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If $f(x)=coslogx$ then $f(x^2).f(y^2)-\large\frac{1}{2}$ $(f(\large\frac{x^2}{y^2})$$+f(x^2y^2))= ?$

(A) -2

(B) -1

(C) 0

(D) 1/2
Can you answer this question?
 
 

1 Answer

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Toolbox:
  • $2cosA.cosB=cos(A+B)+cos(A-B)$
  • $loga+logb=log(ab)$
  • $loga-logb=log(\large\frac{a}{b})$
Given $f(x)=coslog x$
$f(x^2).f(y^2)=coslog x^2.coslog y^2$
$=\large\frac{1}{2}$ $(2coslogx^2.coslogy^2)$
$=\large\frac{1}{2}$ $(cos(logx^2+logy^2)+cos(logx^2-logy^2))$
$=\large\frac{1}{2}$ $\big[cos(log(x^2y^2))+cos(log(\large\frac{x^2}{y^2}))\big]$
$=\large\frac{1}{2}$ $\big[f(x^2y^2)+f(\large\frac{x^2}{y^2})\big]$
$\therefore\:\:f(x^2).f(y^2)-\large\frac{1}{2}$ $\big[f(x^2y^2)+f(\large\frac{x^2}{y^2})\big]$$=0$
answered May 29, 2013 by rvidyagovindarajan_1
 

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