The function {f}(x)=[x(x-2)]² is increasing in the set

Correct Answer is : (0,1)∪(2,∞)(0,1) (2, ∞)(0,1)∪(2,∞)

Correct Answer is : (0,1)∪(2,∞)(0,1) (2, ∞)(0,1)∪(2,∞)

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MHTCET 26 Apr 2025 Shift 2 Paper

Question : 106 of 150

Marks: +1, -0

The function

\mathrm{f}(x)=[x(x-2)]^{2}

is increasing in the set

Solution:

Concept:

A function is increasing when its derivative is positive.

Explanation:

Given

f(x) = [x(x-2)]^2 = x^2(x-2)^2

Differentiate using the product rule.

f'(x) = 2x(x-2)^2 + 2(x-2)x^2

= 2x(x-2)(x-2+x) = 4x(x-1)(x-2)

For

f(x)

to be increasing,

f'(x) > 0

4x(x-1)(x-2) > 0

, which simplifies to

x(x-1)(x-2) > 0

Using the sign scheme for critical points

0

1

, and

2

, the product is positive for

x \in (0,1) \cup (2, \infty)

Answer:

x \in (0,1) \cup (2, \infty)

, i.e. option D.

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