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Functions — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSFunctions1 Mark
MCQ
If $\text{f(x)}=\cos(\log_\text{e}),$ then $\text{f}\Big(\frac{1}{\text{x}}\Big)\text{f}\Big(\frac{1}{\text{y}}\Big)-\frac{1}{2}\Big\{\text{f(xy)}+\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)\Big\}$ is equal to:
  • A
    $\cos(\text{x}-\text{y})$
  • B
    $\log(\cos(\text{x}-\text{y}))$
  • C
    $1$
  • ✓
    $\cos(\text{x}+\text{y})$
✓

Answer

Correct option: D.
$\cos(\text{x}+\text{y})$
$\text{f(x)}=\cos(\log_\text{e}\text{x})$
$\Rightarrow\text{f}\Big(\frac{1}{\text{x}}\Big)=\cos\Big(\log_\text{e}\Big(\frac{1}{\text{x}}\Big)\Big)$
$\Rightarrow\text{f}\Big(\frac{1}{\text{x}}\Big)=\cos\Big(-\log_\text{e}(\text{x})\Big)$
$\Rightarrow\text{f}\Big(\frac{1}{\text{x}}\Big)=\cos\Big(\log_\text{e}(\text{x})\Big)$
Similarly,
$\text{f}\Big(\frac{1}{\text{y}}\Big)=\cos(\log_\text{e}\text{y})$
Now,
$\text{f(xy)}=\cos(\log_\text{e}\text{xy})=\cos\big(\log_\text{e}\text{x}+\log_\text{e}\text{y}\big)$
and
$\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)=\cos\Big(\log_\text{e}\frac{\text{x}}{\text{y}}\Big)=\cos\big(\log_\text{e}\text{x}-\log_\text{e}\text{y}\big)$
$\Rightarrow\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)+\text{f(xy)}=\cos\big(\log_\text{e}\text{x}+\log_\text{e}\text{y}\big)+\cos\big(\log_\text{e}\text{x}-\log_\text{e}\text{y}\big)$
$\Rightarrow\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)+\text{f(xy)}=2\cos\big(\log_\text{e}\text{x}\big)\cos(\log_\text{e}\text{y})$
$\Rightarrow\frac{1}{2}\Big[\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)+\text{f(xy)}\Big]=\cos\big(\log_\text{e}\text{x}\big)\cos\big(\log_\text{e}\text{y}\big)$
$\Rightarrow\text{f}\Big(\frac{1}{\text{x}}\Big)\text{f}\Big(\frac{1}{\text{y}}\Big)-\frac{1}{2}\Big\{\text{f(xy)}+\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)\Big\}=\cos\big(\log_\text{e}\text{x}\big)\cos\big(\log_\text{e}\text{y}\big)\\\ \ -\cos\big(\log_\text{e}\text{x}\big)\cos\big(\log_\text{e}\text{y}\big)=0$
Disclaimer: The question in the book has some error, so none of the options are matching with the solution. The solution is created according to the question given in the book.

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