Q156 / 270 · NDA 2016 (Phase 1) · Mathematics
What is the range of the function $\mathrm{y}=\frac{\mathrm{x}^2}{1+\mathrm{x}^2}$, where $\mathrm{x} \in \mathrm{R}$ ?
- Correct answerA$[0,1)$
- B$[0,1]$
- C$(0,1)$
- D$(0,1]$
Solution
Function $y=\frac{x^2}{1+x^2} \quad x \in R$
$x=0$ $y=0$
$\begin{array}{ll}
x=1,-1 & y=\frac{1}{2} \\
x=2,-2 & y=\frac{4}{5} \\
x=3,-3 & y=\frac{9}{10}
\end{array}$
.
.
.
.
Clearly $0 \leq y \lt 1$
$\Rightarrow y \in[0,1)$
Hence Range of $y=[0,1)$
Hence, the correct answer is option: 'A'.