AP Precalculus -3.9 Inverse Trigonometric Functions- FRQ Exam Style Questions - Effective Fall 2023

AP Precalculus -3.9 Inverse Trigonometric Functions- FRQ Exam Style Questions – Effective Fall 2023

AP Precalculus -3.9 Inverse Trigonometric Functions- FRQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.

AP Precalculus – FRQ Exam Style Questions- All Topics

Question

(A) The functions $g$ and $h$ are given by $g(x)=\log_4(2x)$ and $h(x)=\frac{(e^x)^5}{e^{1/4}}$.
(i) Solve $g(x)=3$ for values of $x$ in the domain of $g$.
(ii) Solve $h(x)=e^{1/2}$ for values of $x$ in the domain of $h$.

(B) The functions $j$ and $k$ are given by $j(x)=\log_{10}(x+1)-5 \log_{10}(2-x)+\log_{10}3$ and $k(x)=\sec x-\cos x$.
(i) Rewrite $j(x)$ as a single logarithm base 10 without negative exponents in any part of the expression. Your result should be of the form $\log_{10}(\text{expression})$.
(ii) Rewrite $k(x)$ as a product involving $\tan x$ and $\sin x$ and no other trigonometric functions.

(C) The function $m$ is given by $m(x)=2 \tan^{-1}(\sqrt{3}\pi x)$.
Find all input values in the domain of $m$ that yield an output value of $\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)$.

Most-appropriate topic codes (AP Precalculus CED):

• 2.13: Exponential and Logarithmic Equations and Inequalities – part A
• 2.12: Logarithmic Function Manipulation – part B(i)
• 3.12: Equivalent Representations of Trigonometric Functions – part B(ii)
• 3.9: Inverse Trigonometric Functions – part C

▶️ Answer/Explanation

A (i)
$g(x)=3 \implies \log_4(2x)=3$.
Convert from logarithmic to exponential form: $4^3=2x$.
$64=2x \implies x=\frac{64}{2}=32$.
✅ Answer: $\boxed{x=32}$

A (ii)
$h(x)=e^{1/2} \implies \frac{e^{5x}}{e^{1/4}}=e^{1/2}$.
Using exponent rules: $e^{5x – 1/4} = e^{1/2}$.
Equate the exponents: $5x – \frac{1}{4} = \frac{1}{2}$.
$5x = \frac{1}{2} + \frac{1}{4} = \frac{3}{4} \implies x=\frac{3}{20}$.
✅ Answer: $\boxed{x=\frac{3}{20}}$

B (i)
Apply logarithmic power and product/quotient properties:
$j(x)=\log_{10}(x+1) – \log_{10}(2-x)^5 + \log_{10}3$.
$j(x)=\log_{10}\left(\frac{3(x+1)}{(2-x)^5}\right)$.
✅ Answer: $\boxed{\log_{10}\left(\frac{3(x+1)}{(2-x)^5}\right)}$

B (ii)
Convert secant to cosine: $k(x)=\frac{1}{\cos x} – \cos x$.
Find a common denominator: $k(x)=\frac{1-\cos^2 x}{\cos x}$.
Use the Pythagorean identity: $k(x)=\frac{\sin^2 x}{\cos x}$.
Factor the fraction: $\left(\frac{\sin x}{\cos x}\right) \cdot \sin x = \tan x \sin x$.
✅ Answer: $\boxed{\tan x \sin x}$

C
First evaluate the target output: $\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{3}$.
Set up the equation: $2 \tan^{-1}(\sqrt{3}\pi x) = \frac{\pi}{3}$.
Divide by 2: $\tan^{-1}(\sqrt{3}\pi x) = \frac{\pi}{6}$.
Take the tangent of both sides: $\sqrt{3}\pi x = \tan\left(\frac{\pi}{6}\right)$.
Since $\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$, we get $\sqrt{3}\pi x = \frac{1}{\sqrt{3}}$.
Divide to isolate $x$: $x = \frac{1}{3\pi}$.
✅ Answer: $\boxed{x=\frac{1}{3\pi}}$

Question

(A) The functions $ g $ and $ h $ are given by
\[ g(x) = 2 \log_3 x \quad]
[\quad h(x) = 4 \cos^2 x. \]
(i) Solve $ g(x) = 4 $ for values of $ x $ in the domain of $ g $.
(ii) Solve $ h(x) = 3 $ for values of $ x $ in the interval $[0, \frac{\pi}{2})$.

(B) The functions $ j $ and $ k $ are given by
\[ j(x) = \log_2 x + 3 \log_2 2 \quad \text{and} \quad k(x) = \frac{6}{\tan x (\csc^2 x – 1)}. \]
(i) Rewrite $ j(x) $ as a single logarithm base 2 without negative exponents in any part of the expression. Your result should be of the form $ \log_2 (\text{expression}) $.
(ii) Rewrite $ k(x) $ as an expression in which $ \tan x $ appears exactly once and no other trigonometric functions are involved.

(C) The function $ m $ is given by $ m(x) = e^{2x} – e^x – 12 $. Find all input values in the domain of $ m $ that yield an output value of 0.

Most-appropriate topic codes (AP Precalculus CED):

• 2.11: Logarithmic Functions — part A(i), B(i)
• 2.13: Exponential and Logarithmic Equations and Inequalities — part A(i), C
• 3.9: Trigonometric Equations and Inequalities — part A(ii)
• 3.4: Trigonometric Identities — part B(ii)

▶️ Answer/Explanation

A. (i) Solving $ g(x) = 4 $:
Given $ g(x) = 2\log_3 x $. Set $ g(x) = 4 $: \[ 2\log_3 x = 4 \] Divide both sides by 2: \[ \log_3 x = 2 \] Convert to exponential form: \[ x = 3^2 = 9 \] Check domain: $ x > 0 $, so $ x = 9 $ is valid. ✅ Answer: $\boxed{x = 9}$

A. (ii) Solving $ h(x) = 3 $ on $ [0, \frac{\pi}{2}) $:
Given $ h(x) = 4\cos^2 x $. Set $ h(x) = 3 $: \[ 4\cos^2 x = 3 \] Divide both sides by 4: \[ \cos^2 x = \frac{3}{4} \] Take square root: \[ \cos x = \pm \frac{\sqrt{3}}{2} \] On $ [0, \frac{\pi}{2}) $, cosine is positive, so \[ \cos x = \frac{\sqrt{3}}{2} \implies x = \frac{\pi}{6} \] ✅ Answer: $\boxed{x = \frac{\pi}{6}}$

B. (i) Rewriting $ j(x) $ as a single logarithm:
Given $ j(x) = \log_2 x + 3\log_2 2 $.
Use power rule: $ 3\log_2 2 = \log_2 (2^3) = \log_2 8 $.
Use product rule: \[ j(x) = \log_2 x + \log_2 8 = \log_2 (8x) \] ✅ Answer: $\boxed{\log_2(8x)}$

B. (ii) Rewriting $ k(x) $ in terms of $ \tan x $ only:
Given $ k(x) = \dfrac{6}{\tan x (\csc^2 x – 1)} $.
Use Pythagorean identity: $ \csc^2 x – 1 = \cot^2 x $.
Thus, \[ k(x) = \frac{6}{\tan x \cdot \cot^2 x} \] Since $ \cot x = \frac{1}{\tan x} $, we have $ \cot^2 x = \frac{1}{\tan^2 x} $.
Substitute: \[ k(x) = \frac{6}{\tan x \cdot \frac{1}{\tan^2 x}} = \frac{6}{\frac{\tan x}{\tan^2 x}} = \frac{6}{\frac{1}{\tan x}} = 6 \tan x \] ✅ Answer: $\boxed{6 \tan x}$

C. Solving $ m(x) = 0 $:
Given $ m(x) = e^{2x} – e^x – 12 $.
Set $ m(x) = 0 $: \[ e^{2x} – e^x – 12 = 0 \] Let $ y = e^x $. Then $ e^{2x} = (e^x)^2 = y^2 $.
Substitute: \[ y^2 – y – 12 = 0 \] Factor: \[ (y – 4)(y + 3) = 0 \] Thus $ y = 4 $ or $ y = -3 $.
Since $ y = e^x > 0 $, discard $ y = -3 $.
So $ e^x = 4 $.
Take natural log: $ x = \ln 4 $.
✅ Answer: $\boxed{x = \ln 4}$

Question

(A) The functions $ g $ and $ h $ are given by:

\[ g(x) = 15 \arcsin x, \quad h(x) = \log_{10}(1 – x) – \log_{10} 4. \]

(i) Solve $ g(x) = 5\pi $ for values of $ x $ in the domain of $ g $.

(ii) Solve $ h(x) = 1 $ for values of $ x $ in the domain of $ h $.

(B) The functions $ j $ and $ k $ are given by:

\[ j(x) = \log_2(x + 4) – 11\log_2(x – 2) + \log_2(x^3), \] \[ k(x) = (\cot x)(\csc x). \]

(i) Rewrite $ j(x) $ as a single logarithm base 2 without negative exponents.

(ii) Rewrite $ k(x) $ as a fraction involving powers of $ \cos x $ and no other trigonometric functions.

(C) The function $ m $ is given by:

\[ m(x) = (2^x)^2 – 3 \cdot 2^x. \]

Find all input values in the domain of $ m $ that yield an output value of 18.

Most-appropriate topic codes (AP Precalculus CED):

• 3.9: Inverse Trigonometric Functions — part (A)(i)
• 2.13: Exponential and Logarithmic Equations and Inequalities — parts (A)(ii), (C)
• 2.12: Logarithmic Function Manipulation — part (B)(i)
• 3.12: Equivalent Representations of Trigonometric Functions — part (B)(ii)

▶️ Answer/Explanation

(A)(i)
Solve $ g(x) = 5\pi $.
$ 15 \arcsin x = 5\pi $
$ \arcsin x = \frac{5\pi}{15} = \frac{\pi}{3} $
$ x = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} $
Domain check: $ \arcsin x $ is defined for $ |x| \leq 1 $. $ \frac{\sqrt{3}}{2} \approx 0.866 $ is valid.
✅ Answer: $ \boxed{\frac{\sqrt{3}}{2}} $

(A)(ii)
Solve $ h(x) = 1 $.
$ \log_{10}(1 – x) – \log_{10}4 = 1 $
$ \log_{10}\left(\frac{1 – x}{4}\right) = 1 $
$ \frac{1 – x}{4} = 10^1 $
$ 1 – x = 40 $
$ x = -39 $
Domain check: $ 1 – x > 0 \Rightarrow x < 1 $. $ x = -39 $ is valid.
✅ Answer: $ \boxed{-39} $

(B)(i)
Rewrite $ j(x) $ as a single logarithm.
$ j(x) = \log_2(x + 4) – 11\log_2(x – 2) + \log_2(x^3) $
$ = \log_2(x + 4) – \log_2((x – 2)^{11}) + \log_2(x^3) $
$ = \log_2\left[ \frac{(x + 4) \cdot x^3}{(x – 2)^{11}} \right] $
✅ Answer: $ \boxed{\log_2\left( \frac{x^3(x + 4)}{(x – 2)^{11}} \right)} $

(B)(ii)
Rewrite $ k(x) $ using $ \cot x = \frac{\cos x}{\sin x} $ and $ \csc x = \frac{1}{\sin x} $.
$ k(x) = \left( \frac{\cos x}{\sin x} \right) \left( \frac{1}{\sin x} \right) = \frac{\cos x}{\sin^2 x} $
Using $ \sin^2 x = 1 – \cos^2 x $:
$ k(x) = \frac{\cos x}{1 – \cos^2 x} = \frac{\cos x}{(1 – \cos x)(1 + \cos x)} $
✅ Answer: $ \boxed{\frac{\cos x}{1 – \cos^2 x}} $ or $ \frac{\cos x}{(1 – \cos x)(1 + \cos x)} $

(C)
Solve $ m(x) = 18 $.
$ (2^x)^2 – 3 \cdot 2^x = 18 $
Let $ u = 2^x $, then $ u^2 – 3u – 18 = 0 $
$ (u – 6)(u + 3) = 0 $
$ u = 6 $ or $ u = -3 $
Since $ 2^x > 0 $ for all real $ x $, discard $ u = -3 $.
$ 2^x = 6 $
$ x = \log_2 6 $
✅ Answer: $ \boxed{\log_2 6} $

Question

Directions:

Unless otherwise specified, the domain of a function $ f $ is assumed to be the set of all real numbers $ x $ for which $ f(x) $ is a real number. Angle measures for trigonometric functions are assumed to be in radians.
Solutions to equations must be real numbers. Determine the exact value of any expression that can be obtained without a calculator. For example, $ \log_2 8 $, $ \cos(\frac{\pi}{2}) $, and $ \sin^{-1}(1) $ can be evaluated without a calculator.
Unless otherwise specified, combine terms using algebraic methods and rules for exponents and logarithms, where applicable. For example, $ 2x + 3x $, $ 5^2 \cdot 5^3 $, $ \frac{x^5}{x^2} $, and $ \ln 3 + \ln 5 $ should be rewritten in equivalent forms.
For each part of the question, show the work that leads to your answers.

Part A
The functions $ g $ and $ h $ are given by

$ g(x) = \log_5(4x – 2) $
$ h(x) = \sin^{-1}(8x) $

(i) Solve $ g(x) = 3 $ for values of $ x $ in the domain of $ g $.

(ii) Solve $ h(x) = \frac{\pi}{4} $ for values of $ x $ in the domain of $ h $.

Part B
The functions $ j $ and $ k $ are given by

$ j(x) = (\sec x)(\cot x) $
$ k(x) = \frac{(16^{3x}) \cdot 4^x}{2} $

(i) Rewrite $ j(x) $ as an expression involving $ \sin x $ and no other trigonometric functions.

(ii) Rewrite $ k(x) $ as an expression of the form $ 4^{(ax+b)} $, where $ a $ and $ b $ are constants.

Part C
The function $ m $ is given by

$ m(x) = \sqrt{3}\tan(x + \frac{\pi}{2}) $.

Find all values in the domain of $ m $ that yield an output value of 1.

▶️ Answer/Explanation

Detailed solution

Part A
(i) Solve $ g(x) = 3 $
Start with the given equation:
$ \log_5(4x – 2) = 3 $
Convert the logarithmic equation to exponential form ($ y = \log_b x \iff x = b^y $):
$ 4x – 2 = 5^3 $
Evaluate the exponent:
$ 4x – 2 = 125 $
Add 2 to both sides:
$ 4x = 127 $
Divide by 4:
$ x = \frac{127}{4} $

(ii) Solve $ h(x) = \frac{\pi}{4} $
Start with the given equation:
$ \sin^{-1}(8x) = \frac{\pi}{4} $
Take the sine of both sides to isolate the argument:
$ 8x = \sin\left(\frac{\pi}{4}\right) $
Substitute the exact value of $ \sin\left(\frac{\pi}{4}\right) $:
$ 8x = \frac{\sqrt{2}}{2} $
Divide by 8:
$ x = \frac{\sqrt{2}}{16} $

Part B
(i) Rewrite $ j(x) $
Start with the function definition:
$ j(x) = (\sec x)(\cot x) $
Substitute the reciprocal and quotient identities ($ \sec x = \frac{1}{\cos x} $ and $ \cot x = \frac{\cos x}{\sin x} $):
$ j(x) = \left(\frac{1}{\cos x}\right) \left(\frac{\cos x}{\sin x}\right) $
Cancel the $ \cos x $ terms:
$ j(x) = \frac{1}{\sin x} $

(ii) Rewrite $ k(x) $
Start with the function definition:
$ k(x) = \frac{(16^{3x}) \cdot 4^x}{2} $
To write in the form $ 4^{(ax+b)} $, convert bases 16 and 2 to base 4.
Since $ 16 = 4^2 $ and $ 2 = \sqrt{4} = 4^{1/2} = 4^{0.5} $:
$ k(x) = \frac{(4^2)^{3x} \cdot 4^x}{4^{0.5}} $
Apply the power of a power rule ($ (a^m)^n = a^{mn} $):
$ k(x) = \frac{4^{6x} \cdot 4^x}{4^{0.5}} $
Apply the product rule for exponents ($ a^m \cdot a^n = a^{m+n} $) in the numerator:
$ k(x) = \frac{4^{6x + x}}{4^{0.5}} = \frac{4^{7x}}{4^{0.5}} $
Apply the quotient rule for exponents ($ \frac{a^m}{a^n} = a^{m-n} $):
$ k(x) = 4^{7x – 0.5} $ (or $ 4^{7x – \frac{1}{2}} $)
Thus, $ a = 7 $ and $ b = -0.5 $.

Part C
Find values where $ m(x) = 1 $
Set the function equal to 1:
$ \sqrt{3}\tan\left(x + \frac{\pi}{2}\right) = 1 $
Isolate the tangent function by dividing by $ \sqrt{3} $:
$ \tan\left(x + \frac{\pi}{2}\right) = \frac{1}{\sqrt{3}} $
Determine the reference angle. We know that $ \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} $.
Set up the general solution for tangent ($ \theta = \text{ref} + n\pi $):
$ x + \frac{\pi}{2} = \frac{\pi}{6} + n\pi $, where $ n $ is any integer.
Solve for $ x $ by subtracting $ \frac{\pi}{2} $ from both sides:
$ x = \frac{\pi}{6} – \frac{\pi}{2} + n\pi $
Find a common denominator (6) to combine fractions:
$ x = \frac{\pi}{6} – \frac{3\pi}{6} + n\pi $
$ x = -\frac{2\pi}{6} + n\pi $
Simplify the fraction:
$ x = -\frac{\pi}{3} + n\pi $

AP Precalculus -3.9 Inverse Trigonometric Functions- FRQ Exam Style Questions - Effective Fall 2023