Q66 / 125 · BITSAT 2015 · Mathematics
The slope of the tangent to the curve $y=e^{x}$ cos $x$ is minimum at $x=\alpha, 0 \leq a \leq 2 \pi$, then the value of $\alpha$ is
- A$\underline{0}$
- Correct answerB$\pi$
- C$2 \pi$
- D$3 \pi 2$
Solution
Let $m$ be the slope of the tangent to the curve
$y=e^{x} \cos x$
Then, $m=\frac{d y}{d x}=e^{x}(\cos x-\sin x)$
Diff. w.r.t 'x'
$$
\begin{array}{r}
\Rightarrow \frac{\mathrm{dm}}{\mathrm{dx}}=\mathrm{e}^{\mathrm{x}}(\cos \mathrm{x}-\sin \mathrm{x})+\mathrm{e}^{\mathrm{x}}(-\sin \mathrm{x}-\cos \mathrm{x}) \\
=-2 \mathrm{e}^{\mathrm{x}} \sin \mathrm{x}
\end{array}
$$
and $\frac{\mathrm{d}^{2} \mathrm{~m}}{\mathrm{dx}^{2}}=-2 \mathrm{e}^{\mathrm{x}}(\sin \mathrm{x}+\cos \mathrm{x})$
Put $\frac{\mathrm{dm}}{\mathrm{dx}}=0 \Rightarrow \sin \mathrm{x}=0 \Rightarrow \mathrm{x}=0, \pi, 2 \pi$
Clearly, $\frac{\mathrm{d}^{2} \mathrm{~m}}{\mathrm{dx}^{2}}>0$ for $\mathrm{x}=\pi$
Thus, $y$ is minimum at $x=\pi$.
Hence the value of $\alpha=\pi$
Hence, the correct answer is option: 'B'.