Medium INTEGER +4 / -1 PYQ · JEE Mains 2024

If the function

$$f(x)= \begin{cases}\frac{1}{|x|}, & |x| \geqslant 2 \\ \mathrm{a} x^2+2 \mathrm{~b}, & |x|<2\end{cases}$$

is differentiable on $\mathbf{R}$, then $48(a+b)$ is equal to __________.

Answer (integer) 15

Solution

<p>$$\mathrm{f}(\mathrm{x})\left\{\begin{array}{c} \frac{1}{\mathrm{x}} ; \mathrm{x} \geq 2 \\ \mathrm{ax}^2+2 \mathrm{~b} ;-2<\mathrm{x}<2 \\ -\frac{1}{\mathrm{x}} ; \mathrm{x} \leq-2 \end{array}\right.$$</p> <p>Continuous at $\mathrm{x}=2 \quad \Rightarrow \frac{1}{2}=\frac{\mathrm{a}}{4}+2 \mathrm{~b}$</p> <p>Continuous at $\mathrm{x}=-2 \quad \Rightarrow \frac{1}{2}=\frac{\mathrm{a}}{4}+2 \mathrm{~b}$</p> <p>Since, it is differentiable at $\mathrm{x}=2$</p> <p>$-\frac{1}{x^2}=2 \mathrm{ax}$</p> <p>Differentiable at $$x=2 \quad \Rightarrow \frac{-1}{4}=4 a \Rightarrow a=\frac{-1}{16}, b =\frac{3}{8}$$</p>

About this question

Subject: Mathematics · Chapter: Limits, Continuity and Differentiability · Topic: Limits and Standard Results

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