Mathematics - Continuity - Differentiability Question with Solution | TestHub

At $x=1,f\left(x\right)$ is continuous,

$\Rightarrow f\left(1^{-}\right)=f\left(1\right)=f\left(1^{+}\right)$

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

$\Rightarrow f\left(1^{+}\right)=\underset{h\to 0}{\lim }2\left(1+h\right)+1$

$\Rightarrow f\left(1^{+}\right)=\underset{h\to 0}{\lim }3+2h=3 ...\left(ii\right)$

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

$\Rightarrow c=0$

At $x=0,f\left(x\right)$ is continuous,

$\Rightarrow f\left(0^{-}\right)=f\left(0\right)=f\left(0^{+}\right) ...\left(iii\right)$

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

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

$\Rightarrow \underset{h\to 0}{\lim }\frac{a-b\cos \left(2h\right)}{h^2}$

$\Rightarrow \underset{h\to 0}{\lim }\frac{a-b\left\{1-\frac{4h^2}{2!}+\frac{16h^4}{4!}+...\right\}}{h^2}$

$\Rightarrow \underset{h\to 0}{\lim }\frac{a-b+b\left\{2h^2-\frac{2}{3}{h}^4...\right\}}{h^2}$

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

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

$\Rightarrow a=b=1$

Checking differentiability at $x=0$.

$\Rightarrow LHD:\underset{h\to 0}{\lim }\frac{\frac{1-\cos 2h}{h^2}-2}{-h}$

$\Rightarrow \underset{h\to 0}{\lim }\frac{1-\left(1-\frac{4h^2}{2!}+\frac{16h^4}{4!}-...\right)-2h^2}{-h^3}=0$

$\Rightarrow RHD:\underset{h\to 0}{\lim }\frac{{\left(0+h\right)}^2+2-2}{h}=0$

Function is differentiable at every point in its domain

$\therefore m=0$

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

$\Rightarrow m+a+b+c=2$