Let f : R \to R be defined by f(x) = x \over 1 + x^2,x \in R . Then the range of f is :

Functions PYQ — JEE Main 2019 · 11 Jan · Shift 1 · Q29 with Solution

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Correct answer: A

We need the range of $f(x)=\frac{x}{1+x^2}, \qquad x\in\mathbb R.$
Since the denominator is always positive, $1+x^2>0 \quad \text{for all } x\in\mathbb R,$ so $f$ is defined for every real $x$ .
To find the range, compute the derivative:

=\frac{1-x^2}{(1+x^2)^2}.$$ 4. Critical points occur when $$1-x^2=0 \implies x=\pm 1.$$ 5. Sign of $f'(x)$: - For $|x|<1$, $1-x^2>0$, so $f'(x)>0$. - For $|x|>1$, $1-x^2<0$, so $f'(x)<0$. Thus: - $f$ decreases on $(-\infty,-1)$, - increases on $(-1,1)$, - decreases on $(1,\infty)$. So $x=-1$ gives the minimum value and $x=1$ gives the maximum value. 6. Evaluate at these points: $$f(1)=\frac{1}{1+1}=\frac12,$$ $$f(-1)=\frac{-1}{1+1}=-\frac12.$$ 7. Also, $$\lim_{x\to\infty} \frac{x}{1+x^2}=0, \qquad \lim_{x\to-\infty} \frac{x}{1+x^2}=0.$$ So the function stays between $-\frac12$ and $\frac12$, and both endpoints are attained. Therefore the range is $$\left[-\frac12,\frac12\right].$$ 8. Check options: - A: $$\left[-\frac12,\frac12\right]$$ ✅ - B, C, D are incorrect. Hence the correct option is **A**.

View written solutionFree

Correct answer: A

We need the range of $f(x)=\frac{x}{1+x^2}, \qquad x\in\mathbb R.$
Since the denominator is always positive, $1+x^2>0 \quad \text{for all } x\in\mathbb R,$ so $f$ is defined for every real $x$ .
To find the range, compute the derivative:

=\frac{1-x^2}{(1+x^2)^2}.$$ 4. Critical points occur when $$1-x^2=0 \implies x=\pm 1.$$ 5. Sign of $f'(x)$: - For $|x|<1$, $1-x^2>0$, so $f'(x)>0$. - For $|x|>1$, $1-x^2<0$, so $f'(x)<0$. Thus: - $f$ decreases on $(-\infty,-1)$, - increases on $(-1,1)$, - decreases on $(1,\infty)$. So $x=-1$ gives the minimum value and $x=1$ gives the maximum value. 6. Evaluate at these points: $$f(1)=\frac{1}{1+1}=\frac12,$$ $$f(-1)=\frac{-1}{1+1}=-\frac12.$$ 7. Also, $$\lim_{x\to\infty} \frac{x}{1+x^2}=0, \qquad \lim_{x\to-\infty} \frac{x}{1+x^2}=0.$$ So the function stays between $-\frac12$ and $\frac12$, and both endpoints are attained. Therefore the range is $$\left[-\frac12,\frac12\right].$$ 8. Check options: - A: $$\left[-\frac12,\frac12\right]$$ ✅ - B, C, D are incorrect. Hence the correct option is **A**.