MHT CET · Maths · Continuity and Differentiability
If
\(f(x)=\left\{\begin{array}{cc}(1+|\sin x|)^{\frac{a}{|\sin x|}}, & -\pi / 6 < x < 0 \ b \end{array}\right.\) \(x=0 \ \frac{\tan 2 x}{e^{\tan 3 x}},0 < x < \pi / 6\)
is continuous at \(x=0\), find the values of \(a\) and \(b\).
- A \(3 / 2, e^{3 / 2}\)
- B \(-2 / 3, e^{-3 / 2}\)
- C \(2 / 3, e^{2 / 3}\)
- D None of these
Answer & Solution
Correct Answer
(C) \(2 / 3, e^{2 / 3}\)
Step-by-step Solution
Detailed explanation
We have, \(\lim _{x \rightarrow 0^{-}} f(x)\)
\(= \lim _{x \rightarrow 0}\{1+|\sin x|\}^{\frac{a}{|\sin x|}} \)
\( = \lim _{x \rightarrow 0}|\sin x| \cdot \frac{a}{|\sin x|}=e^{a} \)
\( \text { and } \quad \lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0} e^{\frac{\tan 2 x}{\tan 3 x}} \)
\( = \lim _{x \rightarrow 0} \frac{\tan 2 x}{2 x} \cdot \frac{3 x}{\tan 3 x} \times \frac{2}{3} \)
\( = e^{2 / 3}\)
For \(f(x)\) to be continuous at \(x=0\), we must have
\(\lim _{x \rightarrow 0^{-}} f(x) =\lim _{x \rightarrow 0^{+}} f(x) \)
\( =f(0) \)
\( \Rightarrow e^{a} =e^{2 / 3}=b \)
\( \Rightarrow a =2 / 3 \)
\( \text { and } b =e^{2 / 3}\)
\(= \lim _{x \rightarrow 0}\{1+|\sin x|\}^{\frac{a}{|\sin x|}} \)
\( = \lim _{x \rightarrow 0}|\sin x| \cdot \frac{a}{|\sin x|}=e^{a} \)
\( \text { and } \quad \lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0} e^{\frac{\tan 2 x}{\tan 3 x}} \)
\( = \lim _{x \rightarrow 0} \frac{\tan 2 x}{2 x} \cdot \frac{3 x}{\tan 3 x} \times \frac{2}{3} \)
\( = e^{2 / 3}\)
For \(f(x)\) to be continuous at \(x=0\), we must have
\(\lim _{x \rightarrow 0^{-}} f(x) =\lim _{x \rightarrow 0^{+}} f(x) \)
\( =f(0) \)
\( \Rightarrow e^{a} =e^{2 / 3}=b \)
\( \Rightarrow a =2 / 3 \)
\( \text { and } b =e^{2 / 3}\)
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