If fx+2f1/x=3x, x≠0, and S=x∈R:fx=f-x, then S

Correct answer: (A) Contains exactly two elements. fx+2. f1/x=3x …1 Replace x by 1/x, we get f1/x+2fx=3/x …2 From 1 and 2 3fx=6/x-3x fx=2/x-x ∵ fx=f-x Therefore 2/x-x= -2/x+x ⇒ 4/x=2x ⇒ 2=x^2 ⇒ x= ± √2 Contains exactly two elements

Q32 / 90 · JEE Main 2016 (03 Apr) · Mathematics

If $f (x) + 2 f (\frac{1}{x}) = 3 x, x \neq 0,$ and $S = \{x \in R : f (x) = f (- x)\}$ , then $S$

Correct answer
A
Contains exactly two elements
B
Contains more than two elements
C
Is an empty set
D
Contains exactly one element

Solution

f (x) + 2 . f (\frac{1}{x}) = 3 x \dots (1)

Replace

x

\frac{1}{x}

, we get

f (\frac{1}{x}) + 2 f (x) = \frac{3}{x} \dots (2)

From

(1)

and

(2)

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

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

∵

f (x) = f (- x)

Therefore

\frac{2}{x} - x = - \frac{2}{x} + x

\Rightarrow \frac{4}{x} = 2 x

\Rightarrow 2 = x^{2}

\Rightarrow x = \pm \sqrt{2}

Contains exactly two elements

Hence, the correct answer is option: 'A'.

If fx+2f1x=3x, x≠0, and S=x∈R:fx=f-x, then S

Solution

Relevant Questions

If $f (x) + 2 f (\frac{1}{x}) = 3 x, x \neq 0,$ and $S = \{x \in R : f (x) = f (- x)\}$ , then $S$