CAT 2024 Set-3 | Quantitative Aptitude | Question: 11

Question

For any non-zero real number $x$, let $f(x)+2 f\left(\frac{1}{x}\right)=3 x$. Then, the sum of all possible values of $x$ for which $f(x)=3$, is

• A. $-3$
• B. $3$
• C. $-2$
• D. $2$

Answer 1

Given f(x)+2f(1/x)=3x       (eq 1)

on substituting 1/x in place of x our equation will be :

f(1/x)+2f(x)=3(1/x)           (eq 2)

(eq1) -(eq2) will give us :

f(1/x)-f(x)=3(x)-3(1/x)

f(1/x)=3[(x)-(1/x)]+f(x)

now substitute the value of f(1/x) in (eq1)

f(x)+2* [3{(x)-(1/x)}+f(x)]=3x

f(x)+6(x)-6(1/x)+2f(x)=3x

3f(x)= -3(x)+6(1/x)

f(x)=2(1/x)-(x)

now given that f(x)=3, so put 3 in place of f(x) in above equation

3=2(1/x)-(x)

3x=2-x^2

x^2+3x-2=0

on solving the above equation for x our roots will be : (-3+√17)/2 , (-3-√17)/2

now on adding both the roots the sum would be -3

Hence the sum of roots for which the value of f(x)=3 is -3.

So option(A) is the correct option.