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JEE Main 20211 SeptemberEvening Shift
Exam
JEE Main
Year
2021
Date
1 September
Shift
Evening Shift
Subject
Mathematics
Chapter
Definite Integration

JEE Main 2021 · 1 September · Evening Shift · Mathematics · Definite Integration

Let J _ n, m = ∫ _ 0 ^ 1 / 2 x ^ n x ^ m - 1 dx, n > m and n, m ∈ N...

Let Jn,m=∫01/2xnxm−1dx,∀n>mand{J}_{n, m} = {∫}_{0}^{1 / 2}\dfrac{{x}^{n}}{{x}^{m} - 1}dx, \forall n > m andn, m \in N .ConsideramatrixConsider a matrixA = {\left[{a}_{ij}\right]}_{3 \times 3}wherewhere{a}_{ij} = \left\{{J6+i,3−Ji+3,3,i≤j0,i>j\begin{cases}{J}_{6 + i, 3} - {J}_{i + 3, 3} & , i \le j \\ 0 & , i > j\end{cases}\right). Then ∣adjA−1∣\left|adj {A}^{- 1}\right| is :

Options

  1. A(15)2×234( 15{)}^{2} \times {2}^{34}
  2. B(15)2×242( 15{)}^{2} \times {2}^{42}
  3. C(105)2×236( 105{)}^{2} \times {2}^{36}
  4. D(105)2×238( 105{)}^{2} \times {2}^{38}

Correct answer

D. (105)2×238( 105{)}^{2} \times {2}^{38}

Step-by-step solution

∵ A = \left[{a11a12a130a22a2300a33\begin{cases}{a}_{11} & {a}_{12} & {a}_{13} \\ 0 & {a}_{22} & {a}_{23} \\ 0 & 0 & {a}_{33}\end{cases}\right] ⇒ \left|A\right| = {a}_{11}{a}_{22}{a}_{33}</p><p></p><p>⇒ | A | = \left({J}_{7, 3} - {J}_{4, 3}\right) \cdot \left({J}_{8, 3} - {J}_{5, 3}\right) \cdot \left({J}_{9, 3} - {J}_{6, 3}\right)</p><p></p><p>= {∫}_{0}^{1 / 2}\left(\dfrac{{x}^{7} - {x}^{4}}{{x}^{3} - 1}\right)dx \cdot {∫}_{0}^{1 / 2}\left(\dfrac{{x}^{8} - {x}^{5}}{{x}^{3} - 1}\right)dx \cdot {∫}_{0}^{1 / 2}\left(\dfrac{{x}^{9} - {x}^{6}}{{x}^{3} - 1}\right)dx</p><p></p><p>= {∫}_{0}^{1 / 2}{x}^{4}dx \cdot {∫}_{0}^{1 / 2}{x}^{5}dx \cdot {∫}_{0}^{1 / 2}{x}^{6}dx</p><p></p><p>= \dfrac{1}{( 210 ) \cdot ( 2{)}^{18}}</p><p>Now</p>∣adjA−1∣=∣A−1∣2=1∣A∣2=((210)⋅(218))2=(105)2⋅(2)38</p><p>Now</p>\left|adj {A}^{- 1}\right| = {\left|{A}^{- 1}\right|}^{2} = \dfrac{1}{| A{|}^{2}} = {\left(( 210 ) \cdot \left({2}^{18}\right)\right)}^{2} = ( 105{)}^{2} \cdot ( 2{)}^{38}

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