Consider the equation $$\int_{1}^{e} \frac{\left(\log _{e} x\right)^{1 / 2}}{x\left(a-\left(\log _{e} x\right)^{3 / 2}\right)^{2}} dx=1, \quad a \in(-\infty, 0) \cup(1, \infty).$$ Which of the following statements is/are TRUE ?

(C) An irrational number $a$ satisfies the above equation; (D) More than one $a$ satisfy the above equation

(A) No $a$ satisfies the above equation

(B) An integer $a$ satisfies the above equation

JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · Definite Integrals

Consider the equation

\int_{1}^{e} \frac{( lo g _{e} x ) ^{1/2}}{x ( a - ( lo g _{e} x ) ^{3/2} ) ^{2}} d x = 1, a \in (- \infty, 0) \cup (1, \infty) .

Which of the following statements is/are TRUE ?

Answer: (C), (D)

Let

t = a - (lo g_{e} x)^{3/2}

. Then

(lo g_{e} x)^{1/2} d x = - \frac{2}{3} d t

.

When

x = 1

t = a

. When

x = e

t = a - 1

\int_{1}^{e} \frac{( lo g _{e} x ) ^{1/2}}{x ( a - ( lo g _{e} x ) ^{3/2} ) ^{2}} d x = \frac{2}{3} \int_{a}^{a - 1} \frac{- d t}{t ^{2}} = \frac{2}{3} [\frac{1}{t}]_{a}^{a - 1}

= \frac{2}{3} (\frac{1}{a - 1} - \frac{1}{a}) = \frac{2}{3} \cdot \frac{1}{a ( a - 1 )} = 1

This gives:

\frac{2}{3 a ( a - 1 )} = 1 \Rightarrow 3 a^{2} - 3 a - 2 = 0

a = \frac{3 \pm 9 + 24}{6} = \frac{3 \pm 33}{6}

Both values are irrational, and both lie in the required domain. So (C) and (D) are correct.

JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · One or more correct options