Mathematics - Continuity - Differentiability Question with Solution | TestHub

For $f(x)$ to be differentiable at $x=1$, it must first be continuous at $x=1$. Given $f(1)=2$, from the first part of the function, $1^2+a(1)+b=2\Rightarrow 1+a+b=2\Rightarrow a+b=1$ (Equation 1). For continuity, ${\lim }_{x\to 1^{+}}f(x)=f(1)$, so $\sqrt{1^2+3}-c=2\Rightarrow \sqrt{4}-c=2\Rightarrow 2-c=2\Rightarrow c=0$. Now, for differentiability, the left-hand derivative and right-hand derivative must be equal. $f^{\prime }(x)=\{\begin{array}{ll}2x+a& ,x<1\\ \frac{x}{\sqrt{x^2+3}}& ,x>1\end{array}$. So, $f^{\prime }(1^{-})=2(1)+a=2+a$. And $f^{\prime }(1^{+})=\frac{1}{\sqrt{1^2+3}}=\frac{1}{\sqrt{4}}=\frac{1}{2}$. Equating them, $2+a=\frac{1}{2}\Rightarrow a=\frac{1}{2}-2=-\frac{3}{2}$. Substitute $a$ into Equation 1: $-\frac{3}{2}+b=1\Rightarrow b=1+\frac{3}{2}=\frac{5}{2}$. Finally, $a+b+c=-\frac{3}{2}+\frac{5}{2}+0=\frac{2}{2}=1$. The value of $a+b+c$ is 1.