AP Precalculus Unit 2 Practice Test

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Question 1 |

### The population of bacteria in an experimental setup triples every 6 years. If the initial population of the bacteria is equal to 200, what would be its population after $t$ years?

$400(3)^{t/6}$ | |

$200(1.2)^{3t}$ | |

$200(3)^{6t}$ | |

$200(3)^{t/6}$ |

Question 1 Explanation:

The correct answer is (D). The population of the bacteria triples every 6 years

If $t$ represents the number of years and $p(t)$ represents the population then,

$p(t)=200(3)^{t/6}$

If $t$ represents the number of years and $p(t)$ represents the population then,

$p(t)=200(3)^{t/6}$

Question 2 |

### Which of the exponential functions below are identical?

### $A: 3^{6x-2}$

### $B: \dfrac{3^{6x}}{9}$

### $C: \dfrac{729^x}{3}$

$A$ and $C$ | |

$A$ and $C$ | |

$A,B$ and $C$ | |

$A$ and $B$ |

Question 2 Explanation:

The correct answer is (D).

Function $A→ 3^{6x−2}=3^{−2} \ .\ 3^{6x}=\dfrac{1}{9} 3^{6x}$

Function $B→ \dfrac{3^{6x}}{9}=\dfrac{1}{9} 3^{6x}$

Function $C→ \dfrac{729^x}{3}=\dfrac{1}{3} (3^6 )^x=\dfrac{1}{3} 3^{6x}$

Thus, function $A$ and $B$ are equal to each other

Function $A→ 3^{6x−2}=3^{−2} \ .\ 3^{6x}=\dfrac{1}{9} 3^{6x}$

Function $B→ \dfrac{3^{6x}}{9}=\dfrac{1}{9} 3^{6x}$

Function $C→ \dfrac{729^x}{3}=\dfrac{1}{3} (3^6 )^x=\dfrac{1}{3} 3^{6x}$

Thus, function $A$ and $B$ are equal to each other

Question 3 |

### What is the y-intercept and the range of the exponential function below?

### $g(x)=2.5^{x+2}-10$

$y\text{-intercept} =- 3.75$ $\text{and range} =(-10,∞)$ | |

$y\text{-intercept} =-10$ $\text{and range} =(-10,∞)$ | |

$y\text{-intercept} =6.25$ $\text{and range} =(0,∞)$ | |

$y\text{-intercept} =- 3.75$ $\text{and range} =(-3.75,∞)$ |

Question 3 Explanation:

The correct answer is (A).

$g(x)=2.5^{x+2}-10$

$\lim\limits_{x→+∞} g(x) $ $ = \lim\limits_{x→+∞} (2.5^{x+2} - 10) $ $ =∞$

And,

$\lim\limits_{x→+∞} g(x) $ $ = \lim\limits_{x→+∞} (2.5^{x+2} - 10) $ $ =0-10=-10$

Range $=(-10,∞)$

$y$-intercept $=g(0)=2.5^{0+2}-10 $ $ =6.25-10=-3.75$

$g(x)=2.5^{x+2}-10$

$\lim\limits_{x→+∞} g(x) $ $ = \lim\limits_{x→+∞} (2.5^{x+2} - 10) $ $ =∞$

And,

$\lim\limits_{x→+∞} g(x) $ $ = \lim\limits_{x→+∞} (2.5^{x+2} - 10) $ $ =0-10=-10$

Range $=(-10,∞)$

$y$-intercept $=g(0)=2.5^{0+2}-10 $ $ =6.25-10=-3.75$

Question 4 |

### If $\log_{10}A=5$ and $\log_{10}B=3$, then what is the value of

### $\log_{10}\sqrt{AB^3} \,$?

$7$ | |

$6$ | |

$2$ | |

$10$ |

Question 4 Explanation:

The correct answer is (A).

$\log_{10}\sqrt{AB^3}$

$=\dfrac{1}{2} \log_{10} \ AB^3$

$=\dfrac{1}{2} \log_{10} \ A+\dfrac{3}{2} \log_{10} \ B$

$=\dfrac{1}{2}.5+\dfrac{3}{2}.3$

$=7$

$\log_{10}\sqrt{AB^3}$

$=\dfrac{1}{2} \log_{10} \ AB^3$

$=\dfrac{1}{2} \log_{10} \ A+\dfrac{3}{2} \log_{10} \ B$

$=\dfrac{1}{2}.5+\dfrac{3}{2}.3$

$=7$

Question 5 |

### If $\log_4(x^2+4)=1.5$, then what is the value of $x$?

$-1,1$ | |

$-3,3$ | |

$-2,2$ | |

$-4,4$ |

Question 5 Explanation:

The correct answer is (C).

$\log_4(x^2+4)=1.5$

$4^{1.5}=x^2+4$

$x^2+4=8$

$x=±2$

$\log_4(x^2+4)=1.5$

$4^{1.5}=x^2+4$

$x^2+4=8$

$x=±2$

Question 6 |

### The per-square-foot rate of houses in a metro city increases by 10% every 6 months. The per-square-foot rate is \$100 in January 2020. What is the cost of a house of area 450 *sq. ft* in June 2021?

### You may use a calculator.

$\$57{,}825$ | |

$\$55{,}395$ | |

$\$59{,}895$ | |

$\$52{,}310$ |

Question 6 Explanation:

The correct answer is (C). Increase in the per square foot cost is 10% every 6 months.

If $t$ represents the number of months and $R(t)$ represents the rate then:

$R(t)=100(1+\frac{10}{100})^{t/6} $ $ =100(1.1)^{t/6}$

Number of months from January 2020 to June 2021 = 18.

$R(18)=100(1.1)^{18/6}=133.1$

Thus, the cost of house of area

$450 \; sq.ft=133.1×450=\$59{,}895$

If $t$ represents the number of months and $R(t)$ represents the rate then:

$R(t)=100(1+\frac{10}{100})^{t/6} $ $ =100(1.1)^{t/6}$

Number of months from January 2020 to June 2021 = 18.

$R(18)=100(1.1)^{18/6}=133.1$

Thus, the cost of house of area

$450 \; sq.ft=133.1×450=\$59{,}895$

Question 7 |

### Which option represents the function for the graph below?

$y(x)=20(2)^{-3x}$ | |

$y(x)=5(2)^{-(x-2)}$ | |

$y(x)=5(2)^{-x}$ | |

$y(x)=0.8(5)^{-(x-2)}$ |

Question 7 Explanation:

The correct answer is (B).

Option (A) $→y(x=2)=\dfrac{20}{64}≠5$ (Incorrect function)

Option (B) $→y(x=0)=20 \,$ and $\, y(x=2)=5$ (Correct function)

Option (C) $→y(x=0)=5≠20$ (Incorrect function)

Option (D) $→y(x=2)=0.8≠5$ (Incorrect function)

Option (A) $→y(x=2)=\dfrac{20}{64}≠5$ (Incorrect function)

Option (B) $→y(x=0)=20 \,$ and $\, y(x=2)=5$ (Correct function)

Option (C) $→y(x=0)=5≠20$ (Incorrect function)

Option (D) $→y(x=2)=0.8≠5$ (Incorrect function)

Question 8 |

### With the help of a calculator, work out the exponential fit for the data given below and thus find the value of $Y$ when $X=6$.

## $X$ |
## $0$ |
## $1$ |
## $2.2$ |
## $3.5$ |
## $4.6$ |
## $5$ |

## $Y$ |
## $5.9$ |
## $12.9$ |
## $45.1$ |
## $166.6$ |
## $495.4$ |
## $744.0$ |

### You may use a calculator.

$2000.2$ | |

$2032.6$ | |

$2009.5$ | |

$2062.4$ |

Question 8 Explanation:

The correct answer is (C). Input the $X$ values as List1(L1) and $Y$ values as List2(L2)

Use the Exponential Regression module on calculator to find the equation,

$Y=5.0018(2.7164)^X$

For $X=6$,

$Y=5.0018(2.7164)^6=2009.5$

Use the Exponential Regression module on calculator to find the equation,

$Y=5.0018(2.7164)^X$

For $X=6$,

$Y=5.0018(2.7164)^6=2009.5$

Question 9 |

### A purely exponentially decaying graph is translated by $4$ units horizontally to the right and $2$ units vertically downwards.

### Which graph represents the function after the transformation?

Question 9 Explanation:

The correct answer is (D).

Option (A) → Horizontal asymptote should be $y=-2$ (Incorrect graph)

Option (B) → Horizontal asymptote should be $y=-2$ (Incorrect graph)

Option (C) → The graph does not represent a decay graph (Incorrect graph)

Option (D) → Best representation of the graph (Correct graph)

Option (A) → Horizontal asymptote should be $y=-2$ (Incorrect graph)

Option (B) → Horizontal asymptote should be $y=-2$ (Incorrect graph)

Option (C) → The graph does not represent a decay graph (Incorrect graph)

Option (D) → Best representation of the graph (Correct graph)

Question 10 |

### Without using a calculator, find the value of $\dfrac{\log_{10}\ 0.001}{\log_{2}\ 0.25}$.

$3$ | |

$1.5$ | |

$2.5$ | |

$1$ |

Question 10 Explanation:

The correct answer is (B).

$\dfrac{\log_{10}\ 0.001}{\log_{2}\ 0.25}$

$=\dfrac{\log_{10}\ 10^{-3}}{\log_{2}\ 2^{-2}}$

$=\dfrac{-3}{-2}$

$=1.5$

$\dfrac{\log_{10}\ 0.001}{\log_{2}\ 0.25}$

$=\dfrac{\log_{10}\ 10^{-3}}{\log_{2}\ 2^{-2}}$

$=\dfrac{-3}{-2}$

$=1.5$

Question 11 |

### Two functions, $h(x)$ and $f(x)$, are defined as $\log_{e}x$ and $x(x+e)$, respectively. What is the value of $h(f(e^2 ))$?

$3+\log_e(e+1)$ | |

$2-\log_e(e-1)$ | |

$3-\log_e(e+1)$ | |

$3+\log_e(1-e)$ |

Question 11 Explanation:

The correct answer is (A).

$h(f(x))$

$=h(x(x+e))$

$=\log_e \ x(x+e)$

Thus, for $x=e^2$

$h(f(e^2 ))$

$=\log_e \ e^2 (e^2+e)$

$=\log_e \ e^3 (e+1)$

$=3\ \log_e \ e+\log_e(e+1) $

$=3+\log_e(e+1)$

$h(f(x))$

$=h(x(x+e))$

$=\log_e \ x(x+e)$

Thus, for $x=e^2$

$h(f(e^2 ))$

$=\log_e \ e^2 (e^2+e)$

$=\log_e \ e^3 (e+1)$

$=3\ \log_e \ e+\log_e(e+1) $

$=3+\log_e(e+1)$

Question 12 |

### Andy works out the inverse of the function $f(x)=2^{2x+1}$. What would the graph of $f^{-1} (x)$ look like when he plots it?

Question 12 Explanation:

The correct answer is (A).

$y=f(x)=2^{2x+1}$

$\log_2 \ y=2x+1$

$x=\dfrac{1}{2} (\log_2 \ y-1)$

In terms of $x$,

$f^{-1} (x)=\dfrac{1}{2} (\log_2 \ x-1)$

Option (A) → Best fit (Correct Graph)

Option (B) → $f^{-1} (2)≠0$ (Incorrect Graph)

Option (C) → $f^{-1} (2)≠0$ (Incorrect Graph)

Option (D) → $f^{-1} (8)≠1$ (Incorrect Graph)

$y=f(x)=2^{2x+1}$

$\log_2 \ y=2x+1$

$x=\dfrac{1}{2} (\log_2 \ y-1)$

In terms of $x$,

$f^{-1} (x)=\dfrac{1}{2} (\log_2 \ x-1)$

Option (A) → Best fit (Correct Graph)

Option (B) → $f^{-1} (2)≠0$ (Incorrect Graph)

Option (C) → $f^{-1} (2)≠0$ (Incorrect Graph)

Option (D) → $f^{-1} (8)≠1$ (Incorrect Graph)

Question 13 |

### Work out the domain of the function $\log_x(x^2-5)$.

$(-\sqrt{5},∞)$ | |

$(\sqrt{5},∞)$ | |

$(-\sqrt{5},\sqrt{5})$ | |

$(6,∞)$ |

Question 13 Explanation:

The correct answer is (B).

$\log_b \ a$ is defined for $b>0, b≠1$ and $a>0$

Thus, for $\log_x(x^2-5)$

$x>0$ and $x≠1$

While, $x^2-5>0→x>\sqrt{5}$

$(\text{rejecting} \; x<-\sqrt{5} \, \text{ as}$ $ \ x>0)$

Thus, $x∈(\sqrt{5},∞) $

$\log_b \ a$ is defined for $b>0, b≠1$ and $a>0$

Thus, for $\log_x(x^2-5)$

$x>0$ and $x≠1$

While, $x^2-5>0→x>\sqrt{5}$

$(\text{rejecting} \; x<-\sqrt{5} \, \text{ as}$ $ \ x>0)$

Thus, $x∈(\sqrt{5},∞) $

Question 14 |

### Determine the inverse of the function $f(x)=\log_6(x+8)$.

$8.6^x$ | |

$\frac{1}{8} 6^x$ | |

$6^x+8$ | |

$6^x-8$ |

Question 14 Explanation:

The correct answer is (D).

$y=f(x)=\log_6(x+8)$

$x+8=6^y$

$x=6^y-8$

In terms of $x$,

$y=f^{-1} (x)=6^x-8$

$y=f(x)=\log_6(x+8)$

$x+8=6^y$

$x=6^y-8$

In terms of $x$,

$y=f^{-1} (x)=6^x-8$

Question 15 |

### The table below shows the function $h(x)$.

## $X$ |
## $1$ |
## $2$ |
## $4$ |
## $6$ |
## $9$ |

## h(x) |
## $5$ |
## $10$ |
## $20$ |
## $100$ |
## $200$ |

### Another function $g(x)$ is equal to $\log_{10}(x)+x^2$. What is the value of $g(h (2))$?

$5$ | |

$10$ | |

$26$ | |

$101$ |

Question 15 Explanation:

The correct answer is (D).

From the table, $h(2)=10$

Thus, $g(h(2))=g(10)$

$g(10)=\log_{10}(10)+10^2 $ $ =1+100=101$

From the table, $h(2)=10$

Thus, $g(h(2))=g(10)$

$g(10)=\log_{10}(10)+10^2 $ $ =1+100=101$

Question 16 |

### The graph below shows the first $5$ terms of an Arithmetic Progression.

### What would be the n^{th} term of a Geometric Progression with the same first term $a$ and with a common ratio $r$ equal to the common difference $d$ of the Arithmetic Progression?

### (Note: $n$ takes on integer values $1,2,3…$ in the formulae below)

$10(3)^{n-1}$ | |

$3(3)^n$ | |

$10(3)^{n+1}$ | |

$30(3)^{n-1}$ |

Question 16 Explanation:

The correct answer is (A).

From the graph, sequence is $10,13,16,19,22,…$

First term $=a=10$

Common ratio $r=$ Common difference $d=13-10=3$

Thus, n

From the graph, sequence is $10,13,16,19,22,…$

First term $=a=10$

Common ratio $r=$ Common difference $d=13-10=3$

Thus, n

^{th}term $=a(r)^{n-1}=10(3)^{n-1}$Question 17 |

### The two arithmetic progressions below have their 5^{th} terms equal.

### $S_1: T_n=10-4(n-1)$

$S_2:T_n=p+2(n-1)$

### What is the sum of the first 5 terms for sequence $S_2$?

$54$ | |

$57$ | |

$50$ | |

$44$ |

Question 17 Explanation:

The correct answer is (C).

The 5

$T_5=10-4(5-1) $ $ =10-16=-6$

For sequence $S_2$,

$T_5=-6=p+8$

Solving for $p$ gives,

$p=-14$

The n

Here, $a=-14$ and $d=2$

Sum of first 5 terms is given by,

$\dfrac{n}{2} [2a+(n-1)d]$

$=\dfrac{5}{2} [-28+4×2]$

$=-5×10 $

$=-50$

The 5

^{th}term of sequence $S_1$ is given by,$T_5=10-4(5-1) $ $ =10-16=-6$

For sequence $S_2$,

$T_5=-6=p+8$

Solving for $p$ gives,

$p=-14$

The n

^{th}term of sequence $S_2 \,$ is: $\, T_n=-14+2(n-1)$Here, $a=-14$ and $d=2$

Sum of first 5 terms is given by,

$\dfrac{n}{2} [2a+(n-1)d]$

$=\dfrac{5}{2} [-28+4×2]$

$=-5×10 $

$=-50$

Question 18 |

### Look at the sequence below.

### $\log_{10}x, \log_{10}x^3 , \log_{10}x^5 , … $

### $S_n (x)$ is defined as the sum of the first $n$ terms of the sequence. Work out the formula for $S_n (10)$.

$n^2$ | |

$n^2-n$ | |

$n^2+n$ | |

$2n^2+1$ |

Question 18 Explanation:

The correct answer is (A). For the sequence given,

First term $=a=\log_{10}x$

Difference of first two terms

$ =d_1=\log_{10}x^3 -\log_{10}x $ $ =2 \ \log_{10}x $

Difference of next two terms

$=d_2=\log_{10}x^5 -\log_{10}x^3 $ $ =2 \ \log_{10}x $

Thus, the sequence is an AP (Arithmetic Progression)

Sum $S_n (x)$

$=\dfrac{n}{2} [2a+(n-1)d]$

$=\dfrac{n}{2}[2 \ \log_{10}x+2(n-1) $ $ \log_{10}x] $

$=n\log_{10}x [1+(n-1)]$

$=n^2\log_{10}x $

$S_n (10)=n^2\log_{10}10=n^2 $

First term $=a=\log_{10}x$

Difference of first two terms

$ =d_1=\log_{10}x^3 -\log_{10}x $ $ =2 \ \log_{10}x $

Difference of next two terms

$=d_2=\log_{10}x^5 -\log_{10}x^3 $ $ =2 \ \log_{10}x $

Thus, the sequence is an AP (Arithmetic Progression)

Sum $S_n (x)$

$=\dfrac{n}{2} [2a+(n-1)d]$

$=\dfrac{n}{2}[2 \ \log_{10}x+2(n-1) $ $ \log_{10}x] $

$=n\log_{10}x [1+(n-1)]$

$=n^2\log_{10}x $

$S_n (10)=n^2\log_{10}10=n^2 $

Question 19 |

### The first 5 terms of a sequence are shown as follows.

### When a smooth line is drawn joining these points, which function would it be proportional to?

$\log_2 \ x$ | |

$2^x $ | |

$x^2$ | |

None of these |

Question 19 Explanation:

The correct answer is (B). The value of the term doubles at every integer.

So, the function must be proportional to $2^x$.

So, the function must be proportional to $2^x$.

Question 20 |

### $ka^x$ and $mb^x$ are two exponential functions such that $1 < a < b < 10$ and $0 < m < k < 1 $. Find the value of $x$ for which $ka^x$ is equal to $mb^x$.

$x=\dfrac{\log_{10}a/b}{\log_{10}m/k}$ | |

$x=\dfrac{2\log_{10}m/k }{\log_{10}a/b}$ | |

$x=\dfrac{\log_{10}(m-k)}{\log_{10}(a-b)} $ | |

$x=\dfrac{\log_{10}m/k}{ \log_{10}a/b}$ |

Question 20 Explanation:

The correct answer is (D).

$ka^x= mb^x$

$\left(\dfrac{a}{b}\right)^x=\dfrac{m}{k} $

Thus,

$x\log_{10}\dfrac{a}{b}=\log_{10}\dfrac{m}{k}$

$x=\dfrac{\log_{10}m/k}{ \log_{10}a/b} $

$ka^x= mb^x$

$\left(\dfrac{a}{b}\right)^x=\dfrac{m}{k} $

Thus,

$x\log_{10}\dfrac{a}{b}=\log_{10}\dfrac{m}{k}$

$x=\dfrac{\log_{10}m/k}{ \log_{10}a/b} $

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