Question

For what value of $a,f\left(x\right)=-x^3+4a{x}^2+2x-5$ is decreasing for all $x$?

• A. $(1,2)$
• B. $(3,4)$
• C. $R$
• D. No value of $a$

Answer 1

The correct option is D No value of $a$

$\because f\left(x\right)=-x^3+4a{x}^2+2x-5$

$\because f^{\prime }\left(x\right)=-3x^2+8ax+2$

Since $f\left(x\right)$ is decreasing,for all $x,$ therefore $f^{\prime }\left(x\right)=<0$

$\Rightarrow -3x^2+8ax+2<0$

$\Rightarrow 3x^2-8ax-2>0$

Now, $D={(-8a)}^2-4\left(3\right)(-2)<0$

$\Rightarrow 64a^2+24<0$

$\Rightarrow a^2<-\frac{3}{8}$

Hence, no value of $a$ exists.