Ncert Solutions Maths class 11th

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

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 = aⁿ - bⁿ   

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

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

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

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

Use [x + n] = n + [x], where $n\in z$  f (x) = [x] – 10 will be minimum for $x\in [0,10]$  break the limits as G.I.F. is discontinuous at integral points.

${\displaystyle \underset{0}{\overset{10}{\int }}f\left(x\right)dx=-10-9-....-1}$    

$=\frac{-10.11}{2}$

${\displaystyle \underset{0}{\overset{10}{\int }}\left|f\left(x\right)\right|dx=10+9+....+1}$

 $\frac{{\left(x+1\right)}^2}{\lambda +5}=\frac{{\left(y+1\right)}^2}{\frac{\lambda +5}{4}}=1$

length of latus rectum = $\frac{2b^2}{a}=\frac{2\left(\frac{\lambda +5}{4}\right)}{\sqrt[]{5+\lambda }}=4$

$\sqrt[]{\lambda +5}=8\Rightarrow \lambda =59$

Major axis = $2\sqrt[]{\lambda +5}=16$

$e_H=\sqrt[]{1+\frac{64}{49}}=\frac{\sqrt[]{113}}{7}$

$\Rightarrow e_H.e_E=\frac{1}{2}$

$\Rightarrow \frac{113}{49}.\frac{\left(64-a^2\right)}{64}=\frac{1}{4}\Rightarrow a^2-64=\frac{32^2}{113}$

$\mathcal{l}=\frac{2a^2}{b}=2\left(64+\frac{32^2}{113}\right).\frac{1}{8}$

$113\mathcal{l}=1552$

The circle $x^2+y^2+6x+8y+16=0$ has centre (3, 4) and radius 3 units

The circle

$x^2+y^2+2\left(3-\sqrt[]{3}\right)x+2\left(4-\sqrt[]{6}\right)y=k+6\sqrt[]{3}+8\sqrt[]{6},k>0$ has centre $\left(\sqrt[]{3}-3,\sqrt[]{6}-4\right)$ and radius $\sqrt[]{k+34}$

$?$ These two circles touch internally hence

$\sqrt[]{3+6}=\left|\sqrt[]{k+34}-3\right|$

here, k = 2 is only possible $\left(?k>0\right)$

Equation of common tangent to two circle is

$2\sqrt[]{3}x+2\sqrt[]{6}y+16+6\sqrt[]{3}+8\sqrt[]{6}+k=0$

$?k=2$ then equation is

$x+\sqrt[]{2}y+3+4\sqrt[]{2}+3\sqrt[]{3}=0$ ….(i)

$?\left(\alpha ,\beta \right)$ are foot of perpendicular from (3, 4) to line (i) then

$\frac{\alpha +3}{1}=\frac{\beta +4}{\sqrt[]{2}}=\frac{-3-4\sqrt[]{2}+3+4\sqrt[]{2}+3+\sqrt[]{3}}{1+2}$

$\therefore {\left(\alpha +\sqrt[]{3}\right)}^2+{\left(\beta +\sqrt[]{6}\right)}^2=25$

 $?a_n={\displaystyle {\int }_{-1}^0\left(1+\frac{x}{2}+\frac{x^2}{2}+....+\frac{x^{n-1}}{n}\right)dx}$

$={\left[x+\frac{x^2}{2^2}+\frac{x^3}{3^2}+.....+\frac{x^n}{n^2}\right]}_{-1}^n$

$a_n=\frac{n+1}{1^2}+\frac{n^2-1}{2^2}=\frac{n^3+1}{3^2}+\frac{n^4-1}{4^2}+.....+\frac{n^n+{\left(-1\right)}^{n+1}}{n^2}$

Here a₁ = 2, $a_2=\frac{2+1}{1}+\frac{2^2-1}{2}=3+\frac{3}{2}=\frac{9}{2}$

$a_4=5+\frac{15}{4}+\frac{65}{9}+\frac{255}{16}>31$

$\therefore$ The required set is {2, 3}

$?a_n\in \left(2,30\right)$

$\therefore$ Sum of elements = 5.

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)

$3=m\pm \sqrt[]{m^2-3}$      

$\therefore m=2$