Let f: R R be defined as : [ f(x)= cases(a-b 2 x)/(x^2) ; & x1cases ] If f is continuous e…

Question

Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be defined as :

$$
f(x)= \begin{cases}\frac{a-b \cos 2 x}{x^2} ; & x1\end{cases}
$$

If $f$ is continuous everywhere in $\mathbf{R}$ and $m$ is the number of points where $f$ is NOT differential then $\mathrm{m}+\mathrm{a}+\mathrm{b}+\mathrm{c}$ equals :

Accepted Answer

At $\mathrm{x}=1, \mathrm{f}(\mathrm{x})$ is continuous therefore,

$\mathrm{f}\left(1^{-}\right)=\mathrm{f}(1)=\mathrm{f}\left(1^{+}\right)$

$f(1)=3+c$ .........(1)

$$
\begin{aligned}
& \mathrm{f}\left(1^{+}\right)=\lim _{h \rightarrow 0} 2(1+\mathrm{h})+1 \\
& \mathrm{f}\left(1^{+}\right)=\lim _{h \rightarrow 0} 3+2 \mathrm{~h}=3 .........(2)
\end{aligned}
$$

from (1) and (2)

$$
\mathrm{c}=0
$$

at $\mathrm{x}=0, \mathrm{f}(\mathrm{x})$ is continuous therefore,

$$
\begin{aligned}
& \mathrm{f}\left(0^{-}\right)=\mathrm{f}(0)=\mathrm{f}\left(0^{+}\right) ........(3) \\
& \mathrm{f}(0)=\mathrm{f}\left(0^{+}\right)=2 ...........(4)
\end{aligned}
$$

$\mathrm{f}\left(0^{-}\right)$has to be equal to 2

$\lim\limits_{h \rightarrow 0} \frac{a-b \cos (2 h)}{h^2}$

= $\lim\limits_{h \rightarrow 0} \frac{a-b\left\{1-\frac{4 h^2}{2 !}+\frac{16 h^4}{4 !}+\ldots\right\}}{h^2}$

= $\lim\limits_{h \rightarrow 0} \frac{a-b+b\left\{2 h^2-\frac{2}{3} h^4 \ldots\right\}}{h^2}$

for limit to exist $a-b=0$ and limit is $2 b$ from (3), (4) and (5)

$$
\mathrm{a}=\mathrm{b}=1
$$

checking differentiability at $\mathrm{x}=0$

$$
\begin{aligned}
& \text { LHD : } \lim _{h \rightarrow 0} \frac{\frac{1-\cos 2 h}{h^2}-2}{-h} \\
& \lim _{h \rightarrow 0} \frac{1-\left(1-\frac{4 h^2}{2 !}+\frac{16 h^4}{4 !} \ldots\right)-2 h^2}{-h^3}=0 \\
& \text { RHD : } \lim _{h \rightarrow 0} \frac{(0+h)^2+2-2}{h}=0
\end{aligned}
$$

Function is differentiable at every point in its domain

$$
\therefore \mathrm{m}=0
$$

m + a + b + c = 0 + 1 + 1 + 0 = 2

Solutions · 1