Relations and Functions — Exercise 2.5

Solve the following equations for $x$:

(i) $[2x - 3] = 5$

(ii) $[2x + 1] = -2$

(iii) $[2 - 3x] = 5$.

Answer

We know that,

$[\cdot]$ denotes the greatest integer function and for an integer $n$,

$[k] = n$ ⇔ $n \le k \lt n + 1$.

(i) $[2x - 3] = 5$

⇒ $5 \le 2x - 3 \lt 6$

⇒ $5 + 3 \le 2x \lt 6 + 3$

⇒ $8 \le 2x \lt 9$

⇒ $4 \le x \lt \dfrac{9}{2}$

Hence, $\bm{x \in \left[4, \dfrac{9}{2}\right)}$.

(ii) $[2x + 1] = -2$

⇒ $-2 \le 2x + 1 \lt -1$

⇒ $-2 - 1 \le 2x \lt -1 - 1$

⇒ $-3 \le 2x \lt -2$

⇒ $-\dfrac{3}{2} \le x \lt -1$

Hence, $\bm{x \in \left[-\dfrac{3}{2}, -1\right)}$.

(iii) $[2 - 3x] = 5$

⇒ $5 \le 2 - 3x \lt 6$

⇒ $5 - 2 \le -3x \lt 6 - 2$

⇒ $3 \le -3x \lt 4$

Dividing throughout by $-3$ and reversing the inequalities,

⇒ $-1 \ge x \gt -\dfrac{4}{3}$

⇒ $-\dfrac{4}{3} \lt x \le -1$

Hence, $\bm{x \in \left(-\dfrac{4}{3}, -1\right]}$.

If $f(x) = \log\left(\dfrac{1 + x}{1 - x}\right)$, then show that $f(x) + f(y) = f\left(\dfrac{x + y}{1 + xy}\right)$.

Answer

Given,

$f(x) = \log\left(\dfrac{1 + x}{1 - x}\right)$

Taking L.H.S.,

$f(x) + f(y) = \log\left(\dfrac{1 + x}{1 - x}\right) + \log\left(\dfrac{1 + y}{1 - y}\right)$

We know that, $\log m + \log n = \log (mn)$.

⇒ $f(x) + f(y) = \log\left(\dfrac{1 + x}{1 - x} \times \dfrac{1 + y}{1 - y}\right)$

⇒ $f(x) + f(y) = \log\left[\dfrac{(1 + x)(1 + y)}{(1 - x)(1 - y)}\right]$ ...........(1)

Taking R.H.S.,

$f\left(\dfrac{x + y}{1 + xy}\right) = \log\left(\dfrac{1 + \dfrac{x + y}{1 + xy}}{1 - \dfrac{x + y}{1 + xy}}\right)$

⇒ $f\left(\dfrac{x + y}{1 + xy}\right) = \log\left(\dfrac{\dfrac{1 + xy + x + y}{1 + xy}}{\dfrac{1 + xy - x - y}{1 + xy}}\right)$

⇒ $f\left(\dfrac{x + y}{1 + xy}\right) = \log\left(\dfrac{1 + xy + x + y}{1 + xy - x - y}\right)$

⇒ $f\left(\dfrac{x + y}{1 + xy}\right) = \log\left[\dfrac{(1 + x) + y(1 + x)}{(1 - x) - y(1 - x)}\right]$

⇒ $f\left(\dfrac{x + y}{1 + xy}\right) = \log\left[\dfrac{(1 + x)(1 + y)}{(1 - x)(1 - y)}\right]$ ...........(2)

From equations (1) and (2), we get:

$f(x) + f(y) = f\left(\dfrac{x + y}{1 + xy}\right)$

Hence, proved that $\bm{f(x) + f(y) = f\left(\dfrac{x + y}{1 + xy}\right)}$.

If $f(x) = \dfrac{9^x}{9^x + 3}$, then show that $f(x) + f(1 - x) = 1$.

Answer

Given,

$f(x) = \dfrac{9^x}{9^x + 3}$

Replacing $x$ by $1 - x$,

$f(1 - x) = \dfrac{9^{1 - x}}{9^{1 - x} + 3}$

⇒ $f(1 - x) = \dfrac{\dfrac{9}{9^x}}{\dfrac{9}{9^x} + 3}$

Multiplying the numerator and the denominator by $9^x$,

⇒ $f(1 - x) = \dfrac{9}{9 + 3 \times 9^x}$

⇒ $f(1 - x) = \dfrac{9}{3(3 + 9^x)}$

⇒ $f(1 - x) = \dfrac{3}{9^x + 3}$

∴ $f(x) + f(1 - x) = \dfrac{9^x}{9^x + 3} + \dfrac{3}{9^x + 3}$

⇒ $f(x) + f(1 - x) = \dfrac{9^x + 3}{9^x + 3}$

⇒ $f(x) + f(1 - x) = 1$

Hence, proved that $\bm{f(x) + f(1 - x) = 1}$.

Find the domain of the following functions:

(i) $\log(4x - 3)$

(ii) $\dfrac{1}{\log(9 - x^2)}$.

Answer

We know that,

$\log k$ is a real number only when $k \gt 0$.

(i) $\log(4x - 3)$

For $\log(4x - 3)$ to be a real number,

⇒ $4x - 3 \gt 0$

⇒ $4x \gt 3$

⇒ $x \gt \dfrac{3}{4}$

Hence, domain $\bm{= \left(\dfrac{3}{4}, \infty\right)}$.

(ii) $\dfrac{1}{\log(9 - x^2)}$

For $\log(9 - x^2)$ to be a real number,

⇒ $9 - x^2 \gt 0$

⇒ $x^2 \lt 9$

⇒ $-3 \lt x \lt 3$

Also, the denominator must not be zero.

⇒ $\log(9 - x^2) \ne 0$

⇒ $9 - x^2 \ne 1$

⇒ $x^2 \ne 8$

⇒ $x \ne \pm 2\sqrt{2}$

Both $2\sqrt{2}$ and $-2\sqrt{2}$ lie between $-3$ and 3, so they must be excluded.

Hence, domain $\bm{= (-3, 3) - {-2\sqrt{2}, 2\sqrt{2}}}$.

Draw the graph of the function $f(x) = |x| + |x - 2|$. Hence find its range.

Answer

Given,

$f(x) = |x| + |x - 2|$

We know that,

$|x| = \begin{cases} x, & \text{if } x \ge 0 \\[1em] -x, & \text{if } x \lt 0 \end{cases}$

and

$|x - 2| = \begin{cases} x - 2, & \text{if } x \ge 2 \\[1em] -(x - 2), & \text{if } x \lt 2 \end{cases}$

When $x \lt 0$, $f(x) = -x + (2 - x) = 2 - 2x$

When $0 \le x \lt 2$, $f(x) = x + (2 - x) = 2$

When $x \ge 2$, $f(x) = x + (x - 2) = 2x - 2$

∴ $f(x)$ can be written as

$f(x) = \begin{cases} 2 - 2x, & x \lt 0 \\[1em] 2, & 0 \le x \lt 2 \\[1em] 2x - 2, & x \ge 2 \end{cases}$

Plotting these three pieces, we get the following graph.

From the graph, the lowest value taken by $f(x)$ is 2 (taken along the whole segment from $x = 0$ to $x = 2$) and $f(x)$ increases without limit as $x$ moves away from this segment on either side.

Hence, range of $\bm{f = [2, \infty)}$.

Draw rough sketch (graph) of the function

$f(x) = 3$

Answer

Here $f(x)$ takes the value 3 for every real value of $x$, so the graph is a straight line parallel to the $x$-axis at a distance of 3 units above it.

Taking a few values of $x$,

when $x = -2$, $f(x) = 3$; when $x = -1$, $f(x) = 3$; when $x = 0$, $f(x) = 3$; when $x = 1$, $f(x) = 3$; when $x = 2$, $f(x) = 3$.

Plotting the points $(-2, 3)$, $(-1, 3)$, $(0, 3)$, $(1, 3)$, $(2, 3)$ and joining them, we get the following graph.

Hence, the graph of $\bm{f(x) = 3}$ is the horizontal straight line drawn above.

Draw rough sketch (graph) of the function

$f(x) = 3x - 2$

Answer

This is of the form $f(x) = mx + c$, so the graph is a straight line.

Taking a few values of $x$,

when $x = -1$, $f(x) = 3(-1) - 2 = -5$; when $x = 0$, $f(x) = -2$; when $x = 1$, $f(x) = 1$; when $x = 2$, $f(x) = 4$.

Plotting the points $(-1, -5)$, $(0, -2)$, $(1, 1)$, $(2, 4)$ and joining them, we get a straight line.

Hence, the graph of $\bm{f(x) = 3x - 2}$ is the straight line drawn above.

Draw rough sketch (graph) of the function

$f(x) = x^2 + 1$

Answer

Taking a few values of $x$,

when $x = -2$, $f(x) = (-2)^2 + 1 = 5$; when $x = -1$, $f(x) = 2$; when $x = 0$, $f(x) = 1$; when $x = 1$, $f(x) = 2$; when $x = 2$, $f(x) = 5$.

Plotting the points $(-2, 5)$, $(-1, 2)$, $(0, 1)$, $(1, 2)$, $(2, 5)$ and joining them by a free-hand smooth curve, we get a parabola opening upwards with its lowest point at $(0, 1)$.

Hence, the graph of $\bm{f(x) = x^2 + 1}$ is the parabola drawn above.

Draw rough sketch (graph) of the function

$f(x) = 2^x$

Answer

Taking a few values of $x$,

when $x = -2$, $f(x) = 2^{-2} = \dfrac{1}{4}$; when $x = -1$, $f(x) = \dfrac{1}{2}$; when $x = 0$, $f(x) = 1$; when $x = 1$, $f(x) = 2$; when $x = 2$, $f(x) = 4$.

Plotting the points $\left(-2, \dfrac{1}{4}\right)$, $\left(-1, \dfrac{1}{2}\right)$, $(0, 1)$, $(1, 2)$, $(2, 4)$ and joining them by a free-hand smooth curve, we get an increasing curve which lies wholly above the $x$-axis and passes through $(0, 1)$.

Hence, the graph of $\bm{f(x) = 2^x}$ is the exponential curve drawn above.

Draw rough sketch (graph) of the function

$f(x) = \dfrac{x}{1 + x^2}$

Answer

Taking a few values of $x$,

when $x = -2$, $f(x) = \dfrac{-2}{1 + 4} = -\dfrac{2}{5}$; when $x = -1$, $f(x) = -\dfrac{1}{2}$; when $x = 0$, $f(x) = 0$; when $x = 1$, $f(x) = \dfrac{1}{2}$; when $x = 2$, $f(x) = \dfrac{2}{5}$.

As $f(-x) = -f(x)$, the curve is symmetrical about the origin. It passes through the origin, rises to the highest point $\left(1, \dfrac{1}{2}\right)$ and then falls towards the $x$-axis.

Hence, the graph of $\bm{f(x) = \dfrac{x}{1 + x^2}}$ is the curve drawn above.