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GRE Subject Test: Math : GRE Subject Test: Math

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Example Questions

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Example Question #5 : Conic Sections

Find the intersection(s) of the two parabolas: y=2x^2+3x+4 , y=-4x^2+4

Possible Answers:

$\left(\frac{1}{2},3\right)$

$\left(-\frac{1}{2},3\right)$

$(0,4) \textup{ and }\left(-\frac{1}{2},3\right)$

(0,4)

$(0,4) \textup{ and }\left(\frac{1}{2},3\right)$

Correct answer:

$(0,4) \textup{ and }\left(-\frac{1}{2},3\right)$

Explanation:

Set both parabolas equal to each other and solve for x.

2x^2+3x+4=-4x^2+4

6x^2+3x=0

3x(2x+1)=0

$x=0,-\frac{1}{2}$

Substitute both values of into either parabola and determine .

y=-4(0)^2+4= 4

$y=-4\left(-\frac{1}{2}\right)^2+4= -4\left(\frac{1}{4}\right)+4= -1+4=3$

The coordinates of intersection are:

(0,4) and $\left(-\frac{1}{2},3\right)$

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Example Question #6 : Conic Sections

Find the points of intersection:

y = x^2 + 40x - 6 ; y = -x^2 + 19 x + 5

Possible Answers:

(11, 555) ; (0.5 , 14.25)

(-11, -325); (0.5, 14.25)

(-11, -567); (0.5, 9.05)

(-0.5, -25.75) ; (-11, -325)

(-11, -567) ; (-0.5, -25.75)

Correct answer:

(-11, -325); (0.5, 14.25)

Explanation:

To solve, set both equations equal to each other:

x^2 + 40x - 6 = -x^2 + 19x + 5

To solve as a quadratic, combine like terms by adding/subtracting all three terms from the right side to the left side:

x^2 + x^2 +40x - 19x - 6 - 5 = 0

This simplifies to

2x^2 + 21 x - 11 = 0

Solving by factoring or the quadratic formula gives the solutions and 0.5 .

Plugging each into either original equation gives us:

(-11)^2 + 40(-11 ) - 6 = 121 - 440 - 6 = -325

(0.5) ^ 2 + 40 (0.5) - 6 = 14.25

Our coordinate pairs are (-11, -325) and (0.5, 14.25 ) .

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Example Question #1 : Trigonometric Functions

Evaluate:

$\int\sin(75x)\cos(52x)dx$

Possible Answers:

$\frac{1}{2}\cos(23x)+\frac{1}{2}\cos(127x)+C$

$-\frac{1}{46}\cos(23x)-\frac{1}{254}\cos(127x)+C$

$-\frac{1}{46}\sin(23x)-\frac{1}{254}\sin(127x)+C$

$\frac{1}{46}\cos(75x)+\frac{1}{254}\cos(52x)+C$

$\frac{1}{52}\sin^2(75x)\cos(52x)+C$

Correct answer:

$-\frac{1}{46}\cos(23x)-\frac{1}{254}\cos(127x)+C$

Explanation:

Evaluating this integral requires use of the "Product to Sum Formulas of Trigonometry":

For:

$\sin(\alpha x)\cos(\beta x)=\frac{1}{2}[\sin((\alpha - \beta)x)+\sin((\alpha + \beta)x)]$

So for our given integral, we can rewrite like so:

$\int \sin(75x)\cos(52x)dx\rightarrow \frac{1}{2}\int[\sin(23x)+\sin(127x)]dx$

This can be rewritten as two separate integrals and solved using a simple substitution.

$\frac{1}{2}\int[\sin(23x)+\sin(127x)]dx\rightarrow \frac{1}{2}\int \sin(23x)dx+\frac{1}{2}\int\sin(127x)dx$

Solving each integral individually, we have:

$\frac{1}{2}\int\sin(23x)dx\rightarrow u=23x, du=23dx$

$\frac{1}{23}du=dx$

Substituting this into the integral results in:

$\frac{1}{2}\int\frac{1}{23}\sin(u)du=-\frac{1}{46}\cos(u)+C\rightarrow -\frac{1}{46}\cos(23x)+C$

The other integral is solved the same way:

$\frac{1}{2}\int\sin(127x)dx\rightarrow u=127x, du=127dx$

$\frac{1}{127}du=dx$

Substituting this into the integral results in:

$\frac{1}{2}\int\frac{1}{127}\sin(u)du=-\frac{1}{254}\cos(u)+C\rightarrow-\frac{1}{254}\cos(127x)+C$

Now combining these two statements together results in one of the answer choices:

$-\frac{1}{46}\cos(23x)-\frac{1}{254}\cos(127x)+C$

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Example Question #2 : Trigonometric Functions

Evaluate:

$\int\sin^4(2x)\cos^5(2x)dx$

Possible Answers:

$\frac{1}{18}\cos^9(2x)-\frac{1}{7}\cos^7(2x)+\frac{1}{10}\cos^5(2x)+C$

$\frac{1}{10}\cos^9(2x)-\frac{1}{7}\cos^7(2x)+\frac{1}{18}\cos^5(2x)+C$

$\frac{1}{10}\sin^9(2x)-\frac{1}{7}\sin^7(2x)+\frac{1}{18}\sin^5(2x)+C$

$\frac{1}{18}\sin^9(2x)-\frac{1}{7}\sin^7(2x)+\frac{1}{10}\sin^5(2x)+C$

$\frac{1}{5}\sin^9(2x)-\frac{2}{7}\sin^7(2x)+\frac{1}{9}\sin^5(2x)+C$

Correct answer:

$\frac{1}{18}\sin^9(2x)-\frac{1}{7}\sin^7(2x)+\frac{1}{10}\sin^5(2x)+C$

Explanation:

This integral can be easily evaluated by following the rules outlined for integrating powers of sine and cosine.

But first a substitution needs to be made:

$\int\sin^4(2x)\cos^5(2x)dx$

$u=2x;du=2dx\rightarrow\frac{1}{2}du=dx$

$\frac{1}{2}\int\sin^4(u)cos^5(u)du$

Now that we've made this substitution, we will use the rules outlined for integrating powers of sine and cosine:

In General:

$\int\sin^m(x)\cos^n(x)dx$

1. If "m" is odd, then we make the substitution $u=\sin(x)$ , and we use the identity $\cos^2(x)=1-\sin^2(x)$ .

2. If "n" is odd, then we make the substitution $u=\cos(x)$ , and we use the identity $\sin^2(x)=1-\cos^2(x)$ .

For our given problem statement we will use the first rule, and alter the integral like so:

$\frac{1}{2}\int\sin^4(u)\cos^5(u)du\rightarrow\frac{1}{2}\int\sin^4(u)\cos^4(u)\cos(u)du$

$\frac{1}{2}\int\sin^4(u)[\cos^2(u)]^2\cos(u)du\rightarrow\frac{1}{2}\int\sin^4(u)[1-\sin^2(u)]^2\cos(u)du$

$v=\sin(u);dv=\cos(u)du$

$\frac{1}{2}\int v^4[1-v^2]^2dv\rightarrow\frac{1}{2}\int v^4[1-2v^2+v^4]dv\rightarrow\frac{1}{2}\int (v^4-2v^6+v^8)dv$

$\frac{1}{2}[\frac{1}{5}v^5-\frac{2}{7}v^7+\frac{1}{9}v^9]+C\rightarrow\frac{1}{10}v^5-\frac{1}{7}v^7+\frac{1}{18}v^9+C$

Now we need to substitute back into v:

$\frac{1}{10}\sin^5(u)-\frac{1}{7}\sin^7(u)+\frac{1}{18}\sin^9(u)+C$

Now we need to substitute back into u, and rearrange to make it look like one of the answer choices:

$\frac{1}{18}\sin^9(2x)-\frac{1}{7}\sin^7(2x)+\frac{1}{10}\sin^5(2x)+C$

Report an Error

Example Question #1 : Trigonometric Functions

Find :

$sin(x^{\circ})=\frac{\sqrt2}{2}$

Possible Answers:

Correct answer:

Explanation:

Step 1: Draw a $45^{\circ}-45^{\circ}-90^{\circ}$ triangle..

The short sides have a length of and the hypotenuse has a length of $\sqrt 2$ .

Step 2: Find Sin (Angle A):

$sin(A)=sin(45^{\circ})=\frac {opp}{hyp}=\frac {1}{\sqrt 2}$

Step 3: Rationalize the root at the bottom:

$sin(45^{\circ})=\frac {1}{\sqrt 2}\cdot\frac {\sqrt 2}{\sqrt 2}=\frac {\sqrt 2}{2}$

x=45

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Example Question #1 : Trigonometric Functions

Evaluate the derivative $\frac{\mathrm{d} }{\mathrm{d} x} sin^{-1}(x)$

Possible Answers:

$\frac{1}{\sqrt{1+x^2}}$

$\frac{1}{\sqrt{1+x}}$

$\frac{1}{\sqrt{1-x}}$

$\frac{1}{\sqrt{1-x^2}}$

Correct answer:

$\frac{1}{\sqrt{1-x^2}}$

Explanation:

The inverse trig functions should be memorized.

$\frac{\mathrm{d} }{\mathrm{d} x} sin^{-1}(x)=\frac{1}{\sqrt{1-x^2}}$

The other common inverse trig functions are

$\frac{\mathrm{d} }{\mathrm{d} x} cos^{-1}(x)=-\frac{1}{\sqrt{1-x^2}}$

$\frac{\mathrm{d} }{\mathrm{d} x} tan^{-1}(x)=\frac{1}{1+x^2}$

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Example Question #1 : Limits

Evaluate: $\lim_{x \to 2} \frac {x-2}{x^2-4}$

Possible Answers:

$\frac {1}{2}$

$-\frac {1}{4}$

Undefined/No Limit

$\frac {1}{4}$

Correct answer:

$\frac {1}{4}$

Explanation:

Step 1: See if we can plug in x=2 into the equation..

We can't because the denominator becomes ..

Step 2: Factor the denominator:

(x^2-4)=(x-2)(x+2) (By the Difference of Perfect Squares Formula)

Step 3: Re-write the function:

$\frac {(x-2)}{[(x-2)(x+2)]}$

Step 4: Divide by (x-2) on both the numerator and denominator because it's common:

We are left with: $\frac {1}{(x+2)}$

Step 5: Plug in x=2 :

$\frac {1}{2+2}\implies \frac {1}{4}$

The limit of this function as x approaches 2 is $\frac {1}{4}$ .

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Example Question #1 : Euler's Method And L'hopital's Rule

Evaluate:

$\lim_{x\rightarrow \infty } \frac{x \sin^{2} x}{2^{x} - 1}$

Possible Answers:

The limit does not exist.

$\frac{1}{2}$

$\infty$

Correct answer:

Explanation:

Let's examine the limit

$\lim_{x\rightarrow \infty } \frac{x }{2^{x} - 1}$

first.

$\lim_{x\rightarrow \infty }x= \lim_{x\rightarrow \infty } \left ( 2^{x} - 1 \right )= \infty$

$\frac{\mathrm{d} }{\mathrm{d} x}x = 1$

and

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( 2^{x} - 1 \right ) = \frac{\mathrm{d} }{\mathrm{d} x}\left ( e^{ x \ln 2} - 1 \right )$

$= \frac{\mathrm{d} }{\mathrm{d} x}e^{ x \ln 2} - \frac{\mathrm{d} }{\mathrm{d} x} 1$

$=\ln 2 \cdot e^{ x \ln 2} - 0 = \ln 2 \cdot 2^{x}$ ,

so by L'Hospital's Rule,

$\lim_{x\rightarrow \infty } \frac{x }{2^{x} - 1} = \lim_{x\rightarrow \infty } \frac{1 }{\ln 2 \cdot 2^{x}}$

Since $\lim_{x\rightarrow \infty } \ln 2 \cdot 2^{x} = \infty$ ,

$\lim_{x\rightarrow \infty } \frac{x }{2^{x} - 1} = \lim_{x\rightarrow \infty } \frac{1 }{\ln 2 \cdot 2^{x}} = 0$

Now, for each x > 0 , $0 \leq \sin^{2} x \leq 1$ ; therefore,

$0 \leq \frac{x \sin^{2} x}{2^{x} - 1} \leq \frac{x }{2^{x} - 1}$

By the Squeeze Theorem,

$0 \leq \lim_{x\rightarrow \infty } \frac{x \sin^{2} x}{2^{x} - 1} \leq \lim_{x\rightarrow \infty } \frac{x }{2^{x} - 1} = 0$

and

$\lim_{x\rightarrow \infty } \frac{x \sin^{2} x}{2^{x} - 1} = 0$

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Example Question #1 : Euler's Method And L'hopital's Rule

Evaluate:

$\lim_{x\rightarrow \infty } \frac{5^{x} - 1}{3^{x} - 1}$

Possible Answers:

$\frac{\ln 5}{\ln 3}$

The limit does not exist.

$\infty$

$\frac{5}{3}$

$\ln \frac{5}{3}$

Correct answer:

$\infty$

Explanation:

$\lim_{x\rightarrow \infty }\left ( 5^{x} - 1 \right ) = \lim_{x\rightarrow \infty }\left ( 3^{x} - 1 \right ) = \infty$

Therefore, by L'Hospital's Rule, we can find $\lim_{x\rightarrow \infty } \frac{5^{x} - 1}{3^{x} - 1}$ by taking the derivatives of the expressions in both the numerator and the denominator:

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( 5^{x} - 1 \right ) = \frac{\mathrm{d} }{\mathrm{d} x}\left ( e^{x \ln 5} - 1 \right )$

$= \frac{\mathrm{d} }{\mathrm{d} x} e^{x \ln 5 } - \frac{\mathrm{d} }{\mathrm{d} x}1$

$=\ln 5\cdot e^{x \ln 5 } - 0 = \ln 5\cdot 5^{x }$

Similarly,

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( 3^{x} - 1 \right ) = \ln 3 \cdot 3 ^{x}$

$\lim_{x\rightarrow \infty } \frac{5^{x} - 1}{3^{x} - 1} = \lim_{x\rightarrow \infty } \frac{\ln 5 \cdot 5^{x} }{\ln 3 \cdot 3^{x} }$

$=\frac{\ln 5 }{\ln 3 } \cdot \lim_{x\rightarrow \infty } \frac{ 5^{x} }{ 3^{x} }= \frac{\ln 5 }{\ln 3 } \cdot \lim_{x\rightarrow \infty }\left ( \frac{ 5 }{ 3 } \right )^{x}$

But $\lim_{x\rightarrow \infty }a^{x} = \infty$ for any a > 1 , so

$\lim_{x\rightarrow \infty } \frac{5^{x} - 1}{3^{x} - 1}= \frac{\ln 5 }{\ln 3 } \cdot \lim_{x\rightarrow \infty }\left ( \frac{ 5 }{ 3 } \right )^{x} = \infty$

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Example Question #1 : Euler's Method And L'hopital's Rule

Evaluate:

$\lim_{x\rightarrow 0 } \frac{7^{x} - 1}{2^{x} - 1}$

Possible Answers:

$\frac{1}{2}\ln 7$

$\infty$

$\ln \frac{7}{2}$

The limit does not exist.

$\log_{2}7$

Correct answer:

$\log_{2}7$

Explanation:

$\lim_{x\rightarrow 0 }\left ( 7^{x} - 1 \right ) = 7^{0} - 1 = 1- 1 = 0$

and

$\lim_{x\rightarrow 0 }\left ( 2^{x} - 1 \right ) = 2^{0} - 1 = 1- 1 = 0$

Therefore, by L'Hospital's Rule, we can find $\lim_{x\rightarrow 0 } \frac{7^{x} - 1}{2^{x} - 1}$ by taking the derivatives of the expressions in both the numerator and the denominator:

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( 7^{x} - 1 \right ) = \frac{\mathrm{d} }{\mathrm{d} x}\left ( e^{x \ln 7} - 1 \right )$

$= \frac{\mathrm{d} }{\mathrm{d} x} e^{x \ln 7 } - \frac{\mathrm{d} }{\mathrm{d} x}1$

$=\ln 7\cdot e^{x \ln 7 } - 0 = \ln 7\cdot 7^{x }$

Similarly,

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( 2^{x} - 1 \right ) = \ln 2 \cdot 2 ^{x}$

$\lim_{x\rightarrow 0 } \frac{7^{x} - 1}{2^{x} - 1} = \lim_{x\rightarrow 0 } \frac{ \ln 7\cdot 7^{x }}{ \ln 2\cdot 2^{x }}$

$= \frac{ \ln 7 \cdot 7^{0 }}{ \ln 2\cdot 2^{0 }} = \frac{ \ln 7\cdot 1}{ \ln 2\cdot 1} = \frac{ \ln 7}{ \ln 2 }= \log_{2}7$

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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 #5 : Conic Sections

Example Question #6 : Conic Sections

Example Question #1 : Trigonometric Functions

Evaluate:

Example Question #2 : Trigonometric Functions

Evaluate:

Example Question #1 : Trigonometric Functions

Example Question #1 : Trigonometric Functions

Example Question #1 : Limits

Example Question #1 : Euler's Method And L'hopital's Rule

Example Question #1 : Euler's Method And L'hopital's Rule

Example Question #1 : Euler's Method And L'hopital's Rule

All GRE Subject Test: Math Resources

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