Solution:
$3\text{f(x)}+5\text{f}\Big(\frac{1}{\text{x}}\Big)=\frac{1}{\text{x}}-3\ ...(\text{i})$
Multiplying (1) by 3,
$15\text{f}\Big(\frac{1}{\text{x}}\Big)+9\text{f(x)}=\frac{3}{\text{x}}-9\ ...(\text{ii})$
Replacing x by $\frac{1}{\text{x}}$ in (i)
$3\text{f}\Big(\frac{1}{\text{x}}\Big)+5\text{f(x)}=\text{x}-3$
Multiplying by 5
$15\text{f}\Big(\frac{1}{\text{x}}\Big)+25\text{f(x)}=5\text{x}-15\ ...(\text{iii})$
Solving (ii) and (iii),
$-16\text{f(x)}=\frac{3}{\text{x}}-5\text{x}+6$
$\Rightarrow\text{f(x)}=\frac{1}{16}\Big(-\frac{3}{\text{x}}+5\text{x}-6\Big)$
Disclaimer: The question in the book has some error, so, none of the options are matching with the solution. The solution is created according to the question given in the book.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Which of the following is not correct?
$\sin\theta=-\frac{1}{5}$
$\cos\theta=1$
$\sec\theta=\frac{1}{2}$
$\tan\theta=20$
If in the expansion of $\Big(\text{x}-\frac{1}{3\text{x}^{3}}\Big)^{9},$ the term independent of x is:
$\text{T}_{3}$
$\text{T}_{4}$
$\text{T}_{5}$
None of these.
If the vertex of the parabola is the point (-3, 0) and the directrix is the line x + 5 = 0, then its equation is:
Equation of horizontal line above x-axis at 5 units from x-axis is: