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# Need explanation for: - Sets, Relations and Functions - BITSAT

if     and

• 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)
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N

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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