Question

There are exactly two distinct linear functions, which map

[- 1, 1]

onto

[0, 3]

, then find the point of intersection of the two functions.

Accepted Answer

$∵$ $f : [- 1, 1] \to [0, 3]$ is an onto function

$\Rightarrow$ Range of $f =$ Co-domain of $f$

$\Rightarrow$ Range of $f = R_{f} = [0, 3]$

And $f$ is linear and there are two such fuctions, one of the function is increasing and other is decreasing.Let functions be $f_{1} & f_{2}$ .

$f_{1} (- 1) = 0, f_{1} (1) = 3$ and $f_{2} (- 1) = 3, f_{2} (1) = 0$

It is given that $f (x)$ is a linear function .

Let $f_{1} (x) = a x + b$

If $f_{1} (- 1) = 0$ and $f_{1} (1) = 3$ $\Rightarrow - a + b = 0$ and $a + b = 3$

$\Rightarrow a = b = \frac{3}{2}$

$∴ f_{1} (x) = \frac{3}{2} x + \frac{3}{2} . . . . . (1)$

If $f_{2} (- 1) = 3$ and $f_{2} (1) = 0$ $\Rightarrow$ $- a + b = 3$ and $a + b = 0$

$\Rightarrow a = - \frac{3}{2}$ $, b = \frac{3}{2}$

$∴ f (x) = - \frac{3}{2} x + \frac{3}{2} . . . . . (2)$

Now the point of intersection of these functions can be evaluated by solving both equations simultaneously.

By equation $(1)$ and $(2),$ we get:

$\frac{3}{2} x + \frac{3}{2} = - \frac{3}{2} x + \frac{3}{2} \Rightarrow x = 0$

Put $x = 0$ in the equation $(1),$ we get:

$y = \frac{3}{2}$

Therefore, the point of intersection of these two functions is $(0, \frac{3}{2})$ .

Amit M Agarwal Solutions for Chapter: Functions, Exercise 7: Target Exercise 3.7

Amit M Agarwal Mathematics Solutions for Exercise - Amit M Agarwal Solutions for Chapter: Functions, Exercise 7: Target Exercise 3.7

Questions from Amit M Agarwal Solutions for Chapter: Functions, Exercise 7: Target Exercise 3.7 with Hints & Solutions

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