PUC Karnataka Science Class 12Department of Pre-University Education, Karnataka
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# 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 _______________ .

• $\frac{1}{4}$

• $- \frac{1}{3}$

• $\frac{1}{6}$

• $\frac{1}{5}$

#### Solution

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

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

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

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

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

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

$\Rightarrow 4 - x + x^2 + x - 2 x^2 = 0$

$\Rightarrow x^2 = 4$

$\Rightarrow x =\pm 2 \not\in \left( - 1, 1 \right)$

$\text { So,}$

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

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

$\text { Hence, the maximum value is } \frac{1}{4} .$

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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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