Let f : R ? R be defined as, If f is continuous everywhere in R and m is the number of points where f is NOT differential, then m + a + b + c equals [2024]

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Solution

Q.
Let $f : R \to R$ be defined as $f (x)$ $= {\begin{matrix} \frac{a - b \cos 2 x}{x^{2}}; & x < 0 \\ x^{2} + c x + 2; & 0 \leq x \leq 1 \\ 2 x + 1; & x > 1 \end{matrix}$

If $f$ is continuous everywhere in $R$ and $m$ is the number of points where $f$ is NOT differential, then $m + a + b + c$ equals [2024]

1 3

2 1

3 4

4 2

Ans.
(4)

$Given, f (x) is continuous everywhere$

$f (0^{-}) = f (0) \Rightarrow 2 b = 2 \Rightarrow b = 1$

$f (1) = f (1^{+}) \Rightarrow 3 + c = 3 \Rightarrow c = 0$

$Since, f (0^{-}) = 2$

$= \lim_{h \to 0} \frac{a - b \cos 2 h}{h^{2}} = \lim_{h \to 0} \frac{a - b {1 - \frac{4 h^{2}}{2!} + \frac{16 h^{4}}{4!} - \dots}}{h^{2}}$

$= \lim_{h \to 0} \frac{(a - b) + b {2 h^{2} - \frac{2}{3} h^{4} \dots}}{h^{2}}$

$=for limit to exist a - b = 0 and limit is 2 b$ $∴ a = b = 1$

$Now, L f' (0) = \lim_{x \to 0} \frac{\frac{1 - \cos 2 h}{h^{2}} - 2}{- h}$

$= \lim_{x \to 0} \frac{1 - (1 - \frac{4 h^{2}}{2!} + \frac{16 h^{2}}{4!} \dots) - 2 h^{2}}{- h^{3}} = 0$

$Also, R f' (0) = \lim_{x \to 0} \frac{{(0 + h)}^{2} + 2 - 2}{h} = 0$

$∴ m = 0$

$∴ m + a + b + c = 0 + 1 + 1 + 0 = 2$

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