Class 11: Functions – Exercise 3.4 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Find $f+g, f-g, cf (c \in R, c \neq 0), fg,$ $\frac{1}{f}$ and $\frac{f}{g}$ in each of the following:

i) $f(x) = x^3 + 1$ and $g(x) = x + 1$ ii) $f(x) = \sqrt{x-1}$ and $g(x) = \sqrt{x+1}$

Answer:

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

$f+g : R \rightarrow R$ is given by

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

$f-g : R \rightarrow R$ is given by

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

$cf : R \rightarrow R$ is given by

$(cf)(x) = c f(x) = c(x^3 + 1)$

$fg : R \rightarrow R$ is given by

$(fg)(x) = f(x).g(x) = (x^3+1)(x+1) = x^4 + x + x^3 + 1$

$\frac{1}{f}$ $: R \rightarrow R$ is given by

$\Big($ $\frac{1}{f}$ $\Big) (x) =$ $\frac{1}{f(x)}$ $=$ $\frac{1}{x^3+1}$

$\frac{f}{g}$ $: R \rightarrow R$ is given by

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

ii) Given $f(x) = \sqrt{x-1}$ and $g(x) = \sqrt{x+1}$

$f+g : [1, \infty ) \rightarrow R$ is given by

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

$f-g : [1, \infty ) \rightarrow R$ is given by

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

$cf : [1, \infty ) \rightarrow R$ is given by

$(cf)(x) = c f(x) = c(\sqrt{x-1})$

$fg : [1, \infty ) \rightarrow R$ is given by

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

$\frac{1}{f}$ $: [1, \infty ) \rightarrow R$ is given by

$\Big($ $\frac{1}{f}$ $\Big) (x) =$ $\frac{1}{f(x)}$ $=$ $\frac{1}{\sqrt{x-1}}$

$\frac{f}{g}$ $: [1, \infty ) \rightarrow R$ is given by

$\Big($ $\frac{f}{g}$ $\Big) (x) =$ $\frac{f(x)}{g(x)}$ $=$ $\frac{\sqrt{x-1}}{\sqrt{x+1}}$

$\\$

Question 2: Let $f(x) = 2x+ 5$ and $g(x) = x^2 + x$ . Describe:

i) $f+g$ ii) $f-g$ iii) $fg$ iv) $\frac{f}{g}$ . Find the domain in each case.

Answer:

We observe $f(x) = 2x+5$ is defined for all $x \in R$ . Hence Domain $(f) = R$

Also $g(x) = x^2 + x$ is defined for all $x \in R$ . Hence Domain $(g) = R$

Therefore Domain $(f) \cap$ Domain $(g) = R$

i) $f+g : R \rightarrow R$ is given by

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

Hence Domain $(f+g) = R$

ii) $f-g : R \rightarrow R$ is given by

$(f-g)(x) = f(x) - g(x) = 2x+5 - x^2 - x = x + 5 -x^2$

Hence Domain $(f-g) = R$

iii) $fg : R \rightarrow R$ is given by

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

Hence Domain $(fg) = R$

iv) $\frac{f}{g}$ $: R \rightarrow R$ is given by

$\Big($ $\frac{f}{g}$ $\Big) (x) =$ $\frac{f(x)}{g(x)}$ $=$ $\frac{2x+5}{x^2+x}$

Now for $x^2 + x \neq 0$ for all $x$ except $x = 0, -1$

Hence Domain $\Big($ $\frac{f}{g}$ $\Big) = R - \{ 0, -1 \}$

$\\$

Question 3: If $f(x)$ be defined on $[-2, 2]$ and is given by

$f(x) = \Bigg\{$ $\begin{array}{ll} -1, \hspace*{1.0cm} -2 \leq x \leq 0 \\ x-1, \hspace*{0.8cm} 0 < x \leq 2 \end{array}$ and $g(x) = f(|x|) + |f(x)|$ . Find $g(x)$ .

Answer:

Given $f(x) = \Bigg\{$ $\begin{array}{ll} -1, \hspace*{1.0cm} -2 \leq x \leq 0 \\ x-1, \hspace*{0.8cm} 0 < x \leq 2 \end{array}$

Now for $g(x) = f(|x|) + g(|x|)$

$= \Bigg\{$ $\begin{array}{lll} x-1+1, \hspace*{1.8cm} -2 \leq x \leq 0 \\ x-1+(-x+1), \hspace*{0.8cm} 0 < x < 1 \\ x-1+x-1, \hspace*{1.3cm} 1 \leq x \leq 2 \end{array}$

$= \Bigg\{$ $\begin{array}{lll} x, \hspace*{1.8cm} -2 \leq x \leq 0 \\ 0, \hspace*{0.8cm} 0 < x < 1 \\ 2x-2, \hspace*{1.3cm} 1 \leq x \leq 2 \end{array}$

$\\$

Question 4: Let $f$ and $g$ be two real functions defined by $f(x) = \sqrt{x+1}$ and $g(x) = \sqrt{9-x^2}$ . Then, describe each of the following functions:

i) $f+g$ ii) $g-f$ iii) $fg$ iv) $\frac{f}{g}$ v) $\frac{g}{f}$ vi) $2f - \sqrt{5} g$ vii) $f^2 + 7f$ viii) $\frac{5}{g}$

Answer:

Given $f(x) = \sqrt{x+1}$ and $g(x) = \sqrt{9-x^2}$

We observe that $f(x) = \sqrt{x+1}$ is defined for $x \geq -1$

Hence Domain $(f) = [-1, \infty)$

Similarly, $g(x) = \sqrt{9-x^2}$ is defined for $9-x^2 \geq 0 \Rightarrow x^2 - 9 \leq 0$

$\Rightarrow (x-3)(x+3) \leq 0$

$\Rightarrow x \in [-3, 3 ]$

Hence Domain $(g) = [-3, 3]$

Now Domain $(f) \cap$ Domain $(g) = [-1, \infty) \cap [-3, 3] = [-1, 3]$

i) $f+g :[-1, 3] \rightarrow R$ is given by

$(f+g)(x) = f(x) + g(x) = \sqrt{x+1}+\sqrt{9-x^2}$

ii) $g-f : [-1, 3] \rightarrow R$ is given by

$(g-f)(x) = g(x)- f(x) = \sqrt{9-x^2} - \sqrt{x+1}$

iii) $(fg) : [-1, 3] \rightarrow R$ is given by

$(fg)(x) = f(x) g(x) = \sqrt{x+1} \sqrt{9-x^2}$

iv) $\frac{f}{g}$ $: [-1, 3 ) \rightarrow R$ is given by

$\Big($ $\frac{f}{g}$ $\Big) (x) =$ $\frac{f(x)}{g(x)}$ $=$ $\frac{\sqrt{x+1}}{\sqrt{9-x^2}}$

v) $\frac{g}{f}$ $: [-1, 3 ) \rightarrow R$ is given by

$\Big($ $\frac{g}{f}$ $\Big) (x) =$ $\frac{g(x)}{f(x)}$ $=$ $\frac{\sqrt{9-x^2}}{\sqrt{x+1}}$

vi) $2f-\sqrt{5} g : [-1, 3) \rightarrow R$ is defined by

$(2f-\sqrt{5} g)(x) = 2f(x) - \sqrt{5} g(x) = 2 \sqrt{x+1} - \sqrt{5} (\sqrt{9-x^2})$

$= 2 \sqrt{x+1} - \sqrt{45-5x^2}$

vii) $f^2 + 7f : [-1, \infty) \rightarrow R$ is given by

$(f^2 + 7f)(x) = [f(x)]^2 + 7f(x) = (\sqrt{x+1})^2 + 7 \sqrt{x+1} = x+1 + 7 \sqrt{x+1}$

viii) $\frac{5}{g(x)}$ $=$ $\frac{5}{\sqrt{9-x^2}}$ will be real for all values of $x$ except for $x \geq 3$ and $x \leq -3$

Therefore $\frac{5}{g}$ $: (-3, 3) \rightarrow R$ is defined by

$\Big($ $\frac{5}{g}$ $\Big) (x) =$ $\frac{5}{\sqrt{9-x^2}}$

$\\$

Question 5: If $f(x) = \log_e (1-x)$ and $g(x) = [x]$ , then determine each of the following functions:

i) $f+g$ ii) $fg$ iii) $g-f$ iv) $\frac{f}{g}$ v) $\frac{g}{f}$

Also find v) $(f+g)(-1), \ \ \ (fg) (0), \ \ \$ $\ \ \ \Big($ $\frac{f}{g}$ $\Big) \Big($ $\frac{1}{2}$ $\Big), \ \ \ \Big($ $\frac{g}{f}$ $\Big) \Big($ $\frac{1}{2}$ $\Big)$

Answer:

Given $f(x) = \log_e (1-x)$ and $g(x) = [x]$

For $f(x) = \log_e (1-x)$ is defined for all $(1-x) > 0 \Rightarrow x < 1 \Rightarrow x \in (-\infty, 1)$

Therefore Domain $(f) = (-\infty, 1)$

For $g(x) = [x]$ is defined for all $x \in R$

Therefore Domain $(g) = R$

Therefore Domain $(f) \cap$ Domain $(g) = (-\infty, 1) \cap R = (-\infty, 1)$

i) $f+g :(-\infty, 1) \rightarrow R$ is given by

$(f+g)(x) = f(x) + g(x) = \log_e (1-x) + g(x) = [x]$

ii) $(fg) : (-\infty, 1) \rightarrow R$ is given by

$(fg)(x) = f(x) g(x) = (\log_e (1-x) ).( g(x) ) = [x]$

iii) Domain $(f) = (-\infty, 1)$

$\frac{1}{g}$ $=$ $\frac{1}{[x]}$ will be real for all $x$ except $[x] = 0$

$\Rightarrow x \notin (0, 1)$

Therefore Domain $\Big($ $\frac{f}{g}$ $\Big) =$ Domain $(f) \cap$ Domain $($ $\frac{1}{g}$ $)$

$= (-\infty, 1) \cap R - \notin (0, 1) = (-\infty, 0)$

$\therefore \Big($ $\frac{f}{g}$ $\Big) : (-\infty, 0) \rightarrow R$ is defined as

$\Big($ $\frac{f}{g}$ $\Big)(x) =$ $\frac{f(x)}{g(x)}$ $=$ $\frac{\log_e (1-x)}{[x] }$

iv) $\frac{1}{f(x)}$ $=$ $\frac{1}{\log_e (1-x)}$

$\therefore$ $\frac{1}{f(x)}$ is real if $\log_e (1-x) \neq 0$

$\Rightarrow 10 x > 0$ and $1-x \neq 0$

$\Rightarrow x < 1$ and $x \neq 1$

$\Rightarrow x \in (-\infty, 0) \cup (0, 1)$

$\therefore$ Domain $($ $\frac{g}{f}$ $) =$ Domain $(g) \cap$ Domain $($ $\frac{1}{f}$ $)$

$= R \cap (-\infty, 0) \cup (0, 1) = (-\infty, 0) \cup (0, 1)$

$\therefore$ $\frac{g}{f}$ $: (-\infty, 0) \cup (0, 1)$ is defined by

$\Big($ $\frac{g}{f}$ $\Big)(x) =$ $\frac{g(x)}{f(x)}$ $=$ $\frac{[x] }{\log_e (1-x)}$

v) $(f+g) (-1) = f(-1) + f(-1) = \log_e (1-(-1)) + [-1] = \log_e 2 - 1$

$(fg)(0) = f(0)g(0) = \log_e (1-0) + [0] = 0$

$\Big($ $\frac{f}{g}$ $\Big) \Big($ $\frac{1}{2}$ $\Big) = \Big($ $\frac{f(\frac{1}{2})}{g(\frac{1}{2})}$ $\Big) =$ $\frac{\log_e (1-\frac{1}{2})}{[\frac{1}{2}]}$ $=$ does not exist because $[\frac{1}{2}]$ $= 0$

$\Big($ $\frac{g}{f}$ $\Big) \Big($ $\frac{1}{2}$ $\Big) = \Big($ $\frac{g(\frac{1}{2})}{f(\frac{1}{2})}$ $\Big) =$ $\frac{[\frac{1}{2}]}{\log_e (1-\frac{1}{2})}$ $= 0$

$\\$

Question 6: If $f, g, h$ are real functions defined by $f(x) = \sqrt{x+1}, \ g(x) =$ $\frac{1}{x}$ and $h(x) = 2x^2 - 3$ , then find the values of $(2f + g- h)(1)$ and $(2f+g-h)(0)$ .

Answer:

Given $f(x) = \sqrt{x+1}, \ g(x) =$ $\frac{1}{x}$ and $h(x) = 2x^2 - 3$

$f(x)$ will be real for all $x \geq -1 \Rightarrow$ Domain $(f) = [-1, \infty)$

$g(x)$ will be real for all $x$ except $x = 0 \Rightarrow$ Domain $(g) = R - \{0 \}$

$h(x)$ will be real for all $x \in R \Rightarrow$ Domain $(h) = R$

$\therefore$ Domain $(f) \cap$ Domain $(g) \cap$ Domain $(h) = [-1, \infty) \cap R - \{0 \} \cap R = [-1, \infty) - \{0 \}$

$\therefore 2f+g - h : [-1, \infty) - \{0 \} \rightarrow R$ is given by

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

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

$(2f+g - h)(0)$ does not exist as $x$ does not lie in domain $[-1, \infty) - \{0 \}$

$\\$

Question 7: The function f is defined by

$f(x) = \Bigg\{$ $\begin{array}{lll} 1-x, \hspace*{1.0cm} x < 0 \\ 1, \hspace*{1.7cm} x = 0 \\ x+1, \hspace*{1.0cm} x > 0 \end{array}$

Draw the graph of $f(x)$

Answer:

Given $f(x) = \Bigg\{$ $\begin{array}{lll} 1-x, \hspace*{1.0cm} x < 0 \\ 1, \hspace*{1.7cm} x = 0 \\ x+1, \hspace*{1.0cm} x > 0 \end{array}$

$\\$

Question 8: Let $f, \ g \colon R \rightarrow R$ be defined respectively by $f(x) = x + 1, \ g(x) = 2x-3$ . Find $f+g, \ f-g$ , and $\frac{f}{g}$ .

Answer:

Given $f(x) = x + 1, \ g(x) = 2x-3$

$f(x)$ will be real for all $x \in R \Rightarrow$ Domain $(f) = R$

$g(x$ ) will be real for all $x \in R \Rightarrow$ Domain $(g) = R$

$\therefore$ Domain $(f) \cap$ Domain $(g) = R$

$f+g : R \rightarrow R$ is defined by $(f+g)(x) = f(x) + g(x) = 3x-2$

$f-g : R \rightarrow R$ is defined by $(f-g)(x) = f(x) - g(x) = -x+4$

$\frac{1}{g}$ $=$ $\frac{1}{2x-3}$ is real for $2x-3 \neq 0 \Rightarrow x \neq$ $\frac{3}{2}$

$\frac{f}{g}$ $: R - \{$ $\frac{3}{2}$ $\} \rightarrow R$ is give by $\Big($ $\frac{f}{g}$ $\Big) (x) =$ $\frac{f(x)}{g(x)}$ $=$ $\frac{x+1}{2x-3}$

$\\$

Question 9: Let $f \colon [ 0, \infty ) \rightarrow R$ and $g \colon R \rightarrow R$ be defined by $f(x) = \sqrt{x}$ and $g(x) = x$ . Find $f+g, \ f-g, fg$ and $\frac{f}{g}$ .

Answer:

Given $f \colon [ 0, \infty ) \rightarrow R$ and $f(x) = \sqrt{x}$

$g \colon R \rightarrow R$ and $g(x) = x$

Domain $(f) \cap$ Domain $(g) = [0, \infty)$

$\therefore f+g : [0, \infty) \rightarrow R$ is defined as $(f+g)(x) = f(x) + g(x) = \sqrt{x}+x$

$f-g : [0, \infty) \rightarrow R$ is defined as $(f-g)(x) = f(x) - g(x) = \sqrt{x}-x$

$\frac{f}{g}$ $: (0, \infty) \rightarrow R$ is defined as $\Big($ $\frac{f}{g}$ $\Big) (x) =$ $\frac{f(x)}{g(x)}$ $=$ $\frac{\sqrt{x}}{x}$ $=$ $\frac{1}{\sqrt{x}}$

$\\$

Question 10: Let $f(x) = x^2$ and $g(x) = 2x+ 1$ be two real functions. Find $(f+g)(x), \ (f-g)(x), \ (fg)(x)$ and $\Big($ $\frac{f}{g}$ $\Big) (x)$

Answer:

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

$f(x)$ will be real for all $x \in R \Rightarrow$ Domain $(f) = R$

Similarly, $g(x)$ will be real for all $x \in R \Rightarrow$ Domain $(g) = R$

$\therefore f+g : R \rightarrow R$ is defined as $(f+g)(x) = f(x) + g(x) = x^2+2x+ 1$

$\therefore f-g : R \rightarrow R$ is defined as $(f-g)(x) = f(x) - g(x) = x^2-2x- 1$

$\therefore fg : R \rightarrow R$ is defined as $(fg)(x) = f(x).g(x) = x^2(2x+ 1)$

$\frac{f}{g}$ $: R - \{$ $\frac{1}{2}$ $\} \rightarrow R$ is defined as $\Big($ $\frac{f}{g}$ $\Big) (x) =$ $\frac{f(x)}{g(x)}$ $=$ $\frac{x^2}{2x+1}$

$\\$

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Class 11: Functions – Exercise 3.4

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