QJEEM24Macc09: JEE Main 2024 Mathematics - Class 12 JEEM24Macc09 Solution: Let be defined as : If is continuous everywhere in and is the…

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