If for non-zero x, af(x) + bfleft( {frac{1}{x | vedclass

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(B) Given the equation: $af(x) + bf\left( {\frac{1}{x}} \right) = \frac{1}{x} - 5$ ... $(i)$Replacing $x$ with $\frac{1}{x}$ in $(i)$,we get: $af\left

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

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(B) Given the equation: $af(x) + bf\left( {\frac{1}{x}} \right) = \frac{1}{x} - 5$ ... $(i)$Replacing $x$ with $\frac{1}{x}$ in $(i)$,we get: $af\left( {\frac{1}{x}} \right) + bf(x) = x - 5$ ... $(ii)$To eliminate $f\left( {\frac{1}{x}} \right)$,multiply $(i)$ by $a$ and $(ii)$ by $b$: $a^2 f(x) + ab f\left( {\frac{1}{x}} \right) = \frac{a}{x} - 5a$$b^2 f(x) + ab f\left( {\frac{1}{x}} \right) = bx - 5b$Subtracting the second from the first: $(a^2 - b^2) f(x) = \frac{a}{x} - bx - 5a + 5b$Integrating both sides from $1$ to $2$: $(a^2 - b^2) \int_1^2 f(x) dx = \int_1^2 \left( \frac{a}{x} - bx - 5(a - b) \right) dx$$= \left[ a \log |x| - \frac{b x^2}{2} - 5(a - b)x \right]_1^2$$= \left( a \log 2 - \frac{b(4)}{2} - 5(a - b)(2) \right) - \left( a \log 1 - \frac{b(1)}{2} - 5(a - b)(1) \right)$$= a \log 2 - 2b - 10a + 10b - 0 + \frac{b}{2} + 5a - 5b$$= a \log 2 - 5a + \frac{7}{2}b$Thus,$\int_1^2 f(x) dx = \frac{1}{a^2 - b^2} \left[ a \log 2 - 5a + \frac{7}{2}b \right]$.

If for non-zero $x,$ $af(x) + bf\left( {\frac{1}{x}} \right) = \frac{1}{x} - 5,$ where $a \ne b,$ then $\int_1^2 {f(x)\,dx = } $

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