JEE Advanced 2020 Paper 2Question 1
Let the function $f:(0,\pi)\to\mathbb{R}$ be defined by $$f(\theta)=(\sin\theta+\cos\theta)^2+(\sin\theta-\cos\theta)^4.$$ Suppose the function $f$ has a local minimum at $\theta$ precisely when $\theta\in\{\lambda_1\pi,\ldots,\lambda_r\pi\}$, where $0<\lambda_1<\cdots<\lambda_r<1$. Then the value of $\lambda_1+\cdots+\lambda_r$ is ____
Differentiation and its Application
Answer$0.5$
Solution
Concept. With $s=\sin2\theta$, $f=s^2-s+2$, which is smallest at $s=\frac12$. Formulas used. (1) $(\sin\theta\pm\cos\theta)^2=1\pm\sin2\theta$Step 1 - Simplify. By (1), $f=(1+s)+(1-s)^2=s^2-s+2$. Step 2 - Derivative. $f'(\theta)=(2s-1)\cdot2\cos2\theta$. Zeros in $(0,\pi)$: $s=\frac12$ ($\theta=\frac{\pi}{12},\frac{5\pi}{12}$) and $\cos2\theta=0$ ($\theta=\frac\pi4,\frac{3\pi}4$). Step 3 - Type. At $s=\frac12$, $f$ takes its least possible value $\frac74$: local minima. At $\theta=\frac\pi4$ ($s=1$, $f=2$) and $\theta=\frac{3\pi}{4}$ ($s=-1$, $f=4$), $f$ is larger than nearby: local maxima. Step 4 - Sum. $\frac{1}{12}+\frac{5}{12}=\frac12$. Answer: 0.5