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