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MHT CET · Maths · Continuity and Differentiability

If \(f(x)=\frac{4^{x-\pi}+4^{\pi-x}-2}{(x-\pi)^2}\) for \(\neq \pi\), is continuous at \(x=\pi\), then \(=\mathrm{k} \quad\) for \(=\pi\)
\(\mathrm{k}=\)

  1. A \(2 \log 2\)
  2. B \((\log 2)^2\)
  3. C \(-4(\log 2)^2\)
  4. D \(8(\log 2)^2\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(-4(\log 2)^2\)

Step-by-step Solution

Detailed explanation

\(f(\pi)=\lim _{x \rightarrow \pi} f(x)=\lim _{x \rightarrow \pi} \frac{4^{x-\pi}+4^{\pi-4}-2}{(x-\pi)^2}
\)
Put \(\mathrm{x}=\pi+\mathrm{h}\). As \(\mathrm{x} \rightarrow \pi, \mathrm{h} \rightarrow 0\)
\(\therefore \mathrm{f}(\pi)=\lim _{\mathrm{h} \rightarrow 0} \frac{4^{\mathrm{h}}+4^{-\mathrm{h}}-1-1}{\mathrm{~h}^2}=\lim _{\mathrm{h} \rightarrow 0} \frac{\left(4^{\mathrm{h}}-1\right)}{\mathrm{h}} \times\) \(\lim _{\mathrm{h} \rightarrow 0} \frac{4^{-\mathrm{h}}-1}{-\mathrm{h}} \times(-1) \)
\( =(\log 4)(-\log 4)=(2 \log 2)(-2 \log 2)=-4(\log 2)^2\)