Question
Let $$f\left( x \right) = {e^x}\sin \,x$$ be the equation of a curve. If at $$x = a,\,0 \leqslant a \leqslant 2\pi ,$$ the slope of the tangent is the maximum then the value of $$a$$ is :
A.
$$\frac{\pi }{2}$$
B.
$$\frac{{3\pi }}{2}$$
C.
$$\pi $$
D.
$$\frac{\pi }{4}$$
Answer :
$$\frac{\pi }{2}$$
Solution :
$$f'\left( x \right) = {e^x}\left( {\sin \,x + \cos \,x} \right).$$ Therefore, the slope $$m$$ of the tangent is given by $$m = {e^x}\left( {\sin \,x + \cos \,x} \right).$$
Now, $$\frac{{dm}}{{dx}} = {e^x}\left\{ {\sin \,x + \cos \,x + \cos \,x - \sin \,x} \right\} = 2{e^x}.\cos \,x$$
and $$\frac{{{d^2}m}}{{d{x^2}}} = 2{e^x}\left( {\cos \,x - \sin \,x} \right)$$
At $$x=a,\,m$$ is the maximum
$$\eqalign{
& \therefore \,\,{\left. {\frac{{dm}}{{dx}}} \right)_{x = a}} = 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\left. {\frac{{{d^2}m}}{{d{x^2}}}} \right)_{x = a}} < 0 \cr
& {\left. {\frac{{dm}}{{dx}}} \right)_{x = a}} = 0\,\,\,\,\,\, \Rightarrow 2{e^a}\cos \,a = 0 \cr
& \Rightarrow a = \frac{\pi }{2},\,\frac{{3\pi }}{2}{\text{ because }}0 \leqslant a \leqslant 2\pi \cr
& {\left. {\frac{{{d^2}m}}{{d{x^2}}}} \right)_{x = a}} < 0\,\,\,\,\,\,\, \Rightarrow 2{e^a}\left( {\cos \,a - \sin \,a} \right) < 0 \cr
& {\text{This is satisfied by }}a = \frac{\pi }{2} \cr} $$