Q235 / 270 · NDA 2024 (Phase 1) · Mathematics
Let $f(x)$ and $g(x)$ be two functions such that $g(x)=x-\frac{1}{x}$ and $f$ o $g(x)=x^3-\frac{1}{x^3}$.
What is $g[f(x)-3 x]$ equal to?
- Correct answerA$x^3-\frac{1}{x^3}$
- B$x^3+\frac{1}{x^3}$
- C$x^2-\frac{1}{x^2}$
- D$x^2+\frac{1}{x^2}$
Solution
$\because f(x)$ and $g(x)$ are two functions
$\begin{aligned}
g(x) & =x-\frac{1}{x} \text { and } f o g(x)=x^3-\frac{1}{x^3} \\
f\left(x-\frac{1}{x}\right) & =x^3-\frac{1}{x^3} \\
\Rightarrow f\left(x-\frac{1}{x}\right) & =\left(x-\frac{1}{x}\right)^3+3\left(x-\frac{1}{x}\right)
\end{aligned}$
$f(x)=x^3+3 x \quad ....(i)$
$\begin{aligned} g[f(x)-3 x] & =f(x)-3 x-\frac{1}{f(x)-3 x} \\ g(f(x)-3 x] & =x^3+3 x-3 x-\frac{1}{x^3+3 x-3 x} \\ & =x^3-\frac{1}{x^3} \\ g(f(x)-3 x] & =x^3-\frac{1}{x^3}\end{aligned}$
Hence, the correct answer is option: 'A'.