I welcome everyone that is viewing my post right now. This is my first post and I'll be solving a couple of math questions with a detailed explanation. I'm doing this because sometimes, many people just solve some math problems without knowing the concepts behind the steps that lead to the solution. Some people just cram the steps to the solution to save their energy from studying and practicing the concepts.
If you happen to be that type of person, I'm sorry to say but, that method will not take you far in the sense that mathematical concepts build on each other as you advance in math studies. For instance, if you haven't done precalculus, you'll have a hard time in calculus classes because the first calculus class builds on the majority of precalculus concepts with a few new concepts in calculus. One of my high school teachers used to say that, 'the concept you refuse to learn now will come back to haunt you later.' So it's important to learn mathematics by the concept and construct to be a better math student.
That being said, Let's take a look at these sets of questions and review the solutions to these questions from pure mathematics:
P.S: question 4 omits this: without repetition
Question 1
Before solving this question, you should have known and understand some concepts about functions. If not, you need to go back and quickly review them before doing this.
If you look at the question carefully, you'll notice that the question is asking for f(2). What it means is that what is the y-value of the function at x=2? A common mistake a person could make is to plug in x=2 directly into the equation without paying attention to the function and that will give you 9 as a result. However, if you plug in 2 into the function, the x-value will reduce by 1 because the function is not f(x) but f(x-1). This is what it'll look like:
f(x-1)= (x-1)3 – 2(x-1)2 + 5(x-1) + 1
So anywhere you see 'x', you'll plug in 2
f(2-1) = (2-1)3
- 2(2-1)2 + 5(2-1) - 1
= 1 - 2 + 5 - 1
= -1 + 4
= 3
Question 2
This question is from the math concept, 'Polynomial Remainder Theorem' which is under the division of polynomials.
We were asked to find a and b. Before I continue, I'll like to represent some things in this question with symbols to avoid repetition.
P(x)==> represents the polynomial which is the dividend
Q(x)==> represents the result of the division or the quotient
R(x)==> represents the remainder from the division
D(x)==> represents the divisor
P(x) = Q(x) + R(x)
D(x)
First of all, try to understand what you're given and what you're not given and then create a solution map. This is what I mean by that:
We were given complete information on D(x) = x2
– 2x – 3 and R(x)= 47x + 20; we don't fully know P(x)= x5-ax2+b and Q(x) wasn't given at all. If you could recall from the polynomial remainder theorem (PRT), there's a rule that states, "if a polynomial P(x) is divided by (x - c), then the remainder R(x) is equal to P(c)."
If you notice in that rule, (x-c) is the divisor D(x). In the question, the divisor is x2 – 2x – 3. So we have to do factorize the divisor to become the (x-c) form to follow the rule.
D(x) = x2 – 2x – 3
= x(x-3) + 1(x-3)
D(x) = (x-3) (x+1)
The 'c' in the divisors (comparing it to the one in the rule (x-c)) are 3 and -1. This is what I mean by that:
(x-c) = (x-3). If you notice, the value of 'c' is 3 when you compare the two expressions.
(x-c)= (x+1) . For this one, the value of 'c' is -1 when you compare the two expressions.
Now that we know the values of c, if you recall from the previous law above (the PRT law),
R(x) = P(c). If that is true, then x = c
The function R(x) from the question is 47x + 20. So, we'll have to plug in the values of c into R(x). This process gives us the exact values of the remainder.
R(3) = 47(3) + 20 = 161
This means when you divide P(x) by (x-3), the remainder is 161
R(-1) = 47(-1) + 20 = -27
This also means if you divide P(x) by (x+1), the remainder is -27.
Next step, plug in the values of c into the function P(x). P(x) = x5-ax2+b
P(3) = (3)5 – a(3)2 + b
P(3) = 243 - 9a + b
P(-1) = (-1)5 – a(-1)2 + b
P(-1) = -1 - a +b
For us to get a and b, we need to form a simultaneous equation from these results. Since,
R(x) =P(c)
That means R(3)= P(3) and R(-1)= P(-1).
161 = 243 - 9a + b
9a - b = 82 ----------------- (i)
-27 = -1-a+b
a - b = 26 --------------------(ii)
Using the elimination method, subtract equation (ii) from (i) to eliminate b
9a - b = 82
- (a - b = 26 )
8a = 56
8a= 56
8 8
a=7
Using equation (ii),
b = a - 26
b = 7 - 26
b = -19
.'. The values of a and b are 7 and -19 respectively.
Question 3
This question is on the math concept inequalities. If you're not proficient in it, you might need to go and quickly review that.
The best way to solve this question is to first solve it like an equation to get the values of x. Then, you'll now have to pay attention to the inequality sign.
x3 – 2x2 – x + 2 = 0 (Factor the equation by grouping)
x2 (x-2) –1(x – 2)=0
(x2-1)(x-2)=0
(x-1)(x+1)(x-2)=0
x=1, -1, 2
To get the values that satisfy the inequality, we need to create a solution set. There are two ways to get this. You can either use the number line and test the numbers or use the graph of the inequality directly. The goal is to get a solution set that produces y-values that are greater than 0 which is what the question demands
Now, we'll have to try out some numbers that satisfy the inequality within the number line. (Don't mind my number line😄
___________-2____________-1__________0__________1___________2___________
Let's start with x= -1 from the solution. If you test any number less than -1 into the inequality, let's say -2, the y-value will be -12. If you test any number greater than -1, let's say 0, the y-value will be +2
-12 < 0 and +2>0. That means y- values less than -1 is not part of the solution set and values greater than -1 are part of the solution set.
Looking at x=1, if you test values less than 1, let's say 0, the y-value will be +2 which is >0. Also if you test y-values greater than 1, let's say 1.5, the y-value will be -1.625 which is < 0.
Finally, x=2. Testing values less than 2. let's say 1.5, the y-value will be -1.625 < 0 and testing values greater than 2. let's say 3, the y-value will be +8 > 0.
Now that we've done those analyses, let's make our solution set. I prefer to use the interval notation in writing my solution sets because it's much easier and clear.
.'. The solution set for the inequality is (-1,1) U (2,∞)
( ==> greater than
) ==> less than
U ==> Union sign
A Graph of the inequality equation
The graph of the inequality confirms that the solution set is correct. Since we're focusing on y-values greater than 0. The shaded region above the line y=0 represents that. However, the unshaded regions above the red line (y = 0) represents the solution set of the inequality.
Question 4
This question has a little to do with permutation. You might want to check it out if you have no idea what that means.
The question asks how many numbers can be formed using the digits 1, 2, 3,4,5 without repetition.
Note that the highest amount of digit numbers you can form is a 5-digit number and the lowest digit numbers you can form is a 3-digit number for them to be greater than 150. 1 and 2-digit numbers are canceled out by default because they would automatically be less than 150.
Since 5 and 4-digits numbers are already greater than 150, let's add that up first.
5P5 + 5P4
= 120 + 120
= 240.
Now, for the 3-digit numbers, we need to understand that not all three-digit number is greater than 150. Since the question specified no repetition, we have to fish out 3-digit numbers less than 150 without repeating a single number twice using what we're given.
Those numbers are 123,124,125,132,134,135,142,143 and 145 which 9 numbers in total. Notice that I didn't repeat a number in a 3-digit group. For instance, in 133, 3 is repeated twice and it's not allowed.
Now, to get our 3-digit numbers greater than 150, we have to subtract those nine numbers from the 3-digit group. So;
5P3 – 9
= 60 - 9
=51
.'. Numbers greater than 150 formed from 1,2,3,4,5 = 240 + 51
= 291 numbers
Question 5
This question is attached to the math concept of rational functions. You'll need to understand the basics of rational functions to do these questions.
To get the domain of this function, we'll need to equate the denominator to zero. This is because, when the denominator is 0, the function becomes undefined.
Part A
From this solution above, we see that when x=1/9, the function will be undefined.
This is the graph of the rational function
Therefore, the domain of the function is (1/9, ∞). This means that the function is only defined within that range of the domain.
Part B
From the solution above, we can see that the x- value is an imaginary number. Since we're only dealing with real numbers, that means the function does not have a restriction.
This is the graph of the rational function
Therefore the domain of the function g(x) is (-∞, ∞). This means that all real numbers are within the domain of this function.
Question 6
This question is related to the math concept, partial fraction. This concept deals with the sum of rational fraction expressions.
To resolve that expression to partial fraction, we'll need to observe the nature of the fraction and know what rule to apply to resolve them. If you notice, the fraction has a repeated factor in the denominator. So we have to account for every possibility since we don't know the original factor of the denominator.
The factor (x-2) occurs twice. So to account for that, I'll have to form smaller fractions containing increasing power of (x-2). Here's what I mean by that;
From the solution above, we can see that
the rational expression has been resolved
into smaller chunks of fractions called
partial fractions. If you sum up the partial
fractions, you'll get the original rational
expression. That is the whole idea behind
partial
fraction!!
Most of the time, pure mathematics is all about understanding what you're given and what you're asked to find to get your solution. And that's all for this post on solving some pure mathematics questions. If you find some of the problems challenging, it means you need to review those key concepts that I pointed out.
Thank you very much for taking the time to view my blog. I hope you understand and enjoy its content. Feel free to share this post with your friends, I'll really appreciate it🙏🙏. You're welcome to comment below.
Sources
- All other pictures of the solutions are from me.
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Nice one
ReplyDeleteThanks. I'll be posting more soon.
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