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### Course: Class 12 math (India)>Unit 1

Lesson 8: Solutions to select NCERT problems

# Select problems from exercise 1.3

solutions to selected NCERT problems
In this article we will look at solutions of a few selected problems from exercise 1.3 of NCERT.
Problem 1:
Show that $f:\left[-1,1\right]\to \mathbb{R}$ given by $f\left(x\right)=\frac{x}{x+2}$ is one-one. Find the inverse of the function .
(Hint: For , $y=f\left(x\right)=\frac{x}{x+2}$, for some $x$ in $\left[-1,1\right]$, i.e. $x=\frac{2y}{\left(1-y\right)}$ )
Solution:
Let's first manipulate the function into a form more easier to read.
$\begin{array}{rl}f\left(x\right)& =\frac{x}{x+2}\\ \\ & =\frac{x+2-2}{x+2}\\ \\ & =\frac{x+2}{x+2}-\frac{2}{x+2}\\ \\ & =1-\frac{2}{x+2}\end{array}$
Note by this manipulation, we now have $x$ at only one place instead of two in the expression.
Now, we will show that $f$ is one-one. Let ${x}_{1}$ and ${x}_{2}$ be two different inputs to $f$ having the same output.
This means the two inputs are the same and our initial assumption is false. For the same output, we cannot have two different inputs. $f$ is one-one.
Let us now find the inverse of $f$. The answer is already given in the question hint but we will look at it in a bit more detail here.
try it out
The first question is, what is the range of $f$? Remember $f:\left[-1,1\right]\to \mathbb{R}$, $f\left(x\right)=1-\frac{2}{x+2}$.

So our function is $f:\left[-1,1\right]\to \left[-1,\frac{1}{3}\right]$.
Let $y=f\left(x\right)=1-\frac{2}{x+2}$.
To find the inverse function, we simply express $x$ in terms of $y$.
$\begin{array}{rl}y& =1-\frac{2}{x+2}\\ \\ \frac{2}{x+2}& =1-y\\ \\ x+2& =\frac{2}{1-y}\\ \\ x& =\frac{2}{1-y}-2\\ \\ x& =\frac{2y}{1-y}\end{array}$
That is it. ${f}^{-1}\left(y\right)=\frac{2y}{1-y}$.
We had $f:\left[-1,1\right]\to \left[-1,\frac{1}{3}\right]$. Can you show that for $-1\le y\le \frac{1}{3}$, $-1\le \frac{2y}{1-y}\le 1$ ?
Problem 2:
Consider $f:{\mathbb{R}}_{+}\to \left[-5,\mathrm{\infty }\right)$ given by $f\left(x\right)=9{x}^{2}+6x-5$. Show that $f$ is invertible with ${f}^{-1}\left(y\right)=\frac{\sqrt{y+6}-1}{3}$ .
Solution:
Let's understand the behaviour of $f\left(x\right)$ better by writing it in a perfect square form.
$f\left(x\right)=9{x}^{2}+6x-5=\left(3x+1{\right)}^{2}-6$
Our domain is ${\mathbb{R}}_{+}$ or $\left(0,\mathrm{\infty }\right)$. See that as $0, $-5. Here is the graph.
Clearly $f\left(x\right)$ is one-one and onto. So, $f$ is invertible.
Let $y=f\left(x\right)=\left(3x+1{\right)}^{2}-6$. To find the inverse, we simply express $x$ in terms of $y$.
$\begin{array}{rl}y& =\left(3x+1{\right)}^{2}-6\\ \\ \left(3x+1{\right)}^{2}& =y+6\\ \\ 3x+1& =\sqrt{y+6}\\ \\ x& =\frac{\sqrt{y+6}-1}{3}\end{array}$
In the calculation above, we took only the positive square root, because $3x+1$ is positive as $x\in {\mathbb{R}}_{+}$.
Finally, ${f}^{-1}\left(y\right)=\frac{\sqrt{y+6}-1}{3}$ .

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