MHT CET · Maths · Continuity and Differentiability
If \(\mathrm{f}(x)=\left\{\begin{array}{ll}\frac{\sqrt{1+\mathrm{m} x}-\sqrt{1-\mathrm{m} x}}{x} & ,-1 \leq x < 0 \ \frac{2 x+1}{x-2} & , 0 \leq x \leq 1\end{array}\right.\) is continuous in the interval \([-1,1]\), then \(\mathrm{m}\) is equal to
- A \(\frac{1}{2}\)
- B \(-\frac{1}{2}\)
- C \(-1\)
- D \(-\frac{1}{4}\)
Answer & Solution
Correct Answer
(B) \(-\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
Since \(\mathrm{f}(x)\) is continuous in \([-1,1]\), it is continuous at \(x=0\).
\(\therefore \lim _{x \rightarrow 0^{-}} \mathrm{f}(x)=\lim _{x \rightarrow 0^{+}} \mathrm{f}(x) \)
\( \Rightarrow \lim _{x \rightarrow 0} \frac{\sqrt{1+\mathrm{m} x}-\sqrt{1-\mathrm{m} x}}{x}=\lim _{x \rightarrow 0} \frac{2 x+1}{x-2} \)
\( \Rightarrow \lim _{x \rightarrow 0} \frac{(1+\mathrm{m} x-1+\mathrm{m} x)}{x(\sqrt{1+\mathrm{m} x}+\sqrt{1-\mathrm{m} x})}=\frac{2(0)+1}{0-2} \)
\( \Rightarrow \frac{2 \mathrm{~m}}{1+1}=\frac{1}{-2} \)
\( \Rightarrow \mathrm{m}=\frac{-1}{2}\)
\(\therefore \lim _{x \rightarrow 0^{-}} \mathrm{f}(x)=\lim _{x \rightarrow 0^{+}} \mathrm{f}(x) \)
\( \Rightarrow \lim _{x \rightarrow 0} \frac{\sqrt{1+\mathrm{m} x}-\sqrt{1-\mathrm{m} x}}{x}=\lim _{x \rightarrow 0} \frac{2 x+1}{x-2} \)
\( \Rightarrow \lim _{x \rightarrow 0} \frac{(1+\mathrm{m} x-1+\mathrm{m} x)}{x(\sqrt{1+\mathrm{m} x}+\sqrt{1-\mathrm{m} x})}=\frac{2(0)+1}{0-2} \)
\( \Rightarrow \frac{2 \mathrm{~m}}{1+1}=\frac{1}{-2} \)
\( \Rightarrow \mathrm{m}=\frac{-1}{2}\)
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