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JEE Main 2009
Exam
JEE Main
Year
2009
Subject
Mathematics
Chapter
Differentiation

JEE Main 2009 · Mathematics · Differentiation

Let y be an implicit function of x defined by x^ 2 x -2 x^x y-1=0 ....

Let yy be an implicit function of xx defined by x2x−2xxcot⁡y−1=0x^{2 x}-2 x^x \cot y-1=0. Then y′(1)y^{\prime}(1) equals

Options

  1. A−1-1
  2. B1
  3. Clog⁡2\log 2
  4. D−log⁡2-\log 2

Correct answer

A. −1-1

Step-by-step solution

x2x−2xxcot⁡y−1=0x^{2 x}-2 x^x \cot y-1=0 Now x=1x=1, 1-2 \cot y-1=0 \Rightarrow \cot y=0 \Rightarrow y=π2\frac{\pi}{2} Now differentiating eq. (1) w.r.t. ' xx ' 2 x^{2 x}(1+log⁡x\log x)-2 \left[x^x \left(-\operatorname{cosec}^2y\right) \frac{d y}{d x}+\cotyxx(1+log⁡xy x^x(1+\log x)\right]=0 Now at (1,π2)\left(1, \frac{\pi}{2}\right) 2(1+log⁡1)−2(1(−1)(dydx)(1,π2)+0)=0⇒2+2(dydx)(1,π2)=0⇒(dydx)(1,π2)=−1\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}
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