Question

The least positive integral value of ${}^{\prime }{\lambda }^{\prime }$ for which $f\left(x\right)=\frac{3x^3}{2}+\frac{\lambda x^2}{3}+x+7$ has a point of maxima

• A. $6$
• B. $7$
• C. $8$
• D. $10$

Answer 1

The correct option is B $7$

$f\left(x\right)=\frac{3}{2}{x}^3+\frac{\lambda x^2}{3}+x+7$
$\Rightarrow f^{\prime }\left(x\right)=\frac{9}{2}{x}^2+\frac{2\lambda x}{3}+1$
or, $f^{\prime }\left(x\right)=\frac{1}{6}(27x^2+4\lambda x+6)$
For $f\left(x\right)$ to have maxima, $D>0\Rightarrow 16{\lambda }^2-\mathrm{4.6.27}>0$
$\Rightarrow 2{\lambda }^2-81>0\Rightarrow |\sqrt{2\lambda }|>9\Rightarrow |\lambda |>\frac{9}{\sqrt{2}}$
$\therefore$ least integral value of $\lambda$ is $7$.