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

यदि \(f( x )=\left\{\begin{array}{cc}\frac{1}{| x |} ; & | x | \geq 1 \\ a x ^{2}+ b ; & | x |<1\end{array}\right.\) अपने प्रांत के प्रत्येक बिंदु पर अवकलनीय है, तो \(a\) तथा \(b\) के मान क्रमशः हैं

  1. A \(\frac{1}{2}, \frac{1}{2}\)
  2. B \(\frac{1}{2},-\frac{3}{2}\)
  3. C \(\frac{5}{2},-\frac{3}{2}\)
  4. D \(-\frac{1}{2}, \frac{3}{2}\)
Verified Solution

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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