The maximum value of ' $a$ ' such that the second derivative of $x^4+a x^3+\frac{3 x^2}{2}+1$ is positive for all real $x$ is

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Aicharya · Accepted Answer

$\begin{aligned} & ext { } \because f(x)=x^4+a x^3+\frac{3 x^2}{2}+1 \ & \Rightarrow f^{\prime \prime}(x)=12 x^2+6 a x+3 \ & \because f^{\prime \prime}(x)>0 \Rightarrow 4 x^2+3 a x+1>0 \ & \Rightarrow\left(2 x+\frac{1}{2} a ight)^2+\left(1-\frac{a^2}{4} ight)>0 \ & \Rightarrow 1-\frac{a^2}{4}>0 \ & \Rightarrow \frac{a^2}{4} < 1 \Rightarrow a^2 < 4 \Rightarrow-2 < a < 2 \ & herefore ext { Maximum value of a is } 2 .\end{aligned}$

The maximum value of ' $a$ ' such that the second derivative of $x^4+a x^3+\frac{3 x^2}{2}+1$ is positive…

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