Solve the Following Question.(5 Marks)

Question 15 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).

Answer

We 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
$2\text{f(x)}=\frac{1}{\text{a}+\text{b}}\Big[\frac{1}{\text{x}}+\text{x}-10\Big]+\frac{1}{\text{a}-\text{b}}\Big[\frac{1}{\text{x}}-\text{x}\Big]$
$\Rightarrow\ 2\text{f(x)}=\frac{(\text{a}-\text{b})\big[\frac{1}{\text{x}}+\text{x}-10\big]+(\text{a}+\text{b})\big[\frac{1}{\text{x}}-\text{x}\big]}{(\text{a}+\text{b})(\text{a}-\text{b})}$
$\Rightarrow\ 2\text{f(x)}=\frac{\frac{\text{a}}{\text{x}}+\text{ax}-10\text{a}-\frac{\text{b}}{\text{x}}-\text{bx}+10\text{b}+\frac{\text{a}}{\text{x}}-\text{ax}+\frac{\text{b}}{\text{x}}-\text{bx}}{\text{a}^2-\text{b}^2}$
$\Rightarrow\ 2\text{f(x)}=\frac{\frac{2\text{a}}{\text{x}}-10\text{a}+10\text{b}-2\text{bx}}{\text{a}^2-\text{b}^2}$
$\Rightarrow\ \text{f(x)}=\frac{1}{\text{a}^2-\text{b}^2}\times\frac{1}{2}\Big[\frac{2\text{a}}{\text{x}}-10\text{a}+10\text{b}-2\text{bx}\Big]$
$\Rightarrow\ \text{f(x)}=\frac{1}{\text{a}^2-\text{b}^2}\Big[\frac{\text{a}}{\text{x}}-5\text{a}+5\text{b}-\text{bx}\Big]$
$\therefore\ \text{f(x)}=\frac{1}{\text{a}^2-\text{b}^2}\Big\{\frac{\text{a}}{\text{x}}-\text{bx}-5\text{a}-5\text{b}\Big\}$
$\Rightarrow\ \text{f(x)}=\frac{1}{\text{a}^2-\text{b}^2}\Big\{\frac{\text{a}}{\text{x}}-\text{bx}\Big\}-\frac{5(\text{a}-\text{b})}{\text{a}^2-\text{b}^2}$
$\Rightarrow\ \text{f(x)}=\frac{1}{\text{a}^2-\text{b}^2}\Big\{\frac{\text{a}}{\text{x}}-\text{bx}\Big\}-\frac{5(\text{a}-\text{b})}{(\text{a}-\text{b})(\text{a}+\text{b})}$
$\Rightarrow\ \text{f(x)}=\frac{1}{\text{a}^2-\text{b}^2}\Big\{\frac{\text{a}}{\text{x}}-\text{bx}\Big\}-\frac{5}{\text{a}+\text{b}}$

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Question 25 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 $\Big(\frac{\text{f}}{\text{g}}\Big)(\text{x})=\frac{1}{\sqrt{\text{x}}}$

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Question 35 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:
$\frac{\text{g}}{\text{f}}$

Answer

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

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Question 45 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:
$2\text{f}-\sqrt{5}\text{g}$

Answer

We 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 $\big(2\text{f}-\sqrt{5}\text{g}\big)(\text{x})=2\sqrt{\text{x}+1}-\sqrt{5}\sqrt{9-\text{x}^2}$
$=2\sqrt{\text{x}+1}-\sqrt{45-5\text{x}^2}$

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Question 55 Marks

If $\text{f(x)}=\log_\text{e}(1-\text{x})$ and $\text{g(x)}=[\text{x}],$ then determine the following functions:
$\frac{\text{f}}{8}$

Answer

We 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 $\Big(\frac{\text{f}}{\text{g}}\Big)(\text{x})=\frac{\log_\text{e}(1-\text{x})}{[\text{x}]}$

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Maths Solve the Following Question.(5 Marks)

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