The range of the function fx=x2x4+1 is

fx=x2x4+1Let y=x2x4+1⇒ x4y+y=x2⇒ x22y=x2+y=0x2=1±1-4y22y is defined iff y≠0 and 1-4y2≥0 ⇒4y2-1≤0 ⇒(2y-1)(2y+1)≤0 ⇒y∈-12,12 but y≥0 ⇒0≤y≤12

The range of the function fx=x2x4+1 is

fx=x2x4+1Let y=x2x4+1⇒ x4y+y=x2⇒ x22y=x2+y=0x2=1±1-4y22y is defined iff y≠0 and 1-4y2≥0 ⇒4y2-1≤0 ⇒(2y-1)(2y+1)≤0 ⇒y∈-12,12 but y≥0 ⇒0≤y≤12

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$T h e r a n g e o f t h e f u n c t i o n f (x) = \frac{x^{2}}{x^{4} + 1} i s$

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$(0, \frac{1}{2})$

$[0, \frac{1}{2}]$

$[0, \infty)$

$[0, 2]$

answer is B.

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Detailed Solution

$\begin{array}{l} f (x) = \frac{x^{2}}{x^{4} + 1} \\ L e t y = \frac{x^{2}}{x^{4} + 1} \\ \Rightarrow x^{4} y + y = x^{2} \\ \Rightarrow {(x^{2})}^{2} y = x^{2} + y = 0 \end{array}$

$x^{2} = \frac{1 \pm \sqrt{1 - 4 y^{2}}}{2 y} i s d e f i n e d i f f y \neq 0 a n d 1 - 4 y^{2} \geq 0 \Rightarrow 4 y^{2} - 1 \leq 0 \Rightarrow (2 y - 1) (2 y + 1) \leq 0 \Rightarrow y \in [\frac{- 1}{2}, \frac{1}{2}] b u t y \geq 0 \Rightarrow 0 \leq y \leq \frac{1}{2}$