PUC Karnataka Science Class 12Department of Pre-University Education, Karnataka
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# Solution for The Maximum Value of F(X) = X 4 + X + X 2 on [ − 1,1] is (A) − 1 4 (B) − 1 3 (C) 1 6 (D) 1 5 - PUC Karnataka Science Class 12 - Mathematics

#### Question

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) $\frac{1}{5}$

#### Solution

(c) $\frac{1}{6}$

$\text { Given }: f\left( x \right) = \frac{x}{4 + x + x^2}$

$\Rightarrow f'\left( x \right) = \frac{4 + x + x^2 - x\left( 1 + 2x \right)}{\left( 4 + x + x^2 \right)^2}$

$\text { For a local maxima or a local minima, we must have }$

$f'\left( x \right) = 0$

$\Rightarrow \frac{4 + x + x^2 - x\left( 1 + 2x \right)}{\left( 4 + x + x^2 \right)^2} = 0$

$\Rightarrow 4 + x + x^2 - x\left( 1 + 2x \right) = 0$

$\Rightarrow 4 - x^2 = 0$

$\Rightarrow x = \pm 2 \not\in \left[ - 1, 1 \right]$

$\text { The values of } f\left( x \right) \text { at extreme points are given by }$

$f\left( 1 \right) = \frac{1}{4 + 1 + 1^2} = \frac{1}{6}$

$f\left( - 1 \right) = \frac{- 1}{4 - 1 + \left( - 1 \right)^2} = \frac{- 1}{4}$

$\text{Thus,}\frac{1}{6}\text{ is the maximum value }$ .

Is there an error in this question or solution?

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Solution The Maximum Value of F(X) = X 4 + X + X 2 on [ − 1,1] is (A) − 1 4 (B) − 1 3 (C) 1 6 (D) 1 5 Concept: Graph of Maxima and Minima.
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