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

Let \(f: R \rightarrow R\) be defined as \(f(x)=\left\{\begin{array}{ccc}\frac{a-b \cos 2 x}{x^2} & ; & x<0 \\ x^2+c x+2 & ; & 0 \leq x \leq 1 \\ 2 x+1 & ; & x>1\end{array}\right.\) If \(f\) is continuous everywhere in \(R\) and \(\mathrm{m}\) is the number of points where \(f\) is \(NOT\) differential then \(\mathrm{m}+\mathrm{a}+\mathrm{b}+\mathrm{c}\) equals :

  1. A \(1\)
  2. B \(4\)
  3. C \(3\)
  4. D \(2\)
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Answer & Solution

Correct Answer

(D) \(2\)

Step-by-step Solution

Detailed explanation

At \(\mathrm{x}=1, \mathrm{f}(\mathrm{x})\) is continuous therefore, \(\mathrm{f}\left(1^{-}\right)=\mathrm{f}(1)=\mathrm{f}\left(1^{+}\right)\) \(\mathrm{f}(1)=3+\mathrm{c}\) \(.....(1)\) \(\mathrm{f}\left(1^{+}\right)=\lim _{h \rightarrow 0} 2(1+\mathrm{h})+1\)…
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