Mathematics Test - 10

$Numberofword=5!\times {}^{7}C_3\times {}^{4}C_2$

$=120\times \frac{7!}{4!3!}\times \frac{4!}{2!2!}=25200$

Let

z=x+iy,

$\overline{z}=x-iy$

$\left|z+\overline{z}\right|=\left|z-\overline{z}\right|$

$\left|\left(x+iy\right)\right|+\left|\left(x-iy\right)\right|=\left|\left(x+iy\right)\right|-\left|\left(x-iy\right)\right|$

$\left|2x\right|=\left|2y\right|$

x=$\pm y$

Total number of balls = (8 + 7 + 6) = 21

Let E = event that the ball drawn is neither blue nor green

=event that the ball drawn is red.

Therefore, n(E) = 8.

P(E) = 8/21.

$\begin{array}{l}A+B+C=\pi \end{array}$

$A+B=\pi -C$

$\begin{array}{l}\cos (A+B)=\cos (\pi -C)\end{array}$

$\cos (A+B)=-\cos C$

$or\cos (\mathrm{A}+\mathrm{B})+\cos \mathrm{C}=0$

$\begin{array}{l}{\sin }^2\left(3\pi \right)+{\cos }^2\left(4\pi \right)+{\tan }^2\left(5\pi \right)\end{array}$

$={\sin }^2\left(3\pi \right)+{\cos }^2(\pi +3\pi )+{\tan }^2\left(5\pi \right)$

${\sin }^2\left(3\pi \right)+{\cos }^2(\pi +3\pi )+{\tan }^2\left(5\pi \right)$

$1+{\tan }^2\pi =se{c}^2\pi =1$

A={1, 2, 3, 4, 5, 6, 7} , Z = (X + Y)

x = set of odd numbers

y = set of even numbers

n (s) = 12

n (E1) = 2

$P\left(Z=5\right)=\frac{n\left(E_1\right)}{n\left(S\right)}=\frac{2}{12}=\frac{1}{6}$

S = Set of all integers and

R = {(a, b), a ,b ∈ S and ab ≥ 0}

For reflexive : aRa ⇒ a.a = a²≥0

For all integers a. a ≥ 0

For symmetric : aRb ⇒ ab ≥ 0 V a, b ∈ S

If a b ≥ 0, then ba ≥ 0 ⇒ bRa

For transitive

If ab ≥ 0, bc ≥ 0, then also ac ≥ 0

Relation R is reflexive, symmetric and transitive.

Therefore relation is equivalence.

$\frac{\left(1+i\right)\left(2+i\right)}{3-i}=\frac{1+3i}{3-i}$

$=\frac{1+3i}{3-i}\times \frac{3+i}{3+i}=\frac{10i}{10}=ior0+i$

Argument,$\theta {\tan }^{-1}\left(\frac{1}{0}\right)={\tan }^{-1}$

$\left(\tan \frac{\pi }{2}\right)=\frac{\pi }{2}$

$x^2+y^2+z^2-4y+3=0$

$x^2+y^2-4y+4-4+z^2+3=0$

$x^2+(y-2{)}^2+z^2=1$

Sphere with centre (0, 2, 0) and radius 1 unit.

$x^2+y^2+z^2+2x+4z-4=0$

$x^2+2x+1-1+y^2+z^2+4z+4-4-4=0$

$(x+1{)}^2+y^2(z+2{)}^2=3^2$

Sphere with centre (-1, 0, -2) and radius 3 units

$C_1{C}_2=\sqrt{(0+1{)}^2+(2-0{)}^2+(0+2{)}^2}=3$