Maths NCERT Exemplar Solutions Class 12th Chapter Three

$y^{2}dx+\left(x^2?xy+y^2\right)dy=0$

$\frac{dx}{dy}+\frac{x^2?xy+y^2}{y^2}=0?\frac{dx}{dy}+{\left(\frac{x}{y}\right)}^2?\left(\frac{x}{y}\right)+1=0$

Put x = vy $?v+y\frac{dv}{dy}+v^2?v+1=0$ $\frac{}{}{}^{}$

$?\frac{ydv}{dy}+v^2+1=0$

$?\frac{?}{6}+\mathcal{l}n\left|y\right|=\frac{?}{4}$

$\mathcal{l}n\left|y\right|=\frac{?}{12}$

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 * $\frac{14}{4}=42$

Since student guesses only two wrong. So there are three possibilities

(i) both wrong in section A

(ii) both wrong in section B

(iii) one wrong in each section A and B.

$\therefore$ Required possibilities =

${=}^4{C}_4{*}^6{C}_4{\left(\frac{3}{4}\right)}^4*{\left(\frac{1}{4}\right)}^4{\left(\frac{3}{4}\right)}^2{+}^4{C}_3{*}^6{C}_5{\left(\frac{3}{4}\right)}^3{\left(\frac{1}{4}\right)}^5*\frac{1}{4}*\frac{3}{4}$ ${+}^4{C}_2{*}^6{C}_6*{\left(\frac{3}{4}\right)}^2{\left(\frac{1}{4}\right)}^2*{\left(\frac{1}{4}\right)}^6$

$=\frac{27}{4^{10}}\left[15*27+24*3+2\right]=\frac{27*479}{4^{10}}$

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

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

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

S = 1 + 3 + 3² + 3³ + ….+ 3²⁰²¹   $=\frac{3^{2022}-1}{2}=\frac{1}{2}\left[a^{1011}-1\right]$

^{$=\frac{1}{2}\left[9^{1011}-1\right]=\frac{1}{2}\left[100k+10110-1-1\right]$}                          

= 50k₁ + 4

Remainder = 4

$\left\{a\in \left(1,2,3,......,100\right):HCF\left(a,24\right)=1\right\}$

HCF of (a, 24) = 1 $\therefore$ a = 1, 5, 7, 11, 13, 17, 19, 23 sum of these numbers = 96

$\therefore$ There are four such blocks and a number 97 is there upto 100.

$\therefore$ complete sum = 96 + (24 * 8 + 96) + (48 * 8 + 96) + (72 * 8 + 96) + 97 = 1633

Sum of all given numbers = 31

Difference between odd and even positions must be 0,11 or 22 but 0 and 22 are not possible.

$\therefore$ Hence 11 is possible.

This is possible only when either 1, 2, 3, 4 if filled in odd places in order and remaining in other order.

Hence 2, 3, 5 or 7, 2, 1 or 4, 5, 1 at even places.

$\therefore$ Total possible ways = (4! * 3!) * 4 = 576

 $S=\{\left(\begin{array}{cc}\hfill 1\hfill & \hfill a\hfill \\ \hfill 0\hfill & \hfill b\hfill \end{array}\right),a,b\in 1,2,3,.....100$

$\therefore A=\left(\begin{array}{cc}\hfill -1\hfill & \hfill a\hfill \\ \hfill 0\hfill & \hfill b\hfill \end{array}\right)$ then even power of A as A = $\left(\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right).$

If b = 1 & $a\in \left\{1,2,3.....100\right\}$ and n (n + 1) is always even

$\therefore T_1,T_2,T_3,........,T_n$ are all 1 for b = 1 and each value of a.

$\therefore {\displaystyle \underset{n=1}{\overset{1000}{\cap }}{T}_n=100}$