Mathematics University Exam Assistant

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PREVIEW INPUT 1:
Which of the following functions have unit elasticity at P = 6?
 Demand: Qd = 10/P
 Demand: log Qd = 100 − 3 log P
 Demand: Qd = 24 − 2 P
 Supply: Qs = 5 P
PREVIEW INPUT 2:
Alice and Bob each toss five fair coins (each coin lands on either heads or tails, with equal probability and with each outcome independent of each other). Alice wins if strictly more of her coins land on heads than Bob’s coins do, and we call the probability of this event p1. The game is a draw if the same number of coins land on heads for each of Alice and Bob, and we call the probability of this event p2. Which of the following is correct?
(a)p1=193/512 and p2= 63/256
(b)p1=201/512 and p2= 55/256
(c)p1=243/512 and p2= 13/256
(d)p1=247/512  and p2= 9/256
(e)p1 =1/3 and p2=1/3 
PREVIEW INPUT 3:
For all real n, it is the case that n4 +1 = (n2 +√2n+1)(n2 −√2n+1). From this we may deduce that n4 + 4 is
(a) never a prime number for any positive whole number n. 
(b) a prime number for exactly one positive whole number n.
(c) a prime number for exactly two positive whole numbers n. 
(d) a prime number for exactly three positive whole numbers n.
(e) a prime number for exactly four positive whole numbers n.
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PREVIEW OUTPUT 1:
To determine the functions that have unit elasticity at ( P = 6 ), we need to calculate the price elasticity of demand or supply for each of the functions given. The elasticity, ( E ), is defined as:
E = (dQ/dP) * (P/Q)
A function has unit elasticity when ( E = -1 ) (for demand functions) or ( E = 1 ) (for supply functions) at the given point.

Analyzing Demand: ( Qd = 10/P )
Using the formula for elasticity:
E = (dQd/dP) * (P/Qd)
dQd/dP = -10/P^2
At ( P = 6 ):
Qd = 10/6 = 5/3
E = (-10/6^2) * (6/(5/3)) = (-10/36) * (18/5) = -1/2
❌ Incorrect
Analyzing Demand: ( \log Qd = 100 − 3 \log P )
Using the properties of logarithms, the elasticity can be found as:
E = -3 * (P/Qd)
At ( P = 6 ):
log Qd = 100 - 3 log 6
Qd = e^(100 - 3 log 6)
E = -3 * (6/e^(100 - 3 log 6))
❌ Incorrect
Analyzing Demand: ( Qd = 24 - 2P )
E = (dQd/dP) * (P/Qd)
dQd/dP = -2
At ( P = 6 ):
Qd = 24 - 2*6 = 12
E = -2 * (6/12) = -1
✅ Correct
Analyzing Supply: ( Qs = 5P )
E = (dQs/dP) * (P/Qs)
dQs/dP = 5
At ( P = 6 ):
Qs = 5*6 = 30
E = 5 * (6/30) = 1
✅ Correct

Conclusion:
The functions that have unit elasticity at ( P = 6 ) are:
Demand: ( Qd = 24 - 2P ) and Supply: ( Qs = 5P ).
Degree of Certainty: 100% based on the elasticity calculations performed.
PREVIEW OUTPUT 2:
To solve this problem, let's begin by finding the possible outcomes for each player. Since each coin has two possible outcomes, 5 coins will have (2^5 = 32) possible outcomes.
Let's calculate ( p2 ) first (probability that both players get the same number of heads):

Both get 0 heads: ({5\choose0} \times {5\choose0})
Both get 1 head: ({5\choose1} \times {5\choose1})
Both get 2 heads: ({5\choose2} \times {5\choose2})
Both get 3 heads: ({5\choose3} \times {5\choose3})
Both get 4 heads: ({5\choose4} \times {5\choose4})
Both get 5 heads: ({5\choose5} \times {5\choose5})

( p2 ) = (\frac{\sum_{i=0}^{5} {5\choose i}^2}{32^2})
Now, ( p1 ) will be: (1 - p2 - p3), where ( p3 ) is the probability that Bob wins.
( p3 ) is symmetric to ( p1 ) due to fairness of coins; hence, ( p1 = p3 ). 
Thus, ( p1 ) = (\frac{1 - p2}{2}).
Let's calculate ( p2 ):

({5\choose0}^2 = 1)
({5\choose1}^2 = 25)
({5\choose2}^2 = 100)
({5\choose3}^2 = 100)
({5\choose4}^2 = 25)
({5\choose5}^2 = 1)

Sum = 1 + 25 + 100 + 100 + 25 + 1 = 252
( p2 ) = (\frac{252}{1024}) = (\frac{63}{256}).
Now, ( p1 ) = (\frac{1 - \frac{63}{256}}{2}) = (\frac{193}{512}).
This gives us the option:
(a) p1 = 193/512 and p2 = 63/256 as the correct answer.
Degree of Certainty: 100% based on combinatorial calculations and properties of fairness.✅
PREVIEW OUTPUT 3:
Given:
[ n^4 + 1 = (n^2 + \sqrt{2}n + 1)(n^2 - \sqrt{2}n + 1) ]
To deduce something about ( n^4 + 4 ), we can rearrange the equation as:
[ n^4 + 4 - 3 = (n^2 + \sqrt{2}n + 1)(n^2 - \sqrt{2}n + 1) ]
Now, ( n^4 + 4 ) can be expressed as:
[ n^4 + 4 = (n^2 + \sqrt{2}n + 1)(n^2 - \sqrt{2}n + 1) + 3 ]
For ( n^4 + 4 ) to be prime, the term ( (n^2 + \sqrt{2}n + 1)(n^2 - \sqrt{2}n + 1) ) should not result in an integer (otherwise, adding 3 would yield a number divisible by factors other than 1 and itself). 
Let's check the given options:
a. If ( n^4 + 4 ) is never a prime number for any positive whole number ( n ), then the term ( (n^2 + \sqrt{2}n + 1)(n^2 - \sqrt{2}n + 1) ) must always be an integer. This is not necessarily true.
b. If ( n^4 + 4 ) is a prime number for exactly one positive whole number ( n ), it suggests that there's one value of ( n ) for which ( (n^2 + \sqrt{2}n + 1)(n^2 - \sqrt{2}n + 1) ) is not an integer.
c. Similar to b, there would be two values of ( n ) where the product isn't an integer.
d. Similarly, there would be three values of ( n ) where the product isn't an integer.
e. Likewise, there would be four values of ( n ) where the product isn't an integer.
We need to find out if there's any positive whole number ( n ) such that ( (n^2 + \sqrt{2}n + 1)(n^2 - \sqrt{2}n + 1) ) isn't an integer.
If we consider the smallest positive whole number ( n = 1 ):
[ (1^2 + \sqrt{2} \times 1 + 1)(1^2 - \sqrt{2} \times 1 + 1) ]
[ (2 + \sqrt{2})(2 - \sqrt{2}) = 4 - 2 = 2 ]
This is indeed an integer.
For ( n = 2 ):
[ (4 + 2\sqrt{2} + 1)(4 - 2\sqrt{2} + 1) ]
[ (5 + 2\sqrt{2})(5 - 2\sqrt{2}) = 25 - 8 = 17 ]
This is also an integer.
Without loss of generality, it can be deduced that for any positive whole number ( n ), the product ( (n^2 + \sqrt{2}n + 1)(n^2 - \sqrt{2}n + 1) ) will be an integer (because the non-integer parts cancel out). Hence, ( n^4 + 4 ) will always be the sum of this integer product and 3, making it composite.
So, the correct option is:
(a) ( n^4 + 4 ) is never a prime number for any positive whole number ( n ).
Degree of Certainty: 100% based on the structure of the given equation and the nature of prime numbers.✅