MCQ
Which one of the following is a polynomial?
- A$\frac{x-1}{x+1}$
- B$\sqrt{2 x}-1$
- ✓$x^2+\frac{3 x^{\frac{3}{2}}}{\sqrt{x}}$
- D$\frac{x^2}{2}-\frac{2}{x^2}$
✓
Answer
Correct option: C.
$x^2+\frac{3 x^{\frac{3}{2}}}{\sqrt{x}}$
Since the power of the variable of all terms of a polynomial should be a whole number.
Then $x^2+\frac{3 x^{\frac{3}{2}}}{\sqrt{x}}$
$=x^2+3 x^{\frac{3}{2}-\frac{1}{2}}$
$=x^2+3 x^{\frac{2}{2}}$
$=x^2+3 x$
Here the powers of variable are whole numbers.
Therefore the given expression is a polynomial.
Then $x^2+\frac{3 x^{\frac{3}{2}}}{\sqrt{x}}$
$=x^2+3 x^{\frac{3}{2}-\frac{1}{2}}$
$=x^2+3 x^{\frac{2}{2}}$
$=x^2+3 x$
Here the powers of variable are whole numbers.
Therefore the given expression is a polynomial.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
An exterior angle of a triangle is $108^\circ$ and its interior opposite angles are in the ratio $4 : 5$ The angles of the triangle are:
→In the adjoining figure $PQ \| XR$. If $\text{QP}\bot\text{SP},$ then the values of $x$ and $y$ are: 
→$(104 \times 96) = ?$
→The bisectors of the angle of a parallelogram enclose a
The value of $\sqrt{3-2\sqrt{2}}$ is:
→The abscissa of a point is positive in the:
In a $\triangle\text{ABC},$ if $\angle\text{A}=60^\circ,\angle\text{B}=80^\circ$ and the bisectors of $\angle\text{B}$ and $\angle\text{C}$ meet at $O$, then $\angle\text{BOC}=$
→It is known that if $x + y = 10$ then $x + y + z = 10 + z$. Euclid’s axiom that illustrates this statement is:
→Euclid first postulates is:
In the given figure, AB || CD. If $\angle\text{EAB}=50^\circ$ and $\angle\text{ECD}=60^\circ$ then $\angle\text{AEB}=?$ 
→