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)
N neha
H Himanshu

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.

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