JEE Main 2020 (Online) 6th September Evening ShiftQuestion 1
The set of all real values of $\lambda$ for which the function $f(x)=(1-\cos^2x)\cdot(\lambda+\sin x)$, $x\in\left(-\frac\pi2,\frac\pi2\right)$, has exactly one maxima and exactly one minima, is :
Differentiation and its Application
Solution
Concept. $f=\sin^2x(\lambda+\sin x)$. With $t=\sin x$ (increasing on the interval), $f$ is the cubic $g(t)=\lambda t^2+t^3$ on $(-1,1)$. Formulas used. (1) $g'(t)=2\lambda t+3t^2=t(3t+2\lambda)$Step 1 - Critical points. By (1), $t=0$ and $t=-\frac{2\lambda}{3}$. Step 2 - One max and one min. We need two distinct sign changes inside $(-1,1)$: $\lambda\ne0$ (else $t=0$ is a double root, no extremum) and $\left|\frac{2\lambda}{3}\right|<1$. Step 3 - Set. $\lambda\in\left(-\frac32,\frac32\right)-\{0\}$. Answer: (C)