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Limits, Continuity and Differentiability: Mathematics | JEE Main

Q: Can we remove the absolute value signs directly?

Yes, because the discriminant of is and the leading coefficient is , meaning for all real . Hence, always.

Q: Is strictly non-decreasing so that min occurs at min of ?

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

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}

Step-by-step solutionView

Correct 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 $f(x) = \frac{3}{4} + [\frac{3}{4}] = \frac{3}{4}$ .

Option analysis

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 $[u] \ge 1$ 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 ·

\frac{3}{4}

None. 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 $f(1/2) = |\frac{3}{4}| + [\frac{3}{4}] = \frac{3}{4} + 0 = \frac{3}{4}$ .

Reviewed route

Solution

StepWorking

01given

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.

02goal

Find the absolute minimum value of $f(x)$ on $[-1, 2]$ .

03approach

Let $u(x) = x^2 - x + 1$ . Check the sign and range of $u(x)$ for $x \in [-1, 2]$ . Since both $|u|$ and $[u]$ are non-decreasing functions of $u$ for $u > 0$ , the minimum of $f(x) = u + [u]$ occurs at the minimum value of $u(x)$ .

04execute

The quadratic expression is $u(x) = x^2 - x + 1 = \left(x - \frac{1}{2}\right)^2 + \frac{3}{4}$ . For all real $x$ , $u(x) \ge \frac{3}{4} > 0$ . The vertex $x = \frac{1}{2}$ lies inside the interval $[-1, 2]$ . Thus, the minimum value of $u(x)$ on $[-1, 2]$ is $u_{\min} = \frac{3}{4}$ .

05execute

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 $f\left(\frac{1}{2}\right) = \frac{3}{4} + \left[\frac{3}{4}\right] = \frac{3}{4} + 0 = \frac{3}{4}$ .

✓verify

Check boundary points: at $x = -1$ , $u(-1) = 1 + 1 + 1 = 3 \implies f(-1) = 3 + [3] = 6$ . At $x = 2$ , $u(2) = 4 - 2 + 1 = 3 \implies f(2) = 6$ . Since $6 > \frac{3}{4}$ , $\frac{3}{4}$ is indeed the absolute minimum.

Hints that build this answer step by step

Let $u(x) = x^2 - x + 1$ . What is the range of $u(x)$ for $x \in [-1, 2]$ ?
$[\frac{3}{4}, 3]$
Since $u \ge \frac{3}{4} > 0$ , the function simplifies to $g(u) = u + [u]$ . What is the value of $g(u)$ at the minimum point $u = \frac{3}{4}$ ?
$\frac{3}{4}$
Could $g(u) = u + [u]$ achieve a value smaller than $\frac{3}{4}$ for any other $u \in [\frac{3}{4}, 3]$ ?
No, because both $u$ and $[u]$ are non-decreasing functions of $u$ , so $g(u)$ is strictly increasing.

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
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 $[u] \ge 1$ 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: $\frac{3}{4}$ - None. 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 $u = \frac{3}{4} \in [0, 1)$ defined on the closed interval $[-1, 2]$ , where $[\cdot]$ represents the greatest integer function.
goal: Find the absolute minimum value of $f(x)$ on $[-1, 2]$ .
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 $[u]$ 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 $u(x) = x^2 - x + 1 = \left(x - \frac{1}{2}\right)^2 + \frac{3}{4}$ . For all real $x$ , $u(x) \ge \frac{3}{4} > 0$ . 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 $f\left(\frac{1}{2}\right) = \frac{3}{4} + \left[\frac{3}{4}\right] = \frac{3}{4} + 0 = \frac{3}{4}$ .
verify: Check boundary points: at $x = -1$ , $u(-1) = 1 + 1 + 1 = 3 \implies f(-1) = 3 + [3] = 6$ . At $x = 2$ , $u(2) = 4 - 2 + 1 = 3 \implies f(2) = 6$ . Since $6 > \frac{3}{4}$ , $\frac{3}{4}$ is indeed the absolute minimum.

Limits, Continuity and Differentiability: Mathematics | JEE Main

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

✓verify

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Limits, Continuity and Differentiability: Mathematics | JEE Main

By expressing the function in terms of u=x2−x+1u = x^2 - x + 1u=x2−x+1, the minimum occurs at the vertex where u=34u = \frac{3}{4}u=43​, yielding f(x)=34+[34]=34f(x) = \frac{3}{4} + [\frac{3}{4}] = \frac{3}{4}f(x)=43​+[43​]=43​.

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Students also ask

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

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

Let $\text{f}, \text{g}$ and $\text{h}$ be the real valued functions defined on…

$\lim_{x\to 0} \frac{48}{x^4} \int_0^x \frac{t^3}{t^6+1} dt$ is equal to