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## Algebra Level 3 Test 8

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*Algebra Level 3 Test 8*. You scored %%SCORE%% out of %%TOTAL%%. You correct answer percentage: %%PERCENTAGE%% . Your performance has been rated as %%RATING%% Your answers are highlighted below.

Question 1 |

The value of The value of 1/(1-x ) +1/(1 +x )+2/ ( 1 +x

^{2}) +4/(1 + x^{6}) 8/(1 – x ^{8}) | |

4x/(1 + x ^{2}) | |

4/(1 – x ^{6}) | |

4/(1 + x ^{4}) |

Question 1 Explanation:

1/(1-x ) +1/(1 +x )+2/ ( 1 +x

= 2/(1 – x

= 4/(1- x

^{2}) +4/(1 + x^{6})= 2/(1 – x

^{2}) + 2/(1 + x^{2}) + 4/(1 + x^{4})= 4/(1- x

^{4}) + 4/(1 + x^{4}) =8/(1 – x^{8}).Question 2 |

Let Y = minimum of {(x+2), (3-x)}. What is the maximum value of Y for 0 ≤ × ≤ 1?

1.0 | |

1.5 | |

3.1 | |

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).

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?

f(2x) = f(x) – 1 | |

x = f(2y)-1 | |

f(1/x) = f(x) | |

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.

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

–2 | |

1 | |

–1 | |

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.

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)

99/100 | |

1/100 | |

100/101 | |

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)

= (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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