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New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

log? (abc) = 6 ⇒ abc = 6?
a = b/r & c = br
⇒ b = 36 and a = 36/r
⇒ r = 2,3,4,6,9,12,18
Also b - a = 36 (1 - 1/r) is a perfect cube.
∴ r = 4 ⇒ a + b + c = 9 + 36 + 144 = 189

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2 months ago

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Vishal Baghel

Contributor-Level 10

β² – 6β + 12 = 0 β − 6 = -12/β
∴ Reqd. Expression is (α – 2)³)? + (-12)¹² / (αβ)¹²) - 1
= (α³ – 8 – 6α (α – 2)? + (12)¹² / (12)¹²) - 1 = (α (α² – 6α + 12) – 8)? = 8? = 2¹²

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2 months ago

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Vishal Baghel

Contributor-Level 10

Solving, x² – 9 = kx² ⇒ x² (k − 1) + 9 = 0 ⇒ x? + x? = 0 and x? = 9 / (k-1)
|x? - x? | = 10 = √ (x? + x? )² - 4x? x? ) ⇒ k = 16/25

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2 months ago

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Vishal Baghel

Contributor-Level 10

A: 3 numbers are in A.P.
B: 3 numbers are even
P (B/A) = n (A ∩ B) / n (A) = (²? C? + ²? C? ) / (? C? +? C? ) = (25 * 24) / (50 * 49) = 12/49

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

New answer posted

2 months ago

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A
alok kumar singh

Contributor-Level 10

k = [ a b c ] + 2 [ a b c ] + [ a b c ] [ a b c ] [ a b c ]            

k = 3

New answer posted

2 months ago

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A
alok kumar singh

Contributor-Level 10

a r e a = 2 * ( 1 2 * 1 * 1 ) = 1 = k

New answer posted

2 months ago

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A
alok kumar singh

Contributor-Level 10

  0 1 . 5 [ x 2 ] d x = 0 1 0 d x + 1 2 1 d x + 2 1 . 5 2 d x = 2 2

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