The function f(x) = 1 + x2 + x4  is strictly increasing for

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Navik GD Mathematics 22 March 2021 (All Shifts) Questions
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  1. x < 0
  2. x ≥ 0
  3. x > 0
  4. None of these

Answer (Detailed Solution Below)

Option 3 : x > 0
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Detailed Solution

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Concept:

  • If f′(x) > 0 then the function is said to be strictly increasing.
  • If f′(x) < 0 then the function is said to be decreasing.

 

Calculation:

Given: f(x) = 1 + x2 + x4

Differentiating with respect to x, we get

⇒ f'(x) = 0 + 2x + 4x3

⇒ f'(x) = 2x + 4x3

For strictly increasing function, f'(x) > 0

⇒ 2x + 4x3 > 0

⇒ 2x(1 + 2x2) > 0

As we know, x2 ≥ 0,  x ∈ R

So, 1 + 2x2 > 0, x ∈ R

Now, 2x > 0

⇒ x > 0

Hence function is strictly increasing for x > 0

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