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CBSE Board Examination 2025 · Section A · 1 mark

Q8: Absolute maximum of x3−3x+2x^3 - 3x + 2 on [0,2][0, 2]

Board 2025Ch 6 · Application of Derivatives
The question

The absolute maximum value of function f(x)=x3−3x+2f(x) = x^3 - 3x + 2 in [0,2][0, 2] is :

(A) 00

(B) 22

(C) 44

(D) 55

Exam answer
f′(x)=3x2−3=0⇒x2=1⇒x=1(x=−1∉[0,2])\begin{aligned} f'(x) &= 3x^2 - 3 = 0 \\ \Rightarrow x^2 &= 1 \\ \Rightarrow x &= 1 \quad (x = -1 \notin [0, 2]) \end{aligned}
Why?

Method for an absolute maximum on a closed interval [a,b][a, b]⁠: it occurs either at a critical point inside the interval, where f′(x)=0f'(x) = 0⁠, or at an end point. So find the critical points, then compare the values.

The derivative of x3x^3 is 3x23x^2⁠, of −3x-3x is −3-3⁠, and of the constant 2 is 0. Then solve:

3x2−3=03x2=3x2=1x=1 or x=−1\begin{aligned} 3x^2 - 3 &= 0 \\ 3x^2 &= 3 \\ x^2 &= 1 \\ x &= 1 \text{ or } x = -1 \end{aligned}

Of x=±1x = \pm 1⁠, only x=1x = 1 lies in [0,2][0, 2]⁠.

f(0)=2f(1)=1−3+2=0f(2)=8−6+2=4\begin{aligned} f(0) &= 2 \\ f(1) &= 1 - 3 + 2 = 0 \\ f(2) &= 8 - 6 + 2 = 4 \end{aligned}
Why?

Evaluate f at both end points, 0 and 2, and at the critical point 1, by putting each value into x3−3x+2x^3 - 3x + 2⁠.

f(0)=(0)3−3(0)+2=0−0+2=2f(1)=(1)3−3(1)+2=1−3+2=0f(2)=(2)3−3(2)+2=8−6+2=4\begin{aligned} f(0) &= (0)^3 - 3(0) + 2 \\ &= 0 - 0 + 2 = 2 \\[4pt] f(1) &= (1)^3 - 3(1) + 2 \\ &= 1 - 3 + 2 = 0 \\[4pt] f(2) &= (2)^3 - 3(2) + 2 \\ &= 8 - 6 + 2 = 4 \end{aligned}

Comparing 2, 0 and 4, the largest is 4.

So the absolute maximum value is 44⁠, at x=2x = 2⁠.

Why?

Watch out: the critical point x=1x = 1 gives the smallest value, 0 (the absolute minimum), not the maximum. Here the maximum is at an end point, which f′(x)=0f'(x) = 0 alone would miss.

Answer: (C) 44

Where this comes from

Chapter 6, Application of Derivatives: Absolute maximum 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 [0,2][0, 2]⁠. f′(x)=3x2−3=0f'(x) = 3x^2 - 3 = 0 gives only x=1x = 1 inside it, so compare f(0)=2f(0) = 2⁠, f(1)=0f(1) = 0 and f(2)=4f(2) = 4⁠.

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