Question

If $2x-7-5x^2$ has maximum value at $x=a$, then $a=..........$

• A. $-1/5$
• B. $1/5$
• C. $34/5$
• D. $-34/5$

Answer 1

The correct option is B $1/5$

$\Rightarrow -\left[\right(\sqrt{5}x{)}^2+2\left(\sqrt{5}x\right)\times \left(\frac{1}{5}\right)+\left(\frac{1}{\sqrt{5}}{)}^2\right]-(\frac{1}{\sqrt{5}}{)}^2+7$

$\Rightarrow -{(\sqrt{5}x+\frac{1}{\sqrt{5}})}^2+\frac{1}{5}-7$

$\Rightarrow \frac{34}{5}-{(\sqrt{5}x+\frac{1}{\sqrt{5}})}^2$

As ${(\sqrt{5}x+\frac{1}{\sqrt{5}})}^2\ge 0$

$\therefore$ max. value $e$ at

$\sqrt{5a}+\frac{1}{\sqrt{5}}=0\Rightarrow a=\frac{1}{5}$