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JEE Main 30 Jan 2024 Shift 1 Paper

Section: Mathematics

Question : 29 of 90

Marks: +1, -0

f(x)=\begin{cases} \frac{1}{|x|} & |x| \ge 2 \\ a x^2+2b & |x| < 2 \end{cases}

R

48(a+b)

Solution: 👈: Video Solution

f(x)=\begin{cases} \frac{1}{x} & x \ge 2 \\ a x^2+2b & -2 < x < 2 \\ -\frac{1}{x} & x \le -2 \end{cases}

Continuous at

x=2 \Rightarrow \frac{1}{2} = \frac{a}{4} + 2b

Continuous at

x=-2 \Rightarrow \frac{1}{2} = \frac{a}{4} + 2b

Since, it is differentiable at

x=2

-\frac{1}{x^2}=2 a x

Differentiable at

x=2 \Rightarrow \frac{-1}{4}=4a \Rightarrow a=\frac{-1}{16}, b=\frac{3}{8}

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