Let $J_{n, m} = \int_{0}^{\frac{1}{2}} \frac{x^{n}}{x^{m} - 1} d x, \forall n > m$ and n, $m \in N$ . Consider a matrix $A = {[a_{i j}]}_{3 \times 3}$ where $a_{i j} = {\begin{matrix} J_{6 + i, 3} -_{i + 3, 3}, & i \leq j \\ 0, & i > j \end{matrix} . T h e n | a d j A^{- 1} | i s :$

Option 1 -

${(15)}^{2} * 2^{34}$

Option 2 -

${(15)}^{2} * 2^{42}$

Option 3 -

${(105)}^{2} * 2^{38}$

Option 4 -

${(105)}^{2} * 2^{36}$

24 Views | Posted 2 months ago

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1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 3

Detailed Solution:

$l_{n, m} = \int_{0}^{\frac{1}{2}} \frac{x^{n}}{x^{m} - 1} d x, m, n \in N, n > m$

$l_{6 + i, 3} - l_{3 + i, 3}$

$A = \frac{1}{2^{5}} [\begin{matrix} \frac{1}{5} & \frac{1}{5} & \frac{1}{5} \\ 0 & \frac{1}{12} & \frac{1}{12} \\ 0 & 0 & \frac{1}{28} \end{matrix}]$ $= \frac{B}{32}$

$| A | = {(\frac{1}{32})}^{3} | B | = \frac{1}{105 . 2^{19}}$

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Let $J_{n, m} = \int_{0}^{\frac{1}{2}} \frac{x^{n}}{x^{m} - 1} d x, \forall n > m$ and n, $m \in N$ . Consider a matrix $A = {[a_{i j}]}_{3 \times 3}$ where $a_{i j} = {\begin{matrix} J_{6 + i, 3} -_{i + 3, 3}, & i \leq j \\ 0, & i > j \end{matrix} . T h e n | a d j A^{- 1} | i s :$

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Let J n , m = ∫ 0 1 2 x n x m − 1 d x , ∀ n > m and n, m ∈ N . Consider a matrix A = [ a i j ] 3 × 3 where a i j = { J 6 + i , 3 − i + 3 , 3 , i ≤ j 0 , i > j . T h e n | a d j A − 1 | i s :

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Let $J_{n, m} = \int_{0}^{\frac{1}{2}} \frac{x^{n}}{x^{m} - 1} d x, \forall n > m$ and n, $m \in N$ . Consider a matrix $A = {[a_{i j}]}_{3 \times 3}$ where $a_{i j} = {\begin{matrix} J_{6 + i, 3} -_{i + 3, 3}, & i \leq j \\ 0, & i > j \end{matrix} . T h e n | a d j A^{- 1} | i s :$