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SAT II Math I : SAT Subject Test in Math I

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Example Questions

Example Question #83 : Single Variable Algebra

Define a function $f(x) =5x- \cos 2x$ .

f(c) = 8 for exactly one value of on the interval (1, 1.5) . Which of the following is true of ?

Possible Answers:

$c \in (1.0, 1.1 )$

$c \in (1.1, 1.2)$

$c \in (1.3, 1.4)$

$c \in (1.2, 1.3 )$

$c\in (1.4, 1.5)$

Correct answer:

$c\in (1.4, 1.5)$

Explanation:

Define $g(x)= f(x) - 8 =5x- \cos 2x - 8$ . Then, if g(c)= f(c) - 8 = 0 , it follows that f(c) = 8 .

By the Intermediate Value Theorem (IVT), if g(x) is a continuous function, and g(a) and g(b) are of unlike sign, then g(c) = 0 for some $c \in (a, b)$ . and $\cos 2x$ are both continuous everywhere, so g(x) is a continuous function, so the IVT applies here.

Evaluate g(x) for each of the following values: $\left \{ 1.0, 1.1, 1.2, 1.3, 1.4, 1.5\right \}$

$g(1) =5 (1)- \cos 2 (1)- 8$

$=5 - \cos 2 - 8$

$\approx 5 - (-0.42) - 8$

$\approx -2.58$

$g(1.1) =5 (1.1)- \cos 2 (1.1)- 8$

$=5.5 - \cos 2.2 - 8$

$\approx 5.5 - (-0.59) - 8$

$\approx -1.91$

$g(1.2) =5 (1.2)- \cos 2 (1.2)- 8$

$=6 - \cos 2.4 - 8$

$\approx 6 - (-0.74) - 8$

$\approx -1.26$

$g(1.3) =5 (1.3)- \cos 2 (1.3)- 8$

$=6.5 - \cos 2.6 - 8$

$\approx 6.5 -(-0.86) - 8$

$\approx -0.64$

$g(1.4) =5 (1.4)- \cos 2 (1.4)- 8$

$=7 - \cos 2.8 - 8$

$\approx 7 - (-0.94) - 8$

$\approx -0.06$

$g(1.5) =5 (1.5)- \cos 2 (1.5)- 8$

$=7.5 - \cos 3 - 8$

$\approx 7.5 -(-0.99)- 8$

$\approx 0.49$

Only in the case of (1.4, 1.5) does it hold that g (x) assumes a different sign at both endpoints - g(1.4) < 0 < g(1.5) . By the IVT, g(c) = 0 , and f(c) = 8 , for some $c\in (1.4, 1.5)$ .

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Example Question #241 : Sat Subject Test In Math I

Consider the absolute value equation

|x + 7| = 13

Which of the following quadratic equations has the same solution set as this one?

Possible Answers:

$x^{2} + 14x - 120 = 0$

$x^{2} - 14x - 120 = 0$

$x^{2} + 26x + 120 = 0$

$x^{2}- 120 = 0$

$x^{2}- 26x + 120 = 0$

Correct answer:

$x^{2} + 14x - 120 = 0$

Explanation:

The absolute value equation |x + 7| = 13 can be rewritten as the compound equation

x+ 7 =- 13 $\or$ x+ 7 = 13

Each simple equation can be solved by subtracting 7 from both sides:

x+ 7 =- 13

x+ 7 - 7 =- 13 - 7

x = -20

x+ 7 = 13

x+ 7 - 7 = 13 - 7

x = 6

The absolute value equation has solution set $\left \{ - 20 , 6 \right \}$ .

Suppose P(x) is a polynomial. Then, by the Factor Theorem, x - c is a factor of P(x) if and only if P(c) = 0 - that is, if is a solution of P(x) = 0 . We are seeking a quadratic equation with solution set the same as that of the absolute value equation, $\left \{ - 20 , 6 \right \}$ , we want polynomial P(x) to have as its linear binomial factors x- (-20) , or x+ 20 , and x - 6 .

The correct polynomial will be the product of these two:

P(x) = (x+20)(x-6)

Using the FOIL technique to rewrite this:

$P(x) = x^{2}-6x + 20x - 120$

$P(x) = x^{2} + 14x - 120$

The correct choice is the equation $x^{2} + 14x - 120 = 0$ .

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Example Question #84 : Single Variable Algebra

Give the set of all real solutions of the equation $8 x^{\frac{2}{3}} -2 x ^{\frac{1}{3}} - 3 = 0$ .

Possible Answers:

$\left \{ - \frac{27}{64} , \frac{1}{8} \right \}$

$\left \{ -\frac{27}{64} , -\frac{1}{8} , \frac{1}{8} , \frac{27}{64} \right \}$

$\left \{ - \frac{27}{64} , \frac{27}{64} \right \}$

$\left \{- \frac{1}{8} , \frac{1}{8} \right \}$

$\left \{ -\frac{1}{8} , \frac{27}{64} \right \}$

Correct answer:

$\left \{ -\frac{1}{8} , \frac{27}{64} \right \}$

Explanation:

Set $y = x^{\frac{1}{3}}$ . Then $x^{\frac{2}{3}} = \left (x^{\frac{1}{3}} \right )^{2} = y ^{2}$ .

$8 x^{\frac{2}{3}} -2 x ^{\frac{1}{3}} - 3 = 0$ can be rewritten as

$8 \left (x^{\frac{1}{3}} \right )^{2} -2 x ^{\frac{1}{3}} - 3 = 0$

Substituting $y^{2}$ for and for $x^{\frac{1}{3}}$ , the equation becomes

$8y^{2} -2 y - 3 = 0$ ,

a quadratic equation in .

This can be solved using the method. We are looking for two integers whose sum is and whose product is 8 (-3) = -24 . Through some trial and error, the integers are found to be and 4, so the above equation can be rewritten, and solved using grouping, as

$8y^{2}-6y + 4 y - 3 = 0$

2y ( 4 y - 3)+1 ( 4 y - 3) = 0

(2y +1) ( 4 y - 3) = 0

By the Zero Product Principle, one of these factors is equal to zero:

Either:

2y + 1 = 0

2y + 1 - 1 = 0 - 1

2y = - 1

$\frac{2y}{2} = \frac{- 1}{2}$

$y = - \frac{1}{2}$

Substituting back:

$x^{\frac{1}{3}} = -\frac{1}{2}$ , or $\sqrt[3]{x} = - \frac{1}{2}$

Cubing both sides:

$(\sqrt[3]{x} )^{3} = \left ( - \frac{1}{2} \right )^{3}$

$x = -\frac{1}{8}$

Or:

4y - 3 = 0

4y - 3+ 3 = 0+ 3

4 y = 3

$\frac{4 y}{4} = \frac{3}{4}$

$y= \frac{3}{4}$

Substituting back:

$x^{\frac{1}{3}} = \frac{3}{4}$ , or $\sqrt[3]{x} = \frac{3}{4}$

Cubing both sides:

$(\sqrt[3]{x} )^{3} = \left (\frac{3}{4} \right )^{3}$

$x = \frac{27}{64}$ .

The solution set is $\left \{ -\frac{1}{8} , \frac{27}{64} \right \}$ .

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Example Question #81 : Single Variable Algebra

Give the set of all real solutions of the equation $8x^{4} - 2x^{2} - 3 = 0$ .

Possible Answers:

$\left \{ -\frac{\sqrt{2}}{2} ,\frac{\sqrt{2}}{2} \right \}$

$\left \{ -\frac{\sqrt{2}}{2} , \frac{\sqrt{3}}{2} \right \}$

$\left \{ -\frac{\sqrt{3}}{2} , \frac{\sqrt{3}}{2} \right \}$

$\left \{ -\frac{\sqrt{3}}{2} , \frac{\sqrt{2}}{2} \right \}$

$\left \{ -\frac{\sqrt{3}}{2}, -\frac{\sqrt{2}}{2} ,\frac{\sqrt{2}}{2} , \frac{\sqrt{3}}{2} \right \}$

Correct answer:

$\left \{ -\frac{\sqrt{3}}{2} , \frac{\sqrt{3}}{2} \right \}$

Explanation:

Set $y = x^{2}$ . Then $x^{4 } = \left (x^{2} \right )^{2} = y ^{2}$ .

$8x^{4} - 2x^{2} - 3 = 0$ can be rewritten as

$8 \left (x^{2} \right )^{2} -2 x ^{2} - 3 = 0$

Substituting $y^{2}$ for $x^{4 }$ and for $x^{2}$ , the equation becomes

$8y^{2} -2 y - 3 = 0$ ,

a quadratic equation in .

$8y^{2}-6y + 4 y - 3 = 0$

2y ( 4 y - 3)+1 ( 4 y - 3) = 0

(2y +1) ( 4 y - 3) = 0

By the Zero Product Principle, one of these factors is equal to zero:

Either:

2y + 1 = 0

2y + 1 - 1 = 0 - 1

2y = - 1

$\frac{2y}{2} = \frac{- 1}{2}$

$y = - \frac{1}{2}$

Substituting back:

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

However, there are no real numbers whose squares are negative. Since we are looking only for real solutions, none are yielded here.

Or:

4y - 3 = 0

4y - 3+ 3 = 0+ 3

4 y = 3

$\frac{4 y}{4} = \frac{3}{4}$

$y= \frac{3}{4}$

Substituting back:

$x^{2} = \frac{3}{4}$

Taking both square roots of both sides and simplifying using the Quotient of Radicals Rule:

$x = \pm \sqrt{ \frac{3}{4}} = \pm \frac{\sqrt{3}}{\sqrt{4}} = \pm \frac{\sqrt{3}}{2}$

The solution set is $\left \{ -\frac{\sqrt{3}}{2} , \frac{\sqrt{3}}{2} \right \}$ .

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Example Question #82 : Single Variable Algebra

Define a function $f(x) = x^{3}+3x$ .

f(c) = -22 for exactly one value of on the interval (-2.5, -2.0) . Which of the following is true of ?

Possible Answers:

$c \in ( -2.4, -2.3)$

$c \in (-2.5, -2.4)$

$c \in ( -2.1, -2.0)$

$c \in ( - 2.3, -2.2)$

$c \in ( -2.2, -2.1)$

Correct answer:

$c \in (-2.5, -2.4)$

Explanation:

Define $g(x) = f(x) + 22 = x^{3}+3x+ 22$ . Then, if g(c)= f(c)+22 = 0 , it follows that f(c) = -22 .

By the Intermediate Value Theorem (IVT), if g(x) is a continuous function, and g(a) and g(b) are of unlike sign, then g(c) = 0 for some $c \in (a, b)$ . As a polynomial, g(x) is a continuous function, so the IVT applies here.

Evaluate g(x) for each of the following values: $\left \{ -2.5, -2.4, -2.3, -2.2, -2.1, -2.0 \right \}$ :

$g(-2.5) = (-2.5)^{3}+3(-2.5)+ 22 =-1.125$

$g(-2.4) = (-2.4)^{3}+3(-2.4)+ 22 =0.976$

$g(-2.3) = (-2.3)^{3}+3(-2.3)+ 22 =2.933$

$g(-2.2) = (-2.2)^{3}+3(-2.2)+ 22 =4.752$

$g(-2.1) = (-2.1)^{3}+3(-2.1)+ 22 =6.439$

$g(-2.0) = (-2.0)^{3}+3(-2.0)+ 22 =8$

Only in the case of (-2.5, -2.4) does it hold that g (x) assumes different signs at the endpoints - g(-2.5) < 0 < g(2.4) . By the IVT, g(c) = 0 , and f(c) = -22 , for some $c \in (-2.5, -2.4)$ .

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Example Question #241 : Sat Subject Test In Math I

Solve: 3x+18 = -3

Possible Answers:

Correct answer:

Explanation:

Subtract 18 on both sides.

3x+18 -18= -3-18

3x=-21

Divide by three on both sides.

$\frac{3x}{3}=\frac{-21}{3}$

The answer is:

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Example Question #87 : Single Variable Algebra

Solve the equation: -9x+7 = -16

Possible Answers:

$\frac{11}{9}$

$\frac{23}{9}$

$-\frac{23}{9}$

Correct answer:

$\frac{23}{9}$

Explanation:

Subtract seven from both sides.

-9x+7-7 = -16-7

-9x = -23

Divide both sides by negative nine.

$\frac{-9x}{-9} =\frac{ -23}{-9}$

The answer is: $\frac{23}{9}$

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Example Question #91 : Single Variable Algebra

Solve the equation: -2x+4 = 18-x

Possible Answers:

$-\frac{14}{3}$

$\frac{14}{3}$

$\frac{22}{3}$

Correct answer:

Explanation:

Add on both sides of the equation.

-2x+4 +2x= 18-x+2x

4=18+x

Subtract 18 on both sides of the equation.

4-18=18+x-18

x=-14

The answer is:

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Example Question #91 : Single Variable Algebra

Solve the following equation: $\frac{2}{3x}+4 = \frac{5}{6x}$

Possible Answers:

$\frac{5}{48}$

$\frac{1}{12}$

$\frac{1}{36}$

$\frac{5}{8}$

$\frac{1}{24}$

Correct answer:

$\frac{1}{24}$

Explanation:

Multiply both sides by the least common denominator . This will eliminate the fractions on both sides of the equation.

$6x(\frac{2}{3x}+4 )= \frac{5}{6x}\cdot 6x$

4+24x = 5

Subtract 4 on both sides.

4+24x -4= 5-4

24x=1

Divide by 24 on both sides.

$\frac{24x}{24}=\frac{1}{24}$

The answer is: $\frac{1}{24}$

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Example Question #92 : Single Variable Algebra

$\textup{A cubic polynomial}$ p(x) $\textup{has lead term is}$ $x^{3}$ $\textup{and rational coefficients}$

p(3) = p(5) = 0

$\textup{Which of the following is this polynomial?}$

Possible Answers:

$p(x)= x^{3} - 5x^{2}- 9x+45$

$\textup{Insufficient information exists to determine the polynomial}$

$p(x)= x^{3} - 3x^{2}- 25 x+75$

$p(x)= x^{3} - 13x^{2}+55 x-75$

$p(x)= x^{3} - 11x^{2}+39 x-45$

Correct answer:

$\textup{Insufficient information exists to determine the polynomial}$

Explanation:

A cubic polynomial has three zeroes, if a zero of multiplicity is counted times. Since its lead term is $x^{3}$ , we know that

$p(x) = (x-b_{1}) (x-b_{2}) (x-b_{3})$ ,

where $b_{1}$ , $b_{2}$ , and $b_{3}$ are its zeroes.

p(3) = p(5) = 0 , so 3 and 5 are zeroes of p(x) , and, by the Factor Theorem, x-3 and x-5 are among the linear factors - that is,

$p(x) = (x- 3) (x-5) (x-b_{3})$

for some $b_{3}$ .

However, we are not given the multiplicity of either 3 or 5, so we do not know which, if either, of the two are equal to $b_{3}$ . Without this information, we cannot determine p(x) for certain.

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SAT II Math I : SAT Subject Test in Math I

Study concepts, example questions & explanations for SAT II Math I

All SAT II Math I Resources

Example Questions

Example Question #83 : Single Variable Algebra

Example Question #241 : Sat Subject Test In Math I

Example Question #84 : Single Variable Algebra

Example Question #81 : Single Variable Algebra

Example Question #82 : Single Variable Algebra

Example Question #241 : Sat Subject Test In Math I

Example Question #87 : Single Variable Algebra

Example Question #91 : Single Variable Algebra

Example Question #91 : Single Variable Algebra

Example Question #92 : Single Variable Algebra

All SAT II Math I Resources

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