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CAT 2024 Slot 3 — QA Question 16

Basics (Functions)Easy

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

Answer & solution

  • A

    3

  • B

    -2

  • C

    2

  • -3

Solution

Medium

The relation f(x)+2f(1/x)=3xf(x)+2f(1/x)=3x pairs xx with 1/x1/x. Replace xx by 1/x1/x to get a second equation, then eliminate f(1/x)f(1/x) to obtain an explicit formula for f(x)f(x). Solve f(x)=3f(x)=3 and sum the roots.

1

Two equations. Write the given relation and its x→1/xx\to 1/x version.

f(x)+2f(1/x)=3x(i)⇒ f(1/x)+2f(x)=3x(ii, replace x→1/x)\begin{aligned} &f(x)+2f(1/x)=3x\quad\text{(i)}\\ &\Rightarrow\ f(1/x)+2f(x)=\frac{3}{x}\quad\text{(ii, replace }x\to1/x\text{)} \end{aligned}
2

Eliminate f(1/x)f(1/x). Take 2×2\times(ii) −- (i).

2 ⁣(f(1/x)+2f(x))−(f(x)+2f(1/x))=6x−3x⇒ 3f(x)=6x−3x(simplify)⇒ f(x)=2x−x\begin{aligned} &2\!\left(f(1/x)+2f(x)\right)-\left(f(x)+2f(1/x)\right)=\frac{6}{x}-3x\\ &\Rightarrow\ 3f(x)=\frac{6}{x}-3x\quad\text{(simplify)}\\ &\Rightarrow\ f(x)=\frac{2}{x}-x \end{aligned}
3

Solve f(x)=3f(x)=3.

2x−x=3⇒ 2−x2=3x(×x)⇒ x2+3x−2=0\begin{aligned} &\frac{2}{x}-x=3\\ &\Rightarrow\ 2-x^2=3x\quad\text{(}\times x\text{)}\\ &\Rightarrow\ x^2+3x-2=0 \end{aligned}
4

Sum of roots. By Vieta's, the sum of the roots of x2+3x−2=0x^2+3x-2=0 is −3-3.

x1+x2=−31=−3\begin{aligned} &x_1+x_2=-\frac{3}{1}=-3 \end{aligned}
−3-3
CAT 2024 Slot 3 QA Q16: For any non-zero real number x, let f (x) + 2 f (1/x) = 3x. Then, the sum of all possible values of x for whic — Solution | TheCATExam