Question

Which one of the following algebraic expressions is a polynomial in variable $x$?

• A. $x^2+\frac{2}{x^2}$
• B. $\sqrt{x}+\frac{1}{\sqrt{x}}$
• C. $x^2+\frac{3x^{{\displaystyle \frac{3}{2}}}}{\sqrt{x}}$
• D. None of these

Answer 1

The correct option is C

$x^2+\frac{3x^{{\displaystyle \frac{3}{2}}}}{\sqrt{x}}$

Explanation of correct option

Identification of polynomials

Option (C):

If we simplify $x^2+\frac{3x^{{\displaystyle \frac{3}{2}}}}{\sqrt{x}}$ we get

$$\begin{gathered} =x^2+\frac{3x^{{\displaystyle \frac{3}{2}}}}{\sqrt{x}} \\ =x^2+\frac{3x^{{\displaystyle \frac{3}{2}}}}{x^{{\displaystyle \frac{1}{2}}}} \\ =x^2+3x^{\frac{3}{2}-\frac{1}{2}} \\ =x^2+3x \end{gathered}$$ $\left[\because \frac{a^x}{a^y}=a^{x-y}\right]$

Hence, it is polynomial.

Explanation of incorrect options

Option (A):

$x^2+\frac{2}{x^2}=x^2+2x^{-2}$ has negative exponent, so it is not a polynomial.

Option (B):

$\sqrt{x}+\frac{1}{\sqrt{x}}=x^{\frac{1}{2}}+\frac{1}{x^{{\displaystyle \frac{1}{2}}}}$has fractional exponent, so it is not a polynomial.

'None of these' can't be an answer because there is already a correct option.

Hence, option C is correct.