Stuck on this function question. I’m not able to figure out how to use f(x) = 3 in the given expression. Do we replace x or 1/x?

Question

For any non-zero real number x , let f(x)+2f(1/x)=3x . Then, the sum of all possible values of x for which f(x)=3 , is
A. 2
B. -3
C. -2
D. 3

V Gupta · Accepted Answer

Faced this exact confusion during practice. The doubt is common &mdash; whether to substitute x or 1/x when f(x) = 3. Here&rsquo;s how to handle it:

Start with f(x) + 2f(1/x) = 3x
Replace x with 1/x in the original equation: f(1/x) + 2f(x) = 3/x
Now use both equations to eliminate f(1/x)
Multiply the second equation by 2 and subtract: You&rsquo;ll get 3f(x) = (6/x) &ndash; 3x
So f(x) = (2/x) &ndash; x
Now apply f(x) = 3: (2/x) &ndash; x = 3
Rearranged: x&sup2; + 3x &ndash; 2 = 0
Use sum of roots formula = &ndash;b/a = &ndash;3

So the final answer is &ndash;3, option B. Just needed to go step by step and build the second equation logically.

T Ronak · Answer

For this type of function question, the trick is to use symmetry. Since the equation has both f(x) and f(1/x), try plugging x and 1/x to create two equations.
Start with the original: f(x) + 2f(1/x) = 3x
Now put x = 1/x, which means you form another equation: f(1/x) + 2f(x) = 3/x
Now solve these two equations together. Multiply the second one by 2 and subtract from the first. That gives you f(x) = (2/x) &ndash; x.
Once you get f(x), now use the condition f(x) = 3. So (2/x) &ndash; x = 3,  which gives a quadratic.
Solving it, you'll get x&sup2; + 3x &ndash; 2 = 0.
Use sum of roots formula: &ndash;b/a = &ndash;3 , So the answer is &ndash;3.