JEE Mains · Maths · STD 12 - 7.2 definite integral
Let \(5 f(x)+4 f\left(\frac{1}{x}\right)=\frac{1}{x}+3, x > 0\). Then \(18 \int \limits_1^2 f(x) d x\) is equal to:
- A \(10 \log _e 2-6\)
- B \(10 \log _e 2+6\)
- C \(5 \log _e 2+3\)
- D \(5 \log _e 2-3\)
Answer & Solution
Correct Answer
(A) \(10 \log _e 2-6\)
Step-by-step Solution
Detailed explanation
\(5 f(x)+4 f\left(\frac{1}{x}\right)=\frac{1}{x}+3\) replace \(x \rightarrow \frac{1}{ x }\) \(5 f\left(\frac{1}{x}\right)+4 f(x)=x+3\) Eq. (1) \(\times 5-\) eq. (2) \(\times 4\) \(f(x)=\frac{1}{9}\left(\frac{5}{x}-4 x+3\right)\)…
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