Let $f: R \rightarrow R$ be defined as : $f(x)= \begin{cases}\frac{ a - b \cos 2 x}{x^2} ; & x1\end{cases}$ If $f$ is continuous everywhere in $R$ and $m$ is the number of points where $f$ is NOT differentiable then $m + a + b + c$ equals : [JEE Main 2024, 1 Feb (Shift 1)]

Question

Let $f: R \rightarrow R$ be defined as : $f(x)= \begin{cases}\frac{ a - b \cos 2 x}{x^2} ; & x<0 \ x^2+ c x+2 ; & 0 \leq x \leq 1 \ 2 x+1 & ; x>1\end{cases}$ If $f$ is continuous everywhere in $R$ and $m$ is the number of points where $f$ is NOT differentiable then $m + a + b + c$ equals : [JEE Main 2024, 1 Feb (Shift 1)]

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Let \(f: R \rightarrow R\) be defined as : \(f(x)= \begin{cases}\frac{ a - b \cos 2 x}{x^2} ; & x<0 \\ x^2+ c x+2…

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Let \(f: R \rightarrow R\) be defined as : \(f(x)= \begin{cases}\frac{ a - b \cos 2 x}{x^2} ; &amp; x&lt;0 \\ x^2+ c x+2…

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Let \(f: R \rightarrow R\) be defined as : \(f(x)= \begin{cases}\frac{ a - b \cos 2 x}{x^2} ; & x<0 \\ x^2+ c x+2…