Ncert Solutions Maths class 11th

$\sim (\sim p\vee \sim q)\to (\sim q\vee r)\Rightarrow p\wedge q\to \sim q\vee r$

$\mathrm{}\mathrm{T}\mathrm{F}\mathrm{T}\Rightarrow \mathrm{F}\to \mathrm{T}\to \mathrm{T}$

$\mathrm{F}\mathrm{T}\mathrm{T}\Rightarrow \mathrm{F}\to \mathrm{T}\to \mathrm{T}$

$\mathrm{T}\mathrm{T}\mathrm{T}\Rightarrow \mathrm{T}\to \mathrm{T}\to \mathrm{T}$

$\mathrm{T}\mathrm{T}\mathrm{F}\Rightarrow \mathrm{T}\to \mathrm{F}\to \mathrm{F}$

Let $z=x+yi\mathrm{}z-2\bar{z}=1\Rightarrow x=-1\mathrm{}y=0$

tanx + tan (x+π/3) + tan (x-π/3) = 3
tanx + (tanx+√3)/ (1-√3tanx) + (tanx-√3)/ (1+√3tanx) = 3
tanx + (8tanx)/ (1-3tan²x) = 3
(tanx (1-3tan²x) + 8tanx)/ (1-3tan²x) = 3
(3 (3tanx – tan³x)/ (1-3tan²x) = 3
⇒ 3tan3x = 3
tan3x = 1

$\sim \left(\sim q\wedge p\right)=\left(\sim \left(\sim q\right)\right)\vee \left(\sim p\right)$  

$=\sim p\vee q$            

$=p\to q$            

$\therefore contrapositiveis\sim q\to \sim p$            

 Hence, if $\sim$ q then $\sim$ p

Case-I : $a-2>0$

$F(-2)>0$

$4(a-2)+4a+a>0$

$9\mathrm{a}-8>0$

$a>\frac{8}{9}$

$F\left(1\right)>0$

$a-2-2a+a>0$

  $-2>0$  (not possible)

Case-II : $a-2<0;a<2$

$F(-2)<0\mathrm{}\Rightarrow \mathrm{}a<\frac{8}{9}$

$F\left(1\right)<0\mathrm{}\Rightarrow \mathrm{}a\in R$

$-2<\frac{-b}{2a}<1\mathrm{}a<\frac{4}{3}\mathrm{}a\in \left[0,\frac{8}{9}\right)\cup \left\{2\right\}$

$D\ge 0\mathrm{}a\ge 0$

|x|³/? = y
⇒ y² – 26y – 27 = 0 ⇒ y = −1 or 27
⇒ y = 27 ⇒ |x|³/? = 3³
|x| = 3? ⇒ x = ±3?
Product of roots = (3? ) (−3? ) = −3¹?

tan α = 1/x, tan β = 2/x, tan γ = 3/x
Now α + β + γ = 180°
⇒ tan α + tan β + tan γ = tan α tan β tan γ
1/x + 2/x + 3/x = (123)/ (xxx)
⇒ x² = 1

x² + y² – 6x + 8y + 24 = 0 is circle having centre (3, −4) & r = √ (9+16-24) = 1
√x² + y² min. is min. distance from origin = 4
∴ minimum value of log? (x² + y²) = log? 16 = 4

Centre of C₃ will lie on the radical axis of C₁ and C₂ which is 10x + 6y + 26 = 0.

Let center of C₃ is (h, k).

Equation of chord of contact through (h, k) to the C₁ may be given as hx + ky = 25 … (I)

Let the mid point of the chord is (x₁, y₁) the equation of the chord with the help of mid point may be given as  

$x{x}_1+y{y}_1=x_1^2+y_1^2$

Since (I) and (II) represents same straight line 10x + 6y + 26 = 0

$\Rightarrow \frac{h}{25}=\frac{x_1}{x_1^2+y_1^2},\frac{k}{25}=\frac{y_1}{x_1^2+y_1^2}$            

The locus of (x₁, y₁) is

$5x+3y+\frac{13}{25}\left(x^2+y^2\right)=0$          

a², b², c² . A.P.
a² + 1, b² + 1, c² + 1 . P.
a² + ab + bc + ca, b² + ab + bc + ca, c² + ab + bc + CA . P.
⇒ (a + b) (a + c), (a + b) (b + c), (b + c) (c + a) . P.
⇒ 1/ (b+c), 1/ (c+a), 1/ (a+b) . A.P.
⇒ b + c, c + a, a + b . P.