JEE Advanced 2022 Paper 1, Mathematics Section 2 Q1: Definite Integration And Fundamental Theorem

JEE Advanced2022Paper 1Mathematics Section 2

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?

A
No $a$ satisfies the above equation
B
An integer $a$ satisfies the above equation
C
An irrational number $a$ satisfies the above equation
D
More than one $a$ satisfy the above equation

Solution

Let $u = (lo g_{e} x)^{3/2}$ , so $d u = \frac{3}{2} \frac{( lo g _{e} x ) ^{1/2}}{x} d x$ . The integrand transforms via $\frac{( lo g _{e} x ) ^{1/2}}{x} d x = \frac{2}{3} d u$ .

With $t = a - u$ , $d t = - d u$ . When $x = 1$ , $u = 0$ , so $t = a$ ; when $x = e$ , $u = 1$ , so $t = a - 1$ .

$I = \int_{a}^{a - 1} \frac{1}{t ^{2}} \cdot (- \frac{2}{3}) d t = \frac{2}{3} [\frac{1}{t}]_{a}^{a - 1} = \frac{2}{3} (\frac{1}{a - 1} - \frac{1}{a}) = \frac{2}{3} \cdot \frac{1}{a ( a - 1 )} .$

Setting $I = 1$ : $\frac{2}{3 a ( a - 1 )} = 1 \Rightarrow 3 a^{2} - 3 a - 2 = 0$ .

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

Both roots are irrational; numerically $a \approx 1.46$ (which lies in $(1, \infty)$ ) and $a \approx - 0.46$ (which lies in $(- \infty, 0)$ ), so both satisfy the domain constraint.

$∴$ Options (C) and (D) are correct.

Concept behind this question

Definite Integration And Fundamental TheoremNotes, formulas and examples →

Practice more Mathematics Section 2

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Advanced 2022 Paper 1, Mathematics Section 2 Q1: Definite Integration And Fundamental Theorem

Concept behind this question

More previous year questions on Definite Integration And Fundamental Theorem

Practice more Mathematics Section 2