JEE Main 2019 (Online) 11th January Morning ShiftQuestion 1
Let $f(x)=\begin{cases}-1,&-2\le x<0\\x^{2}-1,&0\le x\le2\end{cases}$ and $g(x)=|f(x)|+f(|x|)$. Then, in the interval $(-2,2)$, $g$ is :
Differentiation and its Application
Solution
Concept. Write $g$ piece by piece, then compare one-sided slopes at the joins $x=0$ and $x=1$. Formulas used. (1) A function is differentiable at a join iff it is continuous there and the one-sided derivatives agreeStep 1 - $f(|x|)$. $|x|\in[0,2)$, so $f(|x|)=x^2-1$ on $(-2,2)$. Step 2 - $g$ piece by piece. $$(-2,0):\ g=1+(x^2-1)=x^2$$ $$[0,1]:\ g=(1-x^2)+(x^2-1)=0$$ $$[1,2):\ g=(x^2-1)+(x^2-1)=2x^2-2$$ Step 3 - Joins. At $x=0$: slopes $0$ and $0$, so differentiable. At $x=1$: slopes $0$ and $4$, so not differentiable. $g$ is continuous everywhere. Step 4 - Conclude. Not differentiable at exactly one point. Answer: (D)