Question 14 Marks
If for non-zerox, $\text{af(x)}+\text{bf}\Big(\frac{1}{\text{x}}\Big)=\frac{1}{\text{x}}-5,$ where $\text{a}\neq\text{b},$ then find f(x).
AnswerWe have, $\text{af(x)}+\text{bf}\Big(\frac{1}{\text{x}}\Big)=\frac{1}{\text{x}}-5\ ....(\text{i})$ $\Rightarrow\ \text{af}\Big(\frac{1}{\text{x}}\Big)+\text{bf(x)}=\frac{1}{\frac{1}{\text{x}}}-5$ $\Rightarrow\ \text{af}\Big(\frac{1}{\text{x}}\Big)+\text{bf(x)}=\text{x}-5\ ...(\text{ii})$ Adding equations (i) and (ii) we get $\Rightarrow\ \text{af}({\text{x}})+\text{bf(x)}+\text{bf}\Big(\frac{1}{\text{x}}\Big)+\text{af}\Big(\frac{1}{\text{x}}\Big)=\frac{1}{\text{x}}-5+\text{x}-5$ $\Rightarrow\ (\text{a}+\text{b})\text{f(x)}+\text{f}\Big(\frac{1}{\text{x}}\Big)(\text{a}+\text{b})=\frac{1}{\text{x}}+\text{x}-10$ $\Rightarrow\ \text{f(x)}+\text{f}\Big(\frac{1}{\text{x}}\Big)=\frac{1}{\text{a}+\text{b}}\Big[\frac{1}{\text{x}}+\text{x}-10\Big]\ ...(\text{iii})$ Subtracting equation (ii) from equation (i), we get $\text{af(x)}-\text{bf(x)}+\text{bf}\Big(\frac{1}{\text{x}}\Big)-\text{af}\Big(\frac{1}{\text{x}}\Big)=\frac{1}{\text{x}}-5-\text{x}+5$ $\Rightarrow\ (\text{a}-\text{b})\text{f(x)}-\text{f}\Big(\frac{1}{\text{x}}\Big)(\text{a}-\text{b})=\frac{1}{\text{x}}-\text{x}$ $\Rightarrow\ \text{f(x)}-\text{f}\Big(\frac{1}{\text{x}}\Big)=\frac{1}{\text{a}-\text{b}}\Big[\frac{1}{\text{x}}-\text{x}\Big]$ Adding equations (iii) and (iv) we get $2\text{f(x)}=\frac{1}{\text{a}+\text{b}}\Big[\frac{1}{\text{x}}+\text{x}-10\Big]+\frac{1}{\text{a}-\text{b}}\Big[\frac{1}{\text{x}}-\text{x}\Big]$ $\Rightarrow\ 2\text{f(x)}=\frac{(\text{a}-\text{b})\big[\frac{1}{\text{x}}+\text{x}-10\big]+(\text{a}+\text{b})\big[\frac{1}{\text{x}}-\text{x}\big]}{(\text{a}+\text{b})(\text{a}-\text{b})}$ $\Rightarrow\ 2\text{f(x)}=\frac{\frac{\text{a}}{\text{x}}+\text{ax}-10\text{a}-\frac{\text{b}}{\text{x}}-\text{bx}+10\text{b}+\frac{\text{a}}{\text{x}}-\text{ax}+\frac{\text{b}}{\text{x}}-\text{bx}}{\text{a}^2-\text{b}^2}$ $\Rightarrow\ 2\text{f(x)}=\frac{\frac{2\text{a}}{\text{x}}-10\text{a}+10\text{b}-2\text{bx}}{\text{a}^2-\text{b}^2}$ $\Rightarrow\ \text{f(x)}=\frac{1}{\text{a}^2-\text{b}^2}\times\frac{1}{2}\Big[\frac{2\text{a}}{\text{x}}-10\text{a}+10\text{b}-2\text{bx}\Big]$ $\Rightarrow\ \text{f(x)}=\frac{1}{\text{a}^2-\text{b}^2}\Big[\frac{\text{a}}{\text{x}}-5\text{a}+5\text{b}-\text{bx}\Big]$ $\therefore\ \text{f(x)}=\frac{1}{\text{a}^2-\text{b}^2}\Big\{\frac{\text{a}}{\text{x}}-\text{bx}-5\text{a}-5\text{b}\Big\}$ $\Rightarrow\ \text{f(x)}=\frac{1}{\text{a}^2-\text{b}^2}\Big\{\frac{\text{a}}{\text{x}}-\text{bx}\Big\}-\frac{5(\text{a}-\text{b})}{\text{a}^2-\text{b}^2}$ $\Rightarrow\ \text{f(x)}=\frac{1}{\text{a}^2-\text{b}^2}\Big\{\frac{\text{a}}{\text{x}}-\text{bx}\Big\}-\frac{5(\text{a}-\text{b})}{(\text{a}-\text{b})(\text{a}+\text{b})}$ $\Rightarrow\ \text{f(x)}=\frac{1}{\text{a}^2-\text{b}^2}\Big\{\frac{\text{a}}{\text{x}}-\text{bx}\Big\}-\frac{5}{\text{a}+\text{b}}$
View full question & answer→Question 24 Marks
Let $\text{f}:[0,\infty)\rightarrow\text{R}$ and $\text{g}:\text{R}\rightarrow\text{R}$ be defined by $\text{f(x)}=\sqrt{\text{x}}$ and g(x) = x. Find f + g, g - g, fg and $\frac{\text{f}}{\text{g}}$
Answer$\text{f}+\text{g}:[0,\infty)\rightarrow\text{R}$ defined by $(\text{f}+\text{g})(\text{x})=\sqrt{\text{x}}+\text{x}$ $\text{f}-\text{g}:[0,\infty)\rightarrow\text{R}$ defined by $(\text{f}-\text{g})(\text{x})=\sqrt{\text{x}}-\text{x}$ $\text{fg}:[0,\infty)\rightarrow\text{R}$ defined by $\Big(\frac{\text{f}}{\text{g}}\Big)(\text{x})=\frac{1}{\sqrt{\text{x}}}$
View full question & answer→Question 34 Marks
Let f and g be two real functions defined by $\text{f(x)}=\sqrt{\text{x}+1}$ and $\text{g(x)}=\sqrt{9-\text{x}^2}$ Then describe the following functions: $\frac{\text{g}}{\text{f}}$
AnswerWe have, $\text{f(x)}=\sqrt{\text{x}+1}$ and $\text{g(x)}=\sqrt{9-\text{x}^2}$ We observe that $\text{f(x)}=\sqrt{\text{x}+1}$ is defined for all $\text{x}\geq-1$ So, domain $\text{f}=[-1,\infty]$ Clearly, $\text{g(x)}=\sqrt{9-\text{x}^2}$ is defined for $9-\text{x}^2\geq0$ $\Rightarrow\text{x}^2-9\leq0$ $\Rightarrow\text{x}^2-3^2\leq0$ $\Rightarrow\text{x}\in[-3,3]$ $\therefore\ \text{domain(g)}=[-3,3]$ Now, $\text{domain(f)}\cap\text{domain(g)}$ $=[-1,\infty]\cap[-3,3]$ $=[-1,3]$ We have, $\text{f(x)}=\sqrt{\text{x}+1}$ $\therefore\ \sqrt{\text{x}+1}=0$ $\Rightarrow\text{x}+1=0$ $\Rightarrow\text{x}=-1$ So, domain $\Big(\frac{\text{g}}{\text{f}}\Big)=[-1,3]-\{-1\}=[-1,3]$ $\therefore\ \frac{\text{f}}{\text{g}}:[-1,3]\rightarrow\text{R}$ is given by $\frac{\text{g}}{\text{f}}(\text{x})=\frac{\text{g(x)}}{\text{f(x)}}=\frac{\sqrt{9-\text{x}^2}}{\sqrt{\text{x}+1}}$
View full question & answer→Question 44 Marks
Let f and g be two real functions defined by $\text{f(x)}=\sqrt{\text{x}+1}$ and $\text{g(x)}=\sqrt{9-\text{x}^2}$ Then describe the following functions: $2\text{f}-\sqrt{5}\text{g}$
AnswerWe have, $\text{f(x)}=\sqrt{\text{x}+1}$ and $\text{g(x)}=\sqrt{9-\text{x}^2}$ We observe that $\text{f(x)}=\sqrt{\text{x}+1}$ is defined for all $\text{x}\geq-1$ So, domain $\text{f}=[-1,\infty]$ Clearly, $\text{g(x)}=\sqrt{9-\text{x}^2}$ is defined for $9-\text{x}^2\geq0$ $\Rightarrow\text{x}^2-9\leq0$ $\Rightarrow\text{x}^2-3^2\leq0$ $\Rightarrow\text{x}\in[-3,3]$ $\therefore\ \text{domain(g)}=[-3,3]$ Now, $\text{domain(f)}\cap\text{domain(g)}$ $=[-1,\infty]\cap[-3,3]$ $=[-1,3]$ $2\text{f}-\sqrt{5}\text{g}:[-,3]\rightarrow\text{R}$ defined by $\big(2\text{f}-\sqrt{5}\text{g}\big)(\text{x})=2\sqrt{\text{x}+1}-\sqrt{5}\sqrt{9-\text{x}^2}$ $=2\sqrt{\text{x}+1}-\sqrt{45-5\text{x}^2}$
View full question & answer→Question 54 Marks
If $\text{f(x)}=\log_\text{e}(1-\text{x})$ and $\text{g(x)}=[\text{x}],$ then determine the following functions: $\frac{\text{f}}{8}$
AnswerWe have, $\text{f(x)}=\log_\text{e}(1-\text{x})$ and $\text{g(x)}=[\text{x}]$ $\text{f(x)}=\log_\text{e}(1-\text{x})$ is defined, if 1 - x > 0 $\Rightarrow1>\text{x}$ $\Rightarrow\text{x}<1$ $\Rightarrow\text{x}\in(-\infty,1)$ $\therefore\text{ Domain(f)}=(-\infty,1)$ $\text{g(x)}=[\text{x}]$ is defined for all $\text{x}\in\text{R}$ $\therefore\ \text{Domain(g)}=\text{R}$ $\therefore\ \text{Domain(f)}\cap\text{R}\text{ Domain(g)}=(-\infty,1)\cap\text{R}$ $=(-\infty,1)$ $\text{g(x)}=[\text{x}]$ $\therefore\ [\text{x}]=0$ $\Rightarrow\text{x}\in(0,1)$ So, domain $\Big(\frac{\text{f}}{\text{g}}\Big)=\text{domain(f)}\cap\text{domain(g)}-\{\text{x}:\text{g(x)}=0\}$ $\therefore\ \frac{\text{f}}{\text{g}}:(-\infty,0)\rightarrow\text{R}$ defined by $\Big(\frac{\text{f}}{\text{g}}\Big)(\text{x})=\frac{\log_\text{e}(1-\text{x})}{[\text{x}]}$
View full question & answer→Question 64 Marks
Let f and g be two real functions defined by $\text{f(x)}=\sqrt{\text{x}+1}$ and $\text{g(x)}=\sqrt{9-\text{x}^2}$ Then describe the following functions: g - f
AnswerWe have, $\text{f(x)}=\sqrt{\text{x}+1}$ and $\text{g(x)}=\sqrt{9-\text{x}^2}$ We observe that $\text{f(x)}=\sqrt{\text{x}+1}$ is defined for all $\text{x}\geq-1$ So, domain $\text{f}=[-1,\infty]$ Clearly, $\text{g(x)}=\sqrt{9-\text{x}^2}$ is defined for $9-\text{x}^2\geq0$ $\Rightarrow\text{x}^2-9\leq0$ $\Rightarrow\text{x}^2-3^2\leq0$ $\Rightarrow\text{x}\in[-3,3]$ $\therefore\ \text{domain(g)}=[-3,3]$ Now, $\text{domain(f)}\cap\text{domain(g)}$ $=[-1,\infty]\cap[-3,3]$ $=[-1,3]$ f - g : [-3] → R is given by (g - f)(x) = g(x) - g(x) $=\sqrt{9-\text{x}^2}-\sqrt{\text{x}+1}$
View full question & answer→Question 74 Marks
Let f and g be two real functions defined by $\text{f(x)}=\sqrt{\text{x}+1}$ and $\text{g(x)}=\sqrt{9-\text{x}^2}$ Then describe the following functions: $\text{f}^2+7\text{f}$
AnswerWe have, $\text{f(x)}=\sqrt{\text{x}+1}$ and $\text{g(x)}=\sqrt{9-\text{x}^2}$ We observe that $\text{f(x)}=\sqrt{\text{x}+1}$ is defined for all $\text{x}\geq-1$ So, domain $\text{f}=[-1,\infty]$ Clearly, $\text{g(x)}=\sqrt{9-\text{x}^2}$ is defined for $9-\text{x}^2\geq0$ $\Rightarrow\text{x}^2-9\leq0$ $\Rightarrow\text{x}^2-3^2\leq0$ $\Rightarrow\text{x}\in[-3,3]$ $\therefore\ \text{domain(g)}=[-3,3]$ Now, $\text{domain(f)}\cap\text{domain(g)}$ $=[-1,\infty]\cap[-3,3]$ $=[-1,3]$ $\text{f}^2+7\text{f}:[-1,\infty]\rightarrow\text{R}$ defined by $(\text{f}^2+7\text{f})(\text{x})=\text{f}^2(\text{x})+7\text{f}(\text{x})$ $\big[\because\text{D(f)}=[-1,\infty]\big]$ $=\Big(\sqrt{\text{x}+1}\Big)^2+7\sqrt{\text{x}+1}$ $=\text{x}+1+7\sqrt{\text{x}+1}$
View full question & answer→Question 84 Marks
If $\text{f(x)}=\log_\text{e}(1-\text{x})$ and $\text{g(x)}=[\text{x}],$ then determine the following functions: (f + g)(-1)
AnswerWe have, $\text{f(x)}=\log_\text{e}(1-\text{x})$ and $\text{g(x)}=[\text{x}]$ $\text{f(x)}=\log_\text{e}(1-\text{x})$ is defined, if 1 - x > 0 $\Rightarrow1>\text{x}$ $\Rightarrow\text{x}<1$ $\Rightarrow\text{x}\in(-\infty,1)$ $\therefore\text{ Domain(f)}=(-\infty,1)$ $\text{g(x)}=[\text{x}]$ is defined for all $\text{x}\in\text{R}$ $\therefore\ \text{Domain(g)}=\text{R}$ $\therefore\ \text{Domain(f)}\cap\text{R}\text{ Domain(g)}=(-\infty,1)\cap\text{R}$ $=(-\infty,1)$ Now, $(\text{f}+\text{g})(-1)=\text{f}(-1)+\text{g}(-1)$ $=\log_\text{e}(1-(-1))+[-1]$ $=\log_\text{e}2-1$ $(\text{f}+\text{g})(-1)=\log_\text{e}2-1$
View full question & answer→Question 94 Marks
Let $f(x) = x^2$ and $g(x) = 2x + 1$ be two real functions. Find $(f + g)(x), (f - g)(x), (fg)(x)$ and $\Big(\frac{\text{f}}{\text{g}}\Big)\text{x}$
Answer$(\text{f}+\text{g}):\text{R}\rightarrow[0,\infty)$ defined by $(f + g)(x) = x^2 + 2x + 1 = (x + 1)^2 (\text{f}-\text{g}):\text{R}\rightarrow\text{R}$ defined by $(\text{f}-\text{g})(\text{x})=\text{x}^2-2\text{x}-1$ $(\text{fg}):\text{R}\rightarrow\text{R}$ defined by $(fg)(x) = 2x^3 + x^2 \Big(\frac{\text{f}}{\text{g}}\Big):\text{R}\rightarrow\text{R}$ defined by $\Big(\frac{\text{f}}{\text{g}}\Big)(\text{x})=\frac{\text{x}^2}{2\text{x}+1}$
View full question & answer→Question 104 Marks
Let f and g be two real functions defined by $\text{f(x)}=\sqrt{\text{x}+1}$ and $\text{g(x)}=\sqrt{9-\text{x}^2}$ Then describe the following functions: $\frac{5}{\text{g}}$
AnswerWe have, $\text{f(x)}=\sqrt{\text{x}+1}$ and $\text{g(x)}=\sqrt{9-\text{x}^2}$ We observe that $\text{f(x)}=\sqrt{\text{x}+1}$ is defined for all $\text{x}\geq-1$ So, domain $\text{f}=[-1,\infty]$ Clearly, $\text{g(x)}=\sqrt{9-\text{x}^2}$ is defined for $9-\text{x}^2\geq0$ $\Rightarrow\text{x}^2-9\leq0$ $\Rightarrow\text{x}^2-3^2\leq0$ $\Rightarrow\text{x}\in[-3,3]$ $\therefore\ \text{domain(g)}=[-3,3]$ Now, $\text{domain(f)}\cap\text{domain(g)}$ $=[-1,\infty]\cap[-3,3]$ $=[-1,3]$ We have, $\text{g(x)}=\sqrt{9-\text{x}^2}$ $\therefore\ 9-\text{x}^2=0$ $\Rightarrow\text{x}^2-9=0$ $\Rightarrow(\text{x}-3)(\text{x}+3) =0$ $\Rightarrow\text{x}=\pm3$ So, domain $\Big(\frac{1}{\text{g}}\Big)=[-3,3]-\{-3,3\}=(-3,3)$ $\therefore\ \frac{5}{\text{g}}=(-3,3)\rightarrow\text{R}$ defined by $\Big(\frac{5}{\text{g}}\Big)(\text{x})=\frac{5}{\sqrt{9-\text{x}^2}}$
View full question & answer→Question 114 Marks
Let f and g be two real functions defined by $\text{f(x)}=\sqrt{\text{x}+1}$ and $\text{g(x)}=\sqrt{9-\text{x}^2}$ Then describe the following functions: fg
AnswerWe have, $\text{f(x)}=\sqrt{\text{x}+1}$ and $\text{g(x)}=\sqrt{9-\text{x}^2}$ We observe that $\text{f(x)}=\sqrt{\text{x}+1}$ is defined for all $\text{x}\geq-1$ So, domain $\text{f}=[-1,\infty]$ Clearly, $\text{g(x)}=\sqrt{9-\text{x}^2}$ is defined for $9-\text{x}^2\geq0$ $\Rightarrow\text{x}^2-9\leq0$ $\Rightarrow\text{x}^2-3^2\leq0$ $\Rightarrow\text{x}\in[-3,3]$ $\therefore\ \text{domain(g)}=[-3,3]$ Now, $\text{domain(f)}\cap\text{domain(g)}$ $=[-1,\infty]\cap[-3,3]$ $=[-1,3]$ fg : [-1, 3] → R is given by (fg)(x) = f(x) × g(x) $=\sqrt{\text{x}+1}\times\sqrt{9-\text{x}^2}$ $=\sqrt{9+9\text{x}-\text{x}^2-\text{x}^3}$
View full question & answer→Question 124 Marks
If $\text{f(x)}=\log_\text{e}(1-\text{x})$ and $\text{g(x)}=[\text{x}],$ then determine the following functions: $\frac{\text{g}}{\text{f}}$
AnswerWe have, $\text{f(x)}=\log_\text{e}(1-\text{x})$ and $\text{g(x)}=[\text{x}]$ $\text{f(x)}=\log_\text{e}(1-\text{x})$ is defined, if 1 - x > 0 $\Rightarrow1>\text{x}$ $\Rightarrow\text{x}<1$ $\Rightarrow\text{x}\in(-\infty,1)$ $\therefore\text{ Domain(f)}=(-\infty,1)$ $\text{g(x)}=[\text{x}]$ is defined for all $\text{x}\in\text{R}$ $\therefore\ \text{Domain(g)}=\text{R}$ $\therefore\ \text{Domain(f)}\cap\text{R}\text{ Domain(g)}=(-\infty,1)\cap\text{R}$ $=(-\infty,1)$ we have, $\text{f(x)}=\log_\text{e}(1-\text{x})$ $\Rightarrow\frac{1}{\text{f(x)}}=\frac{1}{\log_\text{e}(1-\text{x})}$ $\therefore\ \frac{1}{\text{f(x)}}$ is defined if $\log_\text{e}(1-\text{x})$ is defined and $\log_\text{e}(1-\text{x})\neq0$ $\Rightarrow1-\text{x}>0$ and $1-\text{x}\neq0$ $\Rightarrow\text{x}<1$ and $\text{x}\neq0$ $\Rightarrow\text{x}\in(-\infty,0)\cap(0,1)$ $\therefore\ \text{domain}\Big(\frac{\text{g}}{\text{f}}\Big)=(-\infty,0)\cup(0,1)$ $\frac{\text{g}}{\text{f}}:(-\infty,0)\cap(0,1)\rightarrow\text{R}$ defined by $\Big(\frac{\text{g}}{\text{f}}\Big)(\text{x})=\frac{[\text{x}]}{\log_\text{e}(1-\text{x})}$
View full question & answer→Question 134 Marks
Find $\text{f}+\text{g},\text{ f}-\text{g},\text{ cf}(\text{c}\in\text{R},\text{c}\neq0),\text{ fg},\frac{1}{\text{f}}$ and $\frac{\text{f}}{\text{g}}$ in the following: $\text{f(x)}=\sqrt{\text{x}-1}\text{ and }\text{g(x)}=\sqrt{\text{x}+1}$
AnswerWe have, $\text{f(x)}=\sqrt{\text{x}-1}$ and $\text{g(x)}=\sqrt{\text{x}+1}$ Now, $\text{f}+\text{g}:(1,\infty)\rightarrow\text{R}$ is defined by (f + g)(x) $=\sqrt{\text{x}-1}+\sqrt{\text{x}+1}$ $\text{f}-\text{g}:(1,\infty)\rightarrow\text{R}$ is defined by (f - g)(x) = f(x) - g(x) $=\sqrt{\text{x}-1}-\sqrt{\text{x}+1}$ $\text{cf}:(1,\infty)\rightarrow\text{R}$ is defined by $(\text{cf)(x)}=\text{c}\sqrt{\text{x}-1}$ $(\text{fg}):(1,\infty)\rightarrow\text{R}$ is defined (fg)(x) $=\big(\sqrt{\text{x}-1}\big)\big(\sqrt{\text{x}+1}\big)=\sqrt{\text{x}^2-1}$ $\frac{1}{\text{f}}:(1,\infty)\rightarrow\text{R}$ is defined by $\Big(\frac{1}{\text{f}}\Big)(\text{x})=\frac{1}{\sqrt{\text{x}-1}}$ $\frac{\text{f}}{\text{g}}:\big(1,\infty)\rightarrow\text{R}$ is defined by $\Big(\frac{\text{f}}{\text{g}}\Big)(\text{x})=\sqrt{\frac{\text{x}+1}{\text{x}-1}}$
View full question & answer→Question 144 Marks
If $\text{f(x)}=\log_\text{e}(1-\text{x})$ and $\text{g(x)}=[\text{x}],$ then determine the following functions: fg
AnswerWe have, $\text{f(x)}=\log_\text{e}(1-\text{x})$ and $\text{g(x)}=[\text{x}]$ $\text{f(x)}=\log_\text{e}(1-\text{x})$ is defined, if 1 - x > 0 $\Rightarrow1>\text{x}$ $\Rightarrow\text{x}<1$ $\Rightarrow\text{x}\in-\infty,1)$ $\therefore\text{ Domain(f)}=(-\infty,1)$ $\text{g(x)}=[\text{x}]$ is defined for all $\text{x}\in\text{R}$ $\therefore\ \text{Domain(g)}=\text{R}$ $\therefore\ \text{Domain(f)}\cap\text{R}\text{ Domain(g)}=(-\infty,1)\cap\text{R}$ $=(-\infty,1)$ $\text{f}+\text{g}:(-\infty,1)\rightarrow\text{R}$ defined by $(\text{fg)(x)}=\text{f(x)}\times\text{g(x)}$ $=\log_\text{e}(1-\text{x})\times[\text{x}]$ $=[\text{x}]\log_\text{e}(1-\text{x})$
View full question & answer→Question 154 Marks
Let f and g be two real functions defined by $\text{f(x)}=\sqrt{\text{x}+1}$ and $\text{g(x)}=\sqrt{9-\text{x}^2}$ Then describe the following functions: $\frac{\text{f}}{\text{g}}$
AnswerWe have, $\text{f(x)}=\sqrt{\text{x}+1}$ and $\text{g(x)}=\sqrt{9-\text{x}^2}$ We observe that $\text{f(x)}=\sqrt{\text{x}+1}$ is defined for all $\text{x}\geq-1$ So, domain $\text{f}=[-1,\infty]$ Clearly, $\text{g(x)}=\sqrt{9-\text{x}^2}$ is defined for $9-\text{x}^2\geq0$ $\Rightarrow\text{x}^2-9\leq0$ $\Rightarrow\text{x}^2-3^2\leq0$ $\Rightarrow\text{x}\in[-3,3]$ $\therefore\ \text{domain(g)}=[-3,3]$ Now, $\text{domain(f)}\cap\text{domain(g)}$ $=[-1,\infty]\cap[-3,3]$ $=[-1,3]$ We have, $\text{g(x)}=\sqrt{9-\text{x}^2}$ $\therefore\ 9-\text{x}^2=0$ $\Rightarrow\text{x}^2-9=0$ $\Rightarrow(\text{x}-3)(\text{x}+3)=0$ $\Rightarrow\text{x}=\pm3$ So, domain $\Big(\frac{\text{f}}{\text{g}}\Big)=[-1,3]-[-3,3]=[-1,3]$ $\therefore\ \frac{\text{f}}{\text{g}}:[-1,3]\rightarrow\text{R}$ is given by $\Big(\frac{\text{f}}{\text{g}}\Big)(\text{x})=\frac{\text{f(x)}}{\text{g(x)}}=\frac{\sqrt{\text{x}+1}}{\sqrt{9-\text{x}^2}}$
View full question & answer→Question 164 Marks
If $f(x) = x^2 - 3x + 4$, then find the values of x satisfying the equation $f(x) = f(2x + 1)$.
AnswerWe have, $f(x)=x^2-3 x+4$
Now, $f(2 x+1)=(2 x+1)^2-3(2 x+1)+4=4 x^2+1+4 x-6 x-3+4=4 x^2-2 x+2$ It is given that $f(x)=f(2 x+1)$
$\Rightarrow x^2-3 x+4=4 x^2-2 x+2$
$\Rightarrow 0=4 x^2-x^2-2 x+3 x+2-4$
$\Rightarrow 3 x^2+x-2=0$
$\Rightarrow 3 x^2+3 x-$ $2 x-2=0$
$\Rightarrow 3 x(x+1)-2(x+1)=0$
$\Rightarrow(x+1)(3 x-2)=0$
$\Rightarrow x+1=0$ or $3 x-2=0$
$\Rightarrow x=-1$ or $x=\frac{2}{3}$
View full question & answer→Question 174 Marks
The function f is defined by $\text{f(x)}=\begin{cases}1-\text{x},&\text{x}<0\\1,&\text{x}=0\\\text{x}+1,&\text{x>0}\end{cases}$ Draw the graph of f(x).
AnswerLet, $\text{f(x)}=\begin{cases}1-\text{x},&\text{x}<0\\1,&\text{x}=0\\\text{x}+1,&\text{x>0}\end{cases}$ The graph of f(x) for x < 0 is the part of the line y = 1 - x that lies to the left of origin. The graph of f(x) for x > 0 is the part of the line y = 1 + x that lies to the right of origin. For x = 0, the graph of f(x) represents the point (0, 1) The graph of f(x) is shown below.
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If $\text{f(x)}=\log_\text{e}(1-\text{x})$ and $\text{g(x)}=[\text{x}],$ then determine the following functions: $\Big(\frac{\text{f}}{\text{g}}\Big)\Big(\frac{1}{2}\Big)$
View full question & answer→Question 194 Marks
If $\text{f(x)}=\frac{1+\text{x}}{1-\text{x}},$ show that $\text{f}\big[\text{f}\text{(x)}\big]=\text{x}$
AnswerWe have, $\text{f(x)}=\frac{1+\text{x}}{1-\text{x}}$ Now, $\text{f}\big[\text{f}\text{(x)}\big]=\text{f}\Big(\frac{\text{x}+1}{1-\text{x}}\Big)$ $=\frac{\big(\frac{\text{x}+1}{1-\text{x}}\big)+1}{\big(\frac{\text{x}+1}{1-\text{x}}\big)-1}$ $=\frac{\frac{\text{x}+1+{\text{x}}-1}{{\text{x}}-1}}{\frac{\text{x}+1-1(\text{x}-1)}{\text{x}-1}}$ $=\frac{\frac{2\text{x}}{\text{x}-1}}{\frac{\text{x}+1-\text{x}+1}{\text{x}-1}}$ $=\frac{2\text{x}}{2}$ $=\text{x}$ $\therefore\ \text{f}\big[\text{f}\text{(x)}\big]=\text{x}$ Hence, proved.
View full question & answer→Question 204 Marks
Let f, g : R → R be defined, respectively by f(x) = x + 1 and g(x) = 2x - 3. Find f + g, f - g and $\frac{\text{f}}{\text{g}}$
Answerf, g : R → R defined by (f + g)(x) = 3x - 2 f, g : R → R defined by (f - g)(x) = -x + 4 $\text{f}:\text{R}-\Big\{\frac{3}{2}\Big\}\rightarrow\text{R}$ defined by $\frac{\text{f}}{\text{g}}(\text{x})=\frac{\text{x}+1}{2\text{x}-3}$
View full question & answer→Question 214 Marks
If $\text{f(x)}=\log_\text{e}(1-\text{x})$ and $\text{g(x)}=[\text{x}],$ then determine the following functions: $\Big(\frac{\text{g}}{\text{f}}\Big)\Big(\frac{1}{2}\Big)$
AnswerWe have, $\text{f(x)}=\log_\text{e}(1-\text{x})$ and $\text{g(x)}=[\text{x}]$ $\text{f(x)}=\log_\text{e}(1-\text{x})$ is defined, if 1 - x > 0 $\Rightarrow1>\text{x}$ $\Rightarrow\text{x}<1$ $\Rightarrow\text{x}\in(-\infty,1)$ $\therefore\text{ Domain(f)}=(-\infty,1)$ $\text{g(x)}=[\text{x}]$ is defined for all $\text{x}\in\text{R}$ $\therefore\ \text{Domain(g)}=\text{R}$ $\therefore\ \text{Domain(f)}\cap\text{R}\text{ Domain(g)}=(-\infty,1)\cap\text{R}$ $=(-\infty,1)$ $\Big(\frac{\text{g}}{\text{f}}\Big)\Big(\frac{1}{2}\Big)=\frac{\big[\frac{1}{2}\big]}{\log_\text{e}\big(1-\frac{1}{2}\big)}=0$
View full question & answer→Question 224 Marks
If $\text{f(x)}=\frac{1}{1-\text{x}},$ show that $\text{f}\big[\text{f}\{\text{f(x)}\}\big]=\text{x}$
AnswerWe have, $\text{f(x)}=\frac{1}{1-\text{x}}$ Now, $\text{f}\{\text{f}(\text{x})\}=\text{f}\Big\{\frac{1}{1-\text{x}}\Big\}$ $=\frac{1}{1-\frac{1}{1-\text{x}}}$ $=\frac{1}{\frac{1-\text{x}-1}{1-\text{x}}}$ $=\frac{1-\text{x}}{-\text{x}}$ $=\frac{\text{x}-1}{\text{x}}$ $\therefore\ \text{f}\big[\text{f}\{\text{f(x)}\}\big]=\text{f}\Big\{\frac{1}{1-\text{x}}\Big\}$ $=\frac{1}{1-\big(\frac{\text{x}-1}{\text{x}}\big)}$ $=\frac{1}{\frac{\text{x}-\text{x}+1}{\text{x}}}$ $=\frac{\text{x}}{1}$ $=\text{x}$ $\therefore\ \text{f}\big[\text{f}\{\text{f(x)}\}\big]=\text{x}$ Hence, proved.
View full question & answer→Question 234 Marks
If f, g and h are real functions defined by $\text{f(x)}=\sqrt{\text{x}+1},\text{ g(x)}=\frac{1}{\text{x}}$ and $h(x) = 2x^2 - 3$, find the values of $(2f + g - h)(1)$ and $(2f + g - h) (0)$.
AnswerWe have, $\text{f(x)}=\sqrt{\text{x}+1},\text{ g(x)}=\frac{1}{\text{x}}$ and $h(x) = 2x^2 - 3$ Clearly, f(x) is defined for $\text{x}+1\geq0$
$\Rightarrow\text{x}\geq-1$
$\Rightarrow\text{x}\in[-1,\infty]$ g(x) is defined for $\text{x}\neq0$
$\Rightarrow\text{x}\in\text{R}-\{0\}$ and, h(x) is defined forll all $\text{x}\in\text{R}$
$\therefore\text{ Domain(f)}\cap\text{Domain(g)}\cap\text{Domain(h)}=[-1,\infty]-\{0\}$ Clearly, $2\text{f}+\text{g}-\text{h}:[-1,\infty]-\{0\}\rightarrow\text{R}$ is given by $(2\text{f}+\text{g}-\text{h})(\text{x})=2\text{f(x)}+\text{g(x)}-\text{h(x)}$
$=2\sqrt{\text{x}+1}+\frac{1}{\text{x}}-2\text{x}^2+3$
$\therefore\ (2\text{f}+\text{g}-\text{h})(1)=2\sqrt{1+1}+\frac{1}{1}-2\times(1)^2+3$
$=2\sqrt{2}+1-2+3$
$=2\sqrt{2}+4-2$
$=2\sqrt{2}+2$and $(2\text{f}+\text{g}-\text{h})(0)$ does not exist, it is not list in the domain $\text{x}\in[-1,\infty]-\{0\}$
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