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CBSE Board Examination 2026 · Section A · 1 mark

Q7: Least value of x3−12xx^3 - 12x on [0,3][0, 3]

Board 2026Ch 6 · Application of Derivatives
The question

The least value of f(x)=x3−12x, x∈[0,3]f(x) = x^3 - 12x,\ x \in [0, 3] is

(A) −16-16

(B) −9-9

(C) 00

(D) 1616

Exam answer

f′(x)=3x2−12=0⇒x=2f'(x) = 3x^2 - 12 = 0 \Rightarrow x = 2 (in [0,3][0, 3]⁠).

Why?

On a closed interval, the least value comes from a critical point inside the interval or from an end point. So first find where f′(x)=0f'(x) = 0⁠.

Differentiate term by term:

ddx(x3)=3x2ddx(12x)=12f′(x)=3x2−12\begin{aligned} \frac{d}{dx}(x^3) &= 3x^2 \\ \frac{d}{dx}(12x) &= 12 \\ f'(x) &= 3x^2 - 12 \end{aligned}

Set it equal to 0 and solve:

3x2−12=03x2=12x2=123=4x=±4=±2\begin{aligned} 3x^2 - 12 &= 0 \\ 3x^2 &= 12 \\ x^2 &= \frac{12}{3} = 4 \\ x &= \pm\sqrt{4} = \pm 2 \end{aligned}

Only x=2x = 2 lies in [0,3][0, 3]⁠; x=−2x = -2 is outside and is ignored.

f(0)=0f(0) = 0⁠, f(2)=8−24=−16f(2) = 8 - 24 = -16⁠, f(3)=27−36=−9f(3) = 27 - 36 = -9⁠.

Why?

Substitute into f(x)=x3−12xf(x) = x^3 - 12x⁠:

f(0)=03−12(0)=0−0=0f(2)=23−12(2)=8−24=−16f(3)=33−12(3)=27−36=−9\begin{aligned} f(0) &= 0^3 - 12(0) \\ &= 0 - 0 = 0 \\[4pt] f(2) &= 2^3 - 12(2) \\ &= 8 - 24 = -16 \\[4pt] f(3) &= 3^3 - 12(3) \\ &= 27 - 36 = -9 \end{aligned}

The end points 0 and 3 must be included, because they belong to [0,3][0, 3]⁠.

The least of 00⁠, −16-16⁠, −9-9 is −16-16⁠.

Why?

Watch out: −9-9⁠, option (B), is only the value at the end point x=3x = 3⁠. And 16=f(−2)16 = f(-2)⁠, option (D), comes from the critical point outside the interval.

Answer: (A) −16-16

Where this comes from

Chapter 6, Application of Derivatives: Absolute minimum on a closed interval.

Question type: Absolute maximum and minimum on a closed interval

How to spot it: click to see

Type 9. Absolute maximum and minimum on a closed interval

Spot it: an interval with square brackets, like [0,3][0, 3]⁠, and "maximum value" or "absolute maximum".

Here: The interval is [0,3][0, 3]⁠: f′(x)=3x2−12=0f'(x) = 3x^2 - 12 = 0 gives x=2x = 2 inside it (drop −2-2⁠). Compare f(0)=0f(0) = 0⁠, f(2)=−16f(2) = -16 and f(3)=−9f(3) = -9⁠.

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