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New answer posted

a year ago

0 Follower 9 Views

A
alok kumar singh

Contributor-Level 10

f (0)f (1) ≤ 0
=> 2 (λ² + 1 - 4λ + 2) ≤ 0 => 2 (λ² - 4λ + 3) ≤ 0
=> (λ-1) (λ-3) ≤ 0
=> λ
But at λ=1, both roots are 1 so λ ≠ 1

New answer posted

a year ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

S = 6a² => dS/dt = 12a * da/dt = 3.6
=> 12 (10) da/dt = 3.6
=> da/dt = 0.03
V = a³ => dV/dt = 3a² * da/dt
= 3 (10)² * (3/100) = 9

New answer posted

a year ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

Total = 9 (10? )
Fav. Way =? C? (2? -2) +? C? (2? -1) = 36 (30) + 9 (15) = 1080 + 135
Probability = (36x30+9x15)/ (9x10? ) = (4x30+15)/10? = 135/10?

New answer posted

a year ago

0 Follower 21 Views

A
alok kumar singh

Contributor-Level 10

x³dy + xydx = 2ydx + x²dy
=> (x³-x²)dy = (2-x)ydx
=> dy/y = (2-x)/ (x² (x-1) dx
=> ∫ (dy/y) = ∫ (2-x)/ (x² (x-1)dx
Let (2-x)/ (x² (x-1) = A/x + B/x² + C/ (x-1)
=> 2-x = A (x-1) + B (x-1) + Cx²
=> C=1, B=-2 and A=-1
=> ∫ (dy/y) = ∫ (-1/x - 2/x² + 1/ (x-1)dx
=> lny = -lnx + 2/x + ln|x-1| + C
∴ y (2) = e
=> 1 = -ln2 + 1 + 0 + C
=> C = ln2
=> lny = -lnx + 2/x + ln|x-1| + ln2
at x = 4
=> lny (4) = -ln4 + 1/2 + ln3 + ln2
=> lny (4) = ln (3/4) + 1/2 = ln (3/2)e^ (1/2)
=> y (4) = (3/2)e^ (1/2)

New answer posted

a year ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

k/6 = ∫? ^ (π/6) (x²)/ (1-x²)³/² dx x = sinθ dx = cosθdθ
=> k/6 = ∫? ^ (π/6) (sin²θ)/ (1-sin²θ)³/² * cosθdθ
=> k/6 = ∫? ^ (π/6) (sin²θ)/ (cos³θ) * cosθdθ
=> k/6 = ∫? ^ (π/6) tan²θdθ = ∫? ^ (π/6) (sec²θ-1)dθ
=> k/6 = [tanθ - θ]? ^ (π/6) = (1/√3 - π/6) = (2√3-π)/6
=> k = 2√3 - π

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