Ncert Solutions Maths class 11th

P (x) = 0
x² - x - 2 = 0
(x-2) (x+1) = 0
x = 2, -1 ∴ α = 2
Now lim (x→2? ) (√ (1-cos (x²-x-2) / (x-2)
⇒ lim (x→2? ) (√ (2sin² (x²-x-2)/2) / (x-2)
⇒ lim (x→2? ) (√2 sin (x²-x-2)/2) / (x²-x-2)/2) ⋅ (x²-x-2)/2) ⋅ (1/ (x-2)
⇒ for x→2? , (x²-x-2)/2 → 0?
⇒ lim (x→2? ) √2 ⋅ 1 ⋅ (x-2) (x+1)/ (2 (x-2) = 3/√2

S = tan? ¹ (1/3) + tan? ¹ (1/7) + tan? ¹ (1/13) + . upto 10 term
S = tan? ¹ (2-1)/ (1+1⋅2) + tan? ¹ (3-2)/ (1+2⋅3) + tan? ¹ (4-3)/ (1+3⋅4) + . + tan? ¹ (11-10)/ (1+11⋅10)
S = (tan? ¹2 - tan? ¹1) + (tan? ¹3 - tan? ¹2) + . + (tan? ¹11 - tan? ¹10)
S = tan? ¹11 - tan? ¹1
S = tan? ¹ (11) - π/4
tan (S) = 5/6

∴ x² = |x|² = t let
9t² - 18t + 5 = 0
(3t - 1) (3t - 5) = 0
|x| = 1/3, 5/3
Product of roots = (1/3) (-1/3) (5/3) (-5/3) = 25/81

P: n³ - 1 is even, q: n is odd.
The contrapositive of p → q is ~q → ~p.
~q: n is not odd, i.e., n is even.
~p: n³ - 1 is not even, i.e., n³ - 1 is odd.
⇒ "If n is not odd then n³ - 1 is not even"
⇒ "For an integer n, if n is even, then n³ – 1 is odd."

y² = 4x and x² = 4y
any tangent of y² = 4x is y = mx + 1/m
it also tangent for x² = 4y
1/m = -m² ⇒ m = -1
∴ common tangent is y = -x - 1, it also touches x² + y² = c²
∴ 1 = c² ⋅ (1+1) ⇒ c² = 1/2

Assuming the equation is (x + iy)² = I (x² + y²)
x² - y² + 2ixy = I (x² + y²)
Compare real and imaginary parts
x² - y² = 0 ⇒ x = y or x = -y
2xy = x² + y²
If x=y, then 2x² = x² + x², which is true for all x.
If x=-y, then -2y² = y² + y² = 2y², which implies 4y²=0, so y=0 and x=0.
The non-trivial solution is x = y.

a, b, c are in Andhra Pradesh then
2b = a + c
28 = 3^ (2sin2θ-1) + 3^ (4-2sin2θ)
Put 3^ (2sin2θ) = x
28 = x/3 + 81/x ⇒ x² - 84x + 243 = 0
(x-3) (x-81) = 0
3^ (2sin2θ) = 3 or 3?
2sin2θ = 1 or 4
sin2θ = 1/2
terms are 1, 14, 27,
then T? = 1 + 5 (13)

For the parabola y = x², the tangent at (2,4) is given by (y+4)/2 = 2x, which simplifies to 4x - y - 4 = 0.
The equation of a circle touching the line 4x - y - 4 = 0 at the point (2,4) is
(x-2)² + (y-4)² + λ (4x-y-4) = 0.
It passes through (0,1).
∴ (0-2)² + (1-4)² + λ (4 (0) - 1 - 4) = 0
4 + 9 + λ (-5) = 0 ⇒ 13 = 5λ ⇒ λ = 13/5
∴ the circle is x² - 4x + 4 + y² - 8y + 16 + (13/5) (4x-y-4) = 0
x² + y² + (-4 + 52/5)x + (-8 - 13/5)y + (20 - 52/5) = 0
x² + y² + (32/5)x - (53/5)y + 48/5 = 0
∴ centre is (-g, -f) = (-16/5, 53/10).

S' = 2¹? + 2? ⋅ 3 + 2? ⋅ 3² + . + 2 ⋅ 3? + 3¹?
G.P. → a = 2¹? , r = 3/2, n = 11
S' = 2¹? ⋅ [ (3/2)¹¹ - 1)/ (3/2 - 1)] = 2¹¹ (3¹¹/2¹¹) - 1)
= 3¹¹ - 2¹¹

Negation of x ↔~ y
≡ ~ (x ↔ ~y)
≡ x ↔ ~ (~y)
≡ x ↔ y
≡ (x ∧ y) ∨ (~x ∧ ~y)