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New answer posted

a year ago

0 Follower 9 Views

R
Raj Pandey

Contributor-Level 9

{ (x, y) ∈ R*R, x ≥ 0, 2x² ≤ y ≤ 4 − 2x}.
Required area = ∫? ¹ (4 - 2x - 2x²)dx
= [4x - x² - (2/3)x³]? ¹ = 4-1-2/3 = 7/3

New answer posted

a year ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

A = {2,4,6,8, . . .50} ⇒ 25 element
A = {7,14,21, . . . .49} ⇒ 7 elements
A ∩ B = {14,28,42} = 3 elements
Required number of elements = 25 + 7 - 3 = 29

New answer posted

a year ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

(5/3, 7/3, 17/3)
AD ⋅ PD = 0
((5/3-α)i?+(7/3-7)j?+(17/3-1)k?) . (2/3i?+7/3j?+8/3k?) = 0
(5/3-α)(2/3) + (-14/3)(7/3) + (14/3)(8/3) = 0
A(α, 7, 1) D(7/3, 7/3, 12/3)
⇒ 3α = 12
α = 4

New answer posted

a year ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

Mean = 10 = (3+7+9+12+13+20+x+y)/8
16 = x + y
Variance σ² = 25 = (Σx?²/8) - (mean)²
25 = (3²+7²+9²+12²+13²+20²+x²+y²)/8 = 100
x² + y² = 148
(x+y)² = x² + y² + 2xy
256 = 148 + 2xy
x . y = 54

New answer posted

a year ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

 Δ = |1 1 1| = 0
|1 2 3|
|3 2 λ|
⇒ 1(2λ - 6) - 1(λ - 9) + 1(-4) = 0
⇒ λ = 1
Δx = |6 1 1|
|10 2 3| = 0
|μ 2 λ|
⇒ 2λ + μ = 16
⇒ μ = 14
μ - λ² = 14 - 1 = 13

New answer posted

a year ago

0 Follower 4 Views

R
Raj Pandey

Contributor-Level 9

Sol. Σ? k(k+1)/2 = (1/2)Σ(k²+k) = (1/2)[ (5051101/6) + (5051/2) ]
The solution uses k=1 to 20.
(1/2)[ (20
2141/6) + (2021/2) ] = (1/2)[2870+210] = 1540

New answer posted

a year ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

D≥0
(a-10)² - 4(2)(33/2 - 2a) ≥ 0
a²-4a-32 ≥ 0 ⇒ a∈(-∞, -4] U [8,∞)

New answer posted

a year ago

0 Follower 4 Views

R
Raj Pandey

Contributor-Level 9

0 Red, 1 Red, 2 Red, 3 Red
Number of ways = ?C? + ?C?.?C? + ?C?.?C? + ?C?.?C? = 35+175+210+70=490

New answer posted

a year ago

0 Follower 5 Views

R
Raj Pandey

Contributor-Level 9

P=(x?,y?). 2yy'-6x+y'=0 ⇒ y' = 6x/(2y+1)
(y?-0)/(x?-3/2) = (1+2y?)/(6x?)
9-6y? = 1+2y? ⇒ y?=1. x?=±2. Slope = ±12/3 = ±4. |n|=4.

New answer posted

a year ago

0 Follower 4 Views

R
Raj Pandey

Contributor-Level 9

Let tan?¹x = θ ⇒ x=tanθ ⇒ sinθ = x/√(1+x²)
y = (x/√(1+x²)) + (1/√(1+x²)) = (x+1)/√(1+x²). This is not f(x).
Let's follow the solution:
y = (x+1)²/(1+x²) - 1 = (2x)/(1+x²) = f(x)
Now dy/dx = (1/2√y) * f'(x) = .
The solution seems to take y as a different function. Let's assume y = (x/(√(1+x²))) + (1/√(1+x²)) - 1. No.
Let's assume y's derivative is taken w.r.t to f(x).
y = -tan?¹x + c
given y(√3)=π/6 ⇒ π/6 = -π/3 + c ⇒ c=π/2
y = cot?¹x. Now y(-√3) = cot?¹(-√3) = 5π/6

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