Question

For real values of x, the range of \[\dfrac{{{x}^{2}}+2x+1}{{{x}^{2}}+2x-1}\] is
\[\left( \text{a} \right)\text{ }\left( -\infty ,0 \right)\cup \left( 1,\infty \right)\]
\[\left( \text{b} \right)\text{ }\left[ \dfrac{1}{2},2 \right]\]
\[\left( \text{c} \right)\text{ }\left( -\infty ,\dfrac{-2}{9} \right]\cup \left( 1,\infty \right)\]
\[\left( \text{d} \right)\text{ }\left( -\infty ,-6 \right]\cup \left( -2,\infty \right)\]

Answer 1

Hint: To solve the given question, we will first find out what is meant by the range of any function. Then we will equate the given expression to x. Then, we will cross – multiply the given expression. After doing this, we will find the quadratic equation in x whose coefficients are in terms of y. Then we will take up two cases. In the first case, we will equate the coefficient of \[{{x}^{2}}\] to zero. From here, we will get the value of y as \[\alpha .\] Here, \[y\ne \alpha .\] Then, we will write the quadratic in y by applying \[D\ge 0.\] From here, we will get the values of y. This will be our second case. We will take out the common values of y from both the cases and that will be equal to the range of the given expression.

Complete step-by-step answer:
Before solving the question given, we will find out what meaning of a range of any function is. The range of a function is the set of all values of the dependent variable. Now, we have to find the range of the expression \[\dfrac{{{x}^{2}}+2x+1}{{{x}^{2}}+2x-1}.\] Let this range be equal to y. Thus, we have,
\[y=\dfrac{{{x}^{2}}+2x+1}{{{x}^{2}}+2x-1}\]
Now, we will cross multiply the terms. Thus, we will get,
\[{{x}^{2}}y+2xy-y={{x}^{2}}+2x+1\]
\[\Rightarrow {{x}^{2}}y-{{x}^{2}}+2xy-2x-y-1=0\]
\[\Rightarrow {{x}^{2}}\left( y-1 \right)+x\left( 2y-2 \right)-y-1=0\]
Now, we will form two cases
Case I: The coefficient of \[{{x}^{2}}\] should not be zero, so that the equation remains quadratic. Thus,
\[y-1\ne 0\]
\[\Rightarrow y\ne 1.....\left( i \right)\]
Case II: The quadratic equation formed will take all the real values, so \[D\ge 0.\] Here D is the discriminant. The discriminant of the quadratic equation \[a{{x}^{2}}+bx+c=0\] is \[{{b}^{2}}-4ac.\] Thus, we have,
\[{{\left( 2y-2 \right)}^{2}}-4\left( y-1 \right)\left( -y-1 \right)\ge 0\]
\[\Rightarrow {{\left( 2y-2 \right)}^{2}}+4\left( y-1 \right)\left( y+1 \right)\ge 0\]
\[\Rightarrow 4{{\left( y-1 \right)}^{2}}+4\left( y-1 \right)\left( y+1 \right)\ge 0\]
\[\Rightarrow 4\left[ {{\left( y-1 \right)}^{2}}+\left( y-1 \right)\left( y+1 \right) \right]\ge 0\]
Dividing the inequality by 4, we will get,
\[\Rightarrow {{\left( y-1 \right)}^{2}}+\left( y-1 \right)\left( y+1 \right)\ge 0\]
On taking (y – 1) common, we will get,
\[\Rightarrow \left( y-1 \right)\left[ \left( y-1 \right)+\left( y+1 \right) \right]\ge 0\]
\[\Rightarrow \left( y-1 \right)\left( 2y \right)\ge 0\]
\[\Rightarrow \left( y-1 \right)y\ge 0\]

\[\Rightarrow y\in \left( -\infty ,0 \right]\cup \left[ 1,\infty \right)......\left( ii \right)\]
Now, we will take the common values of y from both the cases. Thus, we will get,
\[y\in \left( -\infty ,0 \right)\cup \left( 1,\infty \right)\]
Hence, option (a) is the right answer.

Note: The alternate method of solving the question is shown below.
\[y=\dfrac{{{x}^{2}}+2x+1}{{{x}^{2}}+2x-1}\]
\[\Rightarrow y=\dfrac{{{x}^{2}}+2x+1}{\left( {{x}^{2}}+2x+1 \right)-2}\]
Now, we will use the identity \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\]
\[\Rightarrow y=\dfrac{{{\left( x+1 \right)}^{2}}}{{{\left( x+1 \right)}^{2}}-2}\]
\[\Rightarrow \dfrac{1}{y}=\dfrac{{{\left( x+1 \right)}^{2}}-2}{{{\left( x+1 \right)}^{2}}}\]
\[\Rightarrow \dfrac{1}{y}=1-\dfrac{2}{{{\left( x+1 \right)}^{2}}}\]
\[\Rightarrow \dfrac{2}{{{\left( x+1 \right)}^{2}}}=1-\dfrac{1}{y}\]
\[\Rightarrow \dfrac{1}{{{\left( x+1 \right)}^{2}}}=\dfrac{y-1}{2y}\]
Taking square root on both the sides, we will get,
\[\Rightarrow \dfrac{1}{\left( x+1 \right)}=\sqrt{\dfrac{y-1}{2y}}\]
\[\Rightarrow x+1=\sqrt{\dfrac{2y}{y-1}}\]
\[\Rightarrow x=\sqrt{\dfrac{2y}{y-1}}-1\]
Now, \[\dfrac{2y}{y-1}\] should be positive or zero and y should not be equal to 1. Thus, we have,
\[\dfrac{2y}{y-1}\ge 0\]
\[\Rightarrow \dfrac{\left( 2y \right)\left( y-1 \right)}{{{\left( y-1 \right)}^{2}}}\ge 0\]
\[\Rightarrow \left( 2y \right)\left( y-1 \right)\ge 0\]
\[\Rightarrow y\left( y-1 \right)\ge 0\]
\[\Rightarrow y\in \left( -\infty ,0 \right]\cup \left[ 1,\infty \right)\]
Thus the final answer is \[y\in \left( -\infty ,0 \right)\cup \left( 1,\infty \right).\]