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Continuity — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsContinuity1 Mark
MCQ
Let $\text{f(x)=}\begin{cases}\frac{\text{x}-4}{|\text{x}-4|}+\text{a},&\text{if }\text{ x} < 4\\\text{a}+\text{b},&\text{if }\text{ x} =4\\\frac{\text{x}-4}{|\text{x}-4|}+\text{b},&\text{if }\text{ x} > 4\end{cases}$ Then, $f(x)$ is continus at $x = 4$ when:
  • A
    $a = 0, b = 0$
  • B
    $a = 1, b = 1$
  • C
    $a = -1, b = 1$
  • ✓
    $a = 1, b = -1$
✓

Answer

Correct option: D.
$a = 1, b = -1$
Given,
$\text{f(x)=}\begin{cases}\frac{\text{x}-4}{|\text{x}-4|}+\text{a},&\text{if }\text{ x} < 4\\\text{a}+\text{b},&\text{if }\text{ x} =4\\\frac{\text{x}-4}{|\text{x}-4|}+\text{b},&\text{if }\text{ x} > 4\end{cases}$
We have
$(\text{LHL at x = 4})=\lim\limits_{\text{x}\rightarrow4^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}(4-\text{h})$
$=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{4- \text{h}-4}{|4-\text{h}-4|}+\text{a}\Big)=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{-\text{h}}{|-\text{h}|}+\text{a}\Big)=\text{a}-1$
$(\text{RHL at x = 4})=\lim\limits_{\text{x}\rightarrow4^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f(4+h)}$
$=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{4+\text{h}-4}{|4+\text{h}-4|}+\text{b}\Big)=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\text{h}}{|\text{h}|}+\text{b}\Big)=\text{b}+1$
Also,
$\text{f}(4)=\text{a+b}$
if $(x)$ is continuous at $x = 4,$ then
$\lim\limits_{\text{x}\rightarrow4^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow4^+}\text{f(x)}=\text{f}(4)$
$\Rightarrow\text{a}-1=\text{b}+1=\text{a + b}$
$\Rightarrow\text{a}-\text{1}=\text{a + b}$ and $\text{b}+1=\text{a + b} $
$\Rightarrow\text{b}=-1$ and $\text{a}=1$

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