JEE MainMathematics

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Minimizing a Function Involving Absolute Value and Greatest Integer Function

Q: Can we remove the absolute value signs directly?

Yes, because the discriminant of x^2 - x + 1 is D = (-1)^2 - 4(1)(1) = -3 0, meaning x^2 - x + 1 > 0 for all real x. Hence, |x^2 - x + 1| = x^2 - x + 1 always.

Q: Is u + [u] strictly non-decreasing so that min occurs at min of u?

Yes, both h_1(u) = u (strictly increasing) and h_2(u) = [u] (monotonically non-decreasing step function) increase with u. Their sum g(u) = u + [u] is strictly increasing everywhere, so its minimum over any set of positive reals is attained at the minimum of u.

The absolute minimum value, of the function

f (x) = ∣ x^{2} - x + 1 ∣ + [x^{2} - x + 1]

, where

[t]

denotes the greatest integer function, in the interval

[- 1, 2]

, is:

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

\frac{1}{4}

\frac{3}{2}

\frac{5}{4}

\frac{3}{4}

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 9 September 2026

Quick checks

Students also ask

Can we remove the absolute value signs directly?

Yes, because the discriminant of $x^{2} - x + 1$ is $D = (- 1)^{2} - 4 (1) (1) = - 3 < 0$ and the leading coefficient is $1 > 0$ , meaning $x^{2} - x + 1 > 0$ for all real $x$ . Hence, $∣ x^{2} - x + 1 ∣ = x^{2} - x + 1$ always.

u + [u]

strictly non-decreasing so that min occurs at min of

u

Yes, both $h_{1} (u) = u$ (strictly increasing) and $h_{2} (u) = [u]$ (monotonically non-decreasing step function) increase with $u$ . Their sum $g (u) = u + [u]$ is strictly increasing everywhere, so its minimum over any set of positive reals is attained at the minimum of $u$ .

Answer

By expressing the function in terms of $u = x^{2} - x + 1$ , the minimum occurs at the vertex where $u = \frac{3}{4}$ , yielding $f (x) = \frac{3}{4} + [\frac{3}{4}] = \frac{3}{4}$ .

Why each option works or fails

A: $\frac{1}{4}$ - Believing that the vertex value of the quadratic part is $\frac{1}{4}$ instead of completing the square properly as $(x - \frac{1}{2})^{2} + \frac{3}{4}$ . Complete the square carefully: $x^{2} - x + 1 = (x - \frac{1}{2})^{2} - \frac{1}{4} + 1 = \frac{3}{4}$ .
B: $\frac{3}{2}$ - Assuming that $[u] \geq 1$ everywhere on the interval, leading to $u + 1 = \frac{3}{4} + 1 = \frac{7}{4}$ , or miscalculating the floor and absolute value values. Check the value of $u$ at the critical point: since $u = \frac{3}{4} \in [0, 1)$ , the floor $[u]$ evaluates strictly to $0$ .
C: $\frac{5}{4}$ - Incorrectly calculating $[u]$ as $1$ or making an arithmetic error when evaluating $u$ at the critical point. Evaluate each component at $u = \frac{3}{4}$ : $∣ u ∣ = \frac{3}{4}$ and $[u] = 0$ , giving a sum of $\frac{3}{4}$ .
D · correct: $\frac{3}{4}$ - This is the correct value. The quadratic $u = x^{2} - x + 1$ achieves its absolute minimum $\frac{3}{4}$ at $x = \frac{1}{2} \in [- 1, 2]$ , so $f (1 / 2) = ∣ \frac{3}{4} ∣ + [\frac{3}{4}] = \frac{3}{4} + 0 = \frac{3}{4}$ .

Step-by-step solution

Given: Function $f (x) = ∣ x^{2} - x + 1 ∣ + [x^{2} - x + 1]$ defined on the closed interval $[- 1, 2]$ , where $[\cdot]$ represents the greatest integer function.
Goal: Find the absolute minimum value of $f (x)$ on $u$ .
Approach: Let $u (x) = x^{2} - x + 1$ . Check the sign and range of $u (x)$ for $x \in [- 1, 2]$ . Since both $∣ u ∣$ and $\frac{3}{4}$ are non-decreasing functions of $u$ for $u > 0$ , the minimum of $f (x) = u + [u]$ occurs at the minimum value of $u (x)$ .
Execute: The quadratic expression is $u (x) = x^{2} - x + 1 = {(x - \frac{1}{2})}^{2} + \frac{3}{4}$ . For all real $x$ , $u (x) \geq \frac{3}{4} > 0$ . The vertex $x = \frac{1}{2}$ lies inside the interval $[- 1, 2]$ . Thus, the minimum value of $u (x)$ on $[- 1, 2]$ is $u_{\min} = \frac{3}{4}$ .
Execute: Since $u > 0$ , $∣ u ∣ = u$ . Thus $f (x) = u + [u]$ . As $g (u) = u + [u]$ is strictly increasing with respect to $u$ , the minimum of $f (x)$ occurs when $u$ is minimal ( $u = \frac{3}{4}$ ). Evaluating $f (\frac{1}{2}) = \frac{3}{4} + [\frac{3}{4}] = \frac{3}{4} + 0 = \frac{3}{4}$ .
Verify: Check boundary points: at $x = - 1$ , $u (- 1) = 1 + 1 + 1 = 3 ⟹ f (- 1) = 3 + [3] = 6$ . At $x = 2$ , $u (2) = 4 - 2 + 1 = 3 ⟹ f (2) = 6$ . Since $6 > \frac{3}{4}$ , $\frac{3}{4}$ is indeed the absolute minimum.

Minimizing a Function Involving Absolute Value and Greatest Integer Function

What feels right?

By expressing the function in terms of $u = x^{2} - x + 1$ , the minimum occurs at the vertex where $u = \frac{3}{4}$ , yielding $f (x) = \frac{3}{4} + [\frac{3}{4}] = \frac{3}{4}$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

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By expressing the function in terms of $u = x^{2} - x + 1$ , the minimum occurs at the vertex where $u = \frac{3}{4}$ , yielding $f (x) = \frac{3}{4} + [\frac{3}{4}] = \frac{3}{4}$ .

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Let $g : R \to R$ be a non constant twice differentiable such that…