AP Precalculus -3.10 Trigonometric Equations and Inequalities- FRQ Exam Style Questions - Effective Fall 2023
AP Precalculus -3.10 Trigonometric Equations and Inequalities- FRQ Exam Style Questions – Effective Fall 2023
AP Precalculus -3.10 Trigonometric Equations and Inequalities- FRQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.
▶️ Answer/Explanation
(A)(i) Rewrite \(g(x)\)
The function is given by \(g(x) = 3 \ln x – \frac{1}{2} \ln x\).
Combine the like terms:
\(g(x) = \left(3 – \frac{1}{2}\right) \ln x\)
\(g(x) = \frac{5}{2} \ln x\)
Apply the power property of logarithms, \(a \ln b = \ln(b^a)\):
\(g(x) = \ln\left(x^{5/2}\right)\)
(A)(ii) Rewrite \(h(x)\)
The function is given by \(h(x) = \frac{\sin^2 x – 1}{\cos x}\).
Recall the Pythagorean identity: \(\sin^2 x + \cos^2 x = 1\).
Rearrange the identity to isolate the numerator expression: \(\sin^2 x – 1 = -\cos^2 x\).
Substitute this into the function:
\(h(x) = \frac{-\cos^2 x}{\cos x}\)
Simplify the expression by canceling one \(\cos x\) term:
\(h(x) = -\cos x\)
(B)(i) Solve \(j(x) = 0\)
Set the function equal to zero: \(2(\sin x)(\cos x) = 0\).
Divide both sides by 2:
\(\sin x \cos x = 0\)
By the zero product property, either \(\sin x = 0\) or \(\cos x = 0\).
Case 1: \(\sin x = 0\). In the interval \(\left[0, \frac{\pi}{2}\right]\), \(x = 0\).
Case 2: \(\cos x = 0\). In the interval \(\left[0, \frac{\pi}{2}\right]\), \(x = \frac{\pi}{2}\).
The solutions are \(x = 0\) and \(x = \frac{\pi}{2}\).
(B)(ii) Solve \(k(x) = 3e\)
Set the function equal to \(3e\):
\(8e^{(3x)} – e = 3e\)
Add \(e\) to both sides:
\(8e^{(3x)} = 4e\)
Divide both sides by 8:
\(e^{(3x)} = \frac{4e}{8}\)
\(e^{(3x)} = \frac{e}{2}\)
Take the natural logarithm (\(\ln\)) of both sides:
\(\ln\left(e^{3x}\right) = \ln\left(\frac{e}{2}\right)\)
Use logarithm properties to simplify (\(\ln(e^a) = a\) and \(\ln(a/b) = \ln a – \ln b\)):
\(3x = \ln e – \ln 2\)
Since \(\ln e = 1\):
\(3x = 1 – \ln 2\)
Divide by 3:
\(x = \frac{1 – \ln 2}{3}\)
(C) Find input values for \(m(x) = \frac{9}{2}\)
Set the function equal to \(\frac{9}{2}\):
\(\cos(2x) + 4 = \frac{9}{2}\)
Subtract 4 from both sides (note that \(4 = \frac{8}{2}\)):
\(\cos(2x) = \frac{9}{2} – \frac{8}{2}\)
\(\cos(2x) = \frac{1}{2}\)
The reference angle for cosine equal to \(\frac{1}{2}\) is \(\frac{\pi}{3}\).
The general solution for \(2x\) is:
\(2x = \frac{\pi}{3} + 2\pi n\) or \(2x = -\frac{\pi}{3} + 2\pi n\) (where \(n\) is an integer).
Solve for \(x\) by dividing by 2:
\(x = \frac{\pi}{6} + \pi n\) or \(x = -\frac{\pi}{6} + \pi n\)
Combining these, the input values are \(x = \pm \frac{\pi}{6} + \pi n\) for any integer \(n\).
▶️ Answer/Explanation
Step 1: Write all logs in base 10 Assume \( \log \) means \( \log_{10} \). Given: \[ g(x) = 2 \log a – 3 \log b + \frac{\log a}{2} + \log_{100} c \] Note: \( \log_{100} c = \frac{\log_{10} c}{\log_{10} 100} = \frac{\log c}{2} \). So: \[ g(x) = 2\log a – 3\log b + \frac{\log a}{2} + \frac{\log c}{2} \] Step 2: Combine like terms Terms with \( \log a \): \( 2\log a + \frac{\log a}{2} = \frac{4\log a}{2} + \frac{\log a}{2} = \frac{5\log a}{2} \). Other terms: \( -3\log b + \frac{\log c}{2} \). Thus: \[ g(x) = \frac{5}{2} \log a – 3\log b + \frac{1}{2} \log c \] Step 3: Express as a single logarithm \[ g(x) = \log\left( a^{5/2} \right) + \log\left( b^{-3} \right) + \log\left( c^{1/2} \right) \] \[ = \log\left( a^{5/2} \cdot b^{-3} \cdot c^{1/2} \right) \] \[ = \log\left( \frac{a^{5/2} c^{1/2}}{b^3} \right) \] \[ = \log\left( \frac{\sqrt{a^5} \sqrt{c}}{b^3} \right) = \log\left( \frac{\sqrt{a^5 c}}{b^3} \right) \] Step 4: Without negative exponents The expression inside log is \( \frac{a^{5/2} c^{1/2}}{b^3} \). No negative exponents inside — all exponents positive in numerator, denominator exponent positive as well (it’s \( b^3 \)). Final: \[ g(x) = \log\left( \frac{a^{5/2} c^{1/2}}{b^3} \right) \quad \text{or} \quad \log\left( \frac{\sqrt{a^5 c}}{b^3} \right) \]
Given: \[ h(x) = \cot x \sec^2 x – \tan x \] Step 1: Write in sines and cosines \[ \cot x = \frac{\cos x}{\sin x}, \quad \sec^2 x = \frac{1}{\cos^2 x}, \quad \tan x = \frac{\sin x}{\cos x} \] So: \[ h(x) = \frac{\cos x}{\sin x} \cdot \frac{1}{\cos^2 x} – \frac{\sin x}{\cos x} \] \[ = \frac{1}{\sin x \cos x} – \frac{\sin x}{\cos x} \] Step 2: Common denominator \( \sin x \cos x \) \[ h(x) = \frac{1 – \sin^2 x}{\sin x \cos x} \] Step 3: Use \( 1 – \sin^2 x = \cos^2 x \) \[ h(x) = \frac{\cos^2 x}{\sin x \cos x} = \frac{\cos x}{\sin x} = \cot x \] Final: \[ h(x) = \cot x \]
Given: \[ j(x) = 3 \cot^2 \theta + 3 \csc \theta \] It appears \( j \) is a function of \( \theta \), so likely \( j(\theta) \). Solve \( j(\theta) = -3 \). \[ 3 \cot^2 \theta + 3 \csc \theta = -3 \] Divide by 3: \[ \cot^2 \theta + \csc \theta = -1 \] Step 1: Express \( \cot^2 \theta \) in terms of \( \csc \theta \) \[ \cot^2 \theta = \csc^2 \theta – 1 \] So: \[ \csc^2 \theta – 1 + \csc \theta = -1 \] \[ \csc^2 \theta + \csc \theta = 0 \] Step 2: Factor \[ \csc \theta (\csc \theta + 1) = 0 \] Thus: \[ \csc \theta = 0 \quad \text{or} \quad \csc \theta = -1 \] But \( \csc \theta = \frac{1}{\sin \theta} \), so \( \csc \theta = 0 \) has no solution. Only: \[ \csc \theta = -1 \implies \sin \theta = -1 \] Step 3: Solve \( \sin \theta = -1 \) on \([0, 2\pi)\) \[ \theta = \frac{3\pi}{2} \] Final: \[ \boxed{\theta = \frac{3\pi}{2}} \] (Note: The problem says \( j(x) \) but uses \( \theta \) — likely a typo. Solving for \( \theta \).)
\[ k(x) = 7e^{7-2x} + 9 = 86 \] \[ 7e^{7-2x} = 77 \] \[ e^{7-2x} = 11 \] Step 1: Take natural log \[ 7-2x = \ln 11 \] \[ 2x = 7 – \ln 11 \] \[ x = \frac{7 – \ln 11}{2} \] Step 2: Domain of \( k \) \( k(x) \) is defined for all real \( x \) (exponential function). So only this one solution. Final: \[ \boxed{x = \frac{7 – \ln 11}{2}} \]
Given: \[ a(\theta) = 3 – 3 \csc 3\theta \] \[ b(\theta) = -1 + \csc 3\theta \] Set \( a(\theta) = b(\theta) \): \[ 3 – 3 \csc 3\theta = -1 + \csc 3\theta \] \[ 3 + 1 = \csc 3\theta + 3 \csc 3\theta \] \[ 4 = 4 \csc 3\theta \] \[ \csc 3\theta = 1 \] \[ \sin 3\theta = 1 \] Step 1: Solve \( \sin 3\theta = 1 \) \[ 3\theta = \frac{\pi}{2} + 2n\pi, \quad n \in \mathbb{Z} \] \[ \theta = \frac{\pi}{6} + \frac{2n\pi}{3} \] Step 2: Domain considerations For \( \csc 3\theta \) to be defined, \( \sin 3\theta \neq 0 \). Here \( \sin 3\theta = 1 \) so fine. We need all \( \theta \) in the domain of both \( a \) and \( b \) — typically all reals except where \( \sin 3\theta = 0 \). Our solutions avoid those points. Step 3: General solution \[ \theta = \frac{\pi}{6} + \frac{2n\pi}{3}, \quad n \in \mathbb{Z} \] Final: \[ \boxed{\theta = \frac{\pi}{6} + \frac{2n\pi}{3}, \quad n \in \mathbb{Z}} \]
▶️ Answer/Explanation
Step 1: Identify parameters
– Clock diameter = 24 inches ⇒ radius \( r = 12 \) inches.
– Clock center height above ground: 5 feet = \( 5 \times 12 = 60 \) inches.
– Bottom of clock: \( 60 – 12 = 48 \) inches above ground.
– Top of clock: \( 60 + 12 = 72 \) inches above ground.
– Cycle period = 30 seconds ⇒ \( b = \frac{2\pi}{30} = \frac{\pi}{15} \).
– At \( t = 0 \), second hand points straight up ⇒ height is maximum: \( h(0) = 72 \).
Step 2:
Form of \( h(t) \) General form: \( h(t) = a\cos(b(t + c)) + d \).
Maximum height = \( d + |a| = 72 \)
Minimum height = \( d – |a| = 48 \)
Solve: \( d = \frac{72+48}{2} = 60 \), \( |a| = 72 – 60 = 12 \).
Since at \( t = 0 \) we have max height, choose cosine with no horizontal shift that gives max at \( t=0 \) ⇒ \( a > 0 \), \( c = 0 \) (since \(\cos(0)=1\) gives max).
Thus \( a = 12 \), \( c = 0 \), \( b = \frac{\pi}{15} \), \( d = 60 \).
Step 3: Final function \[ h(t) = 12\cos\left( \frac{\pi}{15} t \right) + 60 \]
Step 1:
Set up equation \[ 12\cos\left( \frac{\pi}{15} t \right) + 60 = 72 – 6\sqrt{2} \] \[ 12\cos\left( \frac{\pi}{15} t \right) = 12 – 6\sqrt{2} \] \[ \cos\left( \frac{\pi}{15} t \right) = 1 – \frac{\sqrt{2}}{2} \]
Note: \( 1 – \frac{\sqrt{2}}{2} \approx 0.2929 \).
Step 2: General solutions Let \( \theta = \frac{\pi}{15} t \). \(\cos \theta = 1 – \frac{\sqrt{2}}{2} \).
Since cosine is positive, solutions are in QI and QIV (for principal values \( 0 \leq \theta < 2\pi \)).
\(\theta_1 = \cos^{-1}\left(1 – \frac{\sqrt{2}}{2}\right)\) \(\theta_2 = 2\pi – \theta_1\).
Step 3: Find \( t \)
\( t = \frac{15}{\pi} \theta \).
Numerically: \( 1 – \frac{\sqrt{2}}{2} \approx 0.292893 \) \(\theta_1 \approx \cos^{-1}(0.292893) \approx 1.2735 \) rad.
\( t_1 \approx \frac{15}{\pi} \times 1.2735 \approx 6.08 \) s. \(\theta_2 \approx 2\pi – 1.2735 \approx 5.0097 \) rad. \( t_2 \approx \frac{15}{\pi} \times 5.0097 \approx 23.92 \) s.
Both are in \( 0 < t \leq 30 \).
Step 4: Exact form Exact \( \theta_1 = \cos^{-1}\left( 1 – \frac{\sqrt{2}}{2} \right) \), but simpler:
Note \( 1 – \frac{\sqrt{2}}{2} = \frac{2 – \sqrt{2}}{2} \).
Thus: \[ t_1 = \frac{15}{\pi} \cos^{-1}\left( \frac{2 – \sqrt{2}}{2} \right), \quad t_2 = 30 – t_1 \]
because cosine is symmetric about \( \theta = \pi \). Final times: \( t \approx 6.08 \) s and \( t \approx 23.92 \) s.
Step 1: Second derivative
\( h(t) = 12\cos\left( \frac{\pi}{15} t \right) + 60 \) \( h'(t) = -12 \cdot \frac{\pi}{15} \sin\left( \frac{\pi}{15} t \right) \)
\( h”(t) = -12 \cdot \frac{\pi^2}{225} \cos\left( \frac{\pi}{15} t \right) = -\frac{4\pi^2}{75} \cos\left( \frac{\pi}{15} t \right) \).
Step 2: Inflection points occur when \( h”(t) = 0 \) \[ \cos\left( \frac{\pi}{15} t \right) = 0 \] \[ \frac{\pi}{15} t = \frac{\pi}{2} + n\pi, \quad n \in \mathbb{Z} \] \[ t = 7.5 + 15n \]
Step 3: In \( 0 \leq t < 30 \) \( n = 0 \): \( t = 7.5 \) s \( n = 1 \): \( t = 22.5 \) s
So yes, there are two points of inflection at \( t = 7.5 \) s and \( t = 22.5 \) s.
At a point of inflection, the rate of change of height \( h'(t) \) is at a local extremum (maximum or minimum magnitude of vertical speed). Equivalently, the acceleration \( h”(t) \) changes sign, meaning the rate at which the height is changing (speed) stops increasing and starts decreasing, or vice versa.
In this problem:
– At \( t = 7.5 \) s, second hand is horizontal (3 o’clock position), vertical speed is maximum in downward direction, and acceleration changes from negative to positive (decelerating downward motion before starting to slow and reverse).
– At \( t = 22.5 \) s, second hand is horizontal (9 o’clock position), vertical speed is maximum in upward direction, and acceleration changes from positive to negative.
Thus, these are the moments when the vertical motion switches from speeding up to slowing down (or vice versa) while moving toward the extreme top or bottom of its path.
▶️ Answer/Explanation
(A)(i)
\( e^{2x}=10 \)
\( 2x=\ln(10) \)
\( x=\frac{1}{2}\ln(10) \).
(A)(ii)
\( \arcsin(x+3)=\frac{\pi}{4} \)
\( x+3=\sin\!\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2} \)
\( x=\frac{\sqrt{2}}{2}-3 \).
(B)(i)
\( j(x)=\log_{10}\!\left(\frac{8x^9}{2x^3}\right) \)
\( =\log_{10}(4x^6) \).
(B)(ii)
\( k(x)=\frac{\sin x}{\sec x}=\sin x\cos x \)
\( =\frac{\sin x}{\cos x}\cos^2 x=\tan x \).
(C)
\( \cos(\tan(2x-1))=0 \)
\( \tan(2x-1)=\frac{\pi}{2}+k\pi \)
\( 2x-1=\arctan\!\left(\frac{\pi}{2}+k\pi\right) \), where defined.
▶️ Answer/Explanation
Part A: Coordinates of Points
At $t = 0$, $S$ is at its minimum height $-6$ (below $X$).
The maximum height is $20$.
The midline $d = \frac{20 + (-6)}{2} = 7$.
The period is $2$ seconds.
The graph starts at a minimum at $t = 0$, reaches midline at $t = 0.5$, maximum at $t = 1$, midline at $t = 1.5$, and minimum at $t = 2$.
Based on the visual positions in the provided graph:
$F$ (first maximum): $(1, 20)$
$G$ (midline, decreasing): $(1.5, 7)$
$J$ (minimum): $(2, -6)$
$K$ (midline, increasing): $(2.5, 7)$
$P$ (second maximum): $(3, 20)$
Part B: Finding Constants
$a$ (Amplitude) $= \frac{20 – (-6)}{2} = 13$. Since we use $\cos$ and start at a minimum, $a = -13$ (or use a phase shift).
$d$ (Vertical shift/Midline) $= 7$.
$b$ (Frequency factor) $= \frac{2\pi}{\text{period}} = \frac{2\pi}{2} = \pi$.
$c$ (Phase shift): For $h(t) = a \cos(b(t+c)) + d$, if $a = -13$, then at $t=0$, $-13\cos(b(0+c))+7 = -6 \implies \cos(bc)=1 \implies c = 0$.
Final values: $a = -13, b = \pi, c = 0, d = 7$.
Part C: Interval Analysis
(i) At $K$, $h(t)=7$ and is increasing. At $P$, $h(t)=20$ (maximum).
On $(t_1, t_2)$, the height is between $7$ and $20$, so it is positive.
The graph is moving from the midline up to the peak, so it is increasing.
Correct Option: a
(ii) On the interval $(t_1, t_2)$, the graph is concave down as it approaches the maximum.
Therefore, the rate of change of $h$ (the slope) is decreasing.
It starts at its maximum positive value at $K$ and decreases toward zero at $P$.
▶️ Answer/Explanation
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:
\( x = -\frac{\pi}{3} + n\pi \)
▶️ Answer/Explanation
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:
\( x = -\frac{\pi}{3} + n\pi \)
▶️ Answer/Explanation
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:
\( x = -\frac{\pi}{3} + n\pi \)
▶️ Answer/Explanation
(A)(i) Solve \( g(x) = 27 \)
First, simplify the expression for \( g(x) \) using the property of exponents \( a^m \cdot a^n = a^{m+n} \).
\( g(x) = 3^{2x} \cdot 3^{x+4} = 3^{(2x + x + 4)} = 3^{(3x+4)} \)
Set \( g(x) \) equal to 27 and rewrite 27 as a base of 3:
\( 3^{(3x+4)} = 27 \)
\( 3^{(3x+4)} = 3^3 \)
Since the bases are equal, the exponents must be equal:
\( 3x + 4 = 3 \)
\( 3x = 3 – 4 \)
\( 3x = -1 \)
\( x = -\frac{1}{3} \)
(A)(ii) Solve \( h(x) = 5 \) on \( [0, 2\pi) \)
Set the expression for \( h(x) \) equal to 5:
\( 2\tan^2 x – 1 = 5 \)
Add 1 to both sides:
\( 2\tan^2 x = 6 \)
Divide by 2:
\( \tan^2 x = 3 \)
Take the square root of both sides:
\( \tan x = \pm\sqrt{3} \)
The reference angle for \( \tan \theta = \sqrt{3} \) is \( \frac{\pi}{3} \).
Since we have \( \pm\sqrt{3} \), we must consider solutions in all four quadrants within the interval \( [0, 2\pi) \):
Quadrant I: \( x = \frac{\pi}{3} \)
Quadrant II: \( x = \pi – \frac{\pi}{3} = \frac{2\pi}{3} \)
Quadrant III: \( x = \pi + \frac{\pi}{3} = \frac{4\pi}{3} \)
Quadrant IV: \( x = 2\pi – \frac{\pi}{3} = \frac{5\pi}{3} \)
Solution set: \( x = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3} \)
(B)(i) Rewrite \( j(x) \) as a single logarithm
Given: \( j(x) = 2\log_{10}(x+3) – \log_{10} x – \log_{10} 3 \)
Use the power rule \( n\log a = \log(a^n) \):
\( = \log_{10}((x+3)^2) – \log_{10} x – \log_{10} 3 \)
Factor out the negative sign for the last two terms to group them:
\( = \log_{10}((x+3)^2) – (\log_{10} x + \log_{10} 3) \)
Use the product rule \( \log a + \log b = \log(ab) \):
\( = \log_{10}((x+3)^2) – \log_{10}(3x) \)
Use the quotient rule \( \log a – \log b = \log(\frac{a}{b}) \):
\( = \log_{10}\left(\frac{(x+3)^2}{3x}\right) \)
(B)(ii) Rewrite \( k(x) \) involving \( \sec x \)
Given: \( k(x) = \frac{(\tan^2 x)(\cot x)}{\csc x} \)
First, simplify the numerator using \( \tan x \cdot \cot x = 1 \):
\( (\tan^2 x)(\cot x) = \tan x \cdot (\tan x \cdot \cot x) = \tan x \cdot 1 = \tan x \)
Now substitute back into the expression:
\( k(x) = \frac{\tan x}{\csc x} \)
Convert to sine and cosine:
\( = \frac{\frac{\sin x}{\cos x}}{\frac{1}{\sin x}} \)
\( = \frac{\sin x}{\cos x} \cdot \frac{\sin x}{1} = \frac{\sin^2 x}{\cos x} \)
We need the expression in terms of \( \sec x \). Use the identity \( \sin^2 x = 1 – \cos^2 x \):
\( = \frac{1 – \cos^2 x}{\cos x} \)
Substitute \( \cos x = \frac{1}{\sec x} \):
\( = \frac{1 – \left(\frac{1}{\sec x}\right)^2}{\frac{1}{\sec x}} \)
\( = \frac{1 – \frac{1}{\sec^2 x}}{\frac{1}{\sec x}} \)
Find a common denominator for the numerator:
\( = \frac{\frac{\sec^2 x – 1}{\sec^2 x}}{\frac{1}{\sec x}} \)
Multiply by the reciprocal of the denominator:
\( = \frac{\sec^2 x – 1}{\sec^2 x} \cdot \frac{\sec x}{1} \)
\( = \frac{\sec^2 x – 1}{\sec x} \)
(C) Find input values for \( m(x) = \frac{1}{16} \)
First, simplify the expression for \( m(x) \):
\( m(x) = \frac{2^{(5x+3)}}{\left(2^{(x-2)}\right)^3} \)
Simplify the denominator using the power rule \( (a^m)^n = a^{m \cdot n} \):
\( \left(2^{(x-2)}\right)^3 = 2^{3(x-2)} = 2^{3x-6} \)
Now apply the quotient rule \( \frac{a^m}{a^n} = a^{m-n} \):
\( m(x) = \frac{2^{5x+3}}{2^{3x-6}} = 2^{(5x+3) – (3x-6)} \)
\( = 2^{5x + 3 – 3x + 6} \)
\( = 2^{2x + 9} \)
Set \( m(x) = \frac{1}{16} \) and rewrite \( \frac{1}{16} \) as a power of 2:
\( \frac{1}{16} = \frac{1}{2^4} = 2^{-4} \)
Equate the simplified \( m(x) \) to \( 2^{-4} \):
\( 2^{2x + 9} = 2^{-4} \)
Equate the exponents:
\( 2x + 9 = -4 \)
\( 2x = -13 \)
\( x = -\frac{13}{2} \) or \( -6.5 \)
▶️ Answer/Explanation
(A)(i) Rewrite \(g(x)\)
The function is given by \(g(x) = 3 \ln x – \frac{1}{2} \ln x\).
Combine the like terms:
\(g(x) = \left(3 – \frac{1}{2}\right) \ln x\)
\(g(x) = \frac{5}{2} \ln x\)
Apply the power property of logarithms, \(a \ln b = \ln(b^a)\):
\(g(x) = \ln\left(x^{5/2}\right)\)
(A)(ii) Rewrite \(h(x)\)
The function is given by \(h(x) = \frac{\sin^2 x – 1}{\cos x}\).
Recall the Pythagorean identity: \(\sin^2 x + \cos^2 x = 1\).
Rearrange the identity to isolate the numerator expression: \(\sin^2 x – 1 = -\cos^2 x\).
Substitute this into the function:
\(h(x) = \frac{-\cos^2 x}{\cos x}\)
Simplify the expression by canceling one \(\cos x\) term:
\(h(x) = -\cos x\)
(B)(i) Solve \(j(x) = 0\)
Set the function equal to zero: \(2(\sin x)(\cos x) = 0\).
Divide both sides by 2:
\(\sin x \cos x = 0\)
By the zero product property, either \(\sin x = 0\) or \(\cos x = 0\).
Case 1: \(\sin x = 0\). In the interval \(\left[0, \frac{\pi}{2}\right]\), \(x = 0\).
Case 2: \(\cos x = 0\). In the interval \(\left[0, \frac{\pi}{2}\right]\), \(x = \frac{\pi}{2}\).
The solutions are \(x = 0\) and \(x = \frac{\pi}{2}\).
(B)(ii) Solve \(k(x) = 3e\)
Set the function equal to \(3e\):
\(8e^{(3x)} – e = 3e\)
Add \(e\) to both sides:
\(8e^{(3x)} = 4e\)
Divide both sides by 8:
\(e^{(3x)} = \frac{4e}{8}\)
\(e^{(3x)} = \frac{e}{2}\)
Take the natural logarithm (\(\ln\)) of both sides:
\(\ln\left(e^{3x}\right) = \ln\left(\frac{e}{2}\right)\)
Use logarithm properties to simplify (\(\ln(e^a) = a\) and \(\ln(a/b) = \ln a – \ln b\)):
\(3x = \ln e – \ln 2\)
Since \(\ln e = 1\):
\(3x = 1 – \ln 2\)
Divide by 3:
\(x = \frac{1 – \ln 2}{3}\)
(C) Find input values for \(m(x) = \frac{9}{2}\)
Set the function equal to \(\frac{9}{2}\):
\(\cos(2x) + 4 = \frac{9}{2}\)
Subtract 4 from both sides (note that \(4 = \frac{8}{2}\)):
\(\cos(2x) = \frac{9}{2} – \frac{8}{2}\)
\(\cos(2x) = \frac{1}{2}\)
The reference angle for cosine equal to \(\frac{1}{2}\) is \(\frac{\pi}{3}\).
The general solution for \(2x\) is:
\(2x = \frac{\pi}{3} + 2\pi n\) or \(2x = -\frac{\pi}{3} + 2\pi n\) (where \(n\) is an integer).
Solve for \(x\) by dividing by 2:
\(x = \frac{\pi}{6} + \pi n\) or \(x = -\frac{\pi}{6} + \pi n\)
Combining these, the input values are \(x = \pm \frac{\pi}{6} + \pi n\) for any integer \(n\).