Question

$f\left(x\right)=\frac{x^3}{3}+\frac{x^2}{2}+ax+b\forall x\in R$, then find least value of 'a' for which $f\left(x\right)$ is injective function

• A. $\frac{1}{4}$
• B. $1$
• C. $\frac{1}{2}$
• D. $\frac{1}{8}$

Answer 1

Answer: (A)

$\frac{1}{4}$

given function is $f\left(x\right)=\frac{x^3}{3}+\frac{x^2}{2}+ax+b,\forall xϵR$ and also$f\left(x\right)$ is injective.

$\therefore$ $f\left(x\right)$ should be strictly increasing or decreasing function.

1) $f\left(x\right)$ is a strictly increasing function

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

$\Rightarrow$ $x^2+x+a>0$

For a quadratic equation to be positive the coeffcient of $x^2$ should be positive and discriminant of the quadratic equation should be negitive

Here coefficient of $x^2$ is positive.

$\therefore \Delta <0$

$\Rightarrow$ $1-4a$$<0$

$\Rightarrow$ $a$$>\frac{1}{4}$

$\therefore$ the minimum value of $a$ is $\frac{1}{4}$.

2)$f\left(x\right)$ is strictly decreasing function

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

$\Rightarrow$ $x^2+x+a<0$

A quadratic equation can be negitive only if the coeffcient of $x^2$ is negitive,but here it is positive

$\therefore$ this is not a case

Hence the minimum value of $a$ is $\frac{1}{4}$.