Question:

The function f(x)=x2−2xf(x) = x^2 - 2x is strictly decreasing in the interval.

Updated On: Feb 1, 2024
  • (-∞, -1)
  • (-1, ∞)
  • (-∞, 1)
  • (1, ∞)
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The Correct Option is C

Solution and Explanation

f(x)=x2−2xf(x) = x^2-2x
The function f(x)=x2−2xf(x) = x^2 - 2x is strictly decreasing if
f′(x)<0f'(x) < 0
ddx(x2−2x)<0\frac {d}{dx}(x^2-2x) < 0
2x−2<02x-2 < 0
2x<22x < 2
x<1x <1
⇒x∈(−∞,1)⇒ x ∈ (-∞, 1)

So, the correct option is (C): (−∞,1)(-∞, 1)
 

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Concepts Used:

Increasing and Decreasing Functions

Increasing Function:

On an interval I, a function f(x) is said to be increasing, if for any two numbers x and y in I such that x < y, 

⇒ f(x) ≤ f(y)

Decreasing Function:

On an interval I, a function f(x) is said to be decreasing, if for any two numbers x and y in I such that x < y,

⇒ f(x) ≥ f(y)

Strictly Increasing Function:

On an interval I, a function f(x) is said to be strictly increasing, if for any two numbers x and y in I such that x < y,

⇒ f(x) < f(y)

Strictly Decreasing Function:

On an interval I, a function f(x) is said to be strictly decreasing, if for any two numbers x and y in I such that x < y,

⇒ f(x) > f(y)

Graphical Representation of Increasing and Decreasing Functions