Question 33

Mathematics Definite Integrals Medium

If for non-zero \(x\), \( c f(x) + d f\left(\frac{1}{x}\right) = |\log |x|| + 3\), where \(c \neq d\), then the integral \(\int_1^e f(x)\, dx\) equals:

(A) \(\frac{(c-d)(2 e-1)}{c^2-d^2}\)
(B) \(\frac{(c-d)(3 e-2)}{c^2-d^2}\)
(C) \(\frac{(c-d)(3 e+2)}{c^2-d^2}\)
(D) \(\frac{(c-d)(2 e+1)}{c^2-d^2}\)
✓ Correct Answer: Option B Verified Solution

📘 Step-by-Step Solution & Detailed Explanation:

The expression provided is: \[ c^{2} f(x) + c d f\left(\frac{1}{x}\right) = c |\log |x|| + 3c \] \[ d^{2} f(x) + c d f\left(\frac{1}{x}\right) = d |\log |x|| + 3d \] Subtracting equation (ii) from equation (i): \[ (c^{2} - d^{2}) f(x) = (c - d) |\log |x|| + 3(c - d) \] Simplifying further: \[ (c^{2} - d^{2}) f(x) = (c - d) \left(|\log |x|| + 3\right) \] Thus: \[ f(x) = \frac{(c - d)}{(c^{2} - d^{2})} \left( |\log |x|| + 3 \right) \] To evaluate the integral: \[ \frac{(c - d)}{(c^{2} - d^{2})} \int_{1}^{e} \left( |\log |x|| + 3 \right) \] This simplifies to: \[ \frac{(c - d)}{(c^{2} - d^{2})} \left[ \int_{1}^{e} \log x + 3 \, dx \right] \] \[ = \frac{(c - d)}{(c^{2} - d^{2})} \left[ x \log x - x + 3x \right]_{1}^{e} \] Finally: \[ = \frac{(c - d)(3e - 2)}{(c^{2} - d^{2})} \]

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