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New answer posted

6 months ago

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V
Vishal Baghel

Contributor-Level 10

A² - 4A + 4I = 0
(A - 2I)² = 0 ⇒ A-2I ≠ 0
B = 297A - 594I
= 297 (A - 2I)

New answer posted

6 months ago

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V
Vishal Baghel

Contributor-Level 10

? C? +? C? =? C?
Σ (from r=0 to 3)? C? =? C? +? C? +? C? +? C?
=? C? +? C? +? C? +? C?
= ¹²C? = (12*11*10)/6 = 220

New answer posted

6 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

lim (x→0) (ae²? - bcosx + c) / (x sinx)
= lim (x→0) (ae²? - bcosx + c) / x² * lim (x→0) x/sinx
For limit to exist
a - b + c = 0
2a + b = 0
(4a-b)/2 = 1
c = 2, b = 2, a = -1
a+b+c = 3

New answer posted

6 months ago

0 Follower 4 Views

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Vishal Baghel

Contributor-Level 10

∫ (from 0 to π/2) (x/2)|cos (2x)|dx
∫ (from 0 to π/4) (x/2)cos (2x)dx - ∫ (from π/4 to π/2) (x/2)cos (2x)dx
= [x sin (2x)/4 - ∫sin (2x)/4 dx] (from 0 to π/4) - [x sin (2x)/4 - ∫sin (2x)/4 dx] (from π/4 to π/2)
= [x sin (2x)/4 + cos (2x)/8] (from 0 to π/4) - [x sin (2x)/4 + cos (2x)/8] (from π/4 to π/2)
= (π/16 - 1/8) - (-1/8 - π/16) = π/8

New answer posted

6 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

|3a+4b|² = 9|a|² + 16|b|² + 24a·b
But a·b = 0, |a|=|b|=k
|3a+4b| = 5k
|4a-3b|
10k = 20? k = 2 = |a| = |b|

New answer posted

6 months ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-Level 10

Family of circle through (2, 2) and (9, 9), (x-2) (x-9) + (y-2) (y-9) + λ (y-x) = 0
Touches y=0 (x-axis)
(x-2) (x-9) + 18 - λx = 0
x² - 11x - λx + 36 = 0
x² - (11+λ)x + 36 = 0 has repeated roots
D = 0 ⇒ λ = 1, λ = -23
x = 6 or x = -6
Difference = 12

New answer posted

6 months ago

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V
Vishal Baghel

Contributor-Level 10

All even + 2 odd 1 even
¹? C? + ¹? C? * ¹? C?
(10*9*8)/6 + (10*9)/2 * 10
120 + 450 = 570

New answer posted

6 months ago

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V
Vishal Baghel

Contributor-Level 10

2a² + 3b² = 35
a=0, b≠I
a=±1, b²=11 ⇒ b≠I
a=±2, b=9 ⇒ b=3 or -3
a=±3, b²=17/3 ⇒ b≠I
a=±4, b²=1 ⇒ b=±1

New answer posted

6 months ago

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Vishal Baghel

Contributor-Level 10

Differentiable ⇒ continuous
Continuous at x = 2 ⇒ 2a + b = 0
Continuous at x = 3
0 = 9p + 3q + 1
Differentiable at x=2
a = 2*2 - 5 ⇒ a = -1
Differentiable at x=3
2*3 - 5 = 2p*3 + q ⇒ 6p + q = 1
a = -1, b = 2, p = 4/9, q = -5/3

New answer posted

6 months ago

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V
Vishal Baghel

Contributor-Level 10

For point of intersection
1 + 3λ = 3 + μ
2 + λ = 1 + 2μ
5λ = 5 ⇒ λ = 1, μ = 1
Point of intersection (4, 3, 5)
For the greatest distance from origin perpendicular from meet plane at point of intersection
Hence equation r . (4i + 3j + 5k) = 50

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