Question

The curve $f(x)=e^x\sin x$ is defined in the interval $[0,2\pi ]$. The value of $x$ for which the slope of the tangent drawn to the curve at $x$ is maximum, is

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{3}$

Answer 1

Answer: (B)

We have,
$f(x)=e^x\cdot \sin x$
difference w.r.t ' $x^{{}^{\prime }}$
$f^{{}^{\prime }}(x)=e^x\cos x+e^x\sin x$
For $f^{{}^{\prime }}(x)$ is maximum we find $f^{{}^{\prime \prime }}(x)$
$f^{{}^{\prime \prime }}(x)=-e^x\sin x+e^x\cos x+e^x\cos x+e^x\sin x=2e^x\cos x$
$\therefore f^{{}^{\prime }}(x)$ is maximum at $f^{{}^{\prime \prime }}(x)=0$
Now, $f^{{}^{\prime \prime }}(x)=0$
$\therefore 2e^x\cos x=0$
$\Rightarrow x=\frac{\pi }{2}$
So, slope of tangent drawn to curve is maximum
at $x=\frac{\pi }{2}$.