JEE PYQ: Application Of Derivatives Question 42
Question 42 - 2020 (06 Sep Shift 2)
The set of all real values of $\lambda$ for which the function $f(x) = (1 - \cos^2 x)(\lambda + \sin x)$, $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, has exactly one maxima and exactly one minima, is:
(1) $\left{-\frac{1}{2}, \frac{1}{2}\right} - {0}$
(2) $\left(-\frac{3}{2}, \frac{3}{2}\right) - {0}$
(3) $\left(-\frac{1}{2}, \frac{1}{2}\right)$
(4) $\left(-\frac{3}{2}, \frac{3}{2}\right) - {0}$
Show Answer
Answer: (4)
Solution
$f(x) = \sin^2 x(\lambda + \sin x)$. $f’(x) = \sin x \cos x(2\lambda + 3\sin x) = 0$. For exactly one max and one min: $\sin x = -\frac{2\lambda}{3}$ must have a solution in $(-\frac{\pi}{2}, \frac{\pi}{2})$ with sign change, giving $|\lambda| < \frac{3}{2}$, $\lambda \neq 0$.
BETA
