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The function f(x) = 1 + x2 + x4 is strictly increasing for1. x < 02. x ≥ 03. x > 04. None of these |
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Answer» Correct Answer - Option 3 : x > 0 Concept:
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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