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
- A \(\log 2\left[\log \left(\frac{5}{7}\right)\right]\)
- B \(\log 4\left[\log \left(\frac{5}{7}\right)\right]\)
- C \(\log 2\left[\log \left(\frac{7}{5}\right)\right]\)
- D \(\log 4\left[\log \left(\frac{7}{5}\right)\right]\)
Answer & Solution
Correct Answer
(B) \(\log 4\left[\log \left(\frac{5}{7}\right)\right]\)
Step-by-step Solution
Detailed explanation
\(f(0) = \lim_{x \to 0} \frac{10^x+7^x-14^x-5^x}{1-\cos x}\) \( = \lim_{x \to 0} \frac{(2^x-1)(5^x-7^x)}{1-\cos x}\)
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