If for x ≠ 0, $af(x)+bf(\frac{1}{x})=

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Target Exam

CUET

Subject

-- Mathematics - Section A

Chapter

Definite Integration

Question:

If for x ≠ 0, $af(x)+bf(\frac{1}{x})=\frac{1}{x}-5$ where a ≠ b then $\int_1^2xf(x)\, dx$.

Options:

$\frac{b-9a}{9(a^2-b^2)}$

$\frac{b-9a}{b(a^2-b^2)}$

$\frac{b-9a}{6(a^2-b^2)}$

None of these

Correct Answer:

None of these

Explanation:

$af(x)+bf(\frac{1}{x})=\frac{1}{x}-5⇒a^2f(x)+abf(\frac{1}{x})=\frac{a}{x}-5a$

Replace x by $\frac{1}{x}$ $⇒a^2f(x)+baf(\frac{1}{x})=bx-5b$

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