Ncert Solutions Maths class 11th

AB = r, AD = r/2
CD = rsin60° = √3r/2
|0+0-3|/√ (1²+2²) = √3r/2 ⇒ 3/√5 = √3r/2 ⇒ r = 2√3/√5 ⇒ r² = 12/5

α+β = 3/7, αβ = -2/7
α/ (1-α²) + β/ (1-β²) = (α+β) - αβ (α+β) / (1-α²) (1-β²)
= (α+β) - αβ (α+β) / (1 + (αβ)² - (α+β)² - 2αβ)
= (3/7) + (2/7) (3/7) / (1 + (-2/7)² - (3/7)² - 2 (-2/7) = 27/16

Let a, ar, ar² . G.P.
T? + T? + T? = 3 ⇒ ar (1+r+r²) = 3
T? + T? + T? = 243 ⇒ ar? (1+r+r²) = 243
by (i) and (ii)
r? = 81 ⇒ r=3
∴ a = 1/13
S? = a (r? -1)/ (r-1) = (3? -1)/26

Let x = tanθ
y? = tan? ¹ (secθ-1)/tanθ) = tan? ¹ (tan (θ/2) = θ/2 = (1/2)tan? ¹x
x = sinφ, y? = tan? ¹ (2sinφcosφ)/cos2φ)
= tan? ¹ (tan2φ) = 2φ = 2sin? ¹x
dy? /dy? = (dy? /dx)/ (dy? /dx) = (1/2) (1/ (1+x²) / (2/√ (1-x²)
= (√ (1-x²) / (4 (1+x²) = (√ (1-1/4) / (4 (1+1/4) = √3 / 10

Kindly consider the following Image

L = sin (3π/16)sin (-π/16)
= (1/2) (cos (π/4) - cos (π/8)
= (1/2) (1/√2 - cos (π/8)
M = cos (3π/16)cos (-π/16)
= (1/2) (cos (π/4) + cos (π/8)
= (1/2) (1/√2 + cos (π/8)

lim (x→0) (e^ (√ (1+x²+x? )-1)/x - 1) / ( (√ (1+x²+x? )-1)/x )
put (√ (1+x²+x? )-1)/x = t
clearly x→0 ⇒ t→0
∴ given limit = lim (t→0) (e? -1)/t = 1

(-1+i√3)/ (1-i)³? = (2 (cos (2π/3) + isin (2π/3)/ (√2 (cos (-π/4) - isin (-π/4)³?
= (2³? (cos20π + isin20π)/ (2¹? (cos (-15π/2) - isin (-15π/2)
= (2¹? (1 + 0i)/ (0 + i) = -2¹? i

∫? (√ax - x²/a)dx = a²/6
⇒ (2/3)√a b^ (3/2) - b³/3a = a²/6
also area of ΔOQR = 1/2
(1/2)b² = 1/2 ⇒ b=1
Put in (i) ⇒ 4a√a - 2 = a³ ⇒ a? - 12a³ + 4 = 0