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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 : [- 1, 1] \to R

given by

f (x) = \frac{x}{x + 2}

is one-one. Find the inverse of the function

f : [- 1, 1] \to Range f

(Hint: For

y \in Range f

y = f (x) = \frac{x}{x + 2}

, for some

x

[- 1, 1]

, i.e.

x = \frac{2 y}{(1 - y)}

)

Solution:

Let's first manipulate the function into a form more easier to read.

\begin{aligned} f (x) & = \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{aligned}

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.

\begin{aligned} 1 - \frac{2}{x_{1} + 2} & = 1 - \frac{2}{x_{2} + 2} \\ \frac{2}{x_{1} + 2} & = \frac{2}{x_{2} + 2} \\ for x_{1} & \neq - 2 and x_{2} \neq - 2 we can write \\ x_{1} + 2 & = x_{2} + 2 \\ x_{1} & = x_{2} \end{aligned}

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 : [- 1, 1] \to R

f (x) = 1 - \frac{2}{x + 2}

So our function is

f : [- 1, 1] \to [- 1, \frac{1}{3}]

Let

y = f (x) = 1 - \frac{2}{x + 2}

To find the inverse function, we simply express

x

in terms of

y

\begin{aligned} 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{2 y}{1 - y} \end{aligned}

That is it.

f^{- 1} (y) = \frac{2 y}{1 - y}

We had

f : [- 1, 1] \to [- 1, \frac{1}{3}]

. Can you show that for

- 1 \leq y \leq \frac{1}{3}

- 1 \leq \frac{2 y}{1 - y} \leq 1

Problem 2:

Consider

f : R_{+} \to [- 5, \infty)

given by

f (x) = 9 x^{2} + 6 x - 5

. Show that

f

is invertible with

f^{- 1} (y) = \frac{\sqrt{y + 6} - 1}{3}

Solution:

Let's understand the behaviour of

f (x)

better by writing it in a perfect square form.

f (x) = 9 x^{2} + 6 x - 5 = (3 x + 1)^{2} - 6

Our domain is

R_{+}

(0, \infty)

. See that as

0 < x < + \infty

- 5 < f (x) < + \infty

. Here is the graph.

Clearly

f (x)

is one-one and onto. So,

f

is invertible.

Proof that $f$ ‍ is one-one

Let

x_{1}

and

x_{2}

be two different inputs to

f

having the same output.

\begin{aligned} (3 x_{1} + 1)^{2} - 6 & = (3 x_{2} + 1)^{2} - 6 \\ (3 x_{1} + 1)^{2} & = (3 x_{2} + 1)^{2} \\ 3 x_{1} + 1 & = 3 x_{2} + 1 or 3 x_{1} + 1 = - (3 x_{2} + 1) \\ x_{1} & = x_{2} or x_{1} = - x_{2} - \frac{2}{3} \end{aligned}

Now remember,

x_{1}, x_{2} \in R_{+}

. So,

x_{1} = - x_{2} - \frac{2}{3}

is not a possible solution because in this case, both

x_{1}

and

x_{2}

cannot be positive at the same time. So,

x_{1} = x_{2}

meaning inputs are the same. Hence,

f

is one-one.

Proof that $f$ ‍ is onto

f : R_{+} \to [- 5, \infty)

Pick any

y \in [- 5, \infty)

. Take

x = \frac{\sqrt{y + 6} - 1}{3}

Because

y

can be any real number between

- 5

and

+ \infty

x

can be any real number between

0

and

+ \infty

. So

x \in R_{+}

. Remember

f (x) = (3 x + 1)^{2} - 6

\begin{aligned} f (x) & = f (\frac{\sqrt{y + 6} - 1}{3}) \\ = {[3 (\frac{\sqrt{y + 6} - 1}{3}) + 1]}^{2} - 6 \\ = y + 6 - 6 \\ = y \end{aligned}

Hence for every

y \in [- 5, \infty)

, there exists some

x \in R_{+}

such that

f (x) = y

. Therefore,

f

is onto.

Let

y = f (x) = (3 x + 1)^{2} - 6

. To find the inverse, we simply express

x

in terms of

y

\begin{aligned} y & = (3 x + 1)^{2} - 6 \\ (3 x + 1)^{2} & = y + 6 \\ 3 x + 1 & = \sqrt{y + 6} \\ x & = \frac{\sqrt{y + 6} - 1}{3} \end{aligned}

In the calculation above, we took only the positive square root, because

3 x + 1

is positive as

x \in R_{+}

Finally,

f^{- 1} (y) = \frac{\sqrt{y + 6} - 1}{3}

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