Differentiation and its Application: The function f(x)=begincasesfracpi4+tan^-1x, & |x|le1\\… | JEE Main 2020

JEE Main 2020 (Online) 4th September Evening ShiftQuestion 1

The function $f(x)=\begin{cases}\frac\pi4+\tan^{-1}x, & |x|\le1\\ \frac12(|x|-1), & |x|>1\end{cases}$ is :

Differentiation and its Application

both continuous and differentiable on $\mathbf{R}-\{1\}$.continuous on $\mathbf{R}-\{-1\}$ and differentiable on $\mathbf{R}-\{-1,1\}$.continuous on $\mathbf{R}-\{1\}$ and differentiable on $\mathbf{R}-\{-1,1\}$.✓ Correct answerboth continuous and differentiable on $\mathbf{R}-\{-1\}$.

Solution

Concept. Check values and slopes at the joins $x=\pm1$. Formulas used. (1) $\tan^{-1}(\pm1)=\pm\frac\pi4$, $\ \dfrac{d}{dx}\tan^{-1}x=\dfrac{1}{1+x^2}$Step 1 - At $x=1$. Left value $\frac\pi4+\frac\pi4=\frac\pi2$; right limit $0$. Discontinuous. Step 2 - At $x=-1$. Left limit $\frac12(1-1)=0$; value $\frac\pi4-\frac\pi4=0$. Continuous. Slopes: left $-\frac12$, right $\frac12$. Not differentiable. Step 3 - Conclude. Continuous on $\mathbf R-\{1\}$, differentiable on $\mathbf R-\{-1,1\}$. Answer: (C)

Differentiation and its Application - JEE Mathematics question (JEE Main 2020 (Online) 4th September Evening Shift)

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