Q123 / 200 · Manipal (MET) 2010 · Mathematics

If $f(x)=\left\{\begin{array}{cc}a x^2+b, & 0 \leq x \lt 1 \\ x+3, & 1 \lt x \leq 2 \\ 4, & x=1\end{array}\right.$, then the value of $(a, b)$ for which $f(x)$ cannot be continuous at $x=1$ is

  • A
    $(2,2)$
  • B
    $(3,1)$
  • C
    $(4,0)$
  • Correct answer
    D
    $(5,2)$

Solution

We have, $\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0} a(1-h)^2+b=a+b$ $\Rightarrow \lim _{h \rightarrow 0} f(1+h)=\lim _{h \rightarrow 0}(1+h)+3=4$ and $f(1)=4$ $\therefore f(x)$ will not be continuous at $x=1$, if $a+b \neq 4$

Hence, the correct answer is option: 'D'.

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