# Solution to 2014 MMC Division Finals-Orals 4th Year

This blog post will outline the solution to last year’s MMC division finals-orals team category. We will start for the first ten questions, the next set of questions will be added very soon.

15-second mental questions worth 1 point and their solutions.

1. Find the remainder when $x^3-2x+5$ is divided by $x+2$.

Solution: This can be done in many ways. But I prefer to use substitution.

Let $f(x)=x^3-2x+5$, then $f(-2)$ is the remainder. $f(x)=x^3-2x+5$ $f(-2)=(-2)^3-2(-2)+5$ $f(-2)=-8+4+5$ $f(-2)=1$

2. Find the exact radian measure of $330^\circ$

Solution: we need to know that $\pi rad=180^\circ$, hence $330^\circ\times \displaystyle\frac{\pi}{180^\circ}$ $330^\circ=\displaystyle\frac{11\pi}{6}$

3. Find the domain of $f(x)=\log_4 (x-4)$

Solution: To answer the question, take note that in order for logarithmic curve to exist, in its form $y=\log_b a$, b>1 and a>0.

Going back to our problem, the base is fine since it is 4. We just have to focus on x-4. Since it must be positive real number, we have $x-4>0$ $x>4$

The domain is {x|x>4} or (4,+∞)

4. Find the 12th term of an arithmetic series whose first term is 8 and third term is 2.

Solution: Observe that the progression is decreasing and there is one term between 2 and 8.  With the 15-second cramming fever I guess we don’t have much time to use the series to find the common difference. Just observe the following numbers.

8 , x, 2, . . . . What do you think is the number between them to make it an arithmetic series?

yes! it is 5. Therefore, the common difference is -3.

Solving for 12th term: $a_n=a_1+(n-1)d$ $a_{12}=8+(12-1)-3$ $a_{12}=8-33$ $a_{12}=-25$

Therefore, the 12th term is -25.

5. Find the vertex of the parabola $y=-(x+3)^2+8$

Solution: in the vertex form of parabola $y=a(x-h)^2+k$, the vertex is at (h,k). Therefore, in $y=-(x+3)^2+8$, the vertex is at (-3,8). don’t get confused with the negative sign. It is just telling us that the graph is opening downward.

6. Determine all x in so that the ordered pairs (1,3), (2,1), and (x,3) are on the graph of the function.

Solution: first to think is that a function passed the vertical line test. This means that there should be no the same values for the x-coordinates. If x=1, the point will become (1,3) which is the same as the first point. We can still consider the points are on the graph of a function.

If x=2, the set of points will become (1,3), (2,1), and (2,3). Points (2,1) and (2,3) form a vertical line making the graph to not pass the vertical line test. But the question here is to determine all x so that the ordered pairs are on the graph of the function. The answer here are all real numbers except 2 or $x\ne2$.

7. Find the x-intercept of the line $y=7x-4$

Solution: x-intercept is a point at which the given line intersects the x-axis. Thus, by letting y=0, we can get the value of x-intercept.

$latex y=7x-4$ $0=7x-4$ $7x=4$ $x=\displaystyle\frac{4}{7}$

Therefore the x-intercept is 4/7.

8. Find a so that the line $y=ax-2$ is perpendicular to a line with slope 3.

Solution: This problem can be solved quickly if we know that the product of the slope of perpendicular lines is equal to -1. So the value of a is the negative reciprocal of a to attain such condition. Thus, a=-1/3.

9. The parabola $y=3x^2-5x-4$ passes through the point (1,b). Find b.

Solution: If (1,b) is on the graph of the equation, it must satisfy the equation. Thus, by simple substitution, we can solve for the value of b. $y=3x^2-5x-4$ $b=3(1)^2-5(1)-4$ $b=3-5-4$ $b=-6$

10. Suppose u varies directly as the square root of v. If u=3 when v=4, find u when v=10.

Solution: If we know already that this is a direct variation, we can just use ratio and proportion as follows. $\displaystyle\frac{u_1}{v_1}=\displaystyle\frac{u_2}{v_2}$ $\displaystyle\frac{3}{4}=\displaystyle\frac{u_2}{10}$ $u_2=\displaystyle\frac{3\cdot 10}{4}$ $u_2=\displaystyle\frac{15}{2}$