If a b and a f(x)+b f ( 1 x )= 1 x -5 for all x 0 , then _ 1 ^ 2 f(x) d x= 1 a^ 2 -b^ 2…
Mathematics · JEE Advanced · NTA Exams — Integral Calculus
If \(a \neq \mathrm{b}\) and \(a f(x)+b f\left(\frac{1}{x}\right)=\frac{1}{x}-5\)
for all \(x \neq 0\), then
\(\int_{1}^{2} f(x) d x=\frac{1}{a^{2}-b^{2}}\left[a(\log 2-\alpha)+\beta\left(\frac{b}{2}\right)\right]\)
where \(\beta-\alpha\) is equal to
for all \(x \neq 0\), then
\(\int_{1}^{2} f(x) d x=\frac{1}{a^{2}-b^{2}}\left[a(\log 2-\alpha)+\beta\left(\frac{b}{2}\right)\right]\)
where \(\beta-\alpha\) is equal to
- \(5\)
- \(7\)
- \(2\)
Answer
(D) 2
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