Ncert Solutions Maths class 11th

  ${\displaystyle \sum _{i=1}^n{a}_i}=192\Rightarrow a_1+a_2+a_3+......+a_n=192$

$\therefore n\left(a_1+a_n\right)=384..............(i)$

and ${\displaystyle \sum _{i=1}^{n/2}{a}_{2i}}=120\Rightarrow a_2+a_4+a_6+......+a_n=120$  

 n (a₂ + a_n) = 480 ……………… (ii)

n (a₂ – a₁) = 96

$\therefore n=96$

${\left(ta{n}^{-1}x\right)}^3+{\left(co{t}^{-1}x\right)}^3=k{\pi }^3,x\in R$

$\Rightarrow \left(ta{n}^{-1}x+co{t}^{-1}x\right)\left({\left(ta{n}^{-1}x+co{t}^{-1}x\right)}^2-3ta{n}^{-1}xco{t}^{-1}x\right)=k{\pi }^3$

$\Rightarrow \frac{1}{3}\le k<\frac{7}{8}$

$\alpha +\beta =-\frac{\lambda }{3},\alpha \beta =-\frac{1}{3}$

$\therefore \frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}=15\Rightarrow \lambda =\pm 3$

$\therefore 6{\left({\alpha }^3+{\beta }^3\right)}^2=6{\left({\left(\alpha +\beta \right)}^3-3\alpha \beta \left(\alpha +\beta \right)\right)}^2=6*{\left(-\left(1+1\right)\right)}^2=24$

Equation of tangent to the circle at (2, 4) is

(4 + A) x + (8 + B) y + 2A + 2B + 2C = 0……. (i)

Also equation of tangent to parabola y = x² at (2, 4) is

4x – y = 4 ………. (ii)

Comparing (i) and (ii) we get, A + C = 16

(3 * 3 * 3 * ………2022 times) ¸ 5

Remainder = (-1) (-1) (-1) ….1011 times)

Remainder = -1 + 5 = 4

S (4, 4) and V (3, 2)

$\therefore$ point of intersection of directrix with axis of parabola is A (2, 0)

Image of A (2, 0) with respect to line

$\therefore x+2y=6isB\left(x_2,y_2\right)$

$\therefore \frac{x_2-2}{1}=\frac{y_2-0}{2}\Rightarrow \frac{-2\left(2+0-6\right)}{5}$

$\therefore B\left(\frac{18}{5},\frac{16}{5}\right)$

Point B is point of intersection of directrix with axes of parabola P₂.

$\therefore x+2y=\lambda$

$B\left(\frac{18}{5},\frac{16}{5}\right)$ lies on the line x + 2y = $\lambda$  $\therefore \lambda =\frac{18}{5}+\frac{32}{5}=10$

 $\frac{x^2}{a^2}-\frac{y^2}{1}=1$

Length of latus rectum = $\frac{2}{a}$

$and\frac{x^2}{4}+\frac{y^2}{3}=1$

length of latus rectum = $\frac{6}{2}$ = 3

$?\frac{2}{a}=3\Rightarrow a=\frac{2}{3}$

$12\left(e_H^2+e_H^2\right)=12\left[\left(1+\frac{9}{4}\right)+\left(1-\frac{3}{4}\right)\right]=12\left[\frac{13}{4}+\frac{1}{4}\right]$ = 12 * $\frac{14}{4}=42$

OM² = OP² = PM²

^{$\left|\frac{1+r}{\sqrt[]{2}}\right|=r^2-1$}

^{$\therefore r=3$}

^$\therefore$ equation of circle is ${\left(x-1\right)}^2+{\left(y-3\right)}^2=3^2$

$\therefore h+k+r=7$

Required area

= ${\displaystyle {\int }_{-4}^2\left(4-y-\frac{y^2}{2}\right)dy}$

= ${\left(4y-\frac{y^2}{2}-\frac{y^3}{6}\right)}_{-4}^2=18squnits$