Mathematics Problems and Solutions

1.
f : R [0,1]
g : [0,1]  [-1,1]
g(x) = 2y-1
(gof)(x) = 2(f(x)) – 1
= {2| x|
−1−1|x|>1
2| x|−1 |x|<1
(Fog)(x) = {|2 y−1|
−1 |2 y−1|>1
|2 y−1||2 y−1|<1
Case 1:
|2 y−1|> 1
2y -1 >1 or -(2y - 1) >1
2y >2 or -2y>0
y>1 or y<0
y == {-∞, 0} U {1, ∞}
Case 2:
|2y - 1|<1
2y-1 < 1 or -(2y-1) < 1
2y<2 or -2y<0
y<1 or y>0
y == {0,1}
(Fog)(x) = {|2 y−1|−1 y ∈{−∞, 0 }∪ {1 , ∞}
|2 y −1| y ∈ {0,1 }
2.
∑
i=1
n (2i2−1)
i4 < 4−2 n+1
n2
At n=1,

∑
i=1
1 (2i2−1)
i4 =1≤ 4− 2 ( 1 ) + 1
12 ≤ 4−3 ≤1 ..........{Correct Result}
At any positive integer value n = k+1,
∑
i=1
k (2i2−1)
i4 + 2 ( k +1 )2−1
¿¿
≤ 4− ( 2 k +1 ) ( k +1 )4 −2 k2 ( k +1 )2+ k2
k2 ( k +1 ) 4
≤ 4− ( k +1 )2 ( ( 2 k +1 ) ( k2 +2 k +1 ) −2 k2 )+ k 2
k 2 ( k +1 )4
≤ 4− ( k +1 )2 ( 2 k3 +4 k2 +2 k +k2 +2 k +1−2 k2 ) +k2
k 2 ( k +1 )4
≤ 4− ( k +1 )2 ( 2 k3 +2 k2 +4 k +k2 +1 ) + k2
k2 ( k+ 1 )4
≤ 4− 4 k2+ 4 k +1−2 k2
k2 ( k +1 )2
≤ 4− 2 k2+ 4 k +1
k 2 ( k +1 )2
≤ 4−2 (k ¿¿ 2+2 k +1)−2+1
k2 ( k +1 ) 2 ¿
≤ 4− 2 ( k +1 )2−1
k2 ( k +1 )2
Therefore, sum (-1 + 2 i^2)/i^4<=4 - (2 n + 1)/n^2 is not
always true for i>0 and i element Z
3.
V = {a, b, c, d, e}
E = {ab, ac, bc, cd, ce, de}
a d