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