Question

Consider the function f(x)= $\frac{Inx}{8}-ax+x^2$ and $a\ge 0$ is a real constant. f(x) gives a local maxima at

• A. $a=1,x=\frac{1}{4}$
• B. $a=1;x=\frac{a-\sqrt{a^2-1}}{4}$
• C. $0\le a<1$
• D. $a=1;x=\frac{a-\sqrt{a^2+1}}{4}$

Answer 1

The correct option is A $a=1,x=\frac{1}{4}$

We have,

$f\left(x\right)=\frac{\ln x}{8}-ax+x^2$

Differentiation this equation with respect to x and we get,

$f^{{}^{\prime }}\left(x\right)=\frac{1}{8x}-a+2x$

For maximum and minimum that,

$f^{\prime }\left(x\right)=0$

$\frac{1}{8x}-a+2x=0$

$1-ax+16x^2=0$

$16x^2-ax+1=0$

Solving by quadratic formula and we get,

$x=\frac{a\pm \sqrt{a^2-64}}{32}$

Again differentiation and we get,

$f^{\prime \prime }\left(x\right)=-\frac{1}{8x^2}-0+2$

$f^{\prime \prime }\left(x\right)=-\frac{1}{8x^2}+2$

Put the value of x in this equation and we get,

$f^{\prime \prime }\left(x\right)=-\frac{1}{8{\left(\frac{a\pm \sqrt{a^2-64}}{32}\right)}^2}+2$

It is negative sign then,

It is maximum.