Question 1:
What does it mean by heuristic problem solving skills?
Answer:
A heuristic is a thinking strategy, something that can be used to tease out further information about a problem and thus help you figure out what to do when you don't know what to do. According to the definition provided by YourDictionary(heuristics), it says:
A heuristic is a word from the Greek meaning "to discover".
While according to wikipedia(Heuristic), it says:
A heuristic technique sometimes called simply a heuristic, is any approach to problem solving, learning, or discovery that employs a practical methodology not guaranteed to be optimal or perfect, but sufficient for the immediate goals. Where finding an optimal solution is impossible or impractical, heuristic methods can be used to speed up the process of finding a satisfactory solution. Heuristics can be mental shortcuts that ease the cognitive load of making a decision. Examples of this method include using a rule of thumb, an educated guess, an intuitive judgment, stereotyping, profiling, or common sense.
More precisely, heuristics are strategies using readily accessible, though loosely applicable, information to control problem solving in human beings and machines.
But, we have to highlight the hard fact that heuristic skills do not always generate insightful solutions, but what is more matter is we learned invaluably during the thought process while applying the heuristic skills.
Question 2:
Heuristic problem solving skills are something great that
A. We cannot learn.
B. We can learn from the great teachers.
C. It's either we're born inherently good at it or weak at it.
Answer:
Albeit heuristic problem solving skills sound like a greatest scheme that we have for our students, and in many cases it has turned out to be, but the bitter reality tells us not every teacher is capable of teaching it efficiently and passing the the knowledge to the younger generation. That is why education ministers and their workers around the globe are finding ways to train math educators to become well versed in transferring the heuristic problem solving skills to their students slowly but surely.
Therefore, heuristic problem solving skills can be learned from great teachers, i.e. B is the answer.
Question 3:
Heuristic problem solving skills aim for
A. Shortcut that can solve any math problem using the shortest route as possible.
B. Discovery and invention.
C. More accurate answer of the problem.
Answer:
As heuristic is a word from the Greek defined by "to discover", the answer is hence B.
Question 4:
Which of the following is the graph of the function $y=1-2^x$?
The graph of $y=1-2^x$ can be easily constructed if one knows how to draw the simplest exponential function, $y=a^x$, where $a>1$.
Note that:
- The graph passes through the point (0,1)
- The domain is all real numbers
- The range is y>0.
- The graph is increasing
- The graph is asymptotic to the x-axis as x approaches negative infinity
- The graph increases without bound as x approaches positive infinity
- The graph is continuous
In our case, we need to draw for $y=2^x$, and it looks like:
 |
| Graph (1) |
 |
| Graph (2) |
Question 5:
If we have two graphs $y=f(x)$ and $y=g(x)$ as shown in the graph below, what can we say about the function of $y=f(x)·g(x)$?
A. The function of $y=f(x)·g(x)$ has no inflexion points.
B. The function of $y=f(x)·g(x)$ has one inflexion point at $x=0$.
C. The function of $y=f(x)·g(x)$ has one maximum point at $x=0$.
D. The function of $y=f(x)·g(x)$ has one minimum point at $x=0$.
Answer:
From the graph, we can see that:
1. For $x<0$, the function $y=f(x)<0$ whereas $y=g(x)>0$,
2. For $x>0$, the function $y=f(x)>0$ whereas $y=g(x)<0$,
3. At $x=0$, the function $f(x)=g(x)=0$.
We can then conclude that
1. The function of $y=f(x)·g(x)$ is always less than zero in the domain $x<0$ or $x>0$.
2. If we plot it on the coordinate plane, it would give us the /\ shape, that implies we have a maximum point that occurs at $x=0$.
Therefore, C is the answer for this problem.
Question 6:
Based on the previous discoveries in questions 4 and 5, what algebraic manipulation would you use to solve for the equation $x|\cos x|=1-2^x$?
A. Square both sides of the equation to get rid of the absolute value bars.
B. Take the natural log of both sides to get rid of the exponent.
C. Get rid of the absolute value bars by treating the positive x and negative x separately.
D. Multiply both sides by $x$.
Answer:
Question 4 and 5 tells us the graph of function $y=x(1-2^x)$ is always negative, and is zero at $x=0$.
But we're asked to solve for $x|\cos x|=1-2^x$. The first and foremost thing that we need to do is to make sure we have the expression $x(1-2^x)$ on either side of $x|\cos x|=1-2^x$.
That's why we need to multiply both sides of the given equation $x|\cos x|=1-2^x$ by $x$ to get:
$x(x|\cos x|)=x(1-2^x)$
$x^2|\cos x|=x(1-2^x)$
D is hence the answer.
Question 7:
What are the solution(s) for $x|\cos x|=1-2^x$?
From multiplying both sides of the equation $x|\cos x|=1-2^x$ by an $x$, we have, like what we have gotten in question 6 that
$x^2|\cos x|=x(1-2^x)$
But we also gather from question 5 that the function of $y=x(1-2^x)$ has one maximum point at $x=0$. This seems like nothing special an observation, but if we interpret it as the function of $y=x(1-2^x)\le 0$ for all $x$, then, we can see that what we're having now is:
a. $x^2\ge 0$ for all $x$,
b. $|\cos x|\ge 0$ for all $x$ and
c. $y=x(1-2^x)\le 0$ for all $x$
This is saying $x^2|\cos x|=x(1-2^x)$ holds only when both sides equals zero and there is only one real solution for this equation, and that is when $x=0$.
Therefore, the answer for question 7 is $x=0$.
mmm
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