JEE MAIN - Mathematics (2021 - 1st September Evening Shift - No. 7)
Let $${J_{n,m}} = \int\limits_0^{{1 \over 2}} {{{{x^n}} \over {{x^m} - 1}}dx} $$, $$\forall$$ n > m and n, m $$\in$$ N. Consider a matrix $$A = {[{a_{ij}}]_{3 \times 3}}$$ where $${a_{ij}} = \left\{ {\matrix{
{{j_{6 + i,3}} - {j_{i + 3,3}},} & {i \le j} \cr
{0,} & {i > j} \cr
} } \right.$$. Then $$\left| {adj{A^{ - 1}}} \right|$$ is :
(15)2 $$\times$$ 242
(15)2 $$\times$$ 234
(105)2 $$\times$$ 238
(105)2 $$\times$$ 236
Explanation
A = $$\left[ {\matrix{
{a_{11}} & {a_{12}} & {a_{13}} \cr
{{a_{21}}} & {{a_{22}}} & {{a_{23}}} \cr
{{a_{31}}} & {{a_{32}}} & {{a_{33}}} \cr
} } \right]$$
$${J_{6 + i,3}} - {J_{i + 3,3}}; i \le j$$
$$ = \int_0^{{1 \over 2}} {{{{x^{6 + i}}} \over {{x^3} - 1}} - \int_0^{{1 \over 2}} {{{{x^{i + 3}}} \over {{x^3} - 1}}} } $$
$$ =\int_0^{1/2} {{{{x^{i + 3}}({x^3} - 1)} \over {{x^3} - 1}}} $$
$$ = {{{x^{3 + i + 1}}} \over {3 + i + 1}} = \left( {{{{x^{4 + i}}} \over {4 + i}}} \right)_0^{1/2}$$
$$ \therefore $$ $${a_{ij}} = {j_{6 + i,3}} - {j_{i + 3,3}} = {{{{\left( {{1 \over 2}} \right)}^{4 + i}}} \over {4 + i}}$$
$${a_{11}} = {{{{\left( {{1 \over 2}} \right)}^5}} \over 5} = {1 \over {{{5.2}^5}}}$$
$${a_{12}} = {1 \over {{{5.2}^5}}}$$
$${a_{13}} = {1 \over {{{5.2}^5}}}$$
$${a_{22}} = {1 \over {{{6.2}^6}}}$$
$${a_{23}} = {1 \over {{{6.2}^6}}}$$
$${a_{33}} = {1 \over {{{7.2}^7}}}$$
$$A = \left[ {\matrix{ {{1 \over {{{5.2}^5}}}} & {{1 \over {{{5.2}^5}}}} & {{1 \over {{{5.2}^5}}}} \cr 0 & {{1 \over {{{6.2}^6}}}} & {{1 \over {{{6.2}^6}}}} \cr 0 & 0 & {{1 \over {{{7.2}^7}}}} \cr } } \right]$$
$$\left| A \right| = {1 \over {{{5.2}^5}}}\left[ {{1 \over {{{6.2}^6}}} \times {1 \over {{{7.2}^7}}}} \right]$$
$$\left| A \right| = {1 \over {{{210.2}^{18}}}}$$
$$\left| {adj{A^{ - 1}}} \right| = {\left| {{A^{ - 1}}} \right|^{n - 1}} = {\left| {{A^{ - 1}}} \right|^2} = {1 \over {{{\left( {\left| A \right|} \right)}^2}}}$$
$$ = {({210.2^{18}})^2}$$
= $${(105)^2} \times {2^{38}}$$
$${J_{6 + i,3}} - {J_{i + 3,3}}; i \le j$$
$$ = \int_0^{{1 \over 2}} {{{{x^{6 + i}}} \over {{x^3} - 1}} - \int_0^{{1 \over 2}} {{{{x^{i + 3}}} \over {{x^3} - 1}}} } $$
$$ =\int_0^{1/2} {{{{x^{i + 3}}({x^3} - 1)} \over {{x^3} - 1}}} $$
$$ = {{{x^{3 + i + 1}}} \over {3 + i + 1}} = \left( {{{{x^{4 + i}}} \over {4 + i}}} \right)_0^{1/2}$$
$$ \therefore $$ $${a_{ij}} = {j_{6 + i,3}} - {j_{i + 3,3}} = {{{{\left( {{1 \over 2}} \right)}^{4 + i}}} \over {4 + i}}$$
$${a_{11}} = {{{{\left( {{1 \over 2}} \right)}^5}} \over 5} = {1 \over {{{5.2}^5}}}$$
$${a_{12}} = {1 \over {{{5.2}^5}}}$$
$${a_{13}} = {1 \over {{{5.2}^5}}}$$
$${a_{22}} = {1 \over {{{6.2}^6}}}$$
$${a_{23}} = {1 \over {{{6.2}^6}}}$$
$${a_{33}} = {1 \over {{{7.2}^7}}}$$
$$A = \left[ {\matrix{ {{1 \over {{{5.2}^5}}}} & {{1 \over {{{5.2}^5}}}} & {{1 \over {{{5.2}^5}}}} \cr 0 & {{1 \over {{{6.2}^6}}}} & {{1 \over {{{6.2}^6}}}} \cr 0 & 0 & {{1 \over {{{7.2}^7}}}} \cr } } \right]$$
$$\left| A \right| = {1 \over {{{5.2}^5}}}\left[ {{1 \over {{{6.2}^6}}} \times {1 \over {{{7.2}^7}}}} \right]$$
$$\left| A \right| = {1 \over {{{210.2}^{18}}}}$$
$$\left| {adj{A^{ - 1}}} \right| = {\left| {{A^{ - 1}}} \right|^{n - 1}} = {\left| {{A^{ - 1}}} \right|^2} = {1 \over {{{\left( {\left| A \right|} \right)}^2}}}$$
$$ = {({210.2^{18}})^2}$$
= $${(105)^2} \times {2^{38}}$$
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