Ncert Solutions Maths class 11th

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

Ellipse: x²/4 + y²/3 = 1
eccentricity = √ (1 - 3/4) = 1/2
∴ foci = (±1,0)
For hyperbola, given 2a = √2 ⇒ a = 1/√2
∴ hyperbola will be x²/ (1/2) - y²/b² = 1
eccentricity = √ (1 + 2b²)
∴ foci = (±√ ( (1+2b²)/2 ), 0)
∴ Ellipse and hyperbola have same foci
√ ( (1+2b²)/2 ) = 1
⇒ b² = 1/2
∴ Equation of hyperbola: x²/ (1/2) - y²/ (1/2) = 1
⇒ x² - y² = 1/2
Clearly, (√3/2, 1/2) does not lie on it.

α, β are roots of x² + px + 2 = 0
⇒ α² + pα + 2 = 0 and β² + pβ + 2 = 0
⇒ 1/α, 1/β are roots of 2x² + px + 1 = 0
But 1/α, 1/β are roots of 2x² + 2qx + 1 = 0
⇒ p = 2q
Also α + β = -p, αβ = 2
(α - 1/α) (β - 1/β) (α + 1/β) (β + 1/α)
= ( (α²-1)/α ) ( (β²-1)/β ) ( (αβ+1)/β ) ( (αβ+1)/α )
= ( (-pα-3) (-pβ-3) (αβ+1)² ) / ( (αβ)² )
= 9/4 (p²αβ + 3p (α + β) + 9)
= 9/4 (9 - p²) = 9/4 (9 - 4q²)