Maths

Equation of tangent to the parabola at

$P\left(\frac{8}{5},\frac{6}{5}\right)$

$75x.\frac{8}{5}=160\left(y+\frac{6}{5}\right)-192$

$\Rightarrow 120x=160y$

$\Rightarrow 3x=4y$

Equation of circle touching the given parabola at P can be taken as

$\Rightarrow \lambda =\frac{2}{5}or-\frac{8}{5}$

Radius = 1 or 4

Sum of diameter = 10

  $x^2+y^2-2x+2fy+1=0centre=\left(1,-f\right)$

diameter 2px – y = 1 ………. (i)

2x + py = 4p    ……… (ii)

$x=\frac{5P}{2P^2+2}$          

f = 0

$\left[forP=\frac{1}{2}\right]$                

$\frac{5P}{2P^2+2}=1$            

f = 3 [for P = 2]

substitute (2, 3)

$3=m\pm \sqrt[]{m^2-3}$      

$\therefore m=2$

$li{m}_{n\to \infty }{\left(\frac{n+1}{n}\right)}^{k-1}\frac{1}{n}{\displaystyle {\sum }_{r=1}^n\left(k+\frac{r}{n}\right)}$

= $33{\displaystyle {\int }_0^1{x}^{k}dx}$

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

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

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