Ncert Solutions Maths class 11th

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)

$3=m\pm \sqrt[]{m^2-3}$      

$\therefore m=2$

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

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

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.

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

1012 $\le$  Number in the question $\le$  23421

Also, the number has to use digits {2, 3, 4, 5, 6} without repetition and the number has to be divisible by $\frac{55}{11*5}$  

As the number has to be divisible by both 5 and 11,

5 → once place

Let us make 4-digit such numbers first:

{2, 3, 4, 6} (digits are not be repeated)

A number is divisible by 11 it difference of sum of its digits at even places and sum of digits at odd place is 0 or multiple of 11.

x² – x – 4 = 0

$P_n={\alpha }^n-{\beta }^n$

$?=\frac{P_{15}{P}_{16}-P_{14}{P}_{16}-P_{15}^2+P_{14}{P}_{15}}{P_{13}{P}_{14}}$

$=\frac{\left(P_{15}-P_{14}\right)\left(P_{16}-P_{15}\right)}{P_{13}{P}_{14}}$

$=\frac{4P_{13}\cdot 4P_{14}}{P_{13}\cdot P_{14}}=16$

P_n =  αⁿ - βⁿ

$={\alpha }^{n-1}\cdot \alpha -{\beta }^{n-1}\cdot \beta$

$={\alpha }^{n-1}\left({\alpha }^2-4\right)-{\beta }^{n-1}\left(\beta -4\right)$

$P_n={\alpha }^{n+1}-{\beta }^{n+1}-4\left({\alpha }^{n-1}-{\beta }^{n-1}\right)$

$P_n=P_{n+1}-4P_{n-1}\Rightarrow P_{n+1}-P_n=4P_{n-1}$

${\displaystyle \sum _{}^{}{x}_0^1=\frac{3{\left(1-\left(\frac{1}{2}\right)\right)}^{20}}{1-\frac{1}{2}}=6\left(1-\frac{1}{2^{20}}\right)}$

$={\displaystyle \sum _{i=1}^{20}{\left(x_i\right)}^2+{\left(i\right)}^2-2x_{i}i}$

Now, ${\displaystyle \sum _{i=1}^{20}{\left(x_i\right)}^2=\frac{9{\left(1-\left(\frac{1}{4}\right)\right)}^{20}}{\left(1-\frac{1}{4}\right)}=12\left(1-\frac{1}{2^{40}}\right)}$

$\overline{x}=\frac{2858}{20}+\left(\frac{-12}{2^{40}}+\frac{22}{2^{20}}\right)*\frac{1}{20}$

$\left[\overline{x}\right]=142$

$e_E=\sqrt[]{1-\frac{b^2}{a^2}},e_H=\sqrt[]{2}$

$If\Rightarrow e_E=\frac{1}{e_H}\Rightarrow \frac{a^2-b^2}{a^2}=\frac{1}{2}$

$k^2=a^2*\frac{5}{2}+b^2=\frac{3}{2}$

$6b^2=\frac{3}{2}\Rightarrow b^2=\frac{1}{4}and{a}^2=\frac{1}{2}$

$\therefore 4\left(a^2+b^2\right)=3$