Maths

  $\frac{a+20}{2}=3+7r$

a + 20 = 6 + 14r     . (i)

b = -2 + 10r           . (ii)

a = 18r – 2              . (iii)

Solving (i) and (iii) we get

20 + 18r – 2 = 6 + 14r

r = -3

$\therefore$   a = 14 + 14 (-3) = -56 and b = -2 -30 = -32

$\left|a+b\right|\left|-56-32\right|=88$

x – 2y = 1, x – y + kz = -2, ky + 4z = 6

 x – 2y + 0. z – 1 = 0

x – y + kz + 2 = 0

0x + ky + 4z – 6 = 0

0x + ky + 4z – 6 = 0

${\Delta }_1=\left|\begin{array}{ccc}\hfill 1\hfill & \hfill -2\hfill & \hfill 0\hfill \\ \hfill -2\hfill & \hfill -1\hfill & \hfill k\hfill \\ \hfill 6\hfill & \hfill k\hfill & \hfill 4\hfill \end{array}\right|=-\left(k+10\right)\left(k+2\right)$   

For no solution

$\Delta =0,{\Delta }_1\ne 0$       

k = 2

$l={\displaystyle \underset{1}{\overset{3}{\int }}\left[x^2-2x+1-3\right]dx=}{\displaystyle \underset{1}{\overset{3}{\int }}\left[{\left(x-1\right)}^2-3\right]dx={\displaystyle \underset{1}{\overset{3}{\int }}\left[{\left(x-1\right)}^2\right]dx-3{\displaystyle \underset{1}{\overset{3}{\int }}dx}}}$  .(A)

$l_1={\displaystyle \underset{1}{\overset{3}{\int }}\left[{\left(x-1\right)}^2\right]dxPut{\left(x-1\right)}^2=t}$

$l_1=\frac{1}{2}\left[0{\displaystyle \underset{0}{\overset{1}{\int }}\frac{dt}{\sqrt[]{t}}}{\displaystyle \underset{1}{\overset{2}{\int }}\frac{dt}{\sqrt[]{t}}}+2{\displaystyle \underset{2}{\overset{3}{\int }}\frac{dt}{\sqrt[]{t}}}+3{\displaystyle \underset{3}{\overset{4}{\int }}\frac{dt}{\sqrt[]{t}}}\right]=\frac{1}{2}\left\{{\left|\frac{t^{\frac{-1}{2}+1}}{-\frac{1}{2}+1}\right|}_1^2{\left|+2\frac{t^{\frac{-1}{2}+1}}{-\frac{1}{2}+1}\right|}_2^3{+3\frac{t^{\frac{-1}{2}+1}}{-\frac{1}{2}+1}]}_3^4\right\}$      = $5-\sqrt[]{2}-\sqrt[]{3}$  

Hence from (A)

= $5-\sqrt[]{2}-\sqrt[]{3}-\sqrt[]{3}-6=-1-\sqrt[]{2}-\sqrt[]{3}$

2^nd method       

${\displaystyle \underset{1}{\overset{3}{\int }}\left[{\left(x-1\right)}^2\right]dx-6..............\left(A\right)}$

From (A), $l=5-\sqrt[]{2}-\sqrt[]{3}-6=-1-\sqrt[]{2}-\sqrt[]{3}$  

$l={\displaystyle \underset{0}{\overset{2}{\int }}f\left(x\right)dx={\left[xf\left(x\right)\right]}_0^2-{\displaystyle \underset{0}{\overset{2}{\int }}xf\text{'}\left(x\right)dx=2e^2-{\displaystyle \underset{0}{\overset{2}{\int }}xf\text{'}\left(x\right)dx}}}$        .(A)

Put   $l_1={\displaystyle \underset{0}{\overset{2}{\int }}xf\text{'}\left(x\right)dx}$            .(i)

Using properties ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx={\displaystyle \underset{a}{\overset{b}{\int }}f\left(a+b-x\right)dx}}$  

$l_1={\displaystyle \underset{0}{\overset{2}{\int }}\left(2-x\right)f\text{'}\left(2-x\right)dx={\displaystyle \underset{0}{\overset{2}{\int }}\left(2-x\right)f\text{'}\left(x\right)dx}}$    .(ii)

Adding (i) and (ii) we get

  $2l_1=2{\displaystyle \underset{0}{\overset{2}{\int }}f\text{'}\left(x\right)dx\Rightarrow l_1={\left[f\left(x\right)\right]}_0^2}$         

f(2) – f(0) = e² – 1

From (A) l = 2e² – e² + 1 = e² + 1

kx + y + 2z = 1    . (i)

 3x – y – 2z = 2       . (ii)

-2x – 2y – 4z = 3   . (iii)

(ii) * 5 - (i)  $\equiv$ (iii) * 3 -> (15 – k) = -6

K = 21

Let f(x) = x⁶ + ax⁵ + bx⁴ + ax³ + dx² + ex + f

$\underset{x\to 0}{lim}\frac{f\left(x\right)}{x^3}=1$                   Non zero finite

So, d = e = f = 0

f(x)  = x⁶ + ax⁵ + bx⁴ + cx³

$\underset{x\to 0}{lim}\frac{f\left(x\right)}{x^3}=1$          Non zero finite

f'(x) = $6x^5+5a{x}^4+4b{x}^3+3x^2$  

f'(1) = 0

6 + 5a + 4b + 3 = 0

5a + 4b = - 9 .(i)

f'(-1) = 0

-6 + 5a – 4b + 3 = 0 .(ii)

Solving (i) and (ii)

a  -3/5, b = -3/2

$\therefore f\left(x\right)=x^6+\left(\frac{-3}{5}\right)x^5+\left(\frac{-3}{2}\right)x^4+x^3$

$\therefore 5.f\left(2\right)=144$  

K $=\frac{4\sqrt[]{3}}{\sqrt[]{3}x+y}$  

K =    $\frac{\sqrt[]{3}x-y}{4\sqrt[]{3}}$

$\therefore \frac{4\sqrt[]{3}}{\sqrt[]{3}x+y}=\frac{\sqrt[]{3}x-y}{4\sqrt[]{3}}$   

$\frac{x^2}{16}-\frac{y^2}{48}=1$

48 = 16 (e² – 1)

$e=\sqrt[]{4}=2$      

$f\left(x\right)=\frac{5^x}{5^x+5}$

$f\left(2-x\right)=\frac{5^{2-x}}{5^{2-x}+5}=\frac{25/5^x}{\frac{25}{5^x}+5}=\frac{25}{25+5^{x+1}}=\frac{5}{5+5^x}$

$f\left(x\right)+f\left(2-x\right)=1$

$S={\displaystyle \sum _{k=1}^{39}f\left(\frac{k}{20}\right).....\left(i\right)}$

$S={\displaystyle \sum _{k=1}^{39}f}\left(2-\frac{k}{20}\right)..........\left(ii\right)$

(i) + (ii) 2S = ${\displaystyle \sum _{k=1}^{39}\left(f\left(\frac{k}{20}\right)+f\left(2-\frac{k}{20}\right)\right)}$

$S=\frac{39}{2}$

Let a_n be the side of square A_n

$a_n=\sqrt[]{2}{a}_{n+1}$       

a₁ = 12

a_n = 12 $*{\left(\frac{1}{\sqrt[]{2}}\right)}^{n-1}$  

${\left(a_n\right)}^2<1$

$\frac{144}{2^{\left(n-1\right)}}<1$

$\Rightarrow 2^{\left(n-1\right)}>144$   

$\Rightarrow n-1\ge 8$

$\Rightarrow n\ge 9$        

$1+{\alpha }^2=1\Rightarrow \alpha =0$

$\&{\alpha }^2+{\beta }^2=1\Rightarrow {\beta }^2=\pm 1$

$\&\alpha -\alpha \beta =0\Rightarrow \alpha \left(1-\beta \right)=0\Rightarrow \alpha =0or\beta =1$

= 0 & = 1 or = 0 & = 1

4 + 4 = 1