Maths

$f\left(x\right)=\{\begin{array}{l}\left|4x^2-8x+5\right|,if8x^2-6x+1\ge 0\\ \left[4x^2-8x+5\right],if8x^2-6x+1<0\end{array}$

$x=\frac{1}{4},\frac{2-\sqrt[]{2}}{2},\frac{1}{2}$

a₁ = b₁ = 1

$a_n=a_{n-1}+2\left(forn\ge 2\right)b_n=a_n=b_{n-1}$          

Similarly for others

${\displaystyle {\sum }_{n=1}^{11}{a}_n{b}_n={\displaystyle {\sum }_{n=1}^{15}\left(2n-1\right)n^2={\displaystyle {\sum }_{n=1}^{15}2n^3-{\displaystyle {\sum }_{n=1}^{15}{n}^2}}}}$      

$=2{\left[\frac{15*16}{2}\right]}^2-\left[\frac{15*16*31}{6}\right]=27560$

  $?$  Roots of 2ax²  - 8 ax + 1 = 0 are

$\frac{1}{p}and\frac{1}{r}$ and roots of 6bx² + 12bx + 1 = 0 are

  $\frac{1}{q}and\frac{1}{8}$  

Let  $\frac{1}{p},\frac{1}{q},\frac{1}{r},\frac{1}{8}$

as    $\alpha -3\beta ,\alpha -\beta ,\alpha +\beta ,\alpha +3\beta$

$So,\frac{1}{a}-\frac{1}{b}=38$

General term

${=}^{15}{C}_r{\left(t^2{x}^{\frac{1}{5}}\right)}^{15-r}{\left(\frac{{\left(1-x\right)}^{\frac{1}{10}}}{t}\right)}^r$    

for term independent on t

2 (15 – r) – r = 0

$\Rightarrow r=10$          

$\therefore T_{11}{=}^{15}{C}_{10}*\left(1-x\right)$

Maximum value of x (1 – x) occur at

x =  $\frac{1}{2}$

Arranging letter in alphabetical order A   D   I   K   M   N   N for finding rank of MANKIND making arrangements of dictionary we get

A …….->

$\frac{6!}{2!}=360$        

D ………->360

l ………. ->360

K ………->360

MAD …….->

$\frac{4!}{2!}=12$        

$\therefore$ Rank of MANKIND = 1440 + 36 + 12 + 2 + 2 = 1492.

Here,

$A=\left[\begin{array}{ccc}\hfill 2\hfill & \hfill -1\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 0\hfill \end{array}\right]$            

we get A² = A and similarly, for

$B=A-I=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill -1\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill -1\hfill \end{array}\right]$            

We get B² = -B $\Rightarrow$  B³ = B

$\therefore A^n+{\left(\omega B\right)}^n=A+{\left(\omega B\right)}^{n}forn\in N$

For ${\omega }^n$  to be unity n shall be multiple of 3 and for Bⁿ to be B. n shell be 3, 5, 7, ….9

$\therefore n=\left\{3,9,15,.....99\right\}$     

Number of elements = 17.

$\frac{{\left(x-3\right)}^2}{16}+\frac{{\left(y-4\right)}^2}{9}\le 1,x,y\in N,{\left(x-7\right)}^2+{\left(y-4\right)}^2\ge 36,x,y\in R$

Total number of common point = 1 + 5 + 7 + 5 + 5 + 3 + 1 = 27

Given : ${\left|\overrightarrow{a}+\overrightarrow{b}\right|}^2={\left|\overrightarrow{a}\right|}^2+2{\left|\overrightarrow{b}\right|}^2\&\overrightarrow{a}\cdot \overrightarrow{b}=3$

$\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|cos\theta =3$

$\left|\overrightarrow{a}\right|cos\theta =\frac{3}{\sqrt[]{6}}=\sqrt[]{\frac{9}{6}}=\sqrt[]{\frac{3}{2}}$

${\left|\overrightarrow{a}\right|}^2{\left|\overrightarrow{b}\right|}^{2}si{n}^2\theta =75$

$\left|\overrightarrow{a}\right|sin\theta =\sqrt[]{\frac{75}{6}}=\frac{5}{\sqrt[]{2}}$

$\left|\overrightarrow{a}\right|cos\theta =\sqrt[]{\frac{3}{2}}$

${\left|\overrightarrow{a}\right|}^2=\frac{25}{2}+\frac{3}{2}=\frac{28}{2}=14$

Let C be the centre and M be the mid point of AB

$\Delta APC:sin\theta =\frac{13/2}{pc}=\frac{5}{13}\Rightarrow pC=\frac{169}{10}$

$\Delta AMC:cos\theta =\frac{6}{13/2}=\frac{12}{13}$

PC = $\frac{169}{10},MC=\frac{13}{2}sin\theta =\frac{13}{2}\cdot \frac{5}{13}$

PM = PC – MC = $\frac{169}{10}-\frac{5}{2}=\frac{144}{10}$

5PM = 72

y = 5x² + 2x – 25

P(2, -1)

T_(p) : T = 0

Þ y – 1 = 10x(2) + 2(x + 2) – 50

$\Rightarrow y=22x-45$ is also tangent to y = x³ – x² + x at point (a, b)

For y = x³ -x² + x

$\frac{dy}{dx}=3x^2-2x+1>0$               

y = 22x $-\frac{1939}{27}$  which is not tangent to the curve.

$3a^2-2a+1=22$ (slope of tangent)

$\Rightarrow 3a^2-2a-21=0\to a=\frac{2\pm \sqrt[]{4+252}}{6}=\frac{2\pm 16}{6}=3,\frac{-7}{3}$               

b = 27 – 9 + 3 = 21

tangent : y – 21 = 22(x – 3)

 ⇒ y = 22x – 45

a = 3, b = 21

2a + 9b = 6 + 189 = 195

Also, $a^3-a^2+a=b$  

For a = $\frac{-7}{3}$  

b = $\frac{-343}{27}-\frac{49}{9}-\frac{7}{3}$  

= $\frac{-343-147-63}{27}=\frac{-553}{27}$