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Let $$y$$ be an implicit function of $$x$$ defined by $${x^{2x}} - 2{x^x}\cot \,y - 1 = 0$$. Then $$y'(1)$$ equals

JEE · Math · previous-year question

  1. A.$$1$$
  2. B.$$\log \,2$$
  3. C.$$-\log \,2$$
  4. D.$$-1$$correct

Answer

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

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