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1

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

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

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.

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

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

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

$$ \Rightarrow {{dy} \over {dx}} = {y \over x}$$
3

AIEEE 2006

MCQ (Single Correct Answer)
The set of points where $$f\left( x \right) = {x \over {1 + \left| x \right|}}$$ is differentiable is
A
$$\left( { - \infty ,0} \right) \cup \left( {0,\infty } \right)$$
B
$$\left( { - \infty ,1} \right) \cup \left( { - 1,\infty } \right)$$
C
$$\left( { - \infty ,\infty } \right)$$
D
$$\left( {0,\infty } \right)$$

Explanation

$$f\left( x \right) = \left\{ {\matrix{ {{x \over {1 - x}},} & {x < 0} \cr {{x \over {1 + x}},} & {x \ge 0} \cr } } \right.$$

$$ \Rightarrow f'\left( x \right) = \left\{ {\matrix{ {{x \over {{{\left( {1 - x} \right)}^2}}},} & {x < 0} \cr {{x \over {{{\left( {1 + x} \right)}^2}}}} & {x \ge 0} \cr } } \right.$$

$$\therefore$$ $$f'\left( x \right)$$ exist at everywhere.
4

AIEEE 2005

MCQ (Single Correct Answer)
If the roots of the equation $${x^2} - bx + c = 0$$ be two consecutive integers, then $${b^2} - 4c$$ equals
A
$$-2$$
B
$$3$$
C
$$2$$
D
$$1$$

Explanation

Let $$\alpha ,\,\,\alpha + 1\,\,$$ be roots

Then $$\alpha + \alpha + 1 = b = $$ sum of -

roots $$\alpha \left( {\alpha + 1} \right) = c$$

$$=$$ product of roots

$$\therefore$$ $${b^2} - 4c $$

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

$$ = 1.$$

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