Question

If function $f\left(x\right)=\lambda \cos 2x+2\sin x$ has only one extremum point in $[0,\pi ],$ then the value of $\lambda$ can not be

• A. $-\frac{1}{4}$
• B. $\frac{2}{5}$
• C. $\frac{3}{7}$
• D. $\frac{7}{8}$

Answer 1

The correct option is D $\frac{7}{8}$

Given : $f\left(x\right)=\lambda \cos 2x+2\sin x$
$\Rightarrow f^{\prime }\left(x\right)=-2\lambda \sin 2x+2\cos x=2\cos x(1-2\lambda \sin x)$
for critical points : $f^{\prime }\left(x\right)=0$
$\Rightarrow \cos x=0$ or $\sin x=\frac{1}{2\lambda }$
$\Rightarrow \cos x=0$ gives $x=\frac{\pi }{2}$
So, there should not be any solution for $\sin x=\frac{1}{2\lambda }\text{in}[0,\pi ],$ as $f\left(x\right)$ has only one extremum point in $[0,\pi ].$
$\Rightarrow \frac{1}{2\lambda }>1$ or $\frac{1}{2\lambda }<0$
$\Rightarrow 0<\lambda <\frac{1}{2}$ or $\lambda <0$