Question

If $af\left(x\right)+bf\left(\frac{1}{x}\right)=\frac{1}{x}-5,x\ne 0,a\ne b,then{\int }_1^{2}f\left(x\right)dx$ equals

• A. $\frac{(\log 2-5)a+\frac{13}{2}b}{a^2-b^2}$
• B. $\frac{(\log 2-5)a+\frac{7b}{2}}{a^2-b^2}$
• C. $\frac{(5-\log 2)a+\frac{7b}{2}}{a^2-b^2}$
• D. none of these

Answer 1

The correct option is B $\frac{(\log 2-5)a+\frac{7b}{2}}{a^2-b^2}$

Replacing $x\to \frac{1}{x}$ in the given function
We have $af\left(\frac{1}{x}\right)+bf\left(x\right)=x-5$
Eliminating $f\left(\frac{1}{x}\right)$ in the above equation, we get
$(a^2-b^2)f\left(x\right)=\left(\frac{a}{x}\right)-bx-5a+5b\Rightarrow (a^2-b^2){\int }_1^{2}f\left(x\right)dx$
$={\left(a\log \left|x\right|-\frac{{bx}^2}{2}-5(a-b)x\right)}_1^2=a\log 2-5a+\frac{7}{2}b$
$\therefore {\int }_1^{2}f\left(x\right)dx=\frac{1}{a^2-b^2}\left(a\log 2-5a+\frac{7}{2}b\right)$