Relations and Functions — Exercise 2.6

If $f$ and $g$ are real functions defined by $f(x) = x^2 + 7$ and $g(x) = 3x + 5$, then find the value of

$f(3) + g(-5)$

Answer

Given,

$f(x) = x^2 + 7$ and $g(x) = 3x + 5$

⇒ $f(3) = 3^2 + 7 = 9 + 7 = 16$

⇒ $g(-5) = 3(-5) + 5 = -15 + 5 = -10$

∴ $f(3) + g(-5) = 16 + (-10) = 6$

Hence, $\bm{f(3) + g(-5) = 6}$.

If $f$ and $g$ are real functions defined by $f(x) = x^2 + 7$ and $g(x) = 3x + 5$, then find the value of

$f(-2) + g(-1)$

Answer

Given,

$f(x) = x^2 + 7$ and $g(x) = 3x + 5$

⇒ $f(-2) = (-2)^2 + 7 = 4 + 7 = 11$

⇒ $g(-1) = 3(-1) + 5 = -3 + 5 = 2$

∴ $f(-2) + g(-1) = 11 + 2 = 13$

Hence, $\bm{f(-2) + g(-1) = 13}$.

If $f$ and $g$ are real functions defined by $f(x) = x^2 + 7$ and $g(x) = 3x + 5$, then find the value of

$f\left(\dfrac{1}{2}\right) \times g(14)$

Answer

Given,

$f(x) = x^2 + 7$ and $g(x) = 3x + 5$

⇒ $f\left(\dfrac{1}{2}\right) = \left(\dfrac{1}{2}\right)^2 + 7 = \dfrac{1}{4} + 7 = \dfrac{1 + 28}{4} = \dfrac{29}{4}$

⇒ $g(14) = 3(14) + 5 = 42 + 5 = 47$

∴ $f\left(\dfrac{1}{2}\right) \times g(14) = \dfrac{29}{4} \times 47 = \dfrac{1363}{4}$

Hence, $\bm{f\left(\dfrac{1}{2}\right) \times g(14) = \dfrac{1363}{4}}$.

If $f$ and $g$ are real functions defined by $f(x) = x^2 + 7$ and $g(x) = 3x + 5$, then find the value of

$f(t) - f(-2)$

Answer

Given,

$f(x) = x^2 + 7$

⇒ $f(t) = t^2 + 7$

⇒ $f(-2) = (-2)^2 + 7 = 4 + 7 = 11$

∴ $f(t) - f(-2) = t^2 + 7 - 11 = t^2 - 4$

Hence, $\bm{f(t) - f(-2) = t^2 - 4}$.

If $f$ and $g$ are real functions defined by $f(x) = x^2 + 7$ and $g(x) = 3x + 5$, then find the value of

$\dfrac{f(t) - f(5)}{t - 5}$, $t \ne 5$

Answer

Given,

$f(x) = x^2 + 7$

⇒ $f(t) = t^2 + 7$

⇒ $f(5) = 5^2 + 7 = 25 + 7 = 32$

∴ $\dfrac{f(t) - f(5)}{t - 5} = \dfrac{t^2 + 7 - 32}{t - 5}$

⇒ $\dfrac{f(t) - f(5)}{t - 5} = \dfrac{t^2 - 25}{t - 5}$

⇒ $\dfrac{f(t) - f(5)}{t - 5} = \dfrac{(t - 5)(t + 5)}{t - 5}$

⇒ $\dfrac{f(t) - f(5)}{t - 5} = t + 5$ (as $t \ne 5$)

Hence, $\bm{\dfrac{f(t) - f(5)}{t - 5} = t + 5}$.

If $f(x) = e^x$ and $g(x) = \log x$, then find:

$(f - g)(1)$

Answer

Given,

$f(x) = e^x$ and $g(x) = \log x$

We know that,

$(f - g)(x) = f(x) - g(x)$.

⇒ $(f - g)(1) = f(1) - g(1)$

⇒ $(f - g)(1) = e^1 - \log 1$

We know that, $\log 1 = 0$.

⇒ $(f - g)(1) = e - 0 = e$

Hence, $\bm{(f - g)(1) = e}$.

If $f(x) = e^x$ and $g(x) = \log x$, then find:

$(fg)(1)$

Answer

Given,

$f(x) = e^x$ and $g(x) = \log x$

We know that,

$(fg)(x) = f(x), g(x)$.

⇒ $(fg)(1) = f(1) \times g(1)$

⇒ $(fg)(1) = e^1 \times \log 1$

We know that, $\log 1 = 0$.

⇒ $(fg)(1) = e \times 0 = 0$

Hence, $\bm{(fg)(1) = 0}$.

If $f(x) = e^x$ and $g(x) = \log x$, then find:

$\left(\dfrac{f}{g}\right)(3)$

Answer

Given,

$f(x) = e^x$ and $g(x) = \log x$

We know that,

$\left(\dfrac{f}{g}\right)(x) = \dfrac{f(x)}{g(x)}$, where $g(x) \ne 0$.

⇒ $\left(\dfrac{f}{g}\right)(3) = \dfrac{f(3)}{g(3)}$

⇒ $\left(\dfrac{f}{g}\right)(3) = \dfrac{e^3}{\log 3}$

Hence, $\bm{\left(\dfrac{f}{g}\right)(3) = \dfrac{e^3}{\log 3}}$.

If $f$ and $g$ are two real valued functions defined by $f(x) = 2x + 1$ and $g(x) = x^2 + 1$, then find the following function:

$f + g$

Answer

Given,

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

Both $f$ and $g$ are defined for every real value of $x$, so the domain of each of them is $\textbf{R}$.

$(f + g)(x) = f(x) + g(x)$

⇒ $(f + g)(x) = (2x + 1) + (x^2 + 1)$

⇒ $(f + g)(x) = x^2 + 2x + 2$

Hence, $\bm{(f + g)(x) = x^2 + 2x + 2}$, $\bm{x \in \textbf{R}}$.

If $f$ and $g$ are two real valued functions defined by $f(x) = 2x + 1$ and $g(x) = x^2 + 1$, then find the following function:

$f - g$

Answer

Given,

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

Both $f$ and $g$ are defined for every real value of $x$, so the domain of each of them is $\textbf{R}$.

$(f - g)(x) = f(x) - g(x)$

⇒ $(f - g)(x) = (2x + 1) - (x^2 + 1)$

⇒ $(f - g)(x) = 2x + 1 - x^2 - 1$

⇒ $(f - g)(x) = 2x - x^2$

Hence, $\bm{(f - g)(x) = 2x - x^2}$, $\bm{x \in \textbf{R}}$.

If $f$ and $g$ are two real valued functions defined by $f(x) = 2x + 1$ and $g(x) = x^2 + 1$, then find the following function:

$fg$

Answer

Given,

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

Both $f$ and $g$ are defined for every real value of $x$, so the domain of each of them is $\textbf{R}$.

$(fg)(x) = f(x), g(x)$

⇒ $(fg)(x) = (2x + 1)(x^2 + 1)$

⇒ $(fg)(x) = 2x^3 + 2x + x^2 + 1$

⇒ $(fg)(x) = 2x^3 + x^2 + 2x + 1$

Hence, $\bm{(fg)(x) = 2x^3 + x^2 + 2x + 1}$, $\bm{x \in \textbf{R}}$.

If $f$ and $g$ are two real valued functions defined by $f(x) = 2x + 1$ and $g(x) = x^2 + 1$, then find the following function:

$\dfrac{f}{g}$

Answer

Given,

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

Both $f$ and $g$ are defined for every real value of $x$, so the domain of each of them is $\textbf{R}$.

$\left(\dfrac{f}{g}\right)(x) = \dfrac{f(x)}{g(x)}$, where $g(x) \ne 0$.

Here $x^2 + 1 \ge 1$ for every real number $x$, so $g(x)$ is never zero.

⇒ $\left(\dfrac{f}{g}\right)(x) = \dfrac{2x + 1}{x^2 + 1}$

Hence, $\bm{\left(\dfrac{f}{g}\right)(x) = \dfrac{2x + 1}{x^2 + 1}}$, $\bm{x \in \textbf{R}}$.

If $f(x) = x^3 + 1$ and $g(x) = x + 1$ be two real functions, then find the following function:

$f + g$

Answer

Given,

$f(x) = x^3 + 1$ and $g(x) = x + 1$

Both $f$ and $g$ are defined for every real value of $x$, so the domain of each of them is $\textbf{R}$.

$(f + g)(x) = f(x) + g(x)$

⇒ $(f + g)(x) = (x^3 + 1) + (x + 1)$

⇒ $(f + g)(x) = x^3 + x + 2$

Hence, $\bm{(f + g)(x) = x^3 + x + 2}$, $\bm{x \in \textbf{R}}$.

If $f(x) = x^3 + 1$ and $g(x) = x + 1$ be two real functions, then find the following function:

$g - f$

Answer

Given,

$f(x) = x^3 + 1$ and $g(x) = x + 1$

Both $f$ and $g$ are defined for every real value of $x$, so the domain of each of them is $\textbf{R}$.

$(g - f)(x) = g(x) - f(x)$

⇒ $(g - f)(x) = (x + 1) - (x^3 + 1)$

⇒ $(g - f)(x) = x + 1 - x^3 - 1$

⇒ $(g - f)(x) = x - x^3$

Hence, $\bm{(g - f)(x) = x - x^3}$, $\bm{x \in \textbf{R}}$.

If $f(x) = x^3 + 1$ and $g(x) = x + 1$ be two real functions, then find the following function:

$fg$

Answer

Given,

$f(x) = x^3 + 1$ and $g(x) = x + 1$

Both $f$ and $g$ are defined for every real value of $x$, so the domain of each of them is $\textbf{R}$.

$(fg)(x) = f(x), g(x)$

⇒ $(fg)(x) = (x^3 + 1)(x + 1)$

⇒ $(fg)(x) = x^4 + x^3 + x + 1$

Hence, $\bm{(fg)(x) = x^4 + x^3 + x + 1}$, $\bm{x \in \textbf{R}}$.

If $f(x) = x^3 + 1$ and $g(x) = x + 1$ be two real functions, then find the following function:

$\dfrac{f}{g}$

Answer

Given,

$f(x) = x^3 + 1$ and $g(x) = x + 1$

Both $f$ and $g$ are defined for every real value of $x$, so the domain of each of them is $\textbf{R}$.

$\left(\dfrac{f}{g}\right)(x) = \dfrac{f(x)}{g(x)}$, where $g(x) \ne 0$.

Here $g(x) = x + 1 = 0$ when $x = -1$, so $x = -1$ must be excluded.

⇒ $\left(\dfrac{f}{g}\right)(x) = \dfrac{x^3 + 1}{x + 1}$

Hence, $\bm{\left(\dfrac{f}{g}\right)(x) = \dfrac{x^3 + 1}{x + 1}}$, $\bm{x \in \textbf{R}}$, $\bm{x \ne -1}$.

If $f(x) = x^3 + 1$ and $g(x) = x + 1$ be two real functions, then find the following function:

$2g^2 - 3f$.

Answer

Given,

$f(x) = x^3 + 1$ and $g(x) = x + 1$

Both $f$ and $g$ are defined for every real value of $x$, so the domain of each of them is $\textbf{R}$.

$(2g^2 - 3f)(x) = 2[g(x)]^2 - 3f(x)$

⇒ $(2g^2 - 3f)(x) = 2(x + 1)^2 - 3(x^3 + 1)$

⇒ $(2g^2 - 3f)(x) = 2(x^2 + 2x + 1) - 3x^3 - 3$

⇒ $(2g^2 - 3f)(x) = 2x^2 + 4x + 2 - 3x^3 - 3$

⇒ $(2g^2 - 3f)(x) = 2x^2 + 4x - 3x^3 - 1$

Hence, $\bm{(2g^2 - 3f)(x) = 2x^2 + 4x - 3x^3 - 1}$, $\bm{x \in \textbf{R}}$.

If $f(x) = \sqrt{x - 2}$ and $g(x) = \sqrt{x^2 - 1}$ be two real valued functions, then find the following function:

$f + g$

Answer

Given,

$f(x) = \sqrt{x - 2}$ and $g(x) = \sqrt{x^2 - 1}$

For $f(x)$ to be a real number, $x - 2 \ge 0$ i.e. $x \ge 2$.

∴ $D_f = [2, \infty)$

For $g(x)$ to be a real number, $x^2 - 1 \ge 0$ i.e. $x \le -1$ or $x \ge 1$.

∴ $D_g = (-\infty, -1] \cup [1, \infty)$

The set common to both the domains is $[2, \infty)$.

$(f + g)(x) = f(x) + g(x)$

⇒ $(f + g)(x) = \sqrt{x - 2} + \sqrt{x^2 - 1}$

Hence, $\bm{(f + g)(x) = \sqrt{x - 2} + \sqrt{x^2 - 1}}$, $\bm{x \in [2, \infty)}$.

If $f(x) = \sqrt{x - 2}$ and $g(x) = \sqrt{x^2 - 1}$ be two real valued functions, then find the following function:

$g - f$

Answer

Given,

$f(x) = \sqrt{x - 2}$ and $g(x) = \sqrt{x^2 - 1}$

As shown in part (i), $D_f = [2, \infty)$, $D_g = (-\infty, -1] \cup [1, \infty)$ and the set common to both the domains is $[2, \infty)$.

$(g - f)(x) = g(x) - f(x)$

⇒ $(g - f)(x) = \sqrt{x^2 - 1} - \sqrt{x - 2}$

Hence, $\bm{(g - f)(x) = \sqrt{x^2 - 1} - \sqrt{x - 2}}$, $\bm{x \in [2, \infty)}$.

If $f(x) = \sqrt{x - 2}$ and $g(x) = \sqrt{x^2 - 1}$ be two real valued functions, then find the following function:

$fg$

Answer

Given,

$f(x) = \sqrt{x - 2}$ and $g(x) = \sqrt{x^2 - 1}$

As shown in part (i), $D_f = [2, \infty)$, $D_g = (-\infty, -1] \cup [1, \infty)$ and the set common to both the domains is $[2, \infty)$.

$(fg)(x) = f(x), g(x)$

⇒ $(fg)(x) = \sqrt{x - 2} \times \sqrt{x^2 - 1}$

⇒ $(fg)(x) = \sqrt{(x - 2)(x^2 - 1)}$

Hence, $\bm{(fg)(x) = \sqrt{(x - 2)(x^2 - 1)}}$, $\bm{x \in [2, \infty)}$.

If $f(x) = \sqrt{x - 2}$ and $g(x) = \sqrt{x^2 - 1}$ be two real valued functions, then find the following function:

$3f - 2g$

Answer

Given,

$f(x) = \sqrt{x - 2}$ and $g(x) = \sqrt{x^2 - 1}$

As shown in part (i), $D_f = [2, \infty)$, $D_g = (-\infty, -1] \cup [1, \infty)$ and the set common to both the domains is $[2, \infty)$.

$(3f - 2g)(x) = 3f(x) - 2g(x)$

⇒ $(3f - 2g)(x) = 3\sqrt{x - 2} - 2\sqrt{x^2 - 1}$

Hence, $\bm{(3f - 2g)(x) = 3\sqrt{x - 2} - 2\sqrt{x^2 - 1}}$, $\bm{x \in [2, \infty)}$.

If $f(x) = \sqrt{x - 2}$ and $g(x) = \sqrt{x^2 - 1}$ be two real valued functions, then find the following function:

$\dfrac{f}{g}$

Answer

Given,

$f(x) = \sqrt{x - 2}$ and $g(x) = \sqrt{x^2 - 1}$

As shown in part (i), $D_f = [2, \infty)$, $D_g = (-\infty, -1] \cup [1, \infty)$ and the set common to both the domains is $[2, \infty)$.

$\left(\dfrac{f}{g}\right)(x) = \dfrac{f(x)}{g(x)}$, where $g(x) \ne 0$.

For $x \in [2, \infty)$, $x^2 - 1 \ge 3 \gt 0$, so $g(x)$ is never zero.

⇒ $\left(\dfrac{f}{g}\right)(x) = \dfrac{\sqrt{x - 2}}{\sqrt{x^2 - 1}}$

⇒ $\left(\dfrac{f}{g}\right)(x) = \sqrt{\dfrac{x - 2}{x^2 - 1}}$

Hence, $\bm{\left(\dfrac{f}{g}\right)(x) = \sqrt{\dfrac{x - 2}{x^2 - 1}}}$, $\bm{x \in [2, \infty)}$.

If $f(x) = \sqrt{x - 2}$ and $g(x) = \sqrt{x^2 - 1}$ be two real valued functions, then find the following function:

$2f^2 + \sqrt{3}g$

Answer

Given,

$f(x) = \sqrt{x - 2}$ and $g(x) = \sqrt{x^2 - 1}$

As shown in part (i), $D_f = [2, \infty)$, $D_g = (-\infty, -1] \cup [1, \infty)$ and the set common to both the domains is $[2, \infty)$.

$(2f^2 + \sqrt{3}g)(x) = 2[f(x)]^2 + \sqrt{3},g(x)$

⇒ $(2f^2 + \sqrt{3}g)(x) = 2\left(\sqrt{x - 2}\right)^2 + \sqrt{3}\sqrt{x^2 - 1}$

⇒ $(2f^2 + \sqrt{3}g)(x) = 2(x - 2) + \sqrt{3(x^2 - 1)}$

Hence, $\bm{(2f^2 + \sqrt{3}g)(x) = 2(x - 2) + \sqrt{3(x^2 - 1)}}$, $\bm{x \in [2, \infty)}$.

If $f(x) = \sqrt{x - 2}$ and $g(x) = \sqrt{x^2 - 1}$ be two real valued functions, then find the following function:

$\dfrac{1}{f}$

Answer

Given,

$f(x) = \sqrt{x - 2}$ and $g(x) = \sqrt{x^2 - 1}$

As shown in part (i), $D_f = [2, \infty)$, $D_g = (-\infty, -1] \cup [1, \infty)$ and the set common to both the domains is $[2, \infty)$.

$\left(\dfrac{1}{f}\right)(x) = \dfrac{1}{f(x)}$, where $f(x) \ne 0$.

Here $f(x) = \sqrt{x - 2} = 0$ when $x = 2$, so $x = 2$ must be excluded.

⇒ $\left(\dfrac{1}{f}\right)(x) = \dfrac{1}{\sqrt{x - 2}}$

Hence, $\bm{\left(\dfrac{1}{f}\right)(x) = \dfrac{1}{\sqrt{x - 2}}}$, $\bm{x \in (2, \infty)}$.

If $f(x) = \sqrt{x - 2}$ and $g(x) = \sqrt{x^2 - 1}$ be two real valued functions, then find the following function:

$\dfrac{g}{f}$.

Answer

Given,

$f(x) = \sqrt{x - 2}$ and $g(x) = \sqrt{x^2 - 1}$

As shown in part (i), $D_f = [2, \infty)$, $D_g = (-\infty, -1] \cup [1, \infty)$ and the set common to both the domains is $[2, \infty)$.

$\left(\dfrac{g}{f}\right)(x) = \dfrac{g(x)}{f(x)}$, where $f(x) \ne 0$.

Here $f(x) = \sqrt{x - 2} = 0$ when $x = 2$, so $x = 2$ must be excluded.

⇒ $\left(\dfrac{g}{f}\right)(x) = \dfrac{\sqrt{x^2 - 1}}{\sqrt{x - 2}}$

⇒ $\left(\dfrac{g}{f}\right)(x) = \sqrt{\dfrac{x^2 - 1}{x - 2}}$

Hence, $\bm{\left(\dfrac{g}{f}\right)(x) = \sqrt{\dfrac{x^2 - 1}{x - 2}}}$, $\bm{x \in (2, \infty)}$.

If $f(x) = x^3 - \dfrac{1}{x^3}$, prove that $f(x) + f\left(\dfrac{1}{x}\right) = 0$.

Answer

Given,

$f(x) = x^3 - \dfrac{1}{x^3}$, where $x \ne 0$.

Replacing $x$ by $\dfrac{1}{x}$,

$f\left(\dfrac{1}{x}\right) = \left(\dfrac{1}{x}\right)^3 - \dfrac{1}{\left(\dfrac{1}{x}\right)^3}$

⇒ $f\left(\dfrac{1}{x}\right) = \dfrac{1}{x^3} - x^3$

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

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

⇒ $f(x) + f\left(\dfrac{1}{x}\right) = 0$

Hence, proved that $\bm{f(x) + f\left(\dfrac{1}{x}\right) = 0}$.

If $f(x) = x + \dfrac{1}{x}$, prove that $(f(x))^3 = f(x^3) + 3f\left(\dfrac{1}{x}\right)$.

Answer

Given,

$f(x) = x + \dfrac{1}{x}$, where $x \ne 0$.

Replacing $x$ by $x^3$,

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

Replacing $x$ by $\dfrac{1}{x}$,

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

⇒ $f\left(\dfrac{1}{x}\right) = x + \dfrac{1}{x}$ ...........(2)

Taking L.H.S.,

$(f(x))^3 = \left(x + \dfrac{1}{x}\right)^3$

We know that, $(a + b)^3 = a^3 + b^3 + 3ab(a + b)$.

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

⇒ $(f(x))^3 = x^3 + \dfrac{1}{x^3} + 3\left(x + \dfrac{1}{x}\right)$

Using equations (1) and (2),

⇒ $(f(x))^3 = f(x^3) + 3f\left(\dfrac{1}{x}\right)$

Hence, proved that $\bm{(f(x))^3 = f(x^3) + 3f\left(\dfrac{1}{x}\right)}$.

If $y = f(x) = \dfrac{6x - 5}{5x - 6}$, prove that $f(y) = x$, $x \ne \dfrac{6}{5}$.

Answer

Given,

$y = f(x) = \dfrac{6x - 5}{5x - 6}$, where $x \ne \dfrac{6}{5}$.

Replacing $x$ by $y$ in the given function,

$f(y) = \dfrac{6y - 5}{5y - 6}$

Substituting $y = \dfrac{6x - 5}{5x - 6}$,

⇒ $f(y) = \dfrac{6\left(\dfrac{6x - 5}{5x - 6}\right) - 5}{5\left(\dfrac{6x - 5}{5x - 6}\right) - 6}$

⇒ $f(y) = \dfrac{\dfrac{6(6x - 5) - 5(5x - 6)}{5x - 6}}{\dfrac{5(6x - 5) - 6(5x - 6)}{5x - 6}}$

⇒ $f(y) = \dfrac{6(6x - 5) - 5(5x - 6)}{5(6x - 5) - 6(5x - 6)}$

⇒ $f(y) = \dfrac{36x - 30 - 25x + 30}{30x - 25 - 30x + 36}$

⇒ $f(y) = \dfrac{11x}{11}$

⇒ $f(y) = x$

Hence, proved that $\bm{f(y) = x}$.