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:
Step-by-step Solution:
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: \[ = \frac{(c - d)(3e - 2)}{(c^{2} - d^{2})} \]