AP Precalculus Unit 1 Practice Test

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Question 1 |

### Which of the options below shows the domain of the function $f(x)$?

### $f(x)=\dfrac{1}{\sqrt{x-\sqrt{2}}}$

$[\sqrt{2},∞)$ | |

$(\sqrt{2},∞)$ | |

$(-\sqrt{2},∞)$ | |

$[\sqrt{2},∞]$ |

Question 1 Explanation:

The correct answer is (B). To find the domain of the function $f(x)$ work out the values of x for which $f(x)$ is well defined:

$f(x)=\dfrac{1}{\sqrt{x-\sqrt{2}}}$

$x-\sqrt{2}>0 $

$x>\sqrt{2} $

In interval notation, $x ∈(\sqrt{2},∞)$

$f(x)=\dfrac{1}{\sqrt{x-\sqrt{2}}}$

$x-\sqrt{2}>0 $

$x>\sqrt{2} $

In interval notation, $x ∈(\sqrt{2},∞)$

Question 2 |

### Find the average rate of change of the function $g(x)$ from $x=2$ to $x=4$.

### $g(x)=\dfrac{x}{x^2 - 1}$

$-\dfrac{1}{2}$ | |

$-\dfrac{3}{5}$ | |

$-\dfrac{1}{3}$ | |

$-\dfrac{1}{5}$ |

Question 2 Explanation:

The correct answer is (D). Average rate of change

$= \dfrac{g(4) - g(2)}{4 - 2}$

$= \left(\dfrac{4}{15} - \dfrac{2}{3}\right)×\dfrac{1}{2}$

$= -\dfrac{6}{15} × \dfrac{1}{2}$

$= -\dfrac{1}{5}$

$= \dfrac{g(4) - g(2)}{4 - 2}$

$= \left(\dfrac{4}{15} - \dfrac{2}{3}\right)×\dfrac{1}{2}$

$= -\dfrac{6}{15} × \dfrac{1}{2}$

$= -\dfrac{1}{5}$

Question 3 |

### Find all the roots of the polynomial function $p(x)=x^3-3x^2+4$.

$1,2,-1$ | |

$-2,-1,0$ | |

$-2,1,1$ | |

$-1,2,2$ |

Question 3 Explanation:

The correct answer is (D).

Method I: Check each option by substitution

Method II: Factorize

$x^3-3x^2+4$

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

$=x^2 (x+1)-4(x^2-1)$

$=x^2 (x+1)-4(x+1)(x-1)$

$=(x+1)(x^2-4x+4)$

$=(x+1) (x-2)^2$

$=0$

$→x=-1\ \,$ or $\, \ x=2,2 $

Method I: Check each option by substitution

Method II: Factorize

$x^3-3x^2+4$

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

$=x^2 (x+1)-4(x^2-1)$

$=x^2 (x+1)-4(x+1)(x-1)$

$=(x+1)(x^2-4x+4)$

$=(x+1) (x-2)^2$

$=0$

$→x=-1\ \,$ or $\, \ x=2,2 $

Question 4 |

### A function f(x), when plotted on a graph paper, looks as shown below.

### What is $f^{-1} (20)$ equal to?

$15$ | |

$10$ | |

$7.5$ | |

$12.5$ |

Question 4 Explanation:

The correct answer is (B).

$f^{-1} (y)=x→y=f(x)=20$

From the graph, $f(x)$ is equal to 20 when $x$ is 10.

$f^{-1} (y)=x→y=f(x)=20$

From the graph, $f(x)$ is equal to 20 when $x$ is 10.

Question 5 |

### If $h(x)=x^2-5$, then what is the value of $h(h(x))$?

$x^4-20x^2+20$ | |

$x^4-10x^2+25$ | |

$x^4-10x^2+20$ | |

$2x^4-10x^2+20$ |

Question 5 Explanation:

The correct answer is (C).

$h(x)=x^2-5$

Thus,

$h(h(x))$

$=h(x^2-5)$

$=(x^2-5)^2-5$

$=x^4-10x^2+25-5$

$=x^4-10x^2+20$

$h(x)=x^2-5$

Thus,

$h(h(x))$

$=h(x^2-5)$

$=(x^2-5)^2-5$

$=x^4-10x^2+25-5$

$=x^4-10x^2+20$

Question 6 |

### The function $h(x)=x^2$ is moved horizontally by $3$ units to the right, then stretched vertically by a factor of $2$, and then moved vertically up by $6$ units. Which of the below equations shows the function after the transformations?

$h'(x)=2(x-3)^2+12$ | |

$h'(x)=(2x-3)^2+6$ | |

$h'(x)=2(x+6)^2-3$ | |

$h'(x)=2(x-3)^2+6$ |

Question 6 Explanation:

The correct answer is (D).

$h(x)=x^2$

After horizontal translation of 3 units (right):

$(x-3)^2$

After vertical stretch by 2 units:

$2(x-3)^2$

After vertical translation of 6 units (up):

$h'(x)=2(x-3)^2+6$

$h(x)=x^2$

After horizontal translation of 3 units (right):

$(x-3)^2$

After vertical stretch by 2 units:

$2(x-3)^2$

After vertical translation of 6 units (up):

$h'(x)=2(x-3)^2+6$

Question 7 |

### With the help of a calculator, work out the slope of the line of best fit for the points in the table below:

$X$ | $0$ | $2.5$ | $4$ | $5.8$ | $6.8$ | $8$ |

$Y$ | $11$ | $18$ | $24$ | $33$ | $39$ | $43$ |

### You may use a calculator.

$Y=4.18X+9.119$ | |

$Y=4.00X+10.201$ | |

$Y=3.24X+11.212$ | |

$Y=5.12X+9.021$ |

Question 7 Explanation:

Input the $X$ values as List1(L1) and $Y$ values as List2(L2) into the calculator.

Use the Linear Regression module on the calculator to find the line of best fit.

The equation of the line is,

$Y=4.18X+9.119 $

Use the Linear Regression module on the calculator to find the line of best fit.

The equation of the line is,

$Y=4.18X+9.119 $

Question 8 |

### Which graph shows a one-to-one function?

Question 8 Explanation:

The correct answer is (C). To find whether a function is one-to-one, draw a horizontal line and check how many times it cuts the graph:

Option (A) → More than 1 time (Not one-to-one).

Option (B) → More than 1 time (Not one-to-one).

Option (C) → Exactly 1 time (one-to-one).

Option (D) → The graph does not represent a function.

Option (A) → More than 1 time (Not one-to-one).

Option (B) → More than 1 time (Not one-to-one).

Option (C) → Exactly 1 time (one-to-one).

Option (D) → The graph does not represent a function.

Question 9 |

### $f(x)=2x+7$ and $g(x)=ax-b$. What is the value of $a+b$ if $f(x)$ and $g(x)$ are inverses of each other?

$3$ | |

$4$ | |

$2$ | |

$1$ |

Question 9 Explanation:

The correct answer is (B). Since $f(x)$ and $g(x)$ are inverses of each other,

$f(g(x))=x$

Left hand side of the above equation

$=f(ax-b)$

$=2(ax-b)+7$

$=2ax+(-2b+7)$

$=x$

Thus,

$2a=1→a=\dfrac{1}{2}$

And, $2b-7=0→b=\dfrac{7}{2}$

$a+b=\dfrac{1}{2}+\dfrac{7}{2}=4$

$f(g(x))=x$

Left hand side of the above equation

$=f(ax-b)$

$=2(ax-b)+7$

$=2ax+(-2b+7)$

$=x$

Thus,

$2a=1→a=\dfrac{1}{2}$

And, $2b-7=0→b=\dfrac{7}{2}$

$a+b=\dfrac{1}{2}+\dfrac{7}{2}=4$

Question 10 |

### Identify the polynomial from the graph shown below.

$(x-1)^2 (x-2)(x-4)$ | |

$(x-1)(x-2) (x-4)^2$ | |

$(x-1) (x-2)^2 (x-4)^2$ | |

$(x-1)^2 (x-2)^2 (x-4)$ |

Question 10 Explanation:

The correct answer is (B).

Thus, the polynomial must be of the form:

$(x-1)(x-2) (x-4)^2$

$x=1$ | $x=2$ | $x=4$ |

Crosses $x-axis$ | Crosses $x-axis$ | Tangent to $x-axis$ |

Multiplicity should be odd $(1,3,5…)$ | Multiplicity should be odd $(1,3,5…)$ | Multiplicity should be even $(0,2,4,..)$ |

$(x-1)(x-2) (x-4)^2$

Question 11 |

### The population of lions in a sanctuary is equal to 2,060 in 2002. In 2007, the population of lions increased to 2,880. When will the population of lions cross 4,000?

(Assume a linear model for the question)

### You may use a calculator.

$2019$ | |

$2018$ | |

$2013$ | |

$2014$ |

Question 11 Explanation:

The correct answer is (D).

$\text{Slope} = \dfrac{2{,}880 − 2{,}060}{2007 − 2002} $ $ =164$

Thus, the population $p(t)=164t+2{,}060$ (where $t$ is the number of years after 2002)

To find the year in which $p(t)$ crosses 4,000, use: $164t+2,060=4,000$

$t=\dfrac{4{,}000-2{,}060}{164} $ $ = 11.82 ≈12 \text{ years}$

Thus, the population of lions cross 4,000 by: $2002+12=2014$

$\text{Slope} = \dfrac{2{,}880 − 2{,}060}{2007 − 2002} $ $ =164$

Thus, the population $p(t)=164t+2{,}060$ (where $t$ is the number of years after 2002)

To find the year in which $p(t)$ crosses 4,000, use: $164t+2,060=4,000$

$t=\dfrac{4{,}000-2{,}060}{164} $ $ = 11.82 ≈12 \text{ years}$

Thus, the population of lions cross 4,000 by: $2002+12=2014$

Question 12 |

### Find the values of $x$ for which the inequality $6x^2-5x < x^3$ is true.

$(0,1) \cup (5,∞)$ | |

$\left(0,\frac{1}{2}\right)∪ [5,∞)$ | |

$(5,∞)$ | |

$(0,1)$ |

Question 12 Explanation:

The correct answer is (A).

$x^3>6x^2-5x$

$x^3-6x^2+5x>0$

$x(x^2-6x+5)>0$

$x(x-5)(x-1)>0$

Thus, $x>5 \,$ OR $\, 0< x <1$

$→x∈(0,1)∪(5,∞)$

$x^3>6x^2-5x$

$x^3-6x^2+5x>0$

$x(x^2-6x+5)>0$

$x(x-5)(x-1)>0$

Thus, $x>5 \,$ OR $\, 0< x <1$

$→x∈(0,1)∪(5,∞)$

Question 13 |

### A polynomial has degree 10 with the leading coefficient $(a_{10})>0$. Which of the below statements is FALSE regarding the polynomial?

There are 9 local extrema of the polynomial at max | |

There are 10 roots of the polynomial at max | |

The polynomial goes to +∞ as $x$→±∞ | |

The polynomial must have 5 roots each with a multiplicity of 2 |

Question 13 Explanation:

The correct answer is (D).

Option (A) → A polynomial with degree 10 has 10 − 1 = 9 local extrema at max. (True)

Option (B) → A polynomial with degree 10 has 10 roots at max. (True)

Option (C) → Since the leading coefficient $a_{10}$ > 0 the polynomial increases as $x$→±∞. (True)

Option (D) → It may or may not have 5 roots of multiplicity 2. (False)

Option (A) → A polynomial with degree 10 has 10 − 1 = 9 local extrema at max. (True)

Option (B) → A polynomial with degree 10 has 10 roots at max. (True)

Option (C) → Since the leading coefficient $a_{10}$ > 0 the polynomial increases as $x$→±∞. (True)

Option (D) → It may or may not have 5 roots of multiplicity 2. (False)

Question 14 |

### A polynomial $p(x)$ of degree 4 has real coefficients. If $p(x=2)=-6 \,$ and $\, p(x=6)=1$, then which of the statements below is correct?

There is at least 1 root in the interval [−6, 1] | |

There are 3 roots in the interval [−6, 1] | |

There are 1 or 3 roots in the interval [2, 6] | |

There are exactly 3 roots in the interval [2, 6] |

Question 14 Explanation:

The correct answer is (C).

[−6,1] represents the output interval of the polynomial. Thus, option (A) and (B) are ruled out.

Since, $p(x=2)=-6$ and $p(x=6)=1$

Thus, the polynomial must cross 0 at least once in the interval.

The degree of the polynomial is 4, so it can cross the $x$-axis a maximum of 3 times and a minimum of 1 time within the interval (odd number of times because the end values are negative and positive).

Which implies option (C) is correct.

[−6,1] represents the output interval of the polynomial. Thus, option (A) and (B) are ruled out.

Since, $p(x=2)=-6$ and $p(x=6)=1$

Thus, the polynomial must cross 0 at least once in the interval.

The degree of the polynomial is 4, so it can cross the $x$-axis a maximum of 3 times and a minimum of 1 time within the interval (odd number of times because the end values are negative and positive).

Which implies option (C) is correct.

Question 15 |

### Which option shows the graph of the rational function $\dfrac{1}{x - 6}$?

Question 15 Explanation:

The correct answer is (B).

Option (A) → Vertical asymptote at $x$ = 6 is missing (Incorrect graph).

Option (B) → Best fit (correct graph).

Option (C) → Value of the function at $x$ = 8 is bigger than expected value of 0.5 (Incorrect graph).

Option (D) → For $x$ < 6 the graph should be negative (Incorrect graph).

Option (A) → Vertical asymptote at $x$ = 6 is missing (Incorrect graph).

Option (B) → Best fit (correct graph).

Option (C) → Value of the function at $x$ = 8 is bigger than expected value of 0.5 (Incorrect graph).

Option (D) → For $x$ < 6 the graph should be negative (Incorrect graph).

Question 16 |

### Which of the below rational functions have $y=2$ as an asymptote?

$y=\dfrac{x^2 - 3}{x + 10}$ | |

$y=\dfrac{x - 3}{x + 10}$ | |

$y=\dfrac{2x^2 - 3}{x^2 + 10}$ | |

$y=\dfrac{x^3 - 3}{x + 100}$ |

Question 16 Explanation:

The correct answer is (C).

When $\dfrac{P(x)}{Q(x)}$ are polynomials of equal degree there exists a horizontal asymptote of the form $y$=constant.

So, Options (A) and (D) are ruled out.

Option (B)

$ → y= \lim\limits_{x\to ∞} \dfrac{x - 3}{x + 10} $ $ = \lim\limits_{x\to ∞} \dfrac{1-3/x}{1+10/x}$

$=\dfrac{1-0}{1+0}=1$

Option (C)

$ → y= \lim\limits_{x\to ∞} \dfrac{2x^2 - 3}{x^2 + 10} $ $ = \lim\limits_{x\to ∞} \dfrac{2-3/x^2}{1+10/x^2}$

$=\dfrac{2-0}{1+0}=2$

When $\dfrac{P(x)}{Q(x)}$ are polynomials of equal degree there exists a horizontal asymptote of the form $y$=constant.

So, Options (A) and (D) are ruled out.

Option (B)

$ → y= \lim\limits_{x\to ∞} \dfrac{x - 3}{x + 10} $ $ = \lim\limits_{x\to ∞} \dfrac{1-3/x}{1+10/x}$

$=\dfrac{1-0}{1+0}=1$

Option (C)

$ → y= \lim\limits_{x\to ∞} \dfrac{2x^2 - 3}{x^2 + 10} $ $ = \lim\limits_{x\to ∞} \dfrac{2-3/x^2}{1+10/x^2}$

$=\dfrac{2-0}{1+0}=2$

Question 17 |

### Work out the vertical asymptotes for the below function:

$\dfrac{x - 3}{x^2 - 5x - 14}$

$x=−2 \; \; \& \; -7$ | |

$x=−2 \; \; \& \;\; 7$ | |

$x=−2 \; \; \& \; \; 3$ | |

$x=3 \; \; \& \; \; 5$ |

Question 17 Explanation:

The correct answer is (B).

$\dfrac{x - 3}{x^2 − 5x − 14} $ $ =\dfrac{x − 3}{(x − 7)(x + 2)} $ $ =\dfrac{P(x)}{Q(x)}$

For $x=-2 \,$ and $\, x=7$

$P(x)≠0 \,$ and $\, Q(x)=0$

Thus, the vertical asymptotes are: $x =−2 \,$ and $\, 7$

$\dfrac{x - 3}{x^2 − 5x − 14} $ $ =\dfrac{x − 3}{(x − 7)(x + 2)} $ $ =\dfrac{P(x)}{Q(x)}$

For $x=-2 \,$ and $\, x=7$

$P(x)≠0 \,$ and $\, Q(x)=0$

Thus, the vertical asymptotes are: $x =−2 \,$ and $\, 7$

Question 18 |

### Which line describes the end behavior asymptote for the function $g(x)=\dfrac{x^4 - 16}{x^3 - 8}$?

$y=x$ | |

$y=2x$ | |

$y=x+1$ | |

$y=x-2$ |

Question 18 Explanation:

The correct answer is (A). Divide $x^4-16$ by $x^3-8$, using long division:

$x^4-16 $ $ =x(x^3-8)+(8x-16) $

Thus, $\dfrac{x^4 - 16}{x^3 - 8} $ $ =x+\dfrac{8x - 16}{x^3 - 8}$

Hence, end behavior asymptote for the function is $y=x$.

$x^4-16 $ $ =x(x^3-8)+(8x-16) $

Thus, $\dfrac{x^4 - 16}{x^3 - 8} $ $ =x+\dfrac{8x - 16}{x^3 - 8}$

Hence, end behavior asymptote for the function is $y=x$.

Question 19 |

### Work out the intervals on which $\dfrac{1}{x}<\dfrac{2}{2x -9}$.

$(-∞,1)∪(4,∞)$ | |

$(0,\frac{9}{2})$ | |

$(-∞,0)∪(\frac{9}{2},∞)$ | |

$(-∞,0)∪(5,∞)$ |

Question 19 Explanation:

The correct answer is (C).

$\dfrac{2}{2x - 9}>\dfrac{1}{x} $

$\dfrac{2}{2x - 9}-\dfrac{1}{x}>0 $

$\dfrac{2x}{x(2x -9)}-\dfrac{2x-9}{x(2x - 9)} >0 $

$\dfrac{9}{x(2x - 9)}>0$

Thus, $x>\dfrac{9}{2} \,$ OR $\, x<0$

Which implies $x∈(-∞,0)∪(\frac{9}{2},∞)$

$\dfrac{2}{2x - 9}>\dfrac{1}{x} $

$\dfrac{2}{2x - 9}-\dfrac{1}{x}>0 $

$\dfrac{2x}{x(2x -9)}-\dfrac{2x-9}{x(2x - 9)} >0 $

$\dfrac{9}{x(2x - 9)}>0$

Thus, $x>\dfrac{9}{2} \,$ OR $\, x<0$

Which implies $x∈(-∞,0)∪(\frac{9}{2},∞)$

Question 20 |

### What should be the function equation that describes the graph shown below?

$ f(x) =
\begin{cases}
4x-2, &-2≤x≤1 \\
-3x+5,& \; \; \; 1≤x≤5
\end{cases}$ | |

$ f(x) =
\begin{cases}
4x-2, & \; \; \; -2≤x≤1 \\
2x-1,& \; \; \quad 1≤x≤5
\end{cases}$ | |

$ f(x) =
\begin{cases}
-3x+5, &-2≤x≤1 \\
4x+2,& \; \; \; 1≤x≤5
\end{cases}$ | |

$ f(x) =
\begin{cases}
4x-2, & \; \; \; -2≤x≤1 \\
3x+5,& \; \; \quad 1≤x≤5
\end{cases}$ |

Question 20 Explanation:

The correct answer is (A).

In the region $-2≤x≤1$,

Slope = $4$ and y-intercept = $- 2$

Equation is $f(x)=4x-2$

In the region $1≤x≤5$,

Slope = $- 3$ and $(1, 2)$ lies on the graph

Equation is $f(x)=-3x+5$

Thus, the function is,

$ f(x) = \begin{cases} 4x-2, &-2≤x≤1 \\ -3x+5,& \; \; \; 1≤x≤5 \end{cases}$

In the region $-2≤x≤1$,

Slope = $4$ and y-intercept = $- 2$

Equation is $f(x)=4x-2$

In the region $1≤x≤5$,

Slope = $- 3$ and $(1, 2)$ lies on the graph

Equation is $f(x)=-3x+5$

Thus, the function is,

$ f(x) = \begin{cases} 4x-2, &-2≤x≤1 \\ -3x+5,& \; \; \; 1≤x≤5 \end{cases}$

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