Maths

$x={\displaystyle \sum _{n=0}^{\infty }co{s}^{2n}\theta =co{s}^0\theta +co{s}^2\theta +co{s}^4\theta +.....=1+co{s}^2\theta +co{s}^4\theta +.....}$

a = 1, r = cos2

$x=S_{\infty }=\frac{a}{1-r}=\frac{1}{1-co{s}^2\theta }=\frac{1}{si{n}^2\theta }$

Similarly, y = $\frac{1}{co{s}^2\theta }\Rightarrow \frac{1}{y}=co{s}^2\theta$ $\frac{}{{}^{}}\frac{}{}{}^{}$

$z=\frac{xy}{xy-1}\Rightarrow xyz-z=xy...........\left(i\right)$

Also, $\frac{1}{x}+\frac{1}{y}=1\Rightarrow x+y=xy..........\left(ii\right)$ $\frac{}{}\frac{}{}$

(i) & (ii) xyz = xy + z (x + y) z = xy + z

Let

A : Missile hit the target

B : Missile intercepted

P (B) = $\frac{1}{3}P\left(A/\overline{B}\right)=\frac{3}{4}$ $\frac{}{}\stackrel{}{}\frac{}{}$

$P\left(\overline{B}\right)=\frac{2}{3}$

$\Rightarrow P\left(\overline{B}\cap A\right)=\frac{3}{4}*\frac{2}{3}=\frac{1}{2}$

Required Probability = $\frac{2}{3}*\frac{3}{4}*\frac{2}{3}*\frac{3}{4}*\frac{2}{3}*\frac{3}{4}=\frac{1}{8}$

$\frac{a+20}{2}=3+7r$

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)

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

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

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

According to question,

$\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{1}{4}$

$\frac{1}{a}+\frac{1}{b}=\frac{1}{2}........(i)$

Equation of required line is $\frac{x}{a}+\frac{y}{b}=1$ $\frac{}{}\frac{}{}$

Obviously B (2, 2) satisfying condition (i)

$\Delta -\left|\begin{array}{ccc}\hfill 3\hfill & \hfill -2\hfill & \hfill -k\hfill \\ \hfill 2\hfill & \hfill -4\hfill & \hfill -2\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill -1\hfill \end{array}\right|=\left|\begin{array}{ccc}\hfill 3\hfill & \hfill -2\hfill & \hfill -k\hfill \\ \hfill 0\hfill & \hfill -8\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill -1\hfill \end{array}\right|,\left[R_2-R_1-2R_3\right]$

= 8 (3 + k)

For inconsistent $\Delta =0\Rightarrow k=3$

${\Delta }_x=\left|\begin{array}{ccc}\hfill 10\hfill & \hfill -2\hfill & \hfill -k\hfill \\ \hfill 6\hfill & \hfill -4\hfill & \hfill -2\hfill \\ \hfill 5m\hfill & \hfill 2\hfill & \hfill -1\hfill \end{array}\right|=32-40m\ne 0\Rightarrow m\ne \frac{4}{5}$

y = x³

${\frac{dy}{dx}=3x^2\Rightarrow \frac{dy}{dx}|}_{\left(t,t^3\right)}=3t^2$

Equation of tangent y – t³ = 3t² (x – t) 

Let again meet the curve at $Q\left(t_1,t_1^3\right)$

$\Rightarrow t_1^3-t^3=3t^2\left(t_1-t\right)$

$t_1^2+t{t}_1+t^2=3t^2\left[?t_1\ne t\right]$

$t_1^2+t{t}_1-2t^2=0$

=> t₁ = -2t

Required ordinate = $\frac{2t^3+t_1^3}{3}=\frac{2t^3-8t^3}{3}=-2t^3$   

$\frac{dp}{dt}=0.5p-450andP\left(0\right)=850$

$\Rightarrow \frac{dp}{P-900}=0.5dt$

${\displaystyle \underset{850}{\overset{0}{\int }}\frac{dp}{P-900}={\displaystyle \underset{0}{\overset{T}{\int }}0.5dt}}$

$\Rightarrow ln\left(P-900\right){|}_{805}^0=0.5T$

$\frac{T}{2}=ln\left|\frac{900}{50}\right|=ln18$

T = 2 ln 18