Question

If for non-zero x, if \(af\left( x \right) + bf\left( {\frac{1}{x}} \right) = \frac{1}{x} - 25\) where a ≠ b then \(\mathop \smallint \limits_1^2 f\left( x \right)dx\) is
1. \(\frac{1}{{{a^2} - {b^2}}}\left[ {a\left( {ln\;2 - 25} \right) + \frac{{47b}}{2}} \right]\)
2. \(\frac{1}{{{a^2} - {b^2}}}\left[ {a\left( {2\;ln\;2 - 25} \right) - \frac{{47b}}{2}} \right]\)
3. \(\frac{1}{{{a^2} - {b^2}}}\left[ {a\left( {2\;ln\;2 - 25} \right) + \frac{{47b}}{2}} \right]\)
4. \(\frac{1}{{{a^2} - {b^2}}}\left[ {a\left( {ln\;2 - 25} \right) - \frac{{47b}}{2}} \right]\)

Answer 1

Correct Answer - Option 1 : \(\frac{1}{{{a^2} - {b^2}}}\left[ {a\left( {ln\;2 - 25} \right) + \frac{{47b}}{2}} \right]\)

Given x as non- zero,

\(af\left( x \right) + bf\left( {\frac{1}{x}} \right) = \frac{1}{x} - 25\)       ---(1)

Consider x as 1/x

\(af\left( {\frac{1}{x}} \right) + bf\left( {\frac{1}{{\frac{1}{x}}}} \right) = x - 25\)

\(af\left( {\frac{1}{x}} \right) + bf\left( x \right) = x - 25\)       ---(2)

Multiply equation 1 by a and 2 by b and subtract both

\({a^2}f\left( x \right) + abf\left( {\frac{1}{x}} \right) - abf\left( {\frac{1}{x}} \right) - {b^2}f\left( x \right) = \frac{a}{x} - 25a - bx + 25b\)

\({a^2}f\left( x \right) - {b^2}f\left( x \right) = \frac{a}{x} - 25a - bx + 25b\)

\(f\left( x \right) = \frac{a}{{x\left( {{a^2} - {b^2}} \right)}} - \frac{{bx}}{{{a^2} - {b^2}}} - \frac{{5\left( {a - b} \right)}}{{{a^2} - {b^2}}}\)

\(\mathop \smallint \limits_1^2 f\left( x \right)dx = \frac{1}{{{a^2} - {b^2}}}\left[ {a\left( {log2 - 25} \right) + \frac{{47b}}{2}} \right]\)