Question

Given that, for all the real values of x, the expression ${\displaystyle \frac{x^2-2x+4}{x^2+2x+4}}$ lies between ${\displaystyle \frac{1}{3}}\text{and }3$. The values between which the expression ${\displaystyle \frac{{9.3}^{2x}-{6.3}^x+4}{{9.3}^{2x}+{6.3}^x+4}}$ lies are
a. 0 and 2
b. -1 and 1
c. -1 and 0
d. ${\displaystyle \frac{1}{3}}\text{and }3$

Answer 1

Hint: We will first assume $y={\displaystyle \frac{x^2-2x+4}{x^2+2x+4}}$ and on further solving this, we get an equation, $x^2(1-y)-2x(1+y)+4-4y=0$ which is a quadratic equation and we know that discriminant of quadratic equation $b^2-4ac$ must be greater than 0 for any real value of roots. On solving the discriminant, we get to know that the range of $y\in [{\displaystyle \frac{1}{3}},3]$ as its independent of x. Then, in the final step, we will write, ${\displaystyle \frac{{9.3}^{2x}-{6.3}^x+4}{{9.3}^{2x}+{6.3}^x+4}}$ as ${\displaystyle \frac{{\left({3.3}^x\right)}^2-2{\left(3.3\right)}^x+4}{{\left({3.3}^x\right)}^2+2{\left(3.3\right)}^x+4}}$ and find the range of the expression.

Complete step-by-step solution:
It is given in the question that for all the real values of x, the expression ${\displaystyle \frac{x^2-2x+4}{x^2+2x+4}}$ lies between ${\displaystyle \frac{1}{3}}\text{and }3$. Then we have to find the values between which the expression ${\displaystyle \frac{{9.3}^{2x}-{6.3}^x+4}{{9.3}^{2x}+{6.3}^x+4}}$ lies.
Let us first assume the expression $y={\displaystyle \frac{x^2-2x+4}{x^2+2x+4}}$. On cross multiplying both the sides, we get,
$\begin{array}{rl}& y(x^2+2x+4)=x^2-2x+4\\ & x^{2}y+2xy+4y=x^2-2x+4\end{array}$
On transposing all the terms on the LHS to the RHS, we get,
$\begin{array}{rl}& x^2-2x+4-x^{2}y-2xy-4y\\ & x^2-x^{2}y-2x-2xy+4-4y\\ & x^2(1-y)-2x(1+y)+4(1-y)=0\end{array}$
Now, it is a quadratic equation and we know that in a quadratic equation, the discriminant $b^2-4ac$ must be greater than 0 for any real value of roots.
So, from the quadratic equation, we have a = (1 - y), b = -2 (1 + y) and c = 4 (1 – y).
So, we have the discriminant, $d=b^2-4ac$ and we know that d must be greater than 0, so we can write,
$\begin{array}{rl}& d=b^2-4ac\ge 0\\ & ={[-2(1+y)]}^2-4\left[(1-y)\left(4\right)(1-y)\right]\ge 0\\ & =4{(1+y)}^2-16{(1-y)}^2\ge 0\end{array}$
On taking 4 common from both the terms, we get,
$\begin{array}{rl}& =4[{(1+y)}^2-4{(1-y)}^2]\ge 0\\ & =[{(1+y)}^2-4{(1-y)}^2]\ge 0\end{array}$
We know that ${(a+b)}^2=a^2+b^2+2ab$ and ${(a-b)}^2=a^2+b^2-2ab$, so we get,
$\begin{array}{rl}& =1^2+y^2+2\left(1\right)\left(y\right)-4[1^2+y^2-2\left(1\right)\left(y\right)]\ge 0\\ & =1+y^2+2y-4[1+y^2-2y]\ge 0\end{array}$
On opening the bracket, we get,
$=1+y^2+2y-4-4y^2+8y\ge 0$
On further simplifying, we get,
$=-3y^2+10y-3\ge 0$
On multiplying both the sides by (-1), we get,
$=3y^2-10y+3\le 0$
Here, the sign of the inequality changes as we have multiplied the expression with a negative number. Now, we can split -10y as -9y –y, so we get,
$\begin{array}{rl}& =3y^2-9y-y+3\le 0\\ & =3y(y-3)-(y-3)\le 0\\ & =(3y-1)(y-3)\le 0\\ & (3y-1)=0or(y-3)=0\\ & y={\displaystyle \frac{1}{3}}or3\end{array}$
Now, we will plot ${\displaystyle \frac{1}{3}}\text{and }3$ on the number line.

If we put 0 in $(3y-1)(y-3)\le 0$, we get, (-1) (-3) = 3, and it is a positive term.
Again, if we put 1 in $(3y-1)(y-3)\le 0$, we get, (2) (-2) = -4, and it is a negative term.
And we put 4 in $(3y-1)(y-3)\le 0$, we get, (11) (1) = 11, and it is a positive term.
Hence, we can show the number line as,
Thus, we get a region $[{\displaystyle \frac{1}{3}},3]$ where the range is $$y\in [{\displaystyle \frac{1}{3}},3]$$.
Also, ${\displaystyle \frac{x^2-2x+4}{x^2+2x+4}}\in [{\displaystyle \frac{1}{3}},3]$
This means that the range does not depend on x, but it depend on y.
Now, we have been given the expression, ${\displaystyle \frac{{9.3}^{2x}-{6.3}^x+4}{{9.3}^{2x}+{6.3}^x+4}}$. So, we can write,
$\begin{array}{rl}& {\displaystyle \frac{{9.3}^{2x}-{6.3}^x+4}{{9.3}^{2x}+{6.3}^x+4}}={\displaystyle \frac{3^2{.3}^{2x}-2\left({3.3}^x\right)+4}{3^2{.3}^{2x}+2\left({3.3}^x\right)+4}}\\ & ={\displaystyle \frac{{\left({3.3}^x\right)}^2-2{\left(3.3\right)}^x+4}{{\left({3.3}^x\right)}^2+2{\left(3.3\right)}^x+4}}\end{array}$
Let us assume that $\left({3.3}^x\right)$ equal to u, so we get,
${\displaystyle \frac{{\left(u\right)}^2-2u+4}{{\left(u\right)}^2+2u+4}}$
Now, if we compare ${\displaystyle \frac{{\left(u\right)}^2-2u+4}{{\left(u\right)}^2+2u+4}}$ and ${\displaystyle \frac{x^2-2x+4}{x^2+2x+4}}$, we get to know that both are similar. We also know that the range of the expression will depend upon y and not x, so it will also not depend on u.
Thus, the range of the expression, ${\displaystyle \frac{{9.3}^{2x}-{6.3}^x+4}{{9.3}^{2x}+{6.3}^x+4}}$ will lie between $[{\displaystyle \frac{1}{3}},3]$.
Hence, option (d) is the correct answer.

Note: Most of the students make mistake after multiplying the expression, $-3y^2+10y-3\ge 0$ with (-1). They may directly write it as $3y^2-10y+3\ge 0$, but this is wrong and will lead to wrong answers. After multiplying the expression with (-1), the sign of the inequality should change and the expression should be $3y^2-10y+3\le 0$. This question has high chances of calculation errors, so the students are advised to solve this question step by step and carefully.