JEE Main 2020 — Application Of Derivatives Question with Solution

$$f(x)=\left(1-{\cos }^{2}x\right)\left(\lambda +\sin x\right)$$

$$\Rightarrow$$ f(x) = sin² x($$\lambda$$ + sinx) ....(1)

$$\therefore$$ f'(x) = 2sinx cosx ($$\lambda$$ +sinx) + sin²x (cosx)

$$\Rightarrow$$ f'(x) = sin2x($$\frac{2\lambda +3\sin x}{2}$$)

For maixma and minima, f'(x) = 0

$$\therefore$$ sin2x = 0 or 2$$\lambda$$ + 3sinx = 0

when sin2x = 0 $$\Rightarrow$$ x = 0

or when 2$$\lambda$$ + 3sinx = 0

$$\Rightarrow$$ sin x = $$-\frac{2\lambda }{3}$$

As $$x\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$$

$$\therefore$$ -1 < sinx < 1

$$\Rightarrow$$ -1 < $$-\frac{2\lambda }{3}$$ < 1

$$\Rightarrow$$ $$-\frac{3}{2}$$ < $$\lambda$$ < $$\frac{3}{2}$$

$$\therefore$$ $$\lambda$$ $$\in$$ $$\left(-\frac{3}{2},\frac{3}{2}\right)$$ - {0}

Note : If $$\lambda$$ = 0 $$\Rightarrow$$ f(x) = sin³x [from (1)]

Which is monotonic. so no maxima/minima.