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1

AIEEE 2011

MCQ (Single Correct Answer)
$${{{d^2}x} \over {d{y^2}}}$$ equals:
A
$$ - {\left( {{{{d^2}y} \over {d{x^2}}}} \right)^{ - 1}}{\left( {{{dy} \over {dx}}} \right)^{ - 3}}$$
B
$${\left( {{{{d^2}y} \over {d{x^2}}}} \right)^{}}{\left( {{{dy} \over {dx}}} \right)^{ - 2}}$$
C
$$ - \left( {{{{d^2}y} \over {d{x^2}}}} \right){\left( {{{dy} \over {dx}}} \right)^{ - 3}}$$
D
$${\left( {{{{d^2}y} \over {d{x^2}}}} \right)^{ - 1}}$$

Explanation

$${{{d^2}x} \over {d{y^2}}} = {d \over {dy}}\left( {{{dx} \over {dy}}} \right)$$

$$ = {d \over {dx}}\left( {{{dx} \over {dy}}} \right){{dx} \over {dy}}$$

$$ = {d \over {dx}}\left( {{1 \over {dy/dx}}} \right){{dx} \over {dy}}$$

$$ = - {1 \over {{{\left( {{{dy} \over {dx}}} \right)}^2}}}.{{{d^2}y} \over {d{x^2}}}.{1 \over {{{dy} \over {dx}}}}$$

$$ = - {1 \over {{{\left( {{{dy} \over {dx}}} \right)}^3}}}{{{d^2}y} \over {d{x^2}}}$$
2

AIEEE 2010

MCQ (Single Correct Answer)
Let $$f:\left( { - 1,1} \right) \to R$$ be a differentiable function with $$f\left( 0 \right) = - 1$$ and $$f'\left( 0 \right) = 1$$. Let $$g\left( x \right) = {\left[ {f\left( {2f\left( x \right) + 2} \right)} \right]^2}$$. Then $$g'\left( 0 \right) = $$
A
$$-4$$
B
$$0$$
C
$$-2$$
D
$$4$$

Explanation

$$g'\left( x \right) = 2\left( {f\left( {2f\left( x \right) + 2} \right)} \right)$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{d \over {dx}}\left( {f\left( {2f\left( x \right) + 2} \right)} \right)} \right)$$

$$ = 2f\left( {2f\left( x \right) + 2} \right)f'\left( {2f\left( x \right)} \right)$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \left. 2 \right).\left( {2f'\left( x \right)} \right)$$

$$ \Rightarrow g'\left( 0 \right) = 2f\left( {2f\left( 0 \right) + 2} \right).$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,f'\left( {2f\left( 0 \right) + 2} \right).2f'\left( 0 \right)$$

$$ = 4f\left( 0 \right){\left( {f'\left( 0 \right)} \right)^2}$$

$$ = 4\left( { - 1} \right){\left( 1 \right)^2} = - 4$$
3

AIEEE 2009

MCQ (Single Correct Answer)
Let $$y$$ be an implicit function of $$x$$ defined by $${x^{2x}} - 2{x^x}\cot \,y - 1 = 0$$. Then $$y'(1)$$ equals
A
$$1$$
B
$$\log \,2$$
C
$$-\log \,2$$
D
$$-1$$

Explanation

$${x^{2x}} - 2{x^x}\,\cot \,y - 1 = 0$$

$$ \Rightarrow 2\,\cot \,y = {x^x} - {x^{ - x}}$$

$$ \Rightarrow 2\,\cot \,y\, = u - {1 \over u}$$

where $$u = {x^x}$$

Differentiating both sides with respect to $$x,$$

we get $$ \Rightarrow - 2\cos e{c^2}y{{dy} \over {dx}}$$

$$ = \left( {1 + {1 \over {{u^2}}}} \right){{du} \over {dx}}$$

where $$u = {x^x} \Rightarrow \log \,u = x\,\log \,x$$

$$ \Rightarrow {1 \over u}{{du} \over {dx}} = 1 + \log \,x$$

$$ \Rightarrow {{du} \over {dx}} = {x^x}\left( {1 + \log \,x} \right)$$

$$\therefore$$ We get $$ - 2\cos e{c^2}y{{dy} \over {dx}}$$

$$ = \left( {1 + {x^{ - 2x}}} \right){x^x}\left( {1 + \log \,x} \right)$$

$$ \Rightarrow {{dy} \over {dx}} = {{\left( {{x^x} + {x^{ - x}}} \right)\left( {1 + \log x} \right)} \over { - 2\left( {1 + {{\cot }^2}y} \right)}}\,\,\,\,\,\,...\left( i \right)$$

Now when

$$x=1,$$ $${x^{2x}} - 2{x^x}\,\cot \,y - 1 = 0,$$

gives $$1 - 2\,\cot y - 1 = 0$$

$$ \Rightarrow \,\,\cot y\, = 0$$

$$\therefore$$ From equation $$(i),$$ at $$x=1$$

and $$\cot \,y = 0,$$ we get

$$y'\left( 1 \right) = {{\left( {1 + 1} \right)\left( {1 + 0} \right)} \over { - 2\left( {1 + 0} \right)}} = - 1$$
4

AIEEE 2006

MCQ (Single Correct Answer)
If $${x^m}.{y^n} = {\left( {x + y} \right)^{m + n}},$$ then $${{{dy} \over {dx}}}$$ is
A
$${y \over x}$$
B
$${{x + y} \over {xy}}$$
C
$$xy$$
D
$${x \over y}$$

Explanation

$${x^m}.{y^n} = {\left( {x + y} \right)^{m + n}}$$

$$ \Rightarrow m\ln x + n\ln y = \left( {m + n} \right)\ln \left( {x + y} \right)$$

Differentiating both sides.

$$\therefore$$ $${m \over x} + {n \over y}{{dy} \over {dx}} = {{m + n} \over {x + y}}\left( {1 + {{dy} \over {dx}}} \right)$$

$$ \Rightarrow \left( {{m \over x} - {{m + n} \over {x + y}}} \right) = \left( {{{m + n} \over {x + y}} - {n \over y}} \right){{dy} \over {dx}}$$

$$ \Rightarrow {{my - nx} \over {x\left( {x + y} \right)}} = \left( {{{my - nx} \over {y\left( {x + y} \right)}}} \right){{dy} \over {dx}}$$

$$ \Rightarrow {{dy} \over {dx}} = {y \over x}$$

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