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MHT CET · Maths · Limits

If \(\mathrm{f}(x)=\frac{10^x+7^x-14^x-5^x}{1-\cos x}, x \neq 0\) is continuous at \(x=0\), then the value of \(f(0)\) is

  1. A \(\log 2\left[\log \left(\frac{5}{7}\right)\right]\)
  2. B \(\log 4\left[\log \left(\frac{5}{7}\right)\right]\)
  3. C \(\log 2\left[\log \left(\frac{7}{5}\right)\right]\)
  4. D \(\log 4\left[\log \left(\frac{7}{5}\right)\right]\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\log 4\left[\log \left(\frac{5}{7}\right)\right]\)

Step-by-step Solution

Detailed explanation

\(f(0) = \lim_{x \to 0} \frac{10^x+7^x-14^x-5^x}{1-\cos x}\) \( = \lim_{x \to 0} \frac{(2^x-1)(5^x-7^x)}{1-\cos x}\)