InterviewSolution
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InterviewSolution
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If f (x) = cos (log x), then f (x2) . f (y2) - \(\frac{1}{2}\big[f(x^2.y^2)+f\big(\frac{x^2}{y^2}\big)\big]\) equals(a) – 2 (b) – 1 (c) \(\frac{1}{2}\)(d) 0 |
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Answer» Answer : (d) = 0 Given, f (x) = cos (log x) ⇒ f (x2) = cos (log x2) = cos (2 log x) f (y2) = cos (log y2) = cos (2 log y) \(f\big(\frac{x^2}{y^2}\big)\) = \(cos\big(log\frac{x^2}{y^2}\big)\) = cos (log x2 - log y2 ) = cos (2 log x – 2 log y) f (x2y2) = cos (log x2y2) = cos (log x2 + log y2) = cos (2 log x + 2 log y) Now, f(x2). f(y2) - \(\frac{1}{2}\) \(\big[f\big(\frac{x^2}{y^2}\big) +f (x^2y^2)\big]\) = cos (2 log x) . cos (2 log y) – \(\frac{1}{2}\) [cos (2 log x – 2 log y) + cos (2 log x + 2 log y)] = cos (2 log x) . cos (2 log y) - \(\frac{1}{2}\) \(\big[ 2\,cos\big(\frac{2\,log\,x-2\,log\,y+2\,log\,x+2\,log\,y}{2}\big) cos \big( \frac{2\,log\,x+2\,log\,y-2\,log\,x+2\,log\,y}{2}\big)\big]\) \(\big(\because cos\,C + cos\,D = 2 \,cos\big(\frac{C+D}{2}\big)cos\,\big(\frac{D-C}{2}\big)\big)\) = cos (2 log x) . cos (2 log y) – \(\frac{1}{2}\)[2 cos (2 log x) . cos (2 log y)] = 0. |
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