Q94 / 170 · Manipal (MET) 2018 · Mathematics
The function $\sin x(1+\cos x), 0 \leq x \leq \frac{\pi}{2}$ has maximum value, when $x$ is equal to
- A0
- B$\frac{\pi}{2}$
- C$\frac{\pi}{6}$
- Correct answerDNone of these
Solution
Let $z=\sin x(1+\cos x)$
or $z=\sin x+\frac{1}{2} \sin 2 x$
On differentiating w.r.t. $x$, we get
$\frac{d z}{d x}=\cos x+\cos 2 x$
For maximum or minimum, put $\frac{d z}{d x}=0$ $\begin{array}{rr} \Rightarrow & \cos x+\cos 2 x=0 \\ \Rightarrow & 2 \cos ^2 x-1+\cos x=0 \\ \Rightarrow & (\cos x+1)(2 \cos x-1)=0 \\ \Rightarrow & \cos x=-1, \frac{1}{2} \\ \Rightarrow & x \end{array}$
But $0 \leq x \leq \frac{\pi}{2}$, therefore we take only $x=\frac{\pi}{3}$. $\therefore \frac{d^2 z}{d x^2}=-\sin x-2 \sin 2 x=-\mathrm{ve}, \text { for } x=\pi / 3$ $\therefore$ It is maximum at $x=\pi / 3$.
For maximum or minimum, put $\frac{d z}{d x}=0$ $\begin{array}{rr} \Rightarrow & \cos x+\cos 2 x=0 \\ \Rightarrow & 2 \cos ^2 x-1+\cos x=0 \\ \Rightarrow & (\cos x+1)(2 \cos x-1)=0 \\ \Rightarrow & \cos x=-1, \frac{1}{2} \\ \Rightarrow & x \end{array}$
But $0 \leq x \leq \frac{\pi}{2}$, therefore we take only $x=\frac{\pi}{3}$. $\therefore \frac{d^2 z}{d x^2}=-\sin x-2 \sin 2 x=-\mathrm{ve}, \text { for } x=\pi / 3$ $\therefore$ It is maximum at $x=\pi / 3$.
Hence, the correct answer is option: 'D'.