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Free_Kalibri [48]
2 years ago
11

1. If A = {1, 2, 3,4}, B = {3,4,5,6}, C = {1, 2, 4, 6, 7}, then find i) AUB ii) BUC iii) (AUB)UC v) An(BNC) vi) AU(BUC) iv) AN(B

UC)​

Mathematics
2 answers:
Tomtit [17]2 years ago
8 0

Answer:

i can t see the image

Step-by-step explanation:

i cant see the image

notsponge [240]2 years ago
3 0

Answer:

i) AUB={1,2,3,4,5,6}

ii) BUC={1,2,3,4,5,6,7}

iii)(AUB)UC={1,2,3,4,5,6,7}

iv) {1,2,3,4}

v) {4}

vi){1,2,3,4,5,6,7}

Step-by-step explanation

AUB means all the elements belongs to a and b

AnB means elements common to both sets

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Find the partial fraction decomposition of the rational expression with prime quadratic factors in the denominator
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\dfrac{5x^4-7x^3-12x^2+6x+21}{(x-3)(x^2-2)^2}=\dfrac{a_1}{x-3}+\dfrac{a_2x+a_3}{x^2-2}+\dfrac{a_4x+a_5}{(x^2-2)^2}
\implies 5x^4-7x^3-12x^2+6x+21=a_1(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(a_4x+a_5)(x-3)

When x=3, you're left with

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When x=\sqrt2 or x=-\sqrt2, you're left with

\begin{cases}17-8\sqrt2=(\sqrt2a_4+a_5)(\sqrt2-3)&\text{for }x=\sqrt2\\17+8\sqrt2=(-\sqrt2a_4+a_5)(-\sqrt2-3)\end{cases}\implies\begin{cases}-5+\sqrt2=\sqrt2a_4+a_5\\-5-\sqrt2=-\sqrt2a_4+a_5\end{cases}

Adding the two equations together gives -10=2a_5, or a_5=-5. Subtracting them gives 2\sqrt2=2\sqrt2a_4, a_4=1.

Now, you have

5x^4-7x^3-12x^2+6x+21=3(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(x-5)(x-3)
5x^4-7x^3-12x^2+6x+21=3x^4-11x^2-8x+27+(a_2x+a_3)(x-3)(x^2-2)
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By just examining the leading and lagging (first and last) terms that would be obtained by expanding the right side, and matching these with the terms on the left side, you would see that a_2x^4=2x^4 and a_3(-3)(-2)=6a_3=-6. These alone tell you that you must have a_2=2 and a_3=-1.

So the partial fraction decomposition is

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