Ncert Solutions Maths class 11th

x = t² - t + 1 . (i)
y = t² + t + 1 . (ii)
(i) + (ii)
x + y = 2 (t² + 1) . (iii)
(i) – (ii)
x - y = -2t . (iv)
from (iii) and (iv)
x² – 2xy + y² – 2x – 2y + 4 = 0
Here H² = ab and Δ≠ 0
This is parabola, so e = 1

Negation of (p → q) ∨ (p ∨ q)
~ [ (p → q) ∨ (p ∨ q)]
≡ ~ (p → q) ∧ ~ (p ∨ q)
[∴ ~ (p → q) ≡ p ∧ ~q]
≡ (p ∧ ~q) ∧ (~p ∧ ~q)
= F (contradiction)

According to question
aα² + bα + c = 0 . (i)
−aβ² + bβ + c = 0 . (ii)
a/2 γ² + bγ + c = 0 . (iii)
bα + c = -aα² . (iv)
bβ + c = aβ² . (v)
bγ + c = -a/2 γ² . (vi)
By observing (v) and (vi), we get β > γ
By observing (iv) and (vi), we get α > γ
So α < <

By using condition of tangency,
c² = a² (m² + 1)
⇒ c² = 5 [ (2)² + 1]
⇒ c² = 25
⇒ c = ±5

There are 7 letters permutation.
The total number of 4 letters
Out of 2A's, 2K's and 3 different letters O, L, T.
= Coefficient of x? in 4! (1 + x + x²/2!)² (1 + x)³
= Coefficient of x? in 4! (1 + x + x²/2)² (1 + x)³
= Coefficient of x? in 4! [ (1 + x)? + (1 + x)? x² + (1 + x)³ + x? /4]
= 4! [? C? +? C? + ³C? /4] = 4! (5 + 6 + 1/4)
= 24 [11 + 1/4]
= 264 + 6 = 270

2x=tan (π/9)+tan (7π/18)
=sin (π/9+7π/18) / cos (π/9)cos (7π/18)
=sin (π/2) / cos (π/9)cos (7π/18)
=1 / cos (π/9)cos (7π/18)
=1 / cos (π/9)sin (π/2−7π/18)
=1 / cos (π/9)sin (π/9)
⇒x=1 / 2cos (π/9)sin (π/9)
=1 / sin (2π/9)=cosec (2π/9)

Again 2y=tan (π/9)+tan (5π/18)
⇒2y=sin (π/9+5π/18) / cos (π/9)cos (5π/18)
=sin (7π/18) / sin (π/2−π/9)sin (π/2−5π/18)
=sin (7π/18) / sin (7π/18)sin (4π/18) = cosec (2π/9)
⇒|x−2y|=0

Equation of the ellipse
(x−3)²/a² + (y+4)²/b² = 1
a=2
ae=1⇒e=1/2
⇒b²=3
Equation of tangent
y+4=m (x−3)±√4m²+3
⇒mx−y=4+3m±√4m²+3
⇒3m±√4m²+3=0
⇒9m²=4m²+3
⇒5m²=3

Direction ratio of line (1, -1, -6)
Equation of line (x−3)/1 = (y+4)/-1 = (z+5)/-6 =k
x=k+3, y=−k−4, z=−6k−5
Solving with plane k=−2
⇒x=1, y=−2, z=7
⇒Distance=√ (3−1)²+6²+3²=√49=7

N=2¹? 5¹? 11¹¹ 13¹¹
5→4n+1 type→number of choice=11
11→4n+3 type→number of choice=6
13→4n+1 type→number of choice=12
. Number of divisor of 4n+1 type=11*6*12=792

Z = (3+2icosθ)/ (1-3icosθ) = (3−6cos²θ)+i (11cosθ) / (1+9cos²θ)
Real part = 0
⇒3−6cos²θ=0
⇒θ=45?
⇒sin²3θ+cos²θ=1/2+1/2=1