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The maximum possible value of $y = \min\left ( 1/2-3x^{2}/4,5x^{2}/4 \right )$ for the range $0<x<1$ is

1. $1/3$
2. $1/2$
3. $5/27$
4. $5/16$

Given that,

$y = \text{min}( \frac{1}{2} – \frac{3x^{2}}{4}, \frac{5x^{2}}{4}) ; 0<x<1 \longrightarrow (1)$

Minimum values ,we can get when

$\frac{1}{2} – \frac{3x^{2}}{4} =\frac{5x^{2}}{4}$

$\Rightarrow \frac{8x^{2}}{4} = \frac{1}{2}$

$\Rightarrow 2x^{2} = \frac{1}{2}$

$\Rightarrow x^{2} = \frac{1}{4}$

$\Rightarrow x = \sqrt \frac{1}{4}$

$\Rightarrow \boxed{x = \frac{-1}{2}(rejected) \; or \; x= \frac{1}{2}}$

$\Rightarrow \boxed{ x= \frac{1}{2}}$

Now,  $y = \text{min}( \frac{1}{2} – \frac{3x^{2}}{4}, \frac{5x^{2}}{4})$

$\Rightarrow y = \text{min} \left[ \frac{1}{2} – \frac{3(\frac{1}{2})^{2}}{4}, \frac{5(\frac{1}{2})^{2}}{4}\right]$

$\Rightarrow y= \text{min} \left[\frac{1}{2} – \frac{3}{16} , \frac{5}{16}\right]$

$\Rightarrow y= \text{min} (\frac{8-3}{16} , \frac{5}{16})$

$\Rightarrow y= \text{min} (\frac{5}{16} , \frac{5}{16})$

$\Rightarrow \boxed{y= \frac{5}{16}}$

$\therefore$ The minimum possible value of  $y = \text{min}( \frac{1}{2} – \frac{3x^{2}}{4}, \frac{5x^{2}}{4})$ is $\frac{5}{16}$

Correct Answer $:\text{D}$
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