Ncert Solutions Maths class 11th

C? = 2C? S = 2√20-4 = 8

$\left|z_1-1\right|=\mathrm{R}\mathrm{e}?\left(z_1\right)$ Let $z_1{x}_1+i{y}_1$and $z_2=x_2+i{y}_2$

${\left({\mathrm{x}}_1-1\right)}^2+{\mathrm{y}}_1^2={\mathrm{x}}_1^2$

${\mathrm{y}}_1^2-2{\mathrm{x}}_1+1=0$

$\left|z_2-1\right|=\mathrm{R}\mathrm{e}?\left(z_2\right)$

${\left(x_2-1\right)}^2+y_2^2=x_2^2$

${\mathrm{y}}_2^2-2{\mathrm{x}}_2+1=0$

$\left({\mathrm{y}}_1-{\mathrm{y}}_2\right)\left({\mathrm{y}}_1+{\mathrm{y}}_2\right)=2\left({\mathrm{x}}_1-{\mathrm{x}}_2\right)$

${\mathrm{y}}_1+{\mathrm{y}}_2=2\left(\frac{{\mathrm{x}}_1-{\mathrm{x}}_2}{{\mathrm{y}}_1-{\mathrm{y}}_2}\right)$

$\mathrm{a}\mathrm{r}\mathrm{g}?\left(z_1-z_2\right)=\frac{\pi }{6}$

${\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}?\left(\frac{{\mathrm{y}}_1-{\mathrm{y}}_2}{{\mathrm{x}}_1-{\mathrm{x}}_2}\right)=\frac{\pi }{6}$

$\frac{{\mathrm{y}}_1-{\mathrm{y}}_2}{{\mathrm{x}}_1-{\mathrm{x}}_2}=\frac{1}{\sqrt[]{3}}$

$\therefore {\mathrm{y}}_1+{\mathrm{y}}_2=2\sqrt[]{3}$

$\Rightarrow \mathrm{}\mathrm{I}\mathrm{m}?\left(z_1+z_2\right)=2\sqrt[]{3}$

$\mathrm{}488=\frac{n}{2}\left. [2\left(\frac{100}{5}\right)+(n-1)\left(\frac{2}{5}\right)\right]$  

$488=\frac{n}{2}(101-n)$

$\Rightarrow \mathrm{}{\mathrm{n}}^2-101\mathrm{n}+2440=0$

$\Rightarrow \mathrm{n}=61$ or 40

For $n=40\mathrm{}\Rightarrow \mathrm{}{T}_n>0$  

For $\mathrm{n}=61\mathrm{}\Rightarrow \mathrm{}{\mathrm{T}}_{\mathrm{n}}<0$

${\mathrm{T}}_{\mathrm{n}}=\frac{100}{5}+(61-1)\left(-\frac{2}{5}\right)=-4$

( (1+i)/ (1-i) )^ (m/2) = ( (1+i)/ (i-1) )^ (n/3) = 1
⇒ ( (1+i)²/2 )^ (m/2) = ( (1+i)²/ (-2) )^ (n/3) = 1
⇒ (i)^ (m/2) = (-i)^ (n/3) = 1
⇒ m/2 = 4k? and n/3 = 4k?
⇒ m = 8k? and n = 12k?
Least value of m = 8 and n = 12
∴ GCD = 4
∴ GCD = 4

(0.16)^log? (1/3 + 1/3² + . to ∞)
= (4/25)^log? (1/2)
= ( (5/2)? ² )^log? /? (1/2) = (5/2)^ (-2 log? /? (1/2)
= (5/2)^ (log? /? ( (1/2)? ² ) = 4

lim_ (x→0) (1/x? ) {1 - cos (x²/2) - cos (x²/4) + cos (x²/2)cos (x²/4)} = 2?
⇒ lim_ (x→0) ( (1 - cos (x²/2) (1 - cos (x²/4) / x? ) = 2?
⇒ 2? = 2? ⇒ k = 8

∴ center lies on x + y = 2 and in 1st quadrant center = (a, 2-a) where a > 0 and 2-a > 0 ⇒ 0 < a < 2
∴ circle touches x = 3 and y = 2
⇒ |3-a| = |2 - (2-a)| = radius
⇒ |3-a| = |a| ⇒ a = 3/2
∴ radius = a
⇒ Diameter = 2a = 3.

S = (2 . ¹P? - 3 . ²P? + 4 . ³P? upto 51 terms) + (1! - 2! + 3! - . upto 51 terms)
∴ [? ? P_ (n-1) = n!]
= (2! - 3! + 4! + 52!) + (1! - 2! + 3! - 4! + . . + (51)!)
= 1! + 52!

LHL : lim_ (x→0? ) |1-x-x|/|λ-x-1| = 1/|λ-1|
RHL: lim_ (x→0? ) |1-x+x|/|λ-x+0| = 1/|λ|
For existence of limit
LHL = RHL
⇒ 1/|λ-1| = 1/|λ| ⇒ λ = 1/2
∴ L = 1/|λ| = 2