Bn{e^{nx}}.n$
$= Am^2e^{mx} + Bn^2e^{nx}$ ...(iii)
Now, L.H.S.= $\frac{{{d^2}y}}{{d{x^2}}} - \left( {m + n} \right)\frac{{dy}}{{dx}} + mny$
$= Am^2e^{mx} + Bn^2e^{nx} - (m + n)(Ame^{mx} + Bne^{nx}) + mn(Ae^{mx} + Be^{nx})$
$= Am^2e^{mx} + Bn^2e^{nx} - Am^2e^{mx} - Bmne^{nx} - Amne^{mx} - Bn^2e^{nx} + Amne^{mx} + Bmne^{nx}$
$= 0$
= R.H.S. Hence proved.

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Question 85 Marks

If $y = 3 \cos (\log x) + 4 \sin (\log x).$ Show that $x^2y_2 + xy_1 + y = 0$

Answer

$y = 3\cos (\log x) + 4\sin (\log x)$Differentiating both side w.r.t. to $x$
${y_1} = \frac{{ - 3\sin (\log x)}}{x} + \frac{{4\cos (\log x)}}{x}$
$x{y_1} = - 3\sin (\log x) + 4\cos (\log x)$
Again differentiating,
$x{y_2} + {y_1}.1 = \frac{{ - 3\cos (\log x)}}{x} - \frac{{4\sin (\log x)}}{x}$
${x^2}{y_2} + x{y_1} = - 3\cos (\log x) - 4\sin (\log x)$
${x^2}{y_2} + x{y_1} = - y$
${x^2}{y_2} + x{y_1} + y = 0$

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Question 95 Marks

If x and y are connected parametrically by the equation $x=\frac{\sin ^{3} t}{\sqrt{\cos 2 t}}, y=\frac{\cos ^{3} t}{\sqrt{\cos 2 t}}$, without eliminating the parameter, Find $\frac{d y}{d x}$

Answer

It is given that
$\mathrm{x}=\frac{\sin ^{3} \mathrm{t}}{\sqrt{\cos 2 \mathrm{t}}}, \mathrm{y}=\frac{\cos ^{3} \mathrm{t}}{\sqrt{\cos 2 \mathrm{t}}}$
Then, we have
$\frac{d x}{d t}=\frac{d\left(\frac{\sin ^{3} t}{\sqrt{\cos 2 t}}\right)}{d t}$
= $\frac{\sqrt{\cos 2 t} \cdot \frac{d\left(\sin ^{3} t\right)}{d t}-\sin ^{3} t \cdot \frac{d \sqrt{\cos 2 t}}{d t}}{\cos 2 t}$
= $\frac{\sqrt{\cos 2 t} \cdot 3 \sin ^{2} t \cdot \frac{d(\sin t)}{d t}-\sin ^{3} t \times \frac{1}{2 \sqrt{\cos 2 t}} \frac{d(\cos 2 t)}{d t}}{\cos 2 t}$
= $\frac{3 \sqrt{\cos 2 t} \cdot \sin ^{2} t \cdot \cos t-\frac{\sin ^{3} t}{2 \sqrt{\cos 2 t}} \cdot(-2 \sin 2 t)}{\cos 2 t}$
= $\frac{3 \cos 2 t \cdot \sin ^{2} t \cdot \cos t+\sin ^{3} t \sin 2 t}{\cos 2 t \sqrt{\cos 2 t}}$ .........(i)
$\frac{d y}{d t}=\frac{d\left(\frac{\cos ^{3} t}{\sqrt{\cos 2 t}}\right)}{d t}$
= $\frac{\sqrt{\cos 2 t} \cdot \frac{d\left(\cos ^{3} t\right)}{d t}-\cos ^{3} t \cdot \frac{d \sqrt{\cos 2 t}}{d t}}{\cos 2 t}$
= $\frac{\sqrt{\cos 2 t} \cdot 3 \cos ^{2} t \cdot \frac{d(\cos t)}{d t}-\cos ^{3} t \times \frac{1}{2 \sqrt{\cos 2 t}} \frac{d(\cos 2 t)}{d t}}{\cos 2 t}$
= $\frac{3 \sqrt{\cos 2 t} \cos ^{2} t .(-\sin t)-\frac{\cos ^{3} t}{2 \sqrt{\cos 2 t}} \cdot(-2 \sin 2 t)}{\cos 2 t}$
= $\frac{-3 \cos 2 t \cdot \cos ^{2} t \cdot \sin t+\cos ^{3} t \sin 2 t}{\cos 2 t \sqrt{\cos 2 t}}$ .........(ii)
Therefore, from equation (i) and (ii), we get
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}=\frac{-3 \cos 2 \mathrm{t} \cdot \cos ^{2} \mathrm{t} \cdot \operatorname{sint}+\cos ^{3} \mathrm{t} \sin 2 \mathrm{t}}{3 \cos 2 \mathrm{t} \cdot \sin ^{2} \mathrm{t} \cdot \operatorname{cost}+\sin ^{3} \mathrm{tsin} 2 \mathrm{t}}$
= $\frac{-3 \cos 2 t \cdot \cos ^{2} t \cdot \sin t+\cos ^{3} t(2 \sin t \cos t)}{3 \cos 2 t \cdot \sin ^{2} t \cdot \cos t+\sin ^{3} t(2 \sin t \cos t)}$
= $\frac{\sin t \cos t\left[-3 \cos 2 t \cdot \cos t+2 \cos ^{3} t\right]}{\sin t \cos t\left[3 \cos 2 t \cdot \sin t+2 \sin ^{3} t\right]}$
= $\frac{\left[-3\left(2 \cos ^{2} t-1\right) \cos t+2 \cos ^{3} t\right]}{\left[3\left(1-2 \sin ^{2} t\right) \sin t+2 \sin ^{2} t\right]}$ , [Since, cos $2t = (2 \cos^2 t - 1)$ and also $\cos 2t = (1 - 2 \sin^2 t)]$
= $\frac{-4 \cos ^{3} t+3 \cos t}{3 \sin t-4 \sin ^{3} t}$
= $\frac{-\cos 3 t}{\sin 3 t}$ [Since, $\cos 3t = 4 \cos^3 t - 3 \cos t, \sin 3t = 3 \sin t - 4 \sin^3 t$]
= $-\cot 3t$
Hence, the value of $\frac{dy}{dx}$ is -cot 3t

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Question 105 Marks

If x = $ \sqrt { a ^ { \sin ^ { - 1 } t} }$, y = $ \sqrt { a ^ { \cos ^ { - 1 } t}}$, show that $ \frac { d y } { d x } = -\frac { y } { x }$.

Answer

According to the question, x = $ \sqrt { a ^ { \sin ^ { - 1 } t} }$ and y = $ \sqrt { a ^ { \cos ^ { - 1 } t}}$
Consider, $x= \left( a ^ { \sin ^ { - 1 } t } \right) ^ { 1 / 2 }$
Differentiating both sides w.r.t x,
$\Rightarrow \frac { d x } { d t } = \frac { 1 } { 2 } \left( a ^ { \sin ^ { - 1 } t } \right) ^ { - 1 / 2 } \frac { d } { d t } \left( a ^ { \sin ^ { - 1 } t } \right)$[ Using chain rule of derivative]
$ = \frac { 1 } { 2 } \left( a ^ { \mathrm { sin } ^ { - 1 } t } \right) ^ { - 1 / 2 } a ^ { \sin ^ { - 1 } t } \log a \frac { d } { d t } \left( \sin ^ { - 1 } t \right)$
$ = \frac { 1 } { 2 } \left( a ^ { \sin ^ { - 1 } t } \right) ^ { - 1 / 2 } a ^ { \sin ^ { - 1 } t } \log a \cdot \frac { 1 } { \sqrt { 1 - t ^ { 2 } } }$
$ = \frac { 1 } { 2 } \left( a ^ { \sin ^ { - 1 } t } \right) ^ { 1 / 2 } \log a \cdot \frac { 1 } { \sqrt { 1 - t ^ { 2 } } }$
$ \Rightarrow \frac { d x } { d t } = \frac { \frac { 1 } { 2 } \sqrt { a ^ { \sin ^ { - 1 } t } } \cdot \log a } { \sqrt { 1 - t ^ { 2 } } }$..........(i)
Consider , $y= \left( a ^ { \cos ^ { - 1 } t } \right) ^ { 1 / 2 }$
Differentiating both sides w.r.t x,
$ \frac { d y } { d t } = \frac { 1 } { 2 } \left( a ^ { \cos ^ { - 1 } t } \right) ^ { - 1 / 2 } \frac { d } { d t } \left( a ^ { \cos ^ { - 1 } t } \right)$ [Using chain rule of derivative]
$ = \frac { 1 } { 2 } \left( a ^ { \cos ^ { - 1 } t } \right) ^ { - 1 / 2 } a ^ { \cos ^ { - 1 } t } \log a \frac { d } { d t } \left( \cos ^ { - 1 } t \right)$
$ = \frac { 1 } { 2 } \left( a ^ { \cos ^ { - 1 } t } \right) ^ { 1 / 2 } \log a \cdot \frac { ( - 1 ) } { \sqrt { 1 - t ^ { 2 } } }$
$ \Rightarrow \frac { d y } { d t } = \frac { - \frac { 1 } { 2 } \sqrt { a ^ { \cos ^ { - 1 } t } } \cdot \log a } { \sqrt { 1 - t ^ { 2 } } }$..........(ii)
Dividing Eq.(ii) by Eq.(i),
$\Rightarrow \frac{{dy}}{{dx}} = \frac{{\left( {\frac{{dy}}{{dt}}} \right)}}{{\left( {\frac{{dx}}{{dt}}} \right)}} = \frac{{\left( {\frac{{ - \frac{1}{2}\sqrt {{a^{{{\cos }^{ - 1}}t}}} \log a}}{{\sqrt {1 - {t^2}} }}} \right)}}{{\left( {\frac{{\frac{1}{2}\sqrt {{a^{{{\sin }^{ - 1}}t}}\log a} }}{{\sqrt {1 - {t^2}} }}} \right)}}$
$ = - \frac { \sqrt { a ^ { \cos ^ { - 1 } t } } } { \sqrt { a ^ { \sin ^ { - 1 } t } } } = - \frac { y } { x }$
Hence proved

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Question 115 Marks

Differentiate the function ${x^{\sin x}} + {\left( {\sin x} \right)^{\cos x}}$ w.r.t. x.

Answer

Let y = u + v

When $u = {x^{\sin x}}$, $v = {\left( {\sin x} \right)^{\cos x}}$

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

$u = {x^{\sin x}}$

Taking log both side

log $u = \log {x^{\sin x}}$

log $u = \sin x.\log x$

diff. both side w.r. to x

$\frac{1}{u}\frac{{du}}{{dx}} = \sin x.\frac{1}{x} + \log x\cos x$

$\frac{{du}}{{dx}} = u\left[ {\frac{{\sin x}}{x} + \log x.\cos x} \right]$

$\frac{{du}}{{dx}} = {x^{\sin x}}\left[ {\frac{{\sin x + x\log x.\cos x}}{x}} \right]$

$v = {\left( {\sin x} \right)^{\cos x}}$

Taking log both sides

log $v = \log {\left( {\sin x} \right)^{\cos x}}$

$\log v = \cos x.\log \left( {\sin x} \right)$

Differentiating both sides w.r.t to x

$\frac{1}{v}.\frac{{dv}}{{dx}} = \cos x.\frac{1}{{\sin x}}\left( {\cos x} \right) + \log \left( {\sin x} \right)\left( { - \sin x} \right)$

$\frac{{dv}}{{dx}} = v\left[ {\cot x.\cos x - \log \left( {\sin x} \right).\sin x} \right]$

$\frac{{dv}}{{dx}} = {\left( {\sin x} \right)^{\cos x}}\left[ {\cot x.\cos x - \log \left( {\sin x} \right).\sin x} \right]$

Hence

$\frac{{dy}}{{dx}} = {x^{\sin x}}\left[ {\frac{{\sin x + x\log x.\cos x}}{x}} \right] + $${\left( {\sin x} \right)^{\cos x}}\left[ {\cot x\cos x - \log \left( {\sin x} \right).\sin x} \right]$

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Question 125 Marks

Differentiate the function $(\sin x)^x + \sin^{–1}\sqrt{x}$ w.r.t. $x$.

Answer

Given function is: $(\sin x)^{x}+\sin ^{-1} \sqrt{x}$
Let $y=(\sin x)^{x}+\sin ^{-1} \sqrt{x}$
Let $y = u + v$
$\Rightarrow u = (\sin x)^x$ and $\mathrm{v}=\sin ^{-1} \sqrt{\mathrm{x}}$
For, $\mathrm{u}=(\sin \mathrm{x})^{\mathrm{x}}$
Taking log on both sides, we get
$\log u = \log(\sin x)^x$
$\Rightarrow \log u = x.\log \sin x$
Now, differentiate both sides with respect to $x$
$\frac{d}{d x}(\log u)=\frac{d}{d x}[x \cdot \log (\sin x)]$
$\left.\Rightarrow \frac{1}{\mathrm{u}} \frac{\mathrm{du}}{\mathrm{dx}}=\mathrm{x} \cdot \frac{\mathrm{d}}{\mathrm{dx}} \log (\sin \mathrm{x})\right)+\log (\sin \mathrm{x}) \cdot \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})$
$\Rightarrow \frac{\mathrm{du}}{\mathrm{dx}}=\mathrm{u}\left[\mathrm{x} \cdot \frac{1}{\sin \mathrm{x}} \frac{\mathrm{d}}{\mathrm{dx}}(\sin \mathrm{x})+\log (\sin \mathrm{x}) \cdot(1)\right]$
$\Rightarrow \frac{\mathrm{du}}{\mathrm{dx}}=(\sin \mathrm{x})^{\mathrm{x}}\left[\frac{\mathrm{x}}{\sin \mathrm{x}} \cdot \cos \mathrm{x}+\log (\sin \mathrm{x}) \cdot(1)\right]$
$\Rightarrow \frac{d u}{d x}=(\sin x)^{x}[x \cdot \cot x+\log \sin x]$
For, $v = \sin ^{-1} \sqrt{x}$
Now, differentiating both sides with respect $x$
$\frac{d v}{d x}=\frac{d}{d x}\left[\sin ^{-1} \sqrt{x}\right]$
$\Rightarrow \frac{d v}{d x}=\frac{1}{\sqrt{1-(\sqrt{x})^{2}}} \cdot \frac{d}{d x}(\sqrt{x})$
$\Rightarrow \frac{d v}{d x}=\frac{1}{\sqrt{1-x}} \cdot \frac{1}{2(\sqrt{x})}$
$\Rightarrow \frac{d v}{d x}=\frac{1}{2 \sqrt{x-x^{2}}}$
Because, $y = u + v$
$\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$
$\Rightarrow \frac{d y}{d x}=(\sin x)^{x}[x \cdot \cot x+\log \sin x]+\frac{1}{2 \sqrt{x-x^{2}}}$

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Question 135 Marks

Differentiate the function $(\log x)^x + x^{\log x}$ w.r.t. $x$.

Answer

Given: $(\log x)^x + x^{\log x}$
Let $y = (\log x)^x + x^{\log x}$
Let $y = u + v$
$\Rightarrow u = (\log x)^x$ and $v = x^{\log x}$
For, $u = (\log x)^x$
Taking \log on both sides, we get
$\log u = \log (\log x)^x$
$\Rightarrow \log u = x.\log (\log (x))$
Now, differentiate both sides with respect to $x$
$\frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{u})=\frac{\mathrm{d}}{\mathrm{dx}}[\mathrm{x} \cdot \log (\log \mathrm{x})]$
$\left.\Rightarrow \frac{1}{\mathrm{u}} \frac{\mathrm{du}}{\mathrm{dx}}=\mathrm{x} \cdot \frac{\mathrm{d}}{\mathrm{dx}} \log (\log \mathrm{x})\right)+\log (\log \mathrm{x}) \cdot \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})$
$\Rightarrow \frac{\mathrm{du}}{\mathrm{dx}}=\mathrm{u}\left[\mathrm{x} \cdot \frac{1}{\log \mathrm{x}} \frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{x})+\log (\log \mathrm{x}) \cdot(1)\right]$
$\Rightarrow \frac{\mathrm{d} \mathrm{u}}{\mathrm{dx}}=(\log \mathrm{x})^{\mathrm{x}}\left[\frac{\mathrm{x}}{\log \mathrm{x}} \cdot \frac{1}{\mathrm{x}}+\log (\log \mathrm{x}) \cdot(1)\right]$
$\Rightarrow \frac{d u}{d x}=(\log x)^{x}\left[\frac{1+\log (\log x) \cdot(\log x)}{\log x}\right]$
$\Rightarrow \frac{d u}{d x}=(\log x)^{x-1}[1+\log x \cdot \log (\log x)]$
For, $v = x^{\log x}$
Taking log on both sides, we get
$\log v = \log (x^{\log x})$
$\Rightarrow \log v = \log x. \log x$
Now, differentiate both sides with respect to $x$
$\frac{d}{d x}(\log v)=\frac{d}{d x}\left[(\log x)^{2}\right]$
$\Rightarrow \frac{1}{\mathrm{v}} \frac{\mathrm{dv}}{\mathrm{dx}}=2 \cdot \log \mathrm{x} \frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{x})$
$\Rightarrow \frac{d v}{d x}=v\left[2 \cdot \frac{\log x}{x}\right]$
$\Rightarrow \frac{d v}{d x}=x^{\log x}\left[2 \cdot \frac{\log x}{x}\right]$
$\Rightarrow \frac{d v}{d x}=2 \cdot x^{\log x-1} \cdot \log x$
Because,$ y = u + v$
$\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$
$\Rightarrow \frac{d y}{d x}=(\log x)^{x-1}[1+\log x \cdot \log (\log x)]+2 \cdot x^{\log x-1} \cdot \log x$

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Question 145 Marks

Differentiate the function $\left(x+\frac{1}{x}\right)^{x}+x^{\left(1+\frac{1}{x}\right)}$ w.r.t. x.

Answer

Given: $\left(x+\frac{1}{x}\right)^{x}+x^{\left(1+\frac{1}{x}\right)}$
Let y = $\left(x+\frac{1}{x}\right)^{x}+x^{\left(1+\frac{1}{x}\right)}$
Also, let y = u + v
$\Rightarrow \mathrm{u}=\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)^x$ and $\mathrm{v}=\mathrm{x}^{\left(1+\frac{1}{\mathrm{x}}\right)}$
For, $\mathrm{u}=\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)^{\mathrm{x}}$
Taking log on both sides, we get
$\log u=\log \left(x+\frac{1}{x}\right)^{x}$
Now, differentiate both sides with respect to x
$\frac{d}{d x}(\log u)=\frac{d}{d x}\left[x \cdot \log \left(x+\frac{1}{x}\right)\right]$
$\Rightarrow \frac{1}{\mathrm{u}} \frac{\mathrm{du}}{\mathrm{dx}}=\mathrm{x} \cdot \frac{\mathrm{d}}{\mathrm{dx}}\left(\log \left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)\right)+\log \left(\mathrm{x}+\frac{1}{\mathrm{x}}\right) \cdot \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})$
$\Rightarrow \frac{\mathrm{d} \mathrm{u}}{\mathrm{dx}}=\mathrm{u}\left[\mathrm{x} \cdot \frac{1}{\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)} \cdot \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)+\log \left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)\right]$
$\Rightarrow \frac{\mathrm{du}}{\mathrm{dx}}=\mathrm{u}\left[\mathrm{x} \cdot \frac{1}{\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)} \cdot\left(\frac{\mathrm{dx}}{\mathrm{dx}}+\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{1}{\mathrm{x}}\right)\right)+\log \left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)\right]$
$\Rightarrow \frac{d u}{d x}=u\left[\frac{x}{\left(x+\frac{1}{x}\right)} \cdot\left(1-\frac{1}{x^{2}}\right)+\log \left(x+\frac{1}{x}\right)\right]$
$\Rightarrow \frac{d u}{d x}=u\left[\frac{x}{\left(x+\frac{1}{x}\right)} \cdot\left(\frac{x^{2}-1}{x^{2}}\right)+\log \left(x+\frac{1}{x}\right)\right]$
$\Rightarrow \frac{d u}{d x}=\left(x+\frac{1}{x}\right)^{x}\left[\left(\frac{x^{2}-1}{x^{2}+1}\right)+\log \left(x+\frac{1}{x}\right)\right]$
For, $\mathrm{v}=\mathrm{x}^{\left(1+\frac{1}{\mathrm{x}}\right)}$
Taking log on both sides, we get
$\log v=\log {x}^{\left(1+\frac{1}{x}\right)}$
$\Rightarrow \log \mathrm{v}=\left(1+\frac{1}{\mathrm{x}}\right) \cdot \log \mathrm{x}$
Now, differentiate both sides with respect to x
$\frac{d}{d x}(\log v)=\frac{d}{d x}\left[\left(1+\frac{1}{x}\right) \cdot \log x\right]$
$\Rightarrow \frac{1}{\mathrm{v}} \frac{\mathrm{dv}}{\mathrm{dx}}=\log \mathrm{x} \cdot \frac{\mathrm{d}}{\mathrm{dx}}\left(1+\frac{1}{\mathrm{x}}\right)+\left(1+\frac{1}{\mathrm{x}}\right) \cdot \frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{x})$
$\Rightarrow \frac{d V}{d x}=v\left[\log x \cdot\left(0-\frac{1}{x^{2}}\right)+\left(1+\frac{1}{x}\right) \cdot \frac{1}{x}\right]$
$\Rightarrow \frac{d v}{d x}=x^{\left(1+\frac{1}{x}\right)}\left[-\frac{\log x}{x^{2}}+\left(\frac{1}{x}+\frac{1}{x^{2}}\right)\right]$
$\Rightarrow \frac{d v}{d x}=x^{\left(1+\frac{1}{x}\right)}\left[\frac{-\log x+x+1}{x^{2}}\right]$
$\Rightarrow \frac{d v}{d x}=x^{\left(1+\frac{1}{x}\right)}\left[\frac{x+1-\log x}{x^{2}}\right]$
Because, y = u + v
$\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$
$\Rightarrow \frac{d y}{d x}=\left(x+\frac{1}{x}\right)^{x}\left[\left(\frac{x^{2}-1}{x^{2}+1}\right)+\log \left(x+\frac{1}{x}\right)\right]+x^{\left(1+\frac{1}{x}\right)}\left[\frac{x+1-\log x}{x^{2}}\right]$

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Question 155 Marks

Differentiate the function $x^{x \cos x}+\frac{x^{2}+1}{x^{2}-1}$ w.r.t. x.

Answer

Given: $\mathrm{x}^{\mathrm{x} \cos \mathrm{x}}+\frac{\mathrm{x}^{2}+1}{\mathrm{x}^{2}-1}$
Let $y=x^{x \cos x}+\frac{x^{2}+1}{x^{2}-1}$
Let y = u + v
$\Rightarrow u = x^{xcos x}$ and v = $\frac{x^{2}+1}{x^{2}-1}$
For, $\mathrm{u}=\mathrm{x}^{\mathrm{x} \cos x}$
Taking log on both sides, we get
$\log u=\log {x} ^{x \cos x}$
$\Rightarrow \log u=x \cdot \cos x \cdot \log x$
Now, differentiate both sides with respect to x
$\frac{d}{d x}(\log u)=\frac{d}{d x}[x \cdot \cos x \cdot \log x]$
$\Rightarrow \frac{1}{\mathrm{u}} \frac{\mathrm{du}}{\mathrm{dx}}=\cos \mathrm{x} \cdot \log \mathrm{x} \cdot \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})+\mathrm{x} \cdot \log \mathrm{x} \cdot \frac{\mathrm{d}}{\mathrm{dx}}(\cos \mathrm{x})+\mathrm{x} \cdot \cos \mathrm{x} \cdot \frac{\mathrm{d}}{\mathrm{dx}} \log \mathrm{x}$
$\Rightarrow \frac{\mathrm{d} \mathrm{u}}{\mathrm{dx}}=\mathrm{u}\left[\cos \mathrm{x} . \log \mathrm{x}+\mathrm{x} \cdot \log \mathrm{x}(-\sin \mathrm{x})+\mathrm{x} \cdot \cos \mathrm{x} \cdot\left(\frac{1}{\mathrm{x}}\right)\right]$
$\Rightarrow \frac{d u}{d x}=x^{x \cos x}[\cos x \cdot \log x-x \cdot \log x \cdot \sin x+\cos x]$
$\Rightarrow \frac{d u}{d x}=x^{x \cos x}[\cos x(1+\log x)-x \cdot \log x \cdot \sin x]$
For, v = $\frac{x^{2}+1}{x^{2}-1}$
Takinglog on both sides, we get
$\log \mathrm{v}=\log \left(\frac{\mathrm{x}^{2}+1}{\mathrm{x}^{2}-1}\right)$
$\Rightarrow \log v=\log \left(x^{2}+1\right)-\log \left(x^{2}-1\right)$
Now, differentiate both sides with respect to x
$\frac{d}{d x}(\log v)=\frac{d}{d x}\left[\log \left(x^{2}+1\right)-\log \left(x^{2}-1\right)\right]$
$\Rightarrow \frac{1}{\mathrm{v}} \frac{\mathrm{d} \mathrm{v}}{\mathrm{dx}}=\frac{1}{\mathrm{x}^{2}+1} \cdot \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}\right)-\frac{1}{\mathrm{x}^{2}-1} \cdot \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}\right)$
$\Rightarrow \frac{d v}{d x}=v \cdot\left[\frac{1}{x^{2}+1} \cdot(2 x)-\frac{1}{x^{2}-1} \cdot(2 x)\right]$
$\Rightarrow \frac{d v}{d x}=\left(\frac{x^{2}+1}{x^{2}-1}\right) \cdot\left[\frac{2 x\left(x^{2}-1\right)-2 x\left(x^{2}+1\right)}{\left(x^{2}+1\right)\left(x^{2}-1\right)}\right]$
$\Rightarrow \frac{d v}{d x}=\left(\frac{x^{2}+1}{x^{2}-1}\right) \cdot\left[\frac{-4 x}{\left(x^{2}+1\right)\left(x^{2}-1\right)}\right]$
$\Rightarrow \frac{d v}{d x}=\left[\frac{-4 x}{\left(x^{2}-1\right)^{2}}\right]$
Because, y = u + v
$\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$
$\Rightarrow \frac{d y}{d x}=x^{x \cos x}[\cos x(1+\log x)-x \cdot \log x \cdot \sin x]-\left[\frac{4 x}{\left(x^{2}-1\right)^{2}}\right]$

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Question 165 Marks

If $y = 3e^{2x} + 2e^{3x}$, prove that $\frac{d^{2} y}{d x^{2}}-5 \frac{d y}{d x} + 6y = 0$.

Answer

Given that $y = 3e^{2x} + 2e^{3x},$
Then $\frac{d y}{d x} = 6e^{2x} + 6e^{3x}$
$= 6 (e^{2x} + e^{3x})$
Therefore, $\frac{d^{2} y}{d x^{2}}= 12e^{2x} + 18e^{3x}$
$= 6 (2e^{2x} + 3e^{3x})$
Hence $\frac{d^{2} y}{d x^{2}}-5 \frac{d y}{d x} + 6y = 6 (2e^{2x} + 3e^{3x}) – 30 (e^{2x} + e^{3x}) + 6 (3e^{2x} + 2e^{3x}) = 0.$
So, the result is correct.

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Question 175 Marks

If y = A sin x + B cos x, then prove that $\frac{d^{2} y}{d x^{2}}$ + y = 0

Answer

We have
$\frac{d y}{d x}$ = A cos x - B sin x
and $\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}$ (A cos x – B sin x)
= – A sin x – B cos x = – y
Hence $\frac{d^{2} y}{d x^{2}}$ + y = 0

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