Question

If for a non-zero $x$, $af\left(x\right)+bf\left(\frac{1}{x}\right)=\frac{1}{x}-5,$ where $a\ne b$, then ${\int }_1^{2}f\left(x\right)dx=$

• A. $\frac{1}{a^2+b^2}(a\log 2+5a+\frac{7b}{2})$
• B. $\frac{1}{a^2-b^2}(a\log 2-5a+\frac{7b}{2})$
• C. $-\frac{1}{a^2+b^2}(a\log 2+5a-\frac{7b}{2})$
• D. none of these

Answer 1

Answer: (B)

$\frac{1}{a^2-b^2}(a\log 2-5a+\frac{7b}{2})$

Given $af\left(x\right)+bf\left(\frac{1}{x}\right)=\frac{1}{x}-5$ ...(1)
Replacing $x\to \frac{1}{x}$ in (1), we get
$af\left(\frac{1}{x}\right)+bf\left(x\right)=x-5$ ...(2)
On multiplying equation (1) by $a$ and equation (2) by $b$ respectively
$a^{2}f\left(x\right)+abf\left(\frac{1}{x}\right)=a(\frac{1}{x}-5)$ ...(3)
$abf\left(\frac{1}{x}\right)+b^{2}f\left(x\right)=b(x-5)$ ...(4)
On subtracting (4) from (3)
$(a^2-b^2)f\left(x\right)=\frac{a}{x}-bx-5a+5b\Rightarrow f\left(x\right)=\frac{1}{(a^2-b^2)}(\frac{a}{x}-bx-5a+5b)\Rightarrow {\int }_1^{2}f\left(x\right)dx=\frac{1}{(a^2-b^2)}{\int }_1^2(\frac{a}{x}-bx-5a+5b)dx=\frac{1}{(a^2-b^2)}{[a\log \left|x\right|-\frac{b}{2}{x}^2-5(a-b)x]}_1^2=\frac{1}{(a^2-b^2)}[a\log 2-2b-10(a-b)-a\log 1+\frac{b}{2}+5(a-b)]=\frac{1}{(a^2-b^2)}[a\log 2-5a+\frac{7}{2}b]$