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:

Given: f(x) = |x^2 - x + 1| + x^2 - x + 1 for x [-1,2]. Find: The absolute minimum value of f(x). Let g(x) = x^2 - x + 1 Then f(x) = |g(x)| + g(x) Now, g(x) = (x - 12)^2 + 34 So g(x) has minimum value at x = 12. At x = 12, g ( 12) = 34 Hence, |g ( 12) | = 34, g ( 12) = 0 Therefore, f ( 12) = 34 + 0 = 34 So the absolute minimum value is 34. The correct option is A. The solution labels the option as D, but its computed value is 34, which matches option A in the given options. Given: f(x) = |x^2 - x + 1| + x^2 - x + 1 on [-1,2]. Find: The least value of the function. Since x^2 - x + 1 = (x - 12)^2 + 34 we get x^2 - x + 1 34 for all x [-1,2]. Because x^2 - x + 1 is always positive here, its absolute value remains the same: |x^2 - x + 1| = x^2 - x + 1 Also, when the minimum value is 34, x^2 - x + 1 = 0 at that point. Thus at the minimizing point x = 12, f(x) = 34 + 0 = 34 Therefore, the absolute minimum value of the function is 34.

The absolute minimum value of the function f(x) = |x^2 - x + 1| +… — Solution | JEE Mathematics

MCQEasyJEE 2023Applications of Derivatives (Monotonicity, Extrema)

The absolute minimum value of the function $f(x) = |x^2 - x + 1| + \left\lfloor x^2 - x + 1 \right\rfloor, \quad \text{where} \, [t] \, \text{denotes the greatest integer function, in the interval} \, [-1, 2], \, \text{is:}$

A
$\frac{3}{4}$
B
$\frac{3}{2}$
C
$\frac{1}{4}$
D
$\frac{5}{4}$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: $f(x) = |x^2 - x + 1| + \left\lfloor x^2 - x + 1 \right\rfloor$ for $x \in [-1,2]$ .

Find: The absolute minimum value of $f(x)$ .

Let

g(x) = x^2 - x + 1

Then

f(x) = |g(x)| + \lfloor g(x) \rfloor

Now,

g(x) = \left(x - \frac{1}{2}\right)^2 + \frac{3}{4}

So $g(x)$ has minimum value at $x = \frac{1}{2}$ .

At $x = \frac{1}{2}$ ,

g\left(\frac{1}{2}\right) = \frac{3}{4}

Hence,

\left|g\left(\frac{1}{2}\right)\right| = \frac{3}{4}, \qquad \left\lfloor g\left(\frac{1}{2}\right) \right\rfloor = 0

Therefore,

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

So the absolute minimum value is $\frac{3}{4}$ . The correct option is A.

The solution labels the option as D, but its computed value is $\frac{3}{4}$ , which matches option A in the given options.

Vertex-Based Explanation

Given: $f(x) = |x^2 - x + 1| + \left\lfloor x^2 - x + 1 \right\rfloor$ on $[-1,2]$ .

Find: The least value of the function.

Since

x^2 - x + 1 = \left(x - \frac{1}{2}\right)^2 + \frac{3}{4}

we get

x^2 - x + 1 \ge \frac{3}{4}

for all $x \in [-1,2]$ .

Because $x^2 - x + 1$ is always positive here, its absolute value remains the same:

|x^2 - x + 1| = x^2 - x + 1

Also, when the minimum value is $\frac{3}{4}$ ,

\left\lfloor x^2 - x + 1 \right\rfloor = 0

at that point.

Thus at the minimizing point $x = \frac{1}{2}$ ,

f(x) = \frac{3}{4} + 0 = \frac{3}{4}

Therefore, the absolute minimum value of the function is $\frac{3}{4}$ .

Common mistakes

Students may minimize $x^2 - x + 1$ correctly but forget the floor term. This is wrong because $f(x)$ contains both $|g(x)|$ and $\lfloor g(x) \rfloor$ . After finding the minimum of $g(x)$ , evaluate both parts separately.
Students may trust the solution-page option label D without checking the option values. This is wrong because the working gives the minimum as $\frac{3}{4}$ , which corresponds to option A in the listed choices. Always match the computed value with the actual options.
A common error is to think the absolute value changes the expression into a piecewise negative-positive case here. This is wrong because $x^2 - x + 1 = \left(x-\frac{1}{2}\right)^2 + \frac{3}{4} > 0$ for all relevant $x$ , so the modulus does not alter the value.

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Answer

Step-by-step solution

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