Ncert Solutions Maths class 11th

( (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

Sum of 1st 25 terms = sum of its next 15 terms
? (T? + . + T? ) = (T? + . + T? )
? (T? + . + T? ) = 2 (T? + . + T? )
? 40/2 [2*3 + (39d)] = 2 * 25/2 [2*2 + 24 d]
? d = 1/6

p →~ (p ∧ ~ q)
=~ p ∨ ~ (p ∧ ~ q)
=~ p ∨ ~ p ∨ q
=~ p ∨ q

Let P = (3t², 6t); N = (3t²,0)
M = (3t², 3t)
Equation of MQ: y = 3t
∴ Q = (3/4 t², 3t)
Equation of NQ
y = ( 3t / (3/4 t² - 3t²) ) (x - 3t²)
y - intercept of NQ = 4t = 4/3 ⇒ t = 1/3
∴ MQ = 9/4 t² = 1/4
PN = 6t = 2

A: D ≥ 0
⇒ (m + 1)² - 4 (m + 4) ≥ 0
⇒ m² + 2m + 1 - 4m - 16 ≥ 0
⇒ m² - 2m - 15 ≥ 0
⇒ (m - 5) (m + 3) ≥ 0
⇒ m ∈ (-∞, -3] U [5, ∞)
∴ A = (-∞, -3] U [5, ∞)
B = [-3,5)
A − B = (-∞, −3) U [5, ∞)
A ∩ B = {-3}
B - A = (-3,5)
A U B = R