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AP Precalculus -2.10 Inverses of Exponential Functions- FRQ Exam Style Questions - Effective Fall 2023

AP Precalculus -2.10 Inverses of Exponential Functions- FRQ Exam Style Questions – Effective Fall 2023

AP Precalculus -2.10 Inverses of Exponential Functions- FRQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.

AP Precalculus – FRQ Exam Style Questions- All Topics

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:
\( x = -\frac{\pi}{3} + n\pi \)

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