Maths

(t + 1)dx = (2x + (t + 1)³)dt

$\frac{dx}{dt}-\frac{2x}{t+1}={\left(t+1\right)}^2$

I.F. $=e^{{\displaystyle \int -\frac{2}{t+1}dt}}=\frac{1}{{\left(t+1\right)}^2}$  

Solution is

$\frac{x}{{\left(t+1\right)}^2}={\displaystyle \int 1dt}$  

x = (t + c) (t + 1)²

$?$ x (0) = 2 then c = 2

x = (t + 2) (t + 1)²

 x (1) = 12

${\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\frac{8\sqrt[]{2}cosx}{\left(1+e^{sinx}\right)\left(1+si{n}^{4}x\right)}}$ dx

$={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left\{\left(\frac{8\sqrt[]{2}cosx}{\left(1+e^{sinx}\right)\left(1+si{n}^{4}x\right)}+\frac{8\sqrt[]{2}cosx}{\left(1+e^{-sinx}\right)\left(1+si{n}^{4}x\right)}\right)\right\}dx}$            

$=8\sqrt[]{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{cosx}{1+si{n}^{4}x}dx}$            

Let sin x = t

$I=8\sqrt[]{2}{\displaystyle \underset{0}{\overset{1}{\int }}\frac{dt}{1+t^4}}$            

$=4\sqrt[]{2}{\displaystyle \underset{0}{\overset{1}{\int }}\frac{\left(1+\frac{1}{t^2}\right)-\left(1-\frac{1}{t^2}\right)}{t^2+\frac{1}{t^2}}dt}$       

$=4\sqrt[]{2}{\displaystyle \underset{0}{\overset{1}{\int }}\frac{\left(1+\frac{1}{t^2}\right)dt}{{\left(t-\frac{1}{t}\right)}^2+2}}-4\sqrt[]{2}{\displaystyle \underset{0}{\overset{1}{\int }}\frac{\left(1-\frac{1}{t^2}\right)dt}{{\left(t+\frac{1}{t}\right)}^2-2}}$            

$=4\sqrt[]{2}\cdot \frac{1}{\sqrt[]{2}}{\left(ta{n}^{-1}\frac{t-\frac{1}{t}}{\sqrt[]{2}}\right)}_0^1-4\sqrt[]{2}\cdot \frac{1}{2\sqrt[]{2}}{\left[log\left|\frac{t+\frac{1}{t}-\sqrt[]{2}}{t+\frac{1}{t}+\sqrt[]{2}}\right|\right]}_0^1$         

$=2\pi -2log\left|\frac{2-\sqrt[]{2}}{2+\sqrt[]{2}}\right|$        

$=2\pi +2log\left(3+2\sqrt[]{2}\right)$           

a = b = 2

3, 7, 11, 15, 19, 23, 27, . 403 = AP₁

2, 5, 8, 11, 14, 17, 20, 23, . 401 = AP₂

so common terms A.P.

11, 23, 35, ., 395

->395 = 11 + (n – 1) 12

->395 – 11 = 12 (n – 1)

$\frac{384}{12}=n-1$    

32 = n – 1

n = 33

Sum =  $\frac{33}{2}$ [2*11+ (32)12]

=  $\frac{33}{2}$ [22 + 384]

= 6699

|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

$I={\displaystyle \underset{0}{\overset{\pi /4}{\int }}\frac{xdx}{si{n}^4\left(2x\right)+co{s}^4\left(2x\right)}}$

           Let 2x = t then   $dx=\frac{1}{2}dt$

$I={\displaystyle \underset{}{\overset{}{\int }}\frac{\frac{t}{2}\cdot \frac{1}{2}dt}{si{n}^{4}t+co{s}^{4}t}}$

$=\frac{1}{4}{\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{tdt}{si{n}^{4}t+co{s}^{4}t}dt}$            

$I=\frac{1}{4}{\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{\left(\frac{\pi }{2}-t\right)dt}{si{n}^{4}t+co{s}^{4}t}dt}$

$2I=\frac{1}{4}{\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{\frac{\pi }{2}dt}{si{n}^{4}t+co{s}^{4}t}}$

$2I=\frac{1}{4}{\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{\frac{\pi }{2}dt}{si{n}^{4}t+co{s}^{4}t}}$

$2I=\frac{\pi }{8}{\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{si{n}^{4}tdt}{ta{n}^{4}t+1}}$            

Let tan t = y then

$2I=\frac{\pi }{8}{\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{\left(1+y^2\right)dy}{1+y^4}}$             

$=\frac{\pi }{8}{\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{1+\frac{1}{y^2}}{y^2+\frac{1}{y^2}-2+2}dy}$

$=\frac{\pi }{8}{\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{\left(1+\frac{1}{y^2}\right)dy}{2+{\left(y-\frac{1}{y}\right)}^2}}$             

Let $y-\frac{1}{y}=u$  

$2I=\frac{\pi }{8}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{du}{2+u^2}}$

$=\frac{\pi }{8\sqrt[]{2}}{\left[ta{n}^{-1}\frac{4}{\sqrt[]{2}}\right]}_{-\infty }^{\infty }$                  

$I=\frac{{\pi }^2}{16\sqrt[]{2}}$

|A| = 2

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

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

$=2^{2^{2024}}$                                          

$2^{2024}=\left(2^2\right)2^{2022}=4{\left(8\right)}^{674}=4{\left(9-1\right)}^{674}$             

-> $2^{2024}\equiv 4\left(mod9\right)$

->,  $2^{2024}\equiv 9m+4$  m ¬ even

$2^{9m+4}\equiv 16\cdot {\left(2^3\right)}^{3m}\equiv 16\left(mod9\right)$

7

  $?$  R is symmetric relation

⇒   (y, x) ∈ R V (x, y) ∈ R

(x, y) ∈ R ⇒ 2x = 3y and (y, x) ∈ R ⇒ 3x = 2y

Which holds only for (0, 0)

Which does not belongs to R.

∴ Value of n = 0

After giving 2 apples to each child 15 apples left now 15 apples can be distributed in
^15+3–1C₂ = ¹⁷C₂ ways

  $=\frac{17*16}{2}=136$          

${\displaystyle \underset{0}{\overset{\pi }{\int }}\frac{x^{2}sinx\cdot cosx}{si{n}^{4}x+co{s}^{4}x}dx}$

$={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{sinxcosx}{si{n}^{4}x+co{s}^{4}x}\left(x^2-{\left(\pi -x\right)}^2\right)dx}$

$={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{sinx\cdot cosx\left(2\pi x-{\pi }^2\right)}{si{n}^{4}x+co{s}^{4}x}x}$

$=2\pi \cdot \frac{\pi }{4}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{sinxcosx}{si{n}^{4}x+co{s}^{4}x}dx}-{\pi }^2{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{sinxcosx}{si{n}^{4}x+co{s}^{4}x}dx}$

$=-\frac{{\pi }^2}{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{sinxcosx}{si{n}^{4}x+co{s}^{4}x}dx}$

$=-\frac{{\pi }^2}{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\frac{1}{2}sin2x}{1-\frac{1}{2}si{n}^{2}2x}dx}$

$=-\frac{{\pi }^2}{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{sin2x}{2-si{n}^{2}2x}dx}$

$=-\frac{{\pi }^2}{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{sin2x}{1+co{s}^{2}2x}dx}$

Let cot2x = t

$=-\frac{{\pi }^2}{2}{\displaystyle \underset{1}{\overset{-1}{\int }}\frac{-\frac{1}{2}dt}{1+t^2}}$

$=-\frac{{\pi }^2}{4}{\displaystyle \underset{-1}{\overset{1}{\int }}\frac{dt}{1+t^2}}$

$=-\frac{{\pi }^2}{4}\cdot \frac{\pi }{2}=-\frac{{\pi }^2}{8}$

$\frac{120}{{\pi }^3}\left|-\frac{{\pi }^3}{8}\right|=15$