If f(x)= ax^2+b,&0 < 1 4,&x=1 +3,&1<x 2 then the value of (a, b) … - Vidyadip

MCQ

If $\text{f(x)}=\begin{cases}\text{ax}^2+\text{b},&0\leq\text{x} < 1\\4,&\text{x}=1\\\text{x}+3,&1<\text{x}\leq2\end{cases}$ then the value of $(a, b)$ for which $f(x)$ cannot be continuous at $x = 1,$ is :

A
$(2, 2)$
B
$(3, 1)$
C
$(4, 0)$
✓
$(5, 2)$

✓

Answer

Correct option: D.

$(5, 2)$

$\lim\limits_{\text{x}\rightarrow1^-}\text{f(x)}=\text{f}(1)$
$\lim\limits_{\text{x}\rightarrow1}\text{ax}^2+\text{b}=4$
$a + b = 4$
We have possible values as $(2, 2), (3, 1), (4, 0)$
But can not be $(5, 2).$
Function is can not be continuous at $(5, 2).$

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