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The maximum value of f(x) = \(\frac{x}{4+x+x^2}\) on [–1, 1] is : 

A.\(-\frac{1}{4}\)

B.\(-\frac{1}{3}\)

C. \(\frac{1}{6}\)

D. \(\frac{1}{5}\)

1 Answer

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Best answer

Option : (C)

 f(x) = \(\frac{x}{4+x+x^2}\), x ∈ [-1,1]

Differentiating f(x) with respect to x, we get

Since,

4 - x> 0 ∀ x∈[-1,1] and 

(4 + x + x2)> 0 ∀ x ∈ R 

Therefore, 

f’(x) > 0 ∀ x ∈ [-1,1] 

Hence,

f(x) is increasing in [-1,1] and therefore the maximum value of f(x) occurs at x = 1 and 

f(1) = \(\frac{1}{6}\)

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