WBJEE · Maths · Functions
The function \(\mathrm{f}(\mathrm{x})=\sec \left[\log \left(\mathrm{x}+\sqrt{1+\mathrm{x}^2}\right)\right]\) is
- A odd
- B even
- C neither odd nor even
- D constant
Answer & Solution
Correct Answer
(B) even
Step-by-step Solution
Detailed explanation
Hints : \(\mathrm{f}(\mathrm{x})=\sec \left(\ell \ln \left(\mathrm{x}+\sqrt{1+\mathrm{x}^2}\right)\right)=\sec (\) odd function \()=\) even function \(\because\) sec is an even function
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