Maths

ax² + bx + c = 0

D = b² – 4ac

D = 0

b² – 4ac = 0

b² = 4ac

(i) AC = 1, b = 2 (1, 2, 1) is one way

(ii) AC = 4, b = 4

$\begin{array}{l}a=4c=1\\ a=2c=2\\ a=1c=4\end{array}\}3ways$

(iii) AC = 9, b = 6, a = 3, c = 3 is one way

1 + 3 + 1 = 5 way

Required probability = $\frac{5}{216}$   

Let sin = t

${\displaystyle \frac{}{}}$ ${\displaystyle \int \frac{sin\theta \left(2sin\theta .cos\theta \right)\left(si{n}^6\theta +si{n}^4\theta +si{n}^2\theta \right)\sqrt[]{2si{n}^4\theta +3si{n}^2\theta +6}}{2si{n}^2\theta }}$  d

sin = t

cos . d = dt

${\displaystyle \int \frac{u^{1/2}}{12}}du=\frac{u^{3/2}}{18}+C=\frac{{\left(2t^6+3t^4+6t^2\right)}^{3/2}}{18}+C=\frac{{\left(2si{n}^6\theta +3si{n}^4\theta +6si{n}^2\theta \right)}^{3/2}}{18}+C$

l + m – n = 0

l + m = n . (i)

l² + m² = n²

Now from (i)

l² + m² = (l + m)²

=> 2lm = 0

=>lm = 0

l = 0 or m = 0

=> m = n Þ l = n

if we take direction consine of line

$0,\frac{1}{\sqrt[]{2}},\frac{1}{\sqrt[]{2}}and\frac{1}{\sqrt[]{2}},0,\frac{1}{\sqrt[]{2}}$

cos a = $\frac{1}{2}$              

$si{n}^4\alpha +co{s}^4\alpha ={\left(\frac{\sqrt[]{3}}{2}\right)}^4={\left(\frac{1}{2}\right)}^4=\frac{9}{16}+\frac{1}{16}=\frac{10}{16}=\frac{5}{8}$  

xyz = 24

24 = 2³ * 3

Let's distribute 2, 3 among 3 variables. No. of positive integral solution =

No. of ways to distribute =

$\therefore In\Delta AQP$

$tan30°=\frac{PQ}{AQ}$

$\frac{1}{\sqrt[]{3}}=\frac{h}{x+y}$

x + y = $\sqrt[]{3}h.....\left(i\right)$ $\text{exception 25:}$

$\therefore ln\Delta BQP$

$tan45°=\frac{h}{y}$

h = y . (ii)

(i) & (ii) x + y = $\sqrt[]{3}y$ $\text{exception 25:}$

$\Rightarrow x=\left(\sqrt[]{3}-1\right)y.......\left(iii\right)$

Let the speed be S

$\frac{x}{S}=20$

x = 20.S

from (iii)

$\frac{y}{S}=10\left(\sqrt[]{3}+1\right)$

sin 2 + tan 2 > 0

$\frac{2tan\theta }{1+ta{n}^2\theta }+\frac{2tan\theta }{1-ta{n}^2\theta }>0$

Let tan = x

$\frac{2x}{1+x^2}+\frac{2x}{1-x^2}>0$

$tan\theta <-1or0<tan\theta <1$

$\theta \in \left(0,\frac{\pi }{4}\right)\cup \left(\frac{\pi }{2},\frac{3\pi }{4}\right)\cup \left(\pi ,\frac{5\pi }{4}\right)\cup \left(\frac{3\pi }{2},\frac{7\pi }{4}\right)$

(2 – i) z = (2 + i) $\overline{z}$ $\stackrel{}{}$ , put z = x + iy

$y=\frac{x}{2}........\left(i\right)$

(ii) $\left(2+i\right)z+\left(i-2\right)\overline{z}-4i=0$ $\stackrel{}{}$

x + 2y = 2

(iii) $iz+\overline{z}+1+i=0$ $\stackrel{}{}$

Equation of tangent x – y + 1 = 0

Solving (i) and (ii)

$x=1.y=\frac{1}{2}\Rightarrow centre\left(1,\frac{1}{2}\right)$

Perpendicular distance of point $\left(1,\frac{1}{2}\right)$ $\frac{}{}$ from x – y + 1 = 0 is p = r $\Rightarrow \left|\frac{1-\frac{1}{2}+1}{\sqrt[]{2}}\right|=r$ $\frac{\frac{}{}}{\text{exception 25:}}$

$r=\frac{3}{2\sqrt[]{2}}$

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

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

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