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The function $f (x) = \sec [\log (x + \sqrt{1 + x^{2}})] is$

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Correct option is A

Since the function sec x is an even function and log (x+1+x2 is an odd function, therefore the function sec[log (x+1+x2] is an even function.

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The function f(x)=sec⁡log⁡x+1+x2 is

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The function $f (x) = \sec [\log (x + \sqrt{1 + x^{2}})] is$