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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 value of $\text{f(x)}\text{f(4)}-\frac{1}{2}\Big\{\text{f}\Big(\frac{\text{x}}{4}\Big)+\text{f}(4\text{x})\Big\}$ is:
  • A
    $1$
  • B
    $-1$
  • ✓
    $0$
  • D
    $\pm1$
✓

Answer

Correct option: C.
$0$
Given, $\text{f(x)}=\cos(\log\text{x})$
Then, $\text{f(x)}\text{f(4)}-\frac{1}{2}\Big\{\text{f}\Big(\frac{\text{x}}{4}\Big)+\text{f}(4\text{x})\Big\}$
$=\cos(\log\text{x})\cos(\log4)+\frac{1}{2}\Big\{\cos\Big(\log\frac{\text{x}}{4}\Big)+\cos(\log4\text{x})\Big\}$
$=\frac{1}{2}\big[\cos(\log\text{x}+\log4\big)+\cos(\log\text{x}-\log4)\big]-\frac{1}{2}\Big\{\cos\Big(\log\frac{\text{x}}{4}\Big)+\cos(\log4\text{x})\Big\}$
$=\frac{1}{2}\Big\{\cos(\log4\text{x})+\cos\Big(\log\frac{\text{x}}{4}\Big)-\cos\Big(\log\frac{\text{x}}{4}\Big)-\cos(\log4\text{x})\Big\}$
$=\frac{1}{2}\times0$
$=0$

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