JEE MAIN - Mathematics (2024 - 1st February Morning Shift - No. 15)
Let $f: \mathbf{R} \rightarrow \mathbf{R}$ 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 $\mathrm{m}+\mathrm{a}+\mathrm{b}+\mathrm{c}$ equals :
$$ f(x)= \begin{cases}\frac{a-b \cos 2 x}{x^2} ; & x<0 \\\\ x^2+c x+2 ; & 0 \leq x \leq 1 \\\\ 2 x+1 ; & x>1\end{cases} $$
If $f$ is continuous everywhere in $\mathbf{R}$ and $m$ is the number of points where $f$ is NOT differential then $\mathrm{m}+\mathrm{a}+\mathrm{b}+\mathrm{c}$ equals :
1
4
3
2
Explanation
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
$\mathrm{f}\left(1^{-}\right)=\mathrm{f}(1)=\mathrm{f}\left(1^{+}\right)$
$$ f(1)=3+c $$ .........(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
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