Continuity And Differentiability Question 1
Question 1 - 2024 (01 Feb Shift 1)
Let $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ 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 $\mathbf{R}$ and $\mathrm{m}$ is the number of points where $f$ is NOT differential then $\mathrm{m}+\mathrm{a}+\mathrm{b}+\mathrm{c}$ equals :
(1) 1
(2) 4
(3) 3
(4) 2
Show Answer
Answer (4)
Solution
At $\mathrm{x}=1, \mathrm{f}(\mathrm{x})$ is continuous therefore,
$\mathrm{f}\left(1^{-}\right)=\mathrm{f}(1)=\mathrm{f}\left(1^{+}\right)$
$\mathrm{f}(1)=3+\mathrm{c}$
$\mathrm{f}\left(1^{+}\right)=\lim _{h \rightarrow 0} 2(1+\mathrm{h})+1$
$\mathrm{f}\left(1^{+}\right)=\lim _{h \rightarrow 0} 3+2 \mathrm{~h}=3$
from (1) & (2)
$\mathrm{c}=0$
at $\mathrm{x}=0, \mathrm{f}(\mathrm{x})$ is continuous therefore,
$\mathrm{f}\left(0^{-}\right)=\mathrm{f}(0)=\mathrm{f}\left(0^{+}\right)$
$\mathrm{f}(0)=\mathrm{f}\left(0^{+}\right)=2$
$\mathrm{f}\left(0^{-}\right)$has to be equal to 2
$\lim _{h \rightarrow 0} \frac{a-b \cos (2 h)}{h^{2}}$
$\lim _{h \rightarrow 0} \frac{a-b\left{1-\frac{4 h^{2}}{2 !}+\frac{16 h^{4}}{4 !}+\ldots\right}}{h^{2}}$
$\lim _{h \rightarrow 0} \frac{a-b+b\left{2 h^{2}-\frac{2}{3} h^{4} \ldots\right}}{h^{2}}$
for limit to exist $\mathrm{a}-\mathrm{b}=0$ and limit is $2 \mathrm{~b}$
from (3), (4) & (5)
$\mathrm{a}=\mathrm{b}=1$
checking differentiability at $\mathrm{x}=0$
LHD : $\lim _{h \rightarrow 0} \frac{\frac{1-\cos 2 h}{h^{2}}-2}{-h}$
$\lim _{h \rightarrow 0} \frac{1-\left(1-\frac{4 h^{2}}{2 !}+\frac{16 h^{4}}{4 !} \cdot \ldots\right)-2 h^{2}}{-h^{3}}=0$
RHD : $\lim _{h \rightarrow 0} \frac{(0+h)^{2}+2-2}{h}=0$
Function is differentiable at every point in its domain
$\therefore \mathrm{m}=0$
$\mathrm{m}+\mathrm{a}+\mathrm{b}+\mathrm{c}=0+1+1+0=2$
