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.
$$\frac{{\left( {\log \,2 - 5} \right)a + \frac{{13}}{2}b}}{{{a^2} - {b^2}}}$$
B.
$$\frac{{\left( {\log \,2 - 5} \right)a + \frac{{7b}}{2}}}{{{a^2} - {b^2}}}$$
C.
$$\frac{{\left( {5 - \log \,2} \right)a + \frac{{7b}}{2}}}{{{a^2} - {b^2}}}$$
D.
none of these
Answer :
$$\frac{{\left( {\log \,2 - 5} \right)a + \frac{{7b}}{2}}}{{{a^2} - {b^2}}}$$
Solution :
$$\eqalign{
& af\left( x \right) + bf\left( {\frac{1}{x}} \right) = \frac{1}{x} - 5 \cr
& \Rightarrow af\left( {\frac{1}{x}} \right) + bf\left( x \right) = x - 5 \cr
& {\text{From these, }}\left( {{a^2} - {b^2}} \right)f\left( x \right) = a\left( {\frac{1}{x} - 5} \right) - b\left( {x - 5} \right) \cr
& \therefore \,\int_1^2 {f\left( x \right)dx} = \int_1^2 {\frac{1}{{{a^2} - {b^2}}}\left\{ {\frac{a}{x} - bx - 5a + 5b} \right\}dx} \cr
& = \frac{1}{{{a^2} - {b^2}}}\left[ {a\log \,x - b\frac{{{x^2}}}{2} + 5\left( {b - a} \right)x} \right]_1^2 \cr
& = \frac{{a\log \,2 - 2b + 10\left( {b - a} \right) + \frac{b}{2} - 5\left( {b - a} \right)}}{{{a^2} - {b^2}}} \cr} $$