Class 11th

T? = (1/3)T?
⇒ T? , T? , T? G.P.
T = 1/3
T? = 1/243
ar? = 1/243
a (1/3)? = 1/243
a/729 = 1/243 ⇒ a = 3

Sum of infinite series = T? + T? + T?
= (T? ) / (1-r²) = (3 * 1) / (1 - 1/9) = 3 / (8/9) = 27/8

y² = 4x
(x - 3)² + y² = 9
y = mx + 1/m
(3, 0), r = 3

|3m + 1/m| / √ (1+m²) = 3

9m² + 1/m² + 6 = 9 (1+m²)
9m² + 1/m² + 6 = 9 + 9m²
1/m² = 3 ⇒ m = ±1/√3
m = 1/√3 in first quadrant
y = x/√3 + √3 ⇒ √3y = x + 3

b? , b? , b?
a, ar, ar².
a + ar = 1 ⇒ a = 1 / (1+r)

Σ (from k=1 to ∞) b? = 2 ⇒ a / (1-r) = 2
(1 / (1+r) / (1-r) = 2
1 / (1-r²) = 2
1/2 = 1 - r²
r² = 1/2 ⇒ r = ±1/√2

b? < 0 r = -1/2
⇒ a = b? = 2 + √2

$z+\alpha \left|z-1\right|+2i=0$

Let z = x + iy

$\Rightarrow x+i\left(y+2\right)=-\alpha \sqrt[]{{\left(x-1\right)}^2+y^2}$            

for $\alpha \in Ry+2=0\Rightarrow y=-2$

 $x^2={\alpha }^2\left[{\left(x-1\right)}^2+4\right]$

$x^2\left(\frac{1}{{\alpha }^2}-1\right)+2x-5$

$\frac{1}{{\alpha }^2}\ge \frac{4}{5}\Rightarrow {\alpha }^2\le \frac{5}{4}\Rightarrow \frac{-\sqrt[]{5}}{2}\le \alpha \le \frac{\sqrt[]{5}}{2}$

$4\left(p^2+q^2\right)=4\left(\frac{5}{4}+\frac{5}{4}\right)=10$          

= 0

$t_n=\frac{3^n-2^n}{3^n}=1-{\left(\frac{2}{3}\right)}^n$

$S_{100}=100-\frac{\frac{2}{3}\left({\left(\frac{2}{3}\right)}^{100}-1\right)}{\frac{2}{3}-1}=100+2.\frac{\left(2^{100}-3^{100}\right)}{3^{100}}$

$=100-2+\frac{2^{101}}{3^{100}}=98+{\left(\frac{2}{3}\right)}^{100}.2<98+1$

${\alpha }_n=19^n-12^n$

$\frac{31{\alpha }_9-{\alpha }_{10}}{57.{\alpha }_2}=\frac{31\left(19^9-12^9\right)-\left(19^{10}-12^{10}\right)}{57\left(19^8-12^8\right)}$

$=\frac{19^9\left(31-19\right)-12^9\left(31-12\right)}{57\left(19^8-12^8\right)}$

$=\frac{19^9.12-12^9.19}{57\left(19^8-12^8\right)}=4$

$P\left(x_1,y_1\right)$

$Q\left(x_2,y_2\right)$

$x_1+x_2=\frac{r}{2},x_1{x}_2=\frac{P}{2}$

$y_1+y_2=s,y_1,y_2=-9$

$\left(x-x_1\right)\left(x-x_2\right)+\left(y-y_1\right)\left(y-y_2\right)=0$

$2\left(x^2+y^2\right)-rx-2sy+p-2q=0$

r = 11, s = 7, p – 2q = -22

$x_1+x_2=6,13\to 6+6=12$

$\begin{array}{l}{x}_1+x_2=4,11,18\to 4+8+1=13\\ x_1+x_2=2,9,16\to 2+9+3=14\\ x_1+x_2=7,14\to 7+5=12\\ x_1+x_2=5,12\to 5+7=12\end{array}$

= 63

${\displaystyle \sum _{k=1}^{10}\left(2k-1\right).k{.}^{10}{C}_k}$

$=2\sum {k^2}^{10}{C}_k-\sum k{.}^{10}{C}_k$

$x=1\Rightarrow {\displaystyle \sum _{k=0}^{10}k}{.}^{10}{C}_k=10.2^9$

$x=1\Rightarrow \sum k^2{}^{10}{C}_k=90.2^8+10.2^9=2^8\left(90+201\right)=2^8.110$

S = 2⁹ . 110 – 10.2⁹ = 2⁹ . 100

S = 2⁹ . 100

$\frac{2^{11}.25}{2^1-1}$