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Algebra Level 3 Test 8

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Question 1
The value of The value of 1/(1-x ) +1/(1 +x )+2/ ( 1 +x2 ) +4/(1 + x6)
A
8/(1 – x8)
B
4x/(1 + x2)
C
4/(1 – x6)
D
4/(1 + x4)
Question 1 Explanation: 
1/(1-x ) +1/(1 +x )+2/ ( 1 +x2 ) +4/(1 + x6 )
= 2/(1 – x2) + 2/(1 + x2) + 4/(1 + x4)
= 4/(1- x4) + 4/(1 + x4) =8/(1 – x8).
Question 2
Let Y = minimum of {(x+2), (3-x)}. What is the maximum value of Y for 0 ≤ × ≤ 1?
A
1.0
B
1.5
C
3.1
D
2.5
Question 2 Explanation: 
We are given by the minimum value of Y ={(x+2), (3-x)}.
If 0 ≤ × ≤ 1, then 2 ≤ (× +2) ≤ 3 and 3 ≥ (3 − ×) ≥ 2.
So the minimum value among them should also lie between 2 & 3.
Therefore the best option for this that gives you this is (d).
Question 3
If y = f(x) and f(x) = (1-x) / (1 + x), which of the following is true?
A
f(2x) = f(x) – 1
B
x = f(2y)-1
C
f(1/x) = f(x)
D
x = f(y)
Question 3 Explanation: 
In this question we will use the method of simulation.
Let x = 2. Hence f(2) = (1 – 2)/(1 + 2) = -1/3 = y.
Now come to the options and from the options we can
say that only option (d) satisfies the condition.
f(y) = (-1/3) = (1 +1/3)/(1 – 1/3)
= 2 = x.
Question 4
What is the value of k for which the following system of equations has no solution: 2r – 8s = 3 and kr +4s = 10
A
–2
B
1
C
–1
D
2
Question 4 Explanation: 
We know that for a system of two equations :
p1r + q1s = t1 and p2r + q2s = t2 to have no solution,
the following condition should be satisfied:
p1/p2 = q1/q2 ≠ t1/t2.
Hence, in our equations 2/k = –8/4 ≠ 3/10. So, k = –1.
Question 5
What is the sum of the following series: 1/(1× 2) + 1/(2 × 3) + 1/(3 × 4) + ....... + 1/(100 ×101)
A
99/100
B
1/100
C
100/101
D
101/102
Question 5 Explanation: 
1/(1× 2) +1/(2×3)+1/(3× 4)+ − − − +1/(100×101)
= (1−1/ 2)+ (1/ 2 −1/ 3)+ (1/ 3 −1/ 4)+ --- + (1/99 –1/00) + (1/100-1/101) = 1 – 1/101 = 100/101.
So the right answer is option number (c)
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