Vector Subspace Question (GRE 0568 Q3)

This is an interesting question on vector subspaces (a topic from linear algebra):

Question:
If V and W are 2-dimensional subspaces of \mathbb{R}^4, what are the possible dimensions of the subspace V\cap W?

(A) 1 only
(B) 2 only
(C) 0 and 1 only
(D) 0, 1, and 2 only
(E) 0, 1, 2, 3, and 4

To begin this question, we would need this theorem on the dimension of sum and intersection of subspaces (for finite dimensional subspaces):

\dim (M+N)=\dim M+\dim N-\dim (M\cap N)

Note that this looks familiar to the Inclusion-Exclusion principle, which is indeed used in the proof.

Hence, we have \dim(M\cap N)=\dim M+\dim N-\dim (M+N)=4-\dim (M+N).

\dim (M+N), the sum of the subspaces M and N, is at most 4, and at least 2.

Thus, \dim (M\cap N) can take the values of 0, 1, or 2.

Answer: Option D

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GR0568 Q2 Solution (Tangent to Graph)

This is indeed a GRE Math question that even high school students can solve!

Q2) Which of the following is an equation of the line tangent to the graph y=x+e^x at x=0?

(A) y=x
(B) y=x+1
(C) y=x+2
(D) y=2x
(E) y=2x+1

Solution:

Firstly, at x=0, y=0+e^0=1.

\frac{dy}{dx}=1+e^x=1+e^0=2

Substituting these values into y=mx+c, we get

1=2(0)+c, hence c=1.

Thus, the equation is y=2x+1.

Answer: (E)

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GR0568 GRE Math Test Solutions

I have decided to post some solutions to the GRE Mathematics Test Sample Practice Past Year Papers that are available online. Hope it helps!

The GRE (Graduate Record Examinations) Mathematics Test is a standardized test by the company ETS (Educational Testing Service), and meant for students considering PhD in Math in the United States.

Question 1) In the xy-plane, the curve with parametric equations x=\cos t and y=\sin t, 0\leq t\leq\pi, has length

(A) 3
(B) \pi
(C) 3\pi
(D) \frac{3}{2}
(E) \frac{\pi}{2}

Solution:

Using the integral formula for arc length, \displaystyle s=\int_0^\pi \sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}\ dt, we obtain

\displaystyle \begin{aligned}\int_0^\pi \sqrt{(-\sin t)^2+(\cos t)^2}\ dt &=\int_0^\pi 1\ dt\\    &= \pi\end{aligned}

Thus, the answer is option (B).
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