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Common Questions Ask in Post UTME Mathematics Exams. Day1
Mathematics
Chima Sunday (Tutor)
20-06-2016 15:22:00 +0000Try your best to solve these questions. Show working on paper when necessary and include a picture of your hand written Solution in your response. Wish you the best!
1. Find the minimum point for this curve, y = 3x3 - 9x2
A. 2
B. 0
C. 1
D. 18
(uni-uyo)Updated SAolution
If d2y/dx2 < 0; its a maximum point
If d2y/dx2 > 0; its a minimum pint
If d2y/dx2 = 0; its a stationery point.
Thus dy/dx = 9x2 - 18x
d2y/dx2 = 18x - 18 = 18(x - 1)
at x = 0, d2y/dx2 = 18(0 - 1) = -18, thus maximum point
at x = 2, d2y/dx2 = 18(2 - 1) = 18, thus minimum point
So the minimum point is at x = 2
Answer is Option A2. QRS is a triangle with QS = 12m, RQS = 30o and QRS = 45o. Calculate the length of RS
A. 18.2m
B. 12.2m
C. 6.2m
D. 3.2m
(unilag)Updated Solution
f one should sketch the triangle, it will be seen that angle 30 is opposite to RS and 45 is opposite to 12m. By applying sine formular
sin30/RS = sin45/12
by subject formular
RS = 12sin30/sin45 = 12 × 0.5/0.707 = 8.4m
The answer is not in the Option3. Express 2cos(60 + θ) in terms of cosθ and sinθ
A. cosθ + √3sinθ
B. √3cosθ - sinθ
C. cosθ - √3sinθ
D. √3cosθ + sinθ
(uniben)Updated Solution
2cos(60 + θ) = 2[cos60cosθ - sin60sinθ] (i.e difference of double angle expansion for cosine)
cos60 = 1/2; siin60 = √/2
so 2cos(60 + θ) = 2[1/2cosθ - √3/2sinθ]
factoring 2; 2/[cosθ - √3sinθ] = cosθ - √3sinθ
Answer is Option C
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Chima Sunday (Tutor)
Please, just a reminder. This post has been updated with its detailed solution. So try and check it up. Wouldn't forget to commend on the efforts of some exceptional jambites that attempted to solve the question. Your labour of work is not i vain, keep it up and keep the fire burning. I frown to some of us that demand for number when asked to post their hand written solutions to substantiate their answer. Please is not the ethics of the forum, like i said "Show working on paper when necessary and include a picture of your hand written Solution in your response" Once again thanks for attempting. Cheers!
1 22-06-2016 11:35:00 +0000
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