- This is an assessment test.
- To draw maximum benefit, study the concepts for the topic concerned.
- Kindly take the tests in this series with a pre-defined schedule.
Algebra Level 3 Test 8
Congratulations - you have completed 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 +x2 ) +4/(1 + x6)
8/(1 – x8) | |
4x/(1 + x2) | |
4/(1 – x6) | |
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).
= 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?
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)
Once you are finished, click the button below. Any items you have not completed will be marked incorrect.
There are 5 questions to complete.
List |
