JEE Main 2024 — 1 February, Shift 1 — Question 15

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

is continuous therefore, $\mathrm{f}\left(1^{-}\right)=\mathrm{f}(1)=\mathrm{f}\left(1^{+}\right)$

$\mathrm{f}(1)=3+\mathrm{c}$

$f\left(1^{+}\right)=\lim _{h \rightarrow 0} 2(1+h)+1$

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

from $(1) \&(2)$

$\mathrm{c}=0$ at $x=0, f(x)$ 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 b from $(3),(4) \&(5)$

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

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

$\lim _{h \rightarrow 0} \frac{\frac{1-\cos 2 h}{h^{2}}-2}{-h}$

$\lim _{h \rightarrow 0} \frac{1-\left(1-\frac{4 h^{2}}{2!}+\frac{16 h^{4}}{4!} \ldots\right)-2 h^{2}}{-h^{3}}=0$ RHD :

$\lim _{h \rightarrow 0} \frac{(0+h)^{2}+2-2}{h}=0$

Function is differentiable at every point in its domain

$\therefore \mathrm{m}=0$

$\mathrm{m}+\mathrm{a}+\mathrm{b}+\mathrm{c}=0+1+1+0=2$