Mathematics Test - 5

$a=Bc=\sqrt{0^2+(12-0{)}^2}=12$

$b=AC=\sqrt{(0-8{)}^2+(6-0{)}^2}=10$

$c=AB=\sqrt{8^2+6^2}=10$

$(\frac{a{x}_1+b{x}_2+c{x}_3}{a+b+c},\frac{a{y}_1+b{y}_2+c{y}_3}{a+b+c})$

$(\frac{12\times 0+10\times 8+10\times 8}{12+10+10},\frac{12\times 6+10\times 12+10\times 0}{12+10+10})$

$=(\frac{160}{32},\frac{192}{32})=(5,6)$ 

log_n 2 + log_n 3 + log_n 4 + .....+ log_n 2017.

= log_n [2.3.4....2017]

= log_n (2017!)

= log_n n = 1

Given-

The straight line passes through $(1,2)$ $=(x_1,y_1)$

The line is parallel to the line $y=3x+1$.

Slope of the required line $m=3$  (The slope of the given line is $3$)

According to the formula-

$(y-y_1)=m(x-x_1)$

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

$\Rightarrow y-2=3x-3$

$\Rightarrow 3x-y=1$

Given-

$\frac{\sin 5x-\sin 3x}{\cos 5x+\cos 3x}$

$=\frac{2\cos \left(\frac{5x+3x}{2}\right)\sin \left(\frac{5x-3x}{2}\right)}{2\cos \left(\frac{5x+3x}{2}\right)\cos \left(\frac{5x-3x}{2}\right)}$

$=\frac{2\cos 4x\times \sin x}{2\cos 4x\times \cos x}$

$=\frac{\sin x}{\cos x}$

$=\tan x$

Hence the given equation has a rational root and an irrational root.

$\begin{array}{l}\tan (x-y), \tan x, \tan (x+y)\text{ are in G.P. }\\ {\tan }^{2}x=\tan (x-y)\times \tan (x+y)\\ {\tan }^{2}x=\left(\frac{\tan x-\tan y}{1+\tan x\tan y}\right)\times \left(\frac{\tan x+\tan y}{1-\tan x\tan y}\right)\\ {\tan }^{2}x=\frac{{\tan }^{2}x-{\tan }^{2}y}{1-{\tan }^{2}x{\tan }^{2}y}\\ {\tan }^{2}x-{\tan }^{4}x{\tan }^{2}y={\tan }^{2}x-{\tan }^{2}y\\ \Rightarrow {\tan }^{4}x\times {\tan }^{2}y={\tan }^{2}y\\ \Rightarrow {\tan }^{4}x=1\\ \Rightarrow {\tan }^{4}x=1^4\\ \Rightarrow (\tan x{)}^4={\left(\tan \frac{\pi }{4}\right)}^4\\ \Rightarrow x=\frac{\pi }{4}\end{array}$

Given-

$\cos A=\frac{4}{5}$

$\Rightarrow \frac{Base}{Hypotenuse}=\frac{4}{5}$

Let $base=4k$ and $hypotenuse=5k$.

According to Pythagoras Law-

$perpendicular=3k$

$\Rightarrow \tan A=\frac{Perpendicular}{Base}$

$\Rightarrow \tan A=\frac{3k}{4k}$

$\Rightarrow \tan A=\frac{3}{4}$

$\begin{array}{l}\left|\frac{\alpha -\beta }{1-\alpha \overline{\beta }}\right|=\frac{|\alpha -\beta |}{|1-\alpha \overline{\beta }|}\\ \le \frac{|\alpha |+|\beta |}{|1|+|\alpha \overline{\beta }|}\\ =\frac{|\alpha |+|\beta |}{|1|+|\alpha ||\overline{\beta }|}\\ =\frac{|\alpha |+|\beta |}{|1|+|\alpha ||\beta |}\\ =\frac{1+|\beta |}{|1|+|\beta |}[\because |\alpha |=1]\\ =1\end{array}$   

\(\sum_{n=2}^{11}\left(i^{n}+i^{n+1}\right)\)

\(=t^{2}+2 i^{3}+2 i^{4}+2 i^{5}+2 i^{6}+2 i^{7}+2 i^{8}+2 i^{9}+2 i^{10}+2 i^{11}+i^{12}\)

\(=i^{2}+2(i)^{2} i+2\left(i^{2}\right)^{2}+2\left(i^{2}\right)^{2} \cdot i+2\left(i^{2}\right)^{3}+2\left(i^{2}\right)^{3} \cdot i+2\left(i^{2}\right)^{4}+2\left(i^{2}\right)^{4}+2\left(i^{2}\right)^{4} i+2\left(i^{2}\right)^{5}+2\left(i^{2}\right)^{5} i+\) \(\left(i^{2}\right)^{6}\)

\(=-1-2 i+2+2 i-2-2 i+2+2 i-2-2 i+1\)

\(=-2 i\)