If f(x) = cos (log x), then f({x^2})f({y^2}) | vedclass

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Question

(d) The given expression is

$\cos \,(\log {x^2})\cos \,(\log {y^2}) - \frac{1}{2}\,\left[ {\cos \log \frac{{{x^2}}}{2} + \cos \log \frac{{{x^2}}}{{

Vedclass · Accepted Answer

None of these

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(d) The given expression is

$\cos \,(\log {x^2})\cos \,(\log {y^2}) - \frac{1}{2}\,\left[ {\cos \log \frac{{{x^2}}}{2} + \cos \log \frac{{{x^2}}}{{{y^2}}}} \right]$

$ = \frac{1}{2}\left[ {\cos \,(\log {x^2} + \log {y^2}) + \cos \,(\log {x^2} - \log {y^2})} \right]$

$ - \frac{1}{2}\left[ {\cos \,\log \frac{{{x^2}}}{2} + \cos \,(\log {x^2} - \log {y^2})} \right]$

$ = \frac{1}{2}\,\left[ {\cos \log {x^2}{y^2} - \cos \log \frac{{{x^2}}}{2}} \right]$.

If $f(x) = \cos (\log x)$, then $f({x^2})f({y^2}) - \frac{1}{2}\left[ {f\,\left( {\frac{{{x^2}}}{2}} \right) + f\left( {\frac{{{x^2}}}{{{y^2}}}} \right)} \right]$ has the value

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