## IIT JEE Main 2016 Maths Question 32

If f(x) + 2f(1/x) = 3x, x<> 0 and S = {x ∈ R : f(x)=f(-x)}; then S :

(1) contains exactly two elements | (2) contains more than two elements | (3) is an empty set | (4) contains exactly one element |

#### Solution

Solution to the problem requiring to find out the set S as per the relation specified is as follows:

f(x) + 2f(1/x) = 3x …………. (1)

Replace x->(1/x)

=>f(1/x) + 2f(x) = 3/x

=> 2f(1/x) + 4f(x) =6/x ……(2)

=> from (1) and (2)

3f(x) = 6/x – 3x

=> f(x) = (2/x) – x

=> f(-x) = (-2/x) + x

=> f(x) = f(-x)

=> (2/x) – x = (-2/x) + x

=> 4/x = 2x

=>x^{2} = 2

=>x= +/- √2

Hence set S contains two elements, i.e. +√2 , -√2.

#### Answer

Correct answer to given problem No 32 of IIT JEE Main Maths paper of 2016 is option number (1) that is set S contains exactly two elements.