(C) [− 1 4 , ∞) (D)[− 1 4 , 1 4] 1

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Q. (C) [- 1 4 , \infty) (D)[- 1 4 , 1 4] 1

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Concept understanding — Quadratic Range

Range of a Quadratic Function

The range of a function is the complete set of values it can output. For a quadratic $f(x) = ax^2 + bx + c$ (with $a \neq 0$ ) the graph is a parabola with exactly one turning point, so the outputs run in one direction from that turning value. Finding the range is therefore the same as finding the vertex and asking, "does the curve go up from here, or down?"

The core idea

If $a > 0$ the parabola opens upward. It has a lowest point (a minimum) and rises forever above it, so the range is $[\,y_{\min},\, \infty)$ .
If $a < 0$ it opens downward. It has a highest point (a maximum) and falls forever below it, so the range is $(-\infty,\, y_{\max}\,]$ .

The turning point sits at the vertex, whose coordinates are

x = -\frac{b}{2a}, \qquad y = f\!\left(-\frac{b}{2a}\right) = c - \frac{b^2}{4a}.

Why the vertex value is the boundary

Complete the square:

f(x) = a\left(x + \frac{b}{2a}\right)^2 + \left(c - \frac{b^2}{4a}\right).

The squared bracket is never negative, so $f(x)$ can never go below (if $a>0$ ) or above (if $a<0$ ) the constant $c - \dfrac{b^2}{4a}$ . That constant is exactly the extreme output — the edge of the range.

Note

Since $b^2 - 4ac$ is the discriminant $D$ , the extreme value can be written neatly as $-\dfrac{D}{4a}$ .

Worked feel

For $f(x) = 2x^2 - 8x + 5$ : here $a = 2 > 0$ , vertex at $x = 2$ , value $f(2) = -3$ . Range $= [-3, \infty)$ .

For $f(x) = -x^2 + 4x$ : $a = -1 < 0$ , vertex at $x = 2$ , value $4$ . Range $= (-\infty, 4]$ .

Tip

This assumes the domain is all of $\mathbb{R}$ . If the domain is a restricted interval $[p, q]$ , check the vertex and both endpoints — the range is then a closed interval bounded by the largest and smallest of those values.

Finding the range of a quadratic function by completing the square connects the NCERT quadratic equations work with the Class 11 Relations and Functions chapter, and "range of quadratic function formula" is a frequently searched topic ahead of CBSE exams and JEE Main. This vertex-based reasoning is also the same idea used later to find the maximum or minimum value of a quadratic expression in optimization problems.

The range of $f(x) = \frac{x}{1+x^2}$ is found by solving for $x$ in terms of $y$ and applying the discriminant condition for real $x$ . The range is $\left[-\frac{1}{2}, \frac{1}{2}\right]$ , which corresponds to option (B).

The key insight: when a function is a rational expression where the denominator is always positive (here $1+x^2 > 0$ for all real $x$ ), we can treat the equation $y = f(x)$ as a quadratic in $x$ and use the fact that $x$ must be real. The range is then all $y$ for which this quadratic has real solutions.

Let’s walk through it.

Set up the equation. Write $y = \frac{x}{1+x^2}$ . Multiply both sides by $1+x^2$ (which is never zero, so safe):

y(1+x^2) = x \quad \Rightarrow \quad y + yx^2 = x.

Rearrange into standard quadratic form in $x$ :

y x^2 - x + y = 0.

Apply the discriminant condition. For $x$ to be real, the discriminant of this quadratic must be non-negative. Here $a = y$ , $b = -1$ , $c = y$ . The discriminant is:

\Delta = b^2 - 4ac = (-1)^2 - 4(y)(y) = 1 - 4y^2.

The condition $\Delta \geq 0$ gives:

1 - 4y^2 \geq 0 \quad \Rightarrow \quad 4y^2 \leq 1 \quad \Rightarrow \quad y^2 \leq \frac{1}{4}.

Interpret the inequality. $y^2 \leq \frac{1}{4}$ means $|y| \leq \frac{1}{2}$ , i.e.:

-\frac{1}{2} \leq y \leq \frac{1}{2}.

So the range is $\left[-\frac{1}{2}, \frac{1}{2}\right]$ .

Watch out

A common mistake is to forget that $y$ can be zero. When $y=0$ , the quadratic becomes $-x = 0$ , which gives $x=0$ — perfectly valid. So zero is included, and the interval is closed at both ends.

Check the endpoints. At $y = \frac{1}{2}$ , the quadratic is $\frac{1}{2}x^2 - x + \frac{1}{2} = 0$ , which factors as $\frac{1}{2}(x-1)^2 = 0$ , giving $x=1$ . At $y = -\frac{1}{2}$ , we get $-\frac{1}{2}x^2 - x - \frac{1}{2} = 0$ , or $-\frac{1}{2}(x+1)^2 = 0$ , giving $x=-1$ . Both are real, so the endpoints are attained.

Tip

Notice that $f(x) = \frac{x}{1+x^2}$ is an odd function ( $f(-x) = -f(x)$ ), so the range is symmetric about 0. That immediately tells you the range is of the form $[-a, a]$ , and the discriminant method gives $a = \frac{1}{2}$ .

f(x) = \frac{ax+b}{cx^2+dx+e}

✓Final answer

\boxed{\left[-\frac{1}{2}, \frac{1}{2}\right]}

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2020Set 2A2 marks
$2x+5-3x^2$
›Reveal solutionSolution

$2x+5-3x^2$ is a downward-opening quadratic in $x$ ; its maximum occurs at the vertex $x=-b/2a$ .

Write the expression as $f(x) = -3x^2+2x+5$ , a quadratic with $a=-3,\ b=2,\ c=5$ .

Since $a=-3<0$ , the parabola opens downward, so $f$ has a maximum (not minimum) at

$x = -\frac{b}{2a} = -\frac{2}{2(-3)} = \frac{1}{3}$

Substitute back:

$f\left(\frac13\right) = 2\left(\frac13\right)+5-3\left(\frac13\right)^2 = \frac23+5-3\cdot\frac19 = \frac23+5-\frac13$

$= \left(\frac23-\frac13\right)+5 = \frac13+5 = \frac{16}{3}$

So as $x$ varies over $\mathbb{R}$ , the maximum value of $2x+5-3x^2$ is $\dfrac{16}{3}$ .

✓Final answer
$2x+5-3x^2$
CBSE 2018Set ANNUAL2 marks
$f : \mathbb{R} \to \mathbb{R}$
›Reveal solutionSolution

The domain is given as all of $\mathbb R$ ; find the range by bounding $x^2+1$ below and showing every value above the bound is attained.

Domain: $f$ is defined for every real number (no restriction on $x$ ), so

$\text{Domain}(f)=\mathbb R$

Range: For any real $x$ , $x^2\ge0$ , hence

$f(x)=x^2+1\ge0+1=1$

So every value of $f$ is at least $1$ . Conversely, given any $y\ge1$ , set $x=\sqrt{y-1}$ (which is a real number since $y-1\ge0$ ); then $f(x)=(\sqrt{y-1})^2+1=y-1+1=y$ . So every $y\ge1$ is actually attained.

Hence

$\text{Range}(f)=[1,\infty)$

✓Final answer
$=\mathbb R$

Q.(C) [− 1 4 , ∞) (D)[− 1 4 , 1 4] 1

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