Question

The set of values of $a$ for which the function $f:R\to R$ given by $f\left(x\right)=x^3+(a+2)x^2+3ax+5$ is one-one, is

• A. $(-\infty ,1)\cup (4,\infty )$
• B. $(-\infty ,4)$
• C. $(1,\infty )$
• D. $[1,4]$

Answer 1

The correct option is D $[1,4]$

$f\left(x\right)=x^3+(a+2)x^2+3ax+5$
For $x_1,x_2\in R$
Let, $f\left(x_1\right)=f\left(x_2\right)$
$\Rightarrow x_1^3+(a+2)x_1^2+3a{x}_1+5=x_2^3+(a+2)x_2^2+3a{x}_2+5$
$\Rightarrow x_1^3-x_2^3+(a+2)(x_1^2-x_2^2)+3a(x_1-x_2)=0$
$\Rightarrow (x_1-x_2)(x_1^2+x_2^2+x_1{x}_2+(a+2\left)\right(x_1+x_2)+3a)=0$
Since $f$ is one-one,
$\therefore x_1=x_2\cdots \left(1\right)$

Let $y=x_1^2+x_2^2+x_1{x}_2+(a+2)(x_1+x_2)+3a$
Since, $x_1=x_2=x\left(\text{say}\right)$
$\Rightarrow y=3x^2+2x(a+2)+3a$
The equation $y=0$ has two roots i.e, $x_1,x_2.$
So, the function $f\left(x\right)$ should be one-one only when
$x_1=x_2$ or $y\ne 0$

$\therefore D\le 0$
$\Rightarrow 4(a+2{)}^2-4\cdot 3\cdot 3a\le 0$
$\Rightarrow a^2-5a+4\le 0\Rightarrow (a-1)(a-4)\le 0$
$\Rightarrow a\in [1,4]$