Question

The set of all real values of $\lambda$ for which the function $f(x)=\left(1-\cos ^{2} x\right) \cdot(\lambda+\sin x)$, $x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, has exactly one maxima and exactly one minima, is :

• A. $\left(-\frac{1}{2}, \frac{1}{2}\right)-\{0\}$
• B. $\left(-\frac{1}{2}, \frac{1}{2}\right)$
• C. $\left(-\frac{3}{2}, \frac{3}{2}\right)$
• D. $\left(-\frac{3}{2}, \frac{3}{2}\right)-\{0\}$

Answer 1

Answer: (D)

$f(x)=\left(1-\cos ^{2} x\right)(\lambda+\sin x)$
$x \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$
$f(x)=\lambda \sin ^{2} x+\sin ^{3} x$
$f^{\prime}(x)=2 \lambda \sin x \cos x+3 \sin ^{2} x \cos x $
$f^{\prime}(x)=\sin x \cos x(2 \lambda+3 \sin x) $
$\sin x=0, \frac{-2 \lambda}{3},(\lambda \neq 0)$
for exactly one maxima & minima
$\frac{-2 \lambda}{3} \in(-1,1) \Rightarrow \lambda \in\left(\frac{-3}{2}, \frac{3}{2}\right)$
$\lambda \in\left(-\frac{3}{2}, \frac{3}{2}\right)-\{0\}$