Let f:R R be defined as f(x)=2x^2-3x+23x^2+x+3.

Question

Let $f:R	o R$ be defined as $f(x)=\frac{2{x}^{2}-3x+2}{3{x}^{2}+x+3}$. Then $f$ is :

Accepted Answer

To determine the nature of the given function, let us carefully analyze it for both the one-one (injective) and onto (surjective) properties. The function is given by:

$f(x)=\frac{2{x}^{2}-3x+2}{3{x}^{2}+x+3}$

where the domain is the set of real numbers $R$ and the codomain is also $R$.

Checking if the function is one-one

A function $f:R	o R$ is said to be one-one if every element in the domain has a distinct image in the codomain. That is, if $f({x}_{1})=f({x}_{2})$, then we must have ${x}_{1}={x}_{2}$.

Let us check if we can find two different real values of $x$ that give the same value for $f(x)$. Let us set $f(x)=1$ and solve for $x$:

$\frac{2{x}^{2}-3x+2}{3{x}^{2}+x+3}=1$

Cross-multiplying, we obtain:

$2{x}^{2}-3x+2=3{x}^{2}+x+3$

Bringing all terms to one side, we get:

${x}^{2}+4x+1=0$

Using the quadratic formula to find the roots:

$x=\frac{-4\pm \sqrt{16-4(1)(1)}}{2}=\frac{-4\pm \sqrt{12}}{2}=-2\pm \sqrt{3}$

Since we found two distinct real values, ${x}_{1}=-2+\sqrt{3}$ and ${x}_{2}=-2-\sqrt{3}$, such that $f({x}_{1})=f({x}_{2})=1$, the function is not one-one.

Checking if the function is onto

A function is onto if its range is equal to its codomain. Here, the codomain is $R$. To find the range, let us set $f(x)=y$:

$y=\frac{2{x}^{2}-3x+2}{3{x}^{2}+x+3}$

Cross-multiplying gives us:

$y(3{x}^{2}+x+3)=2{x}^{2}-3x+2$

$3y{x}^{2}+yx+3y=2{x}^{2}-3x+2$

Grouping the terms by the powers of $x$, we form a quadratic equation in $x$:

$(3y-2){x}^{2}+(y+3)x+(3y-2)=0$

Since $x$ is a real number, the discriminant ($D$) of this quadratic equation must be greater than or equal to zero ($D\ge 0$).

$D=(y+3{)}^{2}-4(3y-2)(3y-2)\ge 0$

$(y+3{)}^{2}-4(3y-2{)}^{2}\ge 0$

This is in the form of ${a}^{2}-{b}^{2}\ge 0$, which can be factored as $(a-b)(a+b)\ge 0$. Here, $a=y+3$ and $b=2(3y-2)=6y-4$:

$[(y+3)-(6y-4)][(y+3)+(6y-4)]\ge 0$

$(-5y+7)(7y-1)\ge 0$

To solve this inequality, we can multiply by $-1$ (which reverses the inequality sign):

$(5y-7)(7y-1)\le 0$

The critical points are $y=\frac{1}{7}$ and $y=\frac{7}{5}$. Based on the inequality, $y$ must lie between these two values:

$\frac{1}{7}\le y\le \frac{7}{5}$

This means the range of the function is the closed interval $[\frac{1}{7},\frac{7}{5}]$.

Because the range is a small subset of the real numbers and not equal to the entire codomain $R$, the function is not onto.

Conclusion:

Since $f(x)$ is neither one-one nor onto, the correct statement is given by Option D.

Option D

Solutions · 1