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If $\mathrm{f}(\mathrm{x})$
$=K+1 \quad$, for $x=0$
is continuous at $x=0$, then value of $K$ is
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$=K+1 \quad$, for $x=0$
is continuous at $x=0$, then value of $K$ is
Solution:
2551 Upvotes
Verified Answer
The correct answer is:
$0$
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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