MHT CET · Maths · Continuity and Differentiability
If \(\mathrm{f}(\mathrm{x})\) \(=K+1 \quad\), for \(x=0\) is continuous at \(x=0\), then value of \(K\) is
- A \(1\)
- B \(\mathrm{e}^{-1}\)
- C \(0\)
- D e
Answer & Solution
Correct Answer
(C) \(0\)
Step-by-step Solution
Detailed explanation
For continuity at \(\mathrm{x}=0\)
\(\begin{aligned} & \lim _{x \rightarrow 0} f(x)=f(0) \Rightarrow \lim _{x \rightarrow 0} \log \left(\sec ^2 x\right) \cot ^2 x=K+1 \\ & \Rightarrow \lim _{x \rightarrow 0} \log \left(1+\tan ^2 x\right)^{\frac{1}{\tan ^2 x}}=K+1 \\ & \Rightarrow \log _e=K+1 \\ & \Rightarrow 1=K+1 \\ & \Rightarrow K=0\end{aligned}\)
\(\begin{aligned} & \lim _{x \rightarrow 0} f(x)=f(0) \Rightarrow \lim _{x \rightarrow 0} \log \left(\sec ^2 x\right) \cot ^2 x=K+1 \\ & \Rightarrow \lim _{x \rightarrow 0} \log \left(1+\tan ^2 x\right)^{\frac{1}{\tan ^2 x}}=K+1 \\ & \Rightarrow \log _e=K+1 \\ & \Rightarrow 1=K+1 \\ & \Rightarrow K=0\end{aligned}\)
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