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1

WB JEE 2009

MCQ (Single Correct Answer)

The second order derivative of a sin3t with respect to a cos3t at $$t = {\pi \over 4}$$ is

A
2
B
$${1 \over {12a}}$$
C
$${{4\sqrt 2 } \over {3a}}$$
D
$${{3a} \over {4\sqrt 2 }}$$

Explanation

$${{{d^2}y} \over {d{x^2}}} = {1 \over {3a}}{1 \over {{{\cos }^4}t\sin t}}$$ at $$x = {\pi \over 4},{{{d^2}y} \over {d{x^2}}} = {{4\sqrt 2 } \over {3a}}$$.

2

WB JEE 2008

MCQ (Single Correct Answer)

The value of $${{dy} \over {dx}}$$ at $$x = {\pi \over 2}$$, where y is given by $$y = {x^{\sin x}} + \sqrt x $$ is

A
$$1 + {1 \over {\sqrt {2\pi } }}$$
B
1
C
$${1 \over {\sqrt {2\pi } }}$$
D
$$1 - {1 \over {\sqrt {2\pi } }}$$

Explanation

$$y = {x^{\sin x}} + \sqrt x $$

Let $$y = u + v$$

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

Taking log on both sides, $$\log u = \sin x\log x$$

Taking derivative w.r.t. x

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

$${{du} \over {dx}} = u\left( {{{\sin x} \over x} + \cos x\log x} \right)$$ ..... (i)

$$v = \sqrt x $$, $${{dv} \over {dx}} = {1 \over {2\sqrt x }}$$ ...... (ii)

Since, $$y = u + v$$

$${{dy} \over {dx}} = {{du} \over {dx}} + {{dv} \over {dx}}$$

$${{dy} \over {dx}} = {x^{\sin x}}\left( {{{\sin x} \over x} + \cos x\log x} \right) + {1 \over {2\sqrt x }}$$ (from (i) and (ii))

$${\left( {{{dy} \over {dx}}} \right)_{at\,x = \pi /2}} = {\pi \over 2}\left( {{2 \over \pi }} \right) + {{1 \times \sqrt 2 } \over {2\sqrt \pi }} = 1 + {1 \over {\sqrt {2\pi } }}$$.

3

WB JEE 2008

MCQ (Single Correct Answer)

If $$x = {e^t}\sin t$$, $$y = {e^t}\cos t$$ then $${{{d^2}y} \over {d{x^2}}}$$ at x = $$\pi$$ is

A
$$2{e^x}$$
B
$${1 \over 2}{e^x}$$
C
$${1 \over {2{e^x}}}$$
D
$${2 \over {{e^x}}}$$

Explanation

$$x = {e^t}\sin t$$, $$y = {e^t}\cos t$$

$$\therefore$$ $${{dx} \over {dt}} = {d \over {dt}}({e^t}\sin t) = {e^t}{d \over {dt}}\sin t + \sin t{d \over {dt}}{e^t}$$

$$ = {e^t}\cos t + \sin t({e^t})$$

$$ \Rightarrow {{dx} \over {dt}} = {e^t}(\cos t + \sin t)$$ ..... (i)

Similarly, $${{dy} \over {dt}} = {d \over {dt}}{e^t}\cos t = {e^t}{d \over {dt}}\cos t + \cos t.{d \over {dt}}{e^t}$$

$$ = {e^t}( - \sin t) + \cos t({e^t}) = {e^t}(\cos t - \sin t)$$ ...... (ii)

$$\therefore$$ $${{dy} \over {dx}} = {{dy/dt} \over {dx/dt}} = {{{e^t}(\cos t - \sin t)} \over {{e^t}(\cos t + \sin t)}} = {{\cos t - \sin t} \over {\cos t + \sin t}}$$

$$\therefore$$ $${{{d^2}y} \over {d{x^2}}} = {d \over {dx}}\left( {{{dy} \over {dx}}} \right) = {d \over {dx}}\left( {{{\cos t - \sin t} \over {\cos t + \sin t}}} \right) = {d \over {dt}}\left( {{{\cos t - \sin t} \over {\cos t + \sin t}}} \right).{{dt} \over {dx}}$$

$$ = {{(\cos t + \sin t)( - \sin t - \cos t) - (\cos t - \sin t)(\cos t - \sin t)} \over {{{(\cos t + \sin t)}^2}}}.{{dt} \over {dx}}$$

$$ = {{ - {{(\cos t + \sin t)}^2} - {{(\cos t - \sin t)}^2}} \over {{{(\cos t + \sin t)}^2}}}.{1 \over {{e^t}(\cos t + \sin t)}}$$ [$$\because$$ $${{dt} \over {dx}} = {1 \over {dx/dt}}$$]

$$\therefore$$ $${\left( {{{{d^2}y} \over {d{x^2}}}} \right)_{x = \pi }} = {{ - {{(\cos \pi + \sin \pi )}^2} - {{(\cos \pi - \sin \pi )}^2}} \over {{{(\cos \pi + \sin \pi )}^2}}}.{1 \over {{e^\pi }(\cos \pi + \sin \pi )}}$$

$$ = {{ - 1 - 1} \over {{{( - 1)}^2}}} \times {1 \over {{e^\pi }( - 1)}} = {2 \over {{e^\pi }}}$$.

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