Maths

x² – y² cosec²q = 5 $\frac{x^2}{1}-\frac{y^2}{si{n}^2\theta }=5$                        

x² cosec²q + y² = 5  $\frac{x^2}{si{n}^2\theta }+\frac{y^2}{1}=5$        

$e_H=\sqrt[]{7}{e}_e$                  

and $e_H=\sqrt[]{1+\frac{si{n}^2\theta }{1}}$  

-> $\sqrt[]{1+si{n}^2\theta }=\sqrt[]{7}\sqrt[]{1-si{n}^2\theta }$

1 + sin²q = 7 – 7 sin²q

->8sin²q = 6

-> $sin\theta =\sqrt[]{\frac{3}{4}}=\frac{\sqrt[]{3}}{2}$  

-> $\theta =\frac{\pi }{3}$  

5f(x) + 4f $\left(\frac{1}{x}\right)$  = x² – 4           .(1)

Replace x by  $\frac{1}{x}$

5f  $\left(\frac{1}{x}\right)$  + 4f(x) = $\frac{1}{x^2}$  – 4   .(2)

5 * equation (1) – 4 * equation (2)

$9f\left(x\right)=5x^{2-}\frac{4}{x^2}-4$            

$y=9f\left(x\right)\cdot x^2=\frac{5x^4-4-4x^2}{x^2}{x}^2$            

y = 5x⁴ – 4 – 4x²

y = 20x³ – 8x > 0

4x(5x² – 2) > 0

    $x\in \left(-\sqrt[]{\frac{2}{5}},0\right)\cup \left(\sqrt[]{\frac{2}{5}},\infty \right)$

Total ways to partition 5 into 4 parts are:

5 0 

4 1 0  $\frac{5!}{4!}=5$

3 2 0 $\frac{5!}{3!\cdot 2!}=10$

3 1 0 $\frac{5!}{2!2!2!}=15$

2 1 $\frac{5!}{2!*3!}=10$

51 Total way

(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

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

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