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

(D) $\left(-\frac{3}{2}, \frac{3}{2}\right)-\{0\}$

(A) $\left(-\frac{1}{2}, \frac{1}{2}\right)-\{0\}$

(B) $\left(-\frac{3}{2}, \frac{3}{2}\right)$

(C) $\left(-\frac{1}{2}, \frac{1}{2}\right)$

JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Application of Derivatives

The set of all real values of

λ

for which the function

f (x) = (1 - cos^{2} x) \cdot (λ + sin x)

x \in (- \frac{π}{2}, \frac{π}{2})

, has exactly one maxima and exactly one minima, is

Answer: (D)

Write

f (x) = sin^{2} x (λ + sin x)

; its derivative is

f^{'} (x) = sin x cos x (2 λ + 3 sin x)

. On

(- \frac{π}{2}, \frac{π}{2})

the factor

cos x

never vanishes, so the turning points come from

sin x = 0

and

sin x = - \frac{2 λ}{3}

. For exactly one maximum and one minimum we need a second, distinct root, i.e.

- \frac{2 λ}{3} \in (- 1, 1)

but not equal to

0

, which gives

λ \in (- \frac{3}{2}, \frac{3}{2}) - {0}

JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Single correct option