If x - R- and -x2-1-x4 then A- 0 - 2B- 0 - 1C- 0 -1-4D- 0 -1-2

The correct option is D 0 ≤α≤1/2For real x, we have x^2 / (1 + x^4) ≥ 0 Also, 1 + x^4 2x^2 = (1 x^2)^2 ≥ 0 ⇒ 1 + x^4 ≥ 2x^2 ∴ ( x^2/1+x^4 ) ≤1/2 So 0 ≤x^2 ...

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Quantitative Aptitude

Solving Inequalities

If x ∈ R, and...

Question

If x $ϵ$ R, and $α = \frac{x^{2}}{(1 + x^{4})}$ then

0 \leq α \leq 2

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0 \leq α \leq 1

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0 \leq α \leq \frac{1}{4}

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0 \leq α \leq \frac{1}{2}

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Solution

The correct option is D $0 \leq α \leq \frac{1}{2}$
For real x, we have $x^{2} / (1 + x^{4}) \geq 0$
Also, $1 + x^{4} - 2 x^{2} = (1 - x^{2})^{2} \geq 0 \Rightarrow 1 + x^{4} \geq 2 x^{2} ∴ (\frac{x^{2}}{1 + x^{4}}) \leq \frac{1}{2} S o 0 \leq \frac{x^{2}}{1 + x^{4}} \leq \frac{1}{2} \Rightarrow 0 \leq α \leq \frac{1}{2}$

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If x ϵ R, and α=x2(1+x4) then

The correct option is D 0≤α≤12For real x, we have x2/(1+x4)≥0 Also, 1+x4−2x2=(1−x2)2≥0⇒1+x4≥2x2∴(x21+x4)≤12So0≤x21+x4≤12⇒0≤α≤12

If x $ϵ$ R, and $α = \frac{x^{2}}{(1 + x^{4})}$ then