26. Find values of k if area of triangle is 4 sq. units and vertices are

(i) (k, 0), (4, 0), (0, 2)           (ii) (–2, 0), (0, 4), (0, k)

|2A| = 2⁷

8|A| = 2⁷

Now |A| = α²–β² = 2⁴

α² = 16 + β²

α²– β² = 16

(α–β) (α+β) = 16

->α + β = 8 and

α – β = 2

->α = 5 and β = 3

|A| = 3

|B| = 1

->|C| = |ABA^T| = |A|B|A⁷| = |A|²|B|

= 9

->|X| = |A|C|²|A^T|

= 3 * 9² * 3 = 9 * 9² = 729

|A| = 2

$\underset{2024times}{\underset{?}{adj\left(adj\left(adj....\left(a\right)\right)\right)}}={\left|A\right|}^{{\left(n-1\right)}^{2024}}$            

$={\left|A\right|}^{{\left(n-1\right)}^{2024}}$                                             

$=2^{2^{2024}}$                       &nb

${\displaystyle \underset{0}{\overset{2}{\int }}\sqrt[]{2xdx}}-{\displaystyle \underset{0}{\overset{2}{\int }}\sqrt[]{2x-x^{2}dx}}={\displaystyle \underset{0}{\overset{1}{\int }}dy}-{\displaystyle \underset{0}{\overset{1}{\int }}\sqrt[]{1-y^2}dy-{\displaystyle \underset{0}{\overset{2}{\int }}\frac{y^2}{2}dy}+{\displaystyle \underset{1}{\overset{2}{\int }}2dy+I}}$

$\Rightarrow \frac{8}{3}-{\displaystyle \underset{0}{\overset{1}{\int }}\sqrt[]{1-t^2}dt}=1-\frac{8}{6}+2+I$

$I=1-{\displaystyle \underset{0}{\overset{1}{\int }}\sqrt[]{1-t^2}dt}$

${\displaystyle \underset{0}{\overset{2}{\int }}\sqrt[]{2xdx}}-{\displaystyle \underset{0}{\overset{2}{\int }}\sqrt[]{2x-x^{2}dx}}={\displaystyle \underset{0}{\overset{1}{\int }}dy}-{\displaystyle \underset{0}{\overset{1}{\int }}\sqrt[]{1-y^2}dy-{\displaystyle \underset{0}{\overset{2}{\int }}\frac{y^2}{2}dy}+{\displaystyle \underset{1}{\overset{2}{\int }}2dy+I}}$

$\Rightarrow \frac{8}{3}-{\displaystyle \underset{0}{\overset{1}{\int }}\sqrt[]{1-t^2}dt}=1-\frac{8}{6}+2+I$

$I=1-{\displaystyle \underset{0}{\overset{1}{\int }}\sqrt[]{1-t^2}dt}$