Consider the equation int_1^e frac{(log_e x)^ | vedclass

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(A) Let $I = \int_1^e \frac{(\log_e x)^{1/2}}{x(a-(\log_e x)^{3/2})^2} dx = 1$. Substitute $t = a - (\log_e x)^{3/2}$. Then $dt = -\frac{3}{2}(\log_e

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$C, D$

Vedclass · Answer

(A) Let $I = \int_1^e \frac{(\log_e x)^{1/2}}{x(a-(\log_e x)^{3/2})^2} dx = 1$. Substitute $t = a - (\log_e x)^{3/2}$. Then $dt = -\frac{3}{2}(\log_e x)^{1/2} \cdot \frac{1}{x} dx$,which implies $\frac{(\log_e x)^{1/2}}{x} dx = -\frac{2}{3} dt$. When $x = 1$,$t = a - 0 = a$. When $x = e$,$t = a - 1$. Substituting these into the integral: $I = \int_a^{a-1} \frac{-2/3}{t^2} dt = \frac{2}{3} \int_{a-1}^a t^{-2} dt = \frac{2}{3} [-\frac{1}{t}]_{a-1}^a = \frac{2}{3} (\frac{1}{a-1} - \frac{1}{a}) = \frac{2}{3} (\frac{a - (a-1)}{a(a-1)}) = \frac{2}{3a(a-1)}$. Given $I = 1$,we have $\frac{2}{3a(a-1)} = 1$,so $3a^2 - 3a - 2 = 0$. Using the quadratic formula,$a = \frac{3 \pm \sqrt{9 - 4(3)(-2)}}{2(3)} = \frac{3 \pm \sqrt{33}}{6}$. Since $\sqrt{33} \approx 5.74$,$a_1 = \frac{3 + 5.74}{6} \approx 1.45$ and $a_2 = \frac{3 - 5.74}{6} \approx -0.45$. Both values satisfy $a \in (-\infty, 0) \cup (1, \infty)$. Since both values are irrational,statement $C$ is true. Since there are two such values,statement $D$ is true.

Consider the equation $\int_1^e \frac{(\log_e x)^{1/2}}{x(a-(\log_e x)^{3/2})^2} dx = 1$,where $a \in (-\infty, 0) \cup (1, \infty)$. Which of the following statements is/are $TRUE$?

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