Question

If $x$ is real, then greatest and least values of $\frac{[x^2-x+1]}{[x^2+x+1]}$ are.

• A. $3,\frac{-1}{2}$
• B. $3,\frac{1}{3}$
• C. $-3,\frac{1}{3}$
• D. none of these

Answer 1

The correct option is B

$3,\frac{1}{3}$

Explanation for the correct answer:

Ffind the value of $\frac{\left[x^2-x+1\right]}{\left[x^2+x+1\right]}$ :

Let,

$y=\frac{\left[x^2-x+1\right]}{\left[x^2+x+1\right]}$

Differentiate the above equation with respect to $x$ and equate to zero.

Then,

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{\left[(x^2+x+1)(2x–1)–(x^2–x+1)(2x+1)\right]}{{[x^2+x+1]}^2}\\ & \Rightarrow & \frac{2x^2–2}{{\left[x^2+x+1\right]}^2}=0\\ & \Rightarrow & 2x^2–2=0\\ & \Rightarrow & x=–1,1\end{array}$

Now,

$f(-1)=3$ and $f\left(1\right)=1/3$

So

at $x=-1$ the function will occupy maximum value.

$f(-1)=3$

and

at $x=1$ ,the function will occupy minimum value.

$f\left(1\right)=1/3$

Hence, the correct option is B.