Question

If function $f\left(x\right)=2x^2+3x-m\log x$ is monotonic decreasing in the interval $(0,1)$, then the least value of the parameter $m$ is-

• A. $7$
• B. $\frac{15}{2}$
• C. $\frac{31}{4}$
• D. $8$

Answer 1

The correct option is A $7$

Now
$f^{\prime }\left(x\right)<0$ for $xϵ(0,1)$
Hence
$4x+3-\frac{m}{x}<0$
$4x^2+3x-m<0$
Let $x=0$, we get
$-m<0$
Or
$m>0$
And let $x=1$, we get
$7-m<0$
$m>7$
Hence from i and ii, we get
$m>7$
Thus the least value is $7.$