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if $f(x)+2f\left ( \frac{1}{x} \right )=3x,x\neq 0,$ and $S=\left \{ x\, \epsilon \, R : f(x)=f(-x)\right \};then S:$

Option 1)
is an empty set.

Option 2)
contains exactly one element.

Option 3)
contains exactly two elements.

Option 4)
contains more than two elements.

Answers (2)

best_answer

As we learnt in

FUNCTIONS -

A relation f from a set A to a set B is said to be a function if every element of set A has one and only one image in set B.

-

$f(x)+2f \left(\frac{1}{x} \right )=3x$

Put $\frac{1}{x}$ at the place of

$f\left(\frac{1}{x} \right )+2f(x)=\frac{3}{x}$ (i)

$2f\left(\frac{1}{x} \right )+f(x)=3x$ (ii)

Multiplying (i) by 2

$2f\left(\frac{1}{x} \right )+4f(x)=\frac{6}{x}$

$\underline{2f\left(\frac{1}{x} \right )+f(x)=3x}$

$3f(x)=\frac{6}{x}-3x$

$f(x)=\frac{2}{x}-3x$

and $f(-x)=\frac{2}{-x}+x$

$\therefore\ \; \frac{2}{x}-x=-\frac{2}{x}+x$

$\Rightarrow\ \; \frac{4}{x}-2x=0$

$\Rightarrow\ \; \frac{4-2x^{2}}{x}=0$

$\Rightarrow\ \; 4=2x^{2}$

$\Rightarrow\ \; x^{2}=2$

$x=\pm \sqrt{2}, \; x \neq 0$

Correct option is 3.

Option 1)

is an empty set.

Option 2)

contains exactly one element.

Option 3)

contains exactly two elements.

Option 4)

contains more than two elements.

Posted by

Himanshu

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