Ncert Solutions Maths class 11th

z - 2² = z + 2²
⇒ x=0
Hence minimum '0'

  $x^2+y^2-2x-6y+6=0$ centre (1, 3)

$\begin{array}{l}r=\sqrt[]{1+9-6}=2\\ CM=\sqrt[]{1+4}=\sqrt[]{5}\end{array}$           

$r=\sqrt[]{5+4}=3$

~ (p v (~ p v q)
= p ^ ( (~p v q) = ~p ^ (p ^ ~q)

  $\frac{4}{sinx}+\frac{1}{1-sinx}=a\forall x\in \left(0,\frac{\pi }{2}\right)$

Let sin x = t, t  $\in$ (0, 1)

$g\left(t\right)=\frac{4}{t}+\frac{1}{1-t}$         

$g\text{'}\left(t\right)=0\Rightarrow t=\frac{2}{3}$           

$g\text{'}\text{'}\left(\frac{2}{3}\right)>0$           

$\therefore g{\left(t\right)}_{minimum}=\frac{4}{2/3}+\frac{1}{1-2/3}=9$          

Minimum value of a for which solution exist = 9

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

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

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

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