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QJEEM24Mac2c7: JEE Main 2024 Mathematics - Class 12 JEEM24Mac2c7 Solution: If the function is differentiable on , then is equal to…

If the function

f(x)={1∣x∣,amp;∣x∣⩾2ax2+2 b,amp;∣x∣lt;2f(x)= \begin{cases}\frac{1}{|x|}, & |x| \geqslant 2 \\ \mathrm{a} x^2+2 \mathrm{~b}, & |x|<2\end{cases}

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

Step-by-Step Verified Solution

f(x){1x;x≥2ax2+2 b;−2lt;xlt;2−1x;x≤−2\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.

Continuous at x=2⇒12=a4+2 b\mathrm{x}=2 \quad \Rightarrow \frac{1}{2}=\frac{\mathrm{a}}{4}+2 \mathrm{~b}

Continuous at x=−2⇒12=a4+2 b\mathrm{x}=-2 \quad \Rightarrow \frac{1}{2}=\frac{\mathrm{a}}{4}+2 \mathrm{~b}

Since, it is differentiable at x=2\mathrm{x}=2

−1x2=2ax-\frac{1}{x^2}=2 \mathrm{ax}

Differentiable at x=2⇒−14=4a⇒a=−116,b=38x=2 \quad \Rightarrow \frac{-1}{4}=4 a \Rightarrow a=\frac{-1}{16}, b =\frac{3}{8}

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