Question

$Iff(x)=\{\begin{array}{l}\text{if ax2+b,b=0,x≤1}\\ \text{if bx2+ax+c,x>1 }\end{array}$ then
$f(x)$ is continuous and differentiable at $x=1$, if

• A. $c=0,a=2b$
• B. $a=b,c\in R$
• C. $a=b,c=0$
• D. $a=b,c\ne 0$

Answer 1

Answer: (A)

Given, $f(x)=\{\begin{array}{ll}a{x}^2+b,& b\ne 0,x\le 1\\ b{x}^2+ax+c,& x>1\end{array}.$
$\Rightarrow f^{\prime }(x)=\{\begin{array}{ll}2ax,& b\ne 0,x\le 1\\ 2bx+a,& x>1\end{array}$
Since, $f(x)$ is continuous at $x=1$
$\therefore \underset{x\to 1^{-}}{\lim }f(x)=\underset{x\to 1^{+}}{\lim }f(x)$
$\Rightarrow a+b=b+a+c$
$\Rightarrow c=0$
Also, $f(x)$ is differentiable at $x=1$.
$\therefore (LHD$ at $x=1)=(RHD$ at $x=1$
$\Rightarrow 2a=2b(1)+a$
$\Rightarrow a=2b$