JEE Main 2024 (Online) 30th January Morning ShiftQuestion 1
If the function $f(x)=\begin{cases}\frac{1}{|x|}, & |x|\ge2\\ ax^{2}+2b, & |x|<2\end{cases}$ is differentiable on $\mathbf{R}$, then $48(a+b)$ is equal to $\_\_\_\_\_\_$ .
Differentiation and its Application
Answer$15$
Solution
Concept. At $x=2$ both the values and the slopes of the two pieces must match (the case $x=-2$ is symmetric). Formulas used. (1) $\dfrac{d}{dx}\dfrac1x=-\dfrac{1}{x^2}$Step 1 - Slopes. Inside: $2ax$, which is $4a$ at $x=2$. Outside, by (1): $-\frac14$. So $a=-\frac{1}{16}$. Step 2 - Values. $4a+2b=\frac12$, so $2b=\frac12+\frac14=\frac34$ and $b=\frac38$. Step 3 - Value. $a+b=\frac{-1+6}{16}=\frac{5}{16}$, so $48(a+b)=15$. Answer: 15