Class 11th

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New answer posted

2 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

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."

New answer posted

2 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

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

New answer posted

2 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

n (C) = 73, n (T) = 65, n (C ∩ T) = x
n (C ∪ T) ≤ 100
⇒ n (C) + n (T) - n (C ∩ T) ≤ 100
⇒ x ≥ 38
n (C ∩ T) ≤ min (n (C), n (T) ⇒ x ≤ 65
⇒ 38 ≤ x ≤ 65

New answer posted

2 months ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

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.

New answer posted

2 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

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)

New answer posted

2 months ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

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).

New answer posted

2 months ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

The general term is T? = ¹? C? (K/x²)? (√x)¹?
= ¹? C? K? x? ²? x? /² = ¹? C? K? x? /²
For the constant term, the power of x is 0.
5 - 5r/2 = 0 ⇒ r = 2
The term is T? = ¹? C? · K² = 405
45 · K² = 405
⇒ K² = 9 ⇒ |K| = 3

New answer posted

2 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 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

New answer posted

2 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

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

New answer posted

2 months ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

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