Mathematics Test - 3

$\begin{array}{l}{i}^{1000}+i^{1000}\cdot i+i^{1002}+i^{1002}.i\\ {\left(i^2\right)}^{500}+{\left(i^2\right)}^{500}\cdot i+{\left(i^2\right)}^{501}+{\left(i^2\right)}^{501}\cdot i\\ (-1{)}^{500}+(-1{)}^{500}i+(-1{)}^{501}+(-1{)}^{501}\cdot i\\ 1+i-1-i=0\end{array}$

Given that

$x=1-y+y^2-y^3\cdots \left(\mathrm{i}\right)$

$\Rightarrow y-y^2+y^3-\dots \dots =1-x$

$\Rightarrow y(1-y+y^2-y^3+\cdots )=1-x$

$\Rightarrow yx=1-x\left[From\right(i\left)\right]$

$\Rightarrow yx+x=1$

$x=\frac{1}{1+y}$ 

680 students like music.

215 students like dance.

Total = 680 + 215 = 895

The least number of students who like both music and dance = 895 - 850 = 45

By sine rule,

$\begin{array}{l}\frac{a}{\sin A}=\frac{b}{\sin B}\\ \Rightarrow \frac{2}{\frac{2}{3}}=\frac{3}{\sin B}\\ \Rightarrow 2\times \frac{3}{2}=\frac{3}{\sin B}\\ \sin B=1\\ \sin B=\sin \frac{\pi }{2}\\ B=\frac{\pi }{2}\end{array}$

Suppose cos A is given. If only one value of cos (A/2) is possible, then A must be an odd multiple of 180° 

$\begin{array}{rl}& \sin (x+\frac{\pi }{5})+\cos (x+\frac{\pi }{5})\\ & f^{\mathrm{\prime }}\left(x\right)=\cos (x+\frac{\pi }{5})-\sin (x+\frac{\pi }{5})\\ & 0=\cos (x+\frac{\pi }{5})-\sin (x+\frac{\pi }{5})\\ & \cos (x+\frac{\pi }{5})=\sin (x+\frac{\pi }{5})\\ & \tan (x+\frac{\pi }{5})=1\\ & \tan (x+\frac{\pi }{5})=\tan \frac{\pi }{4}\\ & x+\frac{\pi }{5}=\frac{\pi }{4}\\ & x=\frac{\pi }{4}-\frac{\pi }{5}\\ & x=\frac{\pi }{20}\end{array}$

ntercept form of line is xa+yb=1" role="presentation">x/a+y/b=1$\frac{x}{a}+\frac{y}{b}=1$.

We know, the point which divides a line joining two points (x1,y1)" role="presentation">(x1,y1)$(x_1,y_1)$ and (x2,y2)" role="presentation">(x2,y2)$(x_2,y_2)$ in the ratio m:n" role="presentation">m$:n$ is

(mx2+nx1m+n,my2+ny1m+n)" role="presentation">
(mx2+nx1/m+n,my2+ny1/m+n)

Case 1 : m:n=2:3" role="presentation">m:n=2:3

∴(2,3)=((2)(a)+3(0)/2+3,2(0)+3(b)/2+3)

⇒(2,3)=(2a5,3b5)" role="presentation">⇒(2,3)=(2a/5,3b/5)

⇒2a/5=2,3b/5=3

⇒a=5,b=5" role="presentation">⇒a=5,b=5$\Rightarrow a=5,b=5$.

∴" role="presentation">∴ Equation of line is x5+y5=1⇒x+y=5" role="presentation">x/5+y/5=1⇒x+y=5

Case 2 : m:n=3:2" role="presentation">m:n=3:2

∴(2,3)=(3(a)+2(0)/3+2,3(0)+2(b)/3+2)

⇒(2,3)=(3a5,2b5)" role="presentation">⇒(2,3)=(3a/5,2b/5)

⇒3a/5=2,2b/5=3

⇒a=10/3,b=15/2

∴" role="presentation">$\therefore$ Equation of line is x103+y152=1" role="presentation">x/10/3+y/15/2=1

$\frac{3x}{10} + \frac{2y}{15}=1$x103+y152=1" role="presentation">

⇒9x+4y=30

Given that 

⁴³C_2r = 4³C_r+1

Since 2r ≠ r + 1 = 43 

We know that -

2r + r + 1 = 43

3r = 42

r = 14.

Given that 

$A^2-kA+I_2=0 and A=\left[\begin{array}{cc}1& 3\\ 2& 3\end{array}\right]$

$\Rightarrow \left[\begin{array}{ll}1& 2\\ 2& 3\end{array}\right]\left[\begin{array}{ll}1& 2\\ 2& 3\end{array}\right]-k\left[\begin{array}{ll}1& 2\\ 2& 3\end{array}\right]+\left[\begin{array}{ll}1& 0\\ 0& 1\end{array}\right]=0$

$\left[\begin{array}{cc}5& 8\\ 8& 13\end{array}\right]-\left[\begin{array}{cc}\mathrm{k}& 2\mathrm{k}\\ 2\mathrm{k}& 3\mathrm{k}\end{array}\right]+\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=0$

$\left[\begin{array}{cc}5-\mathrm{k}+1& 8-2\mathrm{k}+0\\ 8-2\mathrm{k}+0& 13-3\mathrm{k}+1\end{array}\right]=\left[\begin{array}{ll}0& 0\\ 0& 0\end{array}\right]$

$\Rightarrow 8-2k=0$

$\Rightarrow 8=2k$

$k=4$ 

$\begin{array}{l}\text{ Area }={\int }_0^{\sqrt{6}}(\frac{y^2}{6}+1-\frac{y^2}{3})dy\\ ={[\frac{y^3}{18}+y-\frac{y^3}{9}]}_0^{\sqrt{6}}=\frac{6\sqrt{6}}{18}+\sqrt{6}-\frac{6\sqrt{6}}{9}\\ =\frac{4\sqrt{6}}{3}\end{array}$