JEE Main 2024 — Mathematics Calculus

At $x=1,f(x)$x=1,f(x) is continuous,

$$\Rightarrow f({1}^{-})=f(1)=f({1}^{+})$$⇒f(1−)=f(1)=f(1+)

$$\Rightarrow f(1)=3+c...(i)$$⇒f(1)=3+c...(i)

$$\Rightarrow f({1}^{+})=\underset{h\rightarrow 0}{\mathrm{lim}}2(1+h)+1$$⇒f(1+)=h→0lim​2(1+h)+1

$$\Rightarrow f({1}^{+})=\underset{h\rightarrow 0}{\mathrm{lim}}3+2h=3...(ii)$$⇒f(1+)=h→0lim​3+2h=3...(ii)

Using $(i)\text{and}(ii)$(i)and(ii),

$$\Rightarrow c=0$$⇒c=0

At $x=0,f(x)$x=0,f(x) is continuous,

$$\Rightarrow f({0}^{-})=f(0)=f({0}^{+})...(iii)$$⇒f(0−)=f(0)=f(0+)...(iii)

$$\Rightarrow f(0)=f({0}^{+})=2...(iv)$$⇒f(0)=f(0+)=2...(iv)

So, $f({0}^{-})$f(0−) has to be equal to $2$2.

$$\Rightarrow \underset{h\rightarrow 0}{\mathrm{lim}}\frac{a-b\mathrm{cos}(2h)}{{h}^{2}}$$⇒h→0lim​h2a−bcos(2h)​

$$\Rightarrow \underset{h\rightarrow 0}{\mathrm{lim}}\frac{a-b{1-\frac{4{h}^{2}}{2!}+\frac{16{h}^{4}}{4!}+...}}{{h}^{2}}$$⇒h→0lim​h2a−b1−2!4h2​+4!16h4​+...​

$$\Rightarrow \underset{h\rightarrow 0}{\mathrm{lim}}\frac{a-b+b{2{h}^{2}-\frac{2}{3}{h}^{4}...}}{{h}^{2}}$$⇒h→0lim​h2a−b+b2h2−32​h4...​

For limit to exist $a-b=0$a−b=0 and limit is $2b$2b$...(v)$...(v)

Using, $(iii),(iv)\text{and}(v)$(iii),(iv)and(v)

$$\Rightarrow a=b=1$$⇒a=b=1

Checking differentiability at $x=0$x=0.

$$\Rightarrow LHD:\underset{h\rightarrow 0}{\mathrm{lim}}\frac{\frac{1-\mathrm{cos}2h}{{h}^{2}}-2}{-h}$$⇒LHD:h→0lim​−hh21−cos2h​−2​

$$\Rightarrow \underset{h\rightarrow 0}{\mathrm{lim}}\frac{1-(1-\frac{4{h}^{2}}{2!}+\frac{16{h}^{4}}{4!}-...)-2{h}^{2}}{-{h}^{3}}=0$$⇒h→0lim​−h31−(1−2!4h2​+4!16h4​−...)−2h2​=0

$$\Rightarrow RHD:\underset{h\rightarrow 0}{\mathrm{lim}}\frac{{(0+h)}^{2}+2-2}{h}=0$$⇒RHD:h→0lim​h(0+h)2+2−2​=0

Function is differentiable at every point in its domain

$$\therefore m=0$$∴m=0

$$\Rightarrow m+a+b+c=0+1+1+0$$⇒m+a+b+c=0+1+1+0

$$\Rightarrow m+a+b+c=2$$⇒m+a+b+c=2