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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

  1. A \(1\)
  2. B \(\mathrm{e}^{-1}\)
  3. C \(0\)
  4. D e
Verified Solution

Answer & Solution

Correct Answer

(C) \(0\)

Step-by-step Solution

Detailed explanation

For continuity at \(\mathrm{x}=0\)
\(\begin{aligned} & \lim _{x \rightarrow 0} f(x)=f(0) \Rightarrow \lim _{x \rightarrow 0} \log \left(\sec ^2 x\right) \cot ^2 x=K+1 \\ & \Rightarrow \lim _{x \rightarrow 0} \log \left(1+\tan ^2 x\right)^{\frac{1}{\tan ^2 x}}=K+1 \\ & \Rightarrow \log _e=K+1 \\ & \Rightarrow 1=K+1 \\ & \Rightarrow K=0\end{aligned}\)