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Relations and Functions — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsRelations and Functions1 Mark
MCQ
If $\text{f(x)}=\cos(\log\text{x}),$ then the value of $\text{f(x})\text{f}(\text{y})-\frac{1}{2}\Big\{\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)+\text{f}\big(\text{x}\text{y}\big)\Big\}$ is:
  • A
    $-1$
  • B
    $\frac{1}{2}$
  • C
    $-2$
  • ✓
    None of these.
✓

Answer

Correct option: D.
None of these.
Given,
$\text{f(x)}=\cos(\log\text{x})$
$\therefore\ \text{f(y)}=\cos(\log\text{y})$
Now,
$\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)=\cos\Big(\cos\Big(\frac{\text{x}}{\text{y}}\Big)\Big)=\cos(\log\text{x}-\log\text{y})$
and
$\text{f(xy)}=\cos(\log\text{xy})=\cos(\log\text{x}+\log\text{y})$
$\Rightarrow\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)+\text{f(xy)}=\cos(\log\text{x}-\log\text{y})+\cos(\log\text{x}+\log\text{y})$
$\Rightarrow\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)+\text{f(xy)}=2\cos(\log\text{x})\cos(\log\text{y})$
$\Rightarrow\frac{1}{2}\Big[\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)+\text{f(xy)}\Big]=\cos(\log\text{x})\cos(\log\text{y})$
$\Rightarrow\text{f(x})\text{f}(\text{y})-\frac{1}{2}\Big\{\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)+\text{f}\big(\text{x}\text{y}\big)\Big\}\\=\cos(\log\text{x})\cos(\log\text{y})-\cos(\log\text{x})\cos(\log\text{y})=0$

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