JEE Mains · Maths · STD 12 - 5. continuity and differentiation
If \(f(x)=\left\{\begin{array}{ccc}\frac{1}{|x|} & ; & |x| \geq 1 \\ a x^{2}+b & ; & |x|<1\end{array}\right.\) is differentiable at every point of the domain, then the values of \(a\) and \(b\) are respectively
- A \(\frac{1}{2}, \frac{1}{2}\)
- B \(\frac{1}{2},-\frac{3}{2}\)
- C \(\frac{5}{2},-\frac{3}{2}\)
- D \(-\frac{1}{2}, \frac{3}{2}\)
Answer & Solution
Correct Answer
(D) \(-\frac{1}{2}, \frac{3}{2}\)
Step-by-step Solution
Detailed explanation
\(f(x)=\left\{\begin{array}{cl}\frac{1}{|x|}, & |x| \geq 1 \\ a x^{2}+b, & |x|<1\end{array}\right.\) at \(x =1\) function must be continuous So, \(1= a + b \quad \ldots\) \(.....(1)\) differentiability at \(x =1\) \(\left(-\frac{1}{x^{2}}\right)_{x=1}=(2 a x)_{x=1}\)…
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