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Functions — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsFunctions5 Marks
Question
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).
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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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