Q52 / 90 · JEE Main 2009 · Mathematics

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

  • Correct answer
    A
    $-1$
  • B
    1
  • C
    $\log 2$
  • D
    $-\log 2$

Solution

$x^{2 x}-2 x^x \cot y-1=0$ Now $x=1$, $$ 1-2 \cot y-1=0 \Rightarrow \cot y=0 \Rightarrow y=\frac{\pi}{2} $$ Now differentiating eq. (1) w.r.t. ' $x$ ' $$ 2 x^{2 x}(1+\log x)-2\left[x^x\left(-\operatorname{cosec}^2 y\right) \frac{d y}{d x}+\cot y x^x(1+\log x)\right]=0 $$ Now at $\left(1, \frac{\pi}{2}\right)$ $$ \begin{aligned} & 2(1+\log 1)-2\left(1(-1)\left(\frac{d y}{d x}\right)_{\left(1, \frac{\pi}{2}\right)}+0\right)=0 \\ & \Rightarrow 2+2\left(\frac{d y}{d x}\right)_{\left(1, \frac{\pi}{2}\right)}=0 \Rightarrow\left(\frac{d y}{d x}\right)_{\left(1, \frac{\pi}{2}\right)}=-1 \end{aligned} $$

Hence, the correct answer is option: 'A'.

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