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GRE Subject Test: Math : GRE Subject Test: Math

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Example Questions

Example Question #131 : Linear Algebra

Solve this system of equations.

5x - 2y = 25

4x-2y = 24

Possible Answers:

(-10,-9)

(-10,9)

(-9,-10)

(1,-10)

Correct answer:

(1,-10)

Explanation:

5x -2y = 25

4x-2y=24

To solve this system of equations, subtract the second equation from the first.

5x - 2y = 25

-4x + 2y = -24

____________________________

x=1

Now, substitute in the value for x into one of the equations to solve for the value of .

$\\5x-2y = 25 \\5(1)-2y=25 \\5-2y=25$

Now subtract five from each side.

$\\5-5-2y=25-5 \\-2y=20$

Divide both sides by negative 2:

$\\ \frac{-2y}{-2}=\frac{20}{-2} \\ \\y=-10$

x = 1

(1,-10) is the correct answer.

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Example Question #301 : Algebra

Solve this system of equations:

$y = \frac{1}{5}x +2$

$y = -\frac{1}{5}x -4$

Possible Answers:

(-15,1)

(-1,-15)

(15,-1)

(-15,-1)

Correct answer:

(-15,-1)

Explanation:

To solve this system of equations, set both equations equal to one another.

$\frac{1}{5}x +2 = -\frac{1}{5}x-4$

Add $\frac{1}{5}x$ to both sides of the equation.

$\frac{1}{5}x + \frac{1}{5}x + 2= -\frac{1}{5}x +\frac{1}{5}x -4$

$\frac{2}{5}x + 2 = -4$

Subtract from both sides of the equation.

$\frac{2}{5}x + 2 -2 = -4-2$

$\frac{2}{5}x =-6$

Multiply both sides of the equation by $\frac{5}{2}$ .

$\frac{2}{5}x \times \frac{5}{2} = \frac{-6}{1} \times \frac{5}{2}$

$x = \frac{-30}{2}$

x = -15

Plug the value of , which is into one of the equations to get the value of

$-\frac{1}{5 }\times \frac{-15}{1 } - 4 = y$

3-4 = y

y = -1

(-15,-1) is the correct answer for this system of equations.

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Example Question #131 : Linear Algebra

Find the value of and that satisfy the equations:

(1) x-y=5

and

(2) x+y=7 .

Possible Answers:

x=6,y=-1

x=-6,y=1

x=6,y=1

x=-6,y=-1

Correct answer:

x=6,y=1

Explanation:

Step 1: Write the two equations, one below another and line up the terms.

x-y=5
x+y=7
----------------

Step 2: We see that we have and . We can add these two equations up, which will isolate y and let us solve for x.

$x{\color{Red} -y}=5$
$x+{\color{Blue} y}=7$ We add here.
----------------
2x=12

Step 3: We will isolate x by itself. We need to divide by 2 on both sides to get x.

$\frac {2x}{2}=\frac {12}{2}$
$\rightarrow x=6$

Step 4: We found x, so we can plug in that value into any one of the two equations and solve for y. Let's choose equation (1).

(1)... x-y=5
(6)-y=5 . Isolate y by itself. We are going to subtract 6 from both sides.
6-6-y=5-6 . Simplify the left hand side.
-y=-1

Step 5: We will divide by -1 to get the value of y.

$\frac {-y}{-1}=\frac {-1}{-1}$
$\rightarrow y=1$

The values that solve this system of equations is x=6 and y=1 .

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Example Question #302 : Algebra

4x+3y=11

8x+4y=12

Possible Answers:

x=1

y=-5

x=-1

y=5

Unsolvable

x=5

y=-1

x=-5

y=1

Correct answer:

x=-1

y=5

Explanation:

To answer this question you must first solve for one of the variables. This can be done with either variable with either equation. In this example of how to solve the problem we will solve for y using the second equation

8x+4y=12

subtract 8x from both sides

4y=12-8x

divide both sides by y

y=3-2x

Now we plug this into the first equation for the y variable

4x+3(3-2x)=11

Distribute the 3

4x+9-6x=11

Simplify

-2x+9=11

subtract 9 from both sides

-2x=2

divide by -2 on both sides

x=-1

Using this we solve for y in the second equation

8(-1) +4y=12

simplify

-8+4y=12

add 8 to both sides

4y=20

divide by 4 on both sides

y=5

Final answer x=-1 and y=5

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Example Question #1 : Eigenvalues

Find the eigenvalues of the following matrix, if possible.

$A=\begin{bmatrix} 1&2 \\ -2&1 \end{bmatrix}$

Possible Answers:

The eigenvalues do not exist.

$\lambda_{1,2}=\pm1$

$\lambda=\frac{5}{2}$

$\lambda=\begin{bmatrix} 2i&4i \\ -2i&i \end{bmatrix}$

$\lambda_{1,2}=1\pm2i$

Correct answer:

$\lambda_{1,2}=1\pm2i$

Explanation:

In order to find the eigenvalues of a matrix, apply the following formula:

$det(A-\lambda I) =0$

is the identity matrix.

$\begin{bmatrix} 1&2 \\ -2&1 \end{bmatrix}-\lambda \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix}=\begin{bmatrix} 1-\lambda & 2\\ -2&1-\lambda \end{bmatrix}$

Compute the determinant and set it equal to zero.

$(1-\lambda)(1-\lambda)-(2)(-2)=0$

$1-2\lambda+\lambda^2+4=0$

$\lambda^2-2\lambda+5=0$

Solve for lambda by using the quadratic formula.

$\lambda_{1,2}=1\pm2i$

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Example Question #491 : Gre Subject Test: Math

Find the eigenvalues of the following matrix, if possible.

$A=\begin{bmatrix} 1&2 \\ -2&1 \end{bmatrix}$

Possible Answers:

The eigenvalues do not exist.

$\lambda_{1,2}=1\pm2i$

$\lambda=\frac{5}{2}$

$\lambda_{1,2}=\pm1$

$\lambda=\begin{bmatrix} 2i&4i \\ -2i&i \end{bmatrix}$

Correct answer:

$\lambda_{1,2}=1\pm2i$

Explanation:

In order to find the eigenvalues of a matrix, apply the following formula:

$det(A-\lambda I) =0$

is the identity matrix.

$\begin{bmatrix} 1&2 \\ -2&1 \end{bmatrix}-\lambda \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix}=\begin{bmatrix} 1-\lambda & 2\\ -2&1-\lambda \end{bmatrix}$

Compute the determinant and set it equal to zero.

$(1-\lambda)(1-\lambda)-(2)(-2)=0$

$1-2\lambda+\lambda^2+4=0$

$\lambda^2-2\lambda+5=0$

Solve for lambda by using the quadratic formula.

$\lambda_{1,2}=1\pm2i$

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Example Question #492 : Gre Subject Test: Math

Let $P_{5}(x)$ be the fifth-degree Taylor polynomial approximation for f(x) = sin(x) , centered at x = 0 .

What is the Lagrange error of the polynomial approximation to sin(1) ?

Possible Answers:

$\frac{1}{7!}$

$\frac{1}{6!}$

$\frac{1}{5!}$

Correct answer:

$\frac{1}{7!}$

Explanation:

The fifth degree Taylor polynomial approximating sin(x) centered at x=0 is:

$sin(x) \approx x - \frac{x^3}{3!} + \frac{x^5}{5!}$

The Lagrange error is the absolute value of the next term in the sequence, which is equal to $\left |\frac{x^7}{7!} \right |$ .

We need only evaluate this at and thus we obtain $\frac{1^7}{7!} = \frac{1}{7!}$

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Example Question #72 : Ap Calculus Bc

Which of the following series does not converge?

Possible Answers:

$\sum_{i=0}^{\infty} 0.9999999999999^n$

$\sum_{i=0}^{\infty} \frac{n^2 ln (n)}{n!}$

$\sum_{i=0}^{\infty} \frac{n!}{n^2 cos(n)}$

$\sum_{i=0}^{\infty} \frac{(-1)^n}{n}$

$\sum_{i=0}^{\infty} \frac{n - 1}{ n^3 + 1}$

Correct answer:

$\sum_{i=0}^{\infty} \frac{n!}{n^2 cos(n)}$

Explanation:

We can show that the series $\sum_{i=0}^{\infty} \frac{n!}{n^2 cos(n)}$ diverges using the ratio test.

$L = \lim_{n ->\infty } \frac{a_{n+1}}{a_{n}} = \lim_{n ->\infty } \left[\left(\frac{(n+1)!}{(n+1)^2 cos (n+1)} \div \frac{n!}{n^2 cos (n) }\right)\right]$

$= \lim_{n ->\infty} \frac{n^2(n+1)cos(n)}{(n+1)^2cos(n+1)} = \lim_{n ->\infty} \frac{n^2cos(n)}{(n+1)cos(n+1)}$

n^2 will dominate over (n+1) since it's a higher order term. Clearly, L will not be less than, which is necessary for absolute convergence.

Alternatively, it's clear that is much greater than n^2 , and thus having in the numerator will make the series diverge by the $n^{th}$ limit test (since the terms clearly don't converge to zero).

The other series will converge by alternating series test, ratio test, geometric series, and comparison tests.

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Example Question #1 : Lagrange Multipliers

Find the minimum and maximum of f(x,y)=2x-5y , subject to the constraint x^2+y^2=144 .

Possible Answers:

f(4.46,-11.14) is a maximum

f(-4.46,11.14) is a minimum

There are no maximums or minimums

f(4.46,11.14) is a maximum

f(-4.46,-11.14) is a minimum

f(4.46,11.14) is a maximum

f(4.46,-11.14) is a minimum

f(-4.46,11.14) is a maximum

f(4.46,-11.14) is a minimum

Correct answer:

f(4.46,-11.14) is a maximum

f(-4.46,11.14) is a minimum

Explanation:

First we need to set up our system of equations.

$2=2\lambda x \rightarrow x=\frac{1}{\lambda}$

$-5=2\lambda y \rightarrow y=-\frac{5}{2 \lambda}$

x^2+y^2=144

Now lets plug in these constraints.

$(\frac{1}{\lambda})^2+(-\frac{5}{2\lambda})^2=144$

$\frac{1}{\lambda ^2}+\frac{25}{4\lambda^2}=144$

$\frac{29}{4\lambda^2}=144$

Now we solve for $\lambda$

$\frac{29}{4\lambda^2}=144\rightarrow \lambda^2=\frac{29}{576}\rightarrow\lambda=\pm\sqrt{\frac{29}{576}}$

$\lambda=\sqrt{\frac{29}{576}}$

$x=\frac{1}{\sqrt{\frac{29}{576}}}\approx 4.46$ , $y=-\frac{5}{2\sqrt{\frac{29}{576}}}\approx -11.14$

$\lambda=-\sqrt{\frac{29}{576}}$

$x=\frac{1}{\sqrt{\frac{29}{576}}}\approx -4.46$ , $y=-\frac{5}{2\sqrt{\frac{29}{576}}}\approx 11.14$

Now lets plug in these values of , and into the original equation.

f(4.46,-11.14)=2(4.46)-5(-11.14)=64.62

f(-4.46,11.14)=2(-4.46)-5(11.14)=-64.62

We can conclude from this that f(4.46,-11.14) is a maximum, and f(-4.46,11.14) is a minimum.

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Example Question #3 : Applications Of Partial Derivatives

Find the absolute minimum value of the function f(x,y) = x^2+y^2 subject to the constraint x^2 +2y^2 = 1 .

Possible Answers:

$2\sqrt2$

$\frac{1}{2}$

$\sqrt2$

Correct answer:

$\frac{1}{2}$

Explanation:

Let g = x^2 +2y^2 To find the absolute minimum value, we must solve the system of equations given by

$\triangledown f = \lambda\triangledown g, g =1$ .

So this system of equations is

$f_x = \lambda g_x$ , $f_y = \lambda g_y$ , g =1 .

Taking partial derivatives and substituting as indicated, this becomes

$2x = \lambda(2x), 2y = \lambda(4y), x^2+2y^2 = 1$ .

From the left equation, we see either x=0 or $\lambda = 1$ . If x=0 , then substituting this into the other equations, we can solve for $y, \lambda$ , and get $y = \pm \sqrt{2}/2$ , $\lambda = 1/2$ , giving two extreme candidate points at $(0, \frac{\sqrt{2}}{2}), (0, -\frac{\sqrt{2}}{2})$ .

On the other hand, if instead $\lambda =1$ , this forces y = 0 from the 2nd equation, and $x = \pm 1$ from the 3rd equation. This gives us two more extreme candidate points; (-1,0),(1,0) .

Taking all four of our found points, and plugging them back into , we have

$f(-1,0) = 1, f(1,0) = 1, f(0,\frac{\sqrt{2}}{2}) = \frac{1}{2}, f(0,-\frac{\sqrt{2}}{2}) = \frac{1}{2}$ .

Hence the absolute minimum value is $\frac{1}{2}$ .

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GRE Subject Test: Math : GRE Subject Test: Math

Study concepts, example questions & explanations for GRE Subject Test: Math

All GRE Subject Test: Math Resources

Example Questions

Example Question #131 : Linear Algebra

Example Question #301 : Algebra

Example Question #131 : Linear Algebra

Example Question #302 : Algebra

Example Question #1 : Eigenvalues

Example Question #491 : Gre Subject Test: Math

Example Question #492 : Gre Subject Test: Math

Example Question #72 : Ap Calculus Bc

Example Question #1 : Lagrange Multipliers

Example Question #3 : Applications Of Partial Derivatives

All GRE Subject Test: Math Resources

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