Answer:
A) 2x*2 +6x -3x - 9= 2x*2 +3x -9
B) 2yx*2 +2yx +2y +3x*2 +3x +3
C) 36x*2 y +48yx +3xy*2 + 4y*2
D) 2x*3 - 4x*2 y + 3x -6y
Consider the Set A = {X | X is an even whole number between 0 and 2 } = 
.
Since, whole numbers are the set of numbers starting from zero upto infinity.
Even numbers are the numbers which are exactly divisible by '2'.
So, we have to find the even whole number between 0 and 2.
Since, only '1' is a whole number between 0 and 2 which is not an even number as '1' is not divisible by '2'.
Therefore, there is no even whole number between 0 and 2.
So, this set is empty.
Therefore, A = { X | X is an even whole number between 0 and 2} =
Answer:
12, 17, 24
Step-by-step explanation:
To find the first 3 terms, substitute n = 1, 2, 3 into the rule, that is
T₁ = 1² + 2(1) + 9 = 1 + 2 + 9 = 12
T₂ = 2² + 2(2) + 9 = 4 + 4 + 9 = 17
T₃ = 3² + 2(3) + 9 = 9 + 6 + 9 = 24
Imx->0 (asin2x + b log(cosx))/x4 = 1/2 [0/0 form] ,applying L'Hospital rule ,we get
= > limx->0 (2a*sinx*cosx - (b /cosx)*sinx)/ 4x3 = 1/2 => limx->0 (a*sin2x - b*tanx)/ 4x3 = 1/2 [0/0 form],
applying L'Hospital rule again ,we get,
= > limx->0 (2a*cos2x - b*sec2x) / 12x2 = 1/2
For above limit to exist,Numerator must be zero so that we get [0/0 form] & we can further proceed.
Hence 2a - b =0 => 2a = b ------(A)
limx->0 (b*cos2x - b*sec2x) / 12x2 = 1/2 [0/0 form], applying L'Hospital rule again ,we get,
= > limx->0 b*(-2sin2x - 2secx*secx.tanx) / 24x = 1/2 => limx->0 2b*[-sin2x - (1+tan2x)tanx] / 24x = 1/2
[0/0 form], applying L'Hospital rule again ,we get,
limx->0 2b*[-2cos2x - (sec2x+3tan2x*sec2x)] / 24 = 1/2 = > 2b[-2 -1] / 24 = 1/2 => -6b/24 = 1/2 => b = -2
from (A), we have , 2a = b => 2a = -2 => a = -1
Hence a =-1 & b = -2
