QJEEM09Ma3d2e: JEE Main 2009 Mathematics - Class 12 JEEM09Ma3d2e Solution: Let be an implicit function of defined by . Then equals

Question

Verified Solution • 2025–26 Session moderate

Let

y

be an implicit function of

x

defined by

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

. Then

y'(1)

equals

Accepted Answer

{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

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