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JEE Mains · Maths · STD 12 - 5. continuity and differentiation

જો વિધેય  \(f(x)=\left\{\begin{array}{cl}\frac{1}{|x|} & ,|x| \geq 2 \\ a x^2+2 b, & |x|<2\end{array}\right.\) એ \(R\) પર વિકલનીય હોય, તો \(48(\mathrm{a}+\mathrm{b}) =\) ...........

  1. A \(15\)
  2. B \(16\)
  3. C \(75\)
  4. D \(78\)
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Correct Answer

(A) \(15\)

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Detailed explanation

\(f(x)\left\{\begin{array}{c}\frac{1}{\mathrm{x}} ; \mathrm{x} \geq 2 \\ \mathrm{ax}^2+2 \mathrm{~b} ;-2<\mathrm{x}<2 \\ -\frac{1}{\mathrm{x}} ; \mathrm{x} \leq-2\end{array}\right.\) Continuous at \(\mathrm{x}=2 \quad \Rightarrow \frac{1}{2}=\frac{\mathrm{a}}{4}+2 \mathrm{~b}\)…
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