For nonzero x, af(x)+bf(1/x)=1/x-25, where a≠b. Find 12 f(x)\,dx, assuming the parameters make the equations nonsingular.

$\frac{a(2\ln2-25)-47b/2}{a^2-b^2}$

$\frac{a(2\ln2-25)+47b/2}{a^2-b^2}$

GATE 2015 CS (CS3) – Question 54

Engineering Mathematics · Calculus · 2 marks · Multiple choice

For nonzero x, $af(x)+bf(1/x)=1/x-25$, where a≠b. Find $\int_1^2 f(x)\,dx$, assuming the parameters make the equations nonsingular.

$\frac{a(\ln2-25)+47b/2}{a^2-b^2}$
$\frac{a(2\ln2-25)-47b/2}{a^2-b^2}$
$\frac{a(2\ln2-25)+47b/2}{a^2-b^2}$
$\frac{a(\ln2-25)-47b/2}{a^2-b^2}$

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Correct answer: (A) $\frac{a(\ln2-25)+47b/2}{a^2-b^2}$

Explanation

Replacing x by 1/x gives af(1/x)+bf(x)=x−25. Eliminate f(1/x): $(a^2-b^2)f(x)=a(1/x-25)-b(x-25)$. Integration from one to two gives a(ln2−25)−b(3/2−25)=a(ln2−25)+47b/2. Divide by a²−b² to obtain A. Besides the printed a≠b, a≠−b is needed for this nonsingular formula.

GATE 2015 CS (CS3) – Question 54

Explanation

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