Question

A function f is defined by $f\left(x\right)=x+\frac{1}{x}$.

Consider the following.

$\left(1\right){\left(\mathrm{f}\left(\mathrm{x}\right)\right)}^2=f\left({\mathrm{x}}^2\right)+2$ and

$\left(2\right){\left(f\left(x\right)\right)}^3=f\left(x^3\right)+3f\left(x\right)$.

Which one is correct?

• A. $1$only
• B. $2$ only
• C. Both $1$and $2$
• D. Neither $1$nor $2$

Answer 1

The correct option is C

Both $1$and $2$

The explanation for the correct answer:

Step 1: Finding the expansion for ${\left(f\left(x\right)\right)}^2$:

Given, the function $f\left(x\right)=x+\frac{1}{x}$

$$\begin{gathered} {\left(f\left(x\right)\right)}^2={\left(x+\frac{1}{x}\right)}^2\mathbf{}\left[\because {\left(a+b\right)}^2=a^2+2ab+b^2\right] \\ =x^2+2\left(x\right)\left(\frac{1}{x}\right)+{\left(\frac{1}{x}\right)}^2 \\ =x^2+\frac{1}{x^2}+2 \\ =f\left(x^2\right)+2...................\left(1\right) \end{gathered}$$

Step 2: Finding ${\left(f\left(x\right)\right)}^3$:

$$\begin{gathered} {\left(f\left(x\right)\right)}^3={\left(x+\frac{1}{x}\right)}^3\left[\because {\left(a+b\right)}^3=a^3+3a^{2}b+3a{b}^2+b^3\right] \\ =x^3+3{\left(x\right)}^2\frac{1}{x}+3\left(x\right){\left(\frac{1}{x}\right)}^2+{\left(\frac{1}{x}\right)}^3 \\ =x^3+\frac{1}{x^3}+3x+\frac{3}{x} \\ =\left(f{\left(x\right)}^3\right)+3\left(f\left(x\right)\right).................\left(2\right) \end{gathered}$$

Thus from the equation $\left(1\right)$and $\left(2\right)$ both the statements are true.

Hence, option (C) is the correct answer.