Question

Statement I: If $x=e^{y+e^{{y+e}^{y+\dots \infty }}}$, then $\frac{dy}{dx}=\frac{x}{1-x}$

Statement II: If $y=|\cos x|+|\sin x|$, then the value of $\frac{dy}{dx}$ at $x=\frac{2\pi }{3}$ is $\left(\frac{\sqrt{3}-1}{2}\right)$

Which of the above statement(s) is/are correct?

• A. Statement $I$ only
• B. Statement $II$ only
• C. Both $I$ and $II$
• D. Neither $I$ nor $II$

Answer 1

The correct option is B Statement $II$ only

Statement I: $x=e^{y+e^{y+e^{y+\cdots \infty }}}$
$\Rightarrow x=e^{y+x}$
Differentiating with respect to 'x', we get
$1=e^{y+x}(1+y^{\prime })$
$\Rightarrow \frac{1}{x}-1=y^{\prime }$ (since $e^{x+y}=x$)
Hence, statement I is not correct.
Statement II: $y=\left|\cos x\right|+\left|\sin x\right|$
In the neighborhood of $\frac{2\pi }{3}$, $\cos x$ is negative and $\sin x$ is positive.
Hence, $\left|\cos x\right|=-\cos x$ and $\left|\sin x\right|=\sin x$.
$y=-\cos x+\sin x$ as $x\to \frac{2\pi }{3}$
$\Rightarrow y^{\prime }=\sin x+\cos x$
$\Rightarrow {y^{\prime }|}_{x=\frac{2\pi }{3}}=\frac{\sqrt{3}}{2}-\frac{1}{2}$

Hence, only statement II is correct.