1. Jun 30, 2006

### aalmighty

I'm giving a scholarship exam to study undergrad engineering in Japan. They have a sample test on their site for practice, I've attempted it but have no one to check my answers and they aren't given anywhere, so please check and correct whever nessasary

download the question paper from http://www.studyjapan.go.jp/en/toj/pdf/017e.pdf". Even though the paper is in english, you'll need to get japanese support on your acrobat.

1) 1. x lies between 1 and 2, not including 1 and 2

2. x=1

3. x=2

4. x lies between pi/6 and pi/3 and between 2*pi/3 and 5*pi/6, ends not included

5. x lies between negative infinity to -1 and further between 1 to infinity, ends not included.

6. r=sqrt(3), theta=pi/6

7. a=2, b=13

8. 6

9. 7

10. 0

11. 1/45 and 14/45 respectively

2) 1. (a^3)/8

2. (a/2)^n

3) a=-1, b=0, c=2

4) a. (1-k)x - y + (k^2)/2 = 0

c. Dependent on part b, so couldn't solve.

Last edited by a moderator: Apr 22, 2017
2. Jun 30, 2006

### nazzard

Hello aalmighty,

I only had a look at problem number 4 so far. I get the same equation for the tangent to the curve $$C_1$$. Just try to find the equation for the tangent to the curve $$C_2$$ on the same way and go from there.

I've found two values for k leading to two tangents. If you take a closer look at both curves (maybe sketch them), you'll quickly see which ones they are.

Regards,

nazzard

3. Jun 30, 2006

### HallsofIvy

For 1, part 5, you have a typo: it should be x< -1 or x> 1 (you have x< 1 or x> 1).

For 1, part 6, I get $\frac{2}{5}\sqrt{3}$ not your $\sqrt{3}$.

For 2, the answers, A2 and An, must be matrices not numbers. Did you think they meant the determinant?

4. Jun 30, 2006

### aalmighty

if they were matrices, they could be written as a function of A, which is the original matrix, the question said to write in terms of a, so I assumed it meant the determinant. Isn't that the case? If it isn't, please explain.

5. Jun 30, 2006

### HallsofIvy

That's certainly not how I read the question! Writing them "as a function of A" would just be A2, A3, etc. ! Just go ahead and calculate a few of the products and see what happens.

If
$$A= \left( \begin{array}{cc} a & 0 \\ 1 & a \end{array}\right)$$
then
$$A^2= \left( \begin{array}{cc} a^2 & 0 \\ 2a & a^2 \end{array}\right)$$
$$A^3= \left( \begin{array}{cc} a^3 & 0 \\ 3a^2 & a^3 \end{array}\right)$$

Do you see the pattern?