The maximum possible value of fx-1-x2x2-4 is

Let nbsp;y=f(x)=1 x2x2+4⇒yx2+4=1 x2⇒yx2+4y=1 x2⇒yx2+x2=1 4y⇒x2y+1=1 4y⇒x2=1 4yy+1We nbsp;know nbsp;that nbsp;x2≥0.⇒1 4yy+1≥0⇒4y 1y+1≤0 ∴ 1 lt;x≤14 n ...

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The maximum p...

Question

The maximum possible value of $f (x) = \frac{1 - x^{2}}{x^{2} + 4}$ is

\frac{1}{3}

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\frac{1}{2}

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\frac{1}{4}

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Solution

The correct option is D $\frac{1}{4}$
$Let y = f (x) = \frac{1 - x^{2}}{x^{2} + 4} \Rightarrow y (x^{2} + 4) = 1 - x^{2} \Rightarrow {yx}^{2} + 4 y = 1 - x^{2} \Rightarrow {yx}^{2} + x^{2} = 1 - 4 y \Rightarrow x^{2} (y + 1) = 1 - 4 y \Rightarrow x^{2} = \frac{1 - 4 y}{y + 1} We know that x^{2} \geq 0 . \Rightarrow \frac{1 - 4 y}{y + 1} \geq 0 \Rightarrow \frac{4 y - 1}{y + 1} \leq 0$
$∴ - 1 < x \leq \frac{1}{4} a s x \neq - 1$
The maximum value of f(x) is $\frac{1}{4}$

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The maximum possible value of f(x)=1−x2x2+4 is

The correct option is D 14Let y=f(x)=1−x2x2+4⇒y(x2+4)=1−x2⇒yx2+4y=1−x2⇒yx2+x2=1−4y⇒x2(y+1)=1−4y⇒x2=1−4yy+1We know that x2≥0.⇒1−4yy+1≥0⇒4y−1y+1≤0 ∴−1<x≤14 as x≠−1 The maximum value of f(x) is 14

The maximum possible value of $f (x) = \frac{1 - x^{2}}{x^{2} + 4}$ is