Q45MCQ2 marks, −0.67If for non-zero x, af(x)+bf(1x)=1x−25x, \: af(x) + bf(\frac{1}{x}) = \frac{1}{x} - 25x,af(x)+bf(x1)=x1−25 where a≠b then ∫12f(x)dxa \neq b \text{ then } \int\limits_1^2 f(x)dxa=b then 1∫2f(x)dx is(A)1a2−b2[a(ln2−25)+47b2]\frac{1}{a^2 - b^2} \begin{bmatrix} a(\ln 2 - 25) + \frac{47b}{2} \end{bmatrix}a2−b21[a(ln2−25)+247b](B)1a2−b2[a(2ln2−25)−47b2]\frac{1}{a^2 - b^2} \begin{bmatrix} a(2\ln 2 - 25) - \frac{47b}{2} \end{bmatrix}a2−b21[a(2ln2−25)−247b](C)1a2−b2[a(2ln2−25)+47b2]\frac{1}{a^2 - b^2} \begin{bmatrix} a(2\ln 2 - 25) + \frac{47b}{2} \end{bmatrix}a2−b21[a(2ln2−25)+247b](D)1a2−b2[a(ln2−25)−47b2]\frac{1}{a^2 - b^2} \begin{bmatrix} a(\ln 2 - 25) - \frac{47b}{2} \end{bmatrix}a2−b21[a(ln2−25)−247b]Sign in to save attempts.Show answer← PreviousNext →SolutionsDiscussionLoading…Sign in with Google