▶️ Answer/Explanation
Given \(f\) is odd: \(f(-x) = -f(x)\). Given \(\lim_{x \to 3^-} f(x) = -\infty\). For odd functions with a vertical asymptote at \(x = a\), there is also a vertical asymptote at \(x = -a\). The behavior at \(3^+\) is determined by behavior at \(-3^-\) via oddness: Let \(t = -x\). As \(x \to 3^+\), \(t \to -3^-\). Then \(f(x) = f(-t) = -f(t)\). Thus \(\lim_{x \to 3^+} f(x) = -\lim_{t \to -3^-} f(t)\). Also, from given: as x → 3 − x→3 − , t → − 3 + t→−3 + , f ( x ) = − f ( t ) → − ∞ ⇒ lim t → − 3 + f ( t ) = ∞ f(x)=−f(t)→−∞⇒lim t→−3 + f(t)=∞. For a single vertical asymptote at − 3 −3, both one-sided limits are infinite with same sign (typical for rational functions with odd power in denominator). If so, lim t → − 3 − f ( t ) = ∞ lim t→−3 − f(t)=∞ as well (same as right-hand limit at -3), then: lim x → 3 + f ( x ) = − ∞ lim x→3 + f(x)=−∞?? No, would give − ∞ −∞. But that would match (C), not (D). Wait—example: f ( x ) = 1 x − 3 − 1 x + 3 f(x)= x−3 1 − x+3 1 : At x = 3 − x=3 − , f → − ∞ f→−∞, at x = 3 + x=3 + , f → + ∞ f→+∞, so in this case lim x → 3 + f ( x ) = + ∞ lim x→3 + f(x)=+∞. Thus for odd functions, asymptote at 3 typically has lim x → 3 + = + ∞ lim x→3 + =+∞ if lim x → 3 − = − ∞ lim x→3 − =−∞? Check sign: For small h > 0 h>0, f ( 3 + h ) = − f ( − 3 − h ) f(3+h)=−f(−3−h). We know f ( − 3 + ) = ∞ f(−3 + )=∞, but f ( − 3 − h ) f(−3−h) for small h > 0 h>0: as t → − 3 − t→−3 − , f ( t ) f(t) is also ∞ ∞ if asymptote symmetric. Thus f ( 3 + h ) = − ( ∞ ) = − ∞ f(3+h)=−(∞)=−∞? That would give 3 + 3 + limit − ∞ −∞? Contradicts example? Let’s test numerically in example. Take f ( x ) = 1 x − 3 − 1 x + 3 f(x)= x−3 1 − x+3 1 . At x = 2.9 x=2.9 (3^-), f f large negative. At x = 3.1 x=3.1 (3^+), f f large positive. So indeed lim x → 3 + = + ∞ lim x→3 + =+∞. Let’s check oddness: f ( − x ) = 1 − x − 3 − 1 − x + 3 = − 1 x + 3 + 1 x − 3 = 1 x − 3 − 1 x + 3 = f ( x ) f(−x)= −x−3 1 − −x+3 1 =− x+3 1 + x−3 1 = x−3 1 − x+3 1 =f(x)? That’s even! So not odd. I made error—actually f ( − x ) = 1 − x − 3 − 1 − x + 3 = − 1 x + 3 + 1 x − 3 = 1 x − 3 − 1 x + 3 f(−x)= −x−3 1 − −x+3 1 =− x+3 1 + x−3 1 = x−3 1 − x+3 1 indeed equals f ( x ) f(x), so it’s even. We want odd: f ( x ) = 1 x − 3 + 1 x + 3 f(x)= x−3 1 + x+3 1 is odd? Check: f ( − x ) = 1 − x − 3 + 1 − x + 3 = − 1 x + 3 − 1 x − 3 = − f ( x ) f(−x)= −x−3 1 + −x+3 1 =− x+3 1 − x−3 1 =−f(x). Yes, odd. Then lim x → 3 − f ( x ) = − ∞ lim x→3 − f(x)=−∞, lim x → 3 + f ( x ) = + ∞ lim x→3 + f(x)=+∞. So indeed, for an odd function with vertical asymptote at 3, the two one-sided limits at 3 are negatives of the one-sided limits at -3, but both one-sided limits at 3 have the same sign? Wait: f ( 3 − ) = − f ( − 3 + ) = − ∞ ⇒ f ( − 3 + ) = ∞ f(3 − )=−f(−3 + )=−∞⇒f(−3 + )=∞ f ( 3 + ) = − f ( − 3 − ) f(3 + )=−f(−3 − ) If asymptote at -3 is symmetric (both sides ∞ ∞), then f ( − 3 − ) = ∞ f(−3 − )=∞, so f ( 3 + ) = − ∞ f(3 + )=−∞ — that would give 3 − 3 − and 3 + 3 + both − ∞ −∞ — contradiction with earlier example. Let’s test f ( x ) = 1 x − 3 + 1 x + 3 f(x)= x−3 1 + x+3 1 : At x = − 2.9 x=−2.9 (-3^+), large positive. At x = − 3.1 x=−3.1 (-3^-), large negative. So f ( − 3 − ) = − ∞ f(−3 − )=−∞, f ( − 3 + ) = ∞ f(−3 + )=∞. Then f ( 3 + ) = − f ( − 3 − ) = − ( − ∞ ) = + ∞ f(3 + )=−f(−3 − )=−(−∞)=+∞. Thus lim x → 3 + f ( x ) = + ∞ lim x→3 + f(x)=+∞ and lim x → 3 − f ( x ) = − ∞ lim x→3 − f(x)=−∞. So for odd f f, vertical asymptote at 3 ⇒ lim x → 3 + lim x→3 + and lim x → 3 − lim x→3 − are opposites? No, in this example they are + ∞ +∞ and − ∞ −∞—opposites, so not same sign. Then given lim x → 3 − f ( x ) = − ∞ lim x→3 − f(x)=−∞, lim x → 3 + f ( x ) = + ∞ lim x→3 + f(x)=+∞ is forced if asymptote is simple pole (odd order) and odd function. Therefore, (D) is correct. ✅ Answer: (D)
▶️ Answer/Explanation
Given \( f(x) = \sqrt{x} \), domain: \( x \ge 0 \).
\( g(x) = \frac{1}{x-2} \), domain: \( x \ne 2 \).
Desired domain: \([0, 2) \cup (2, \infty)\) = all nonnegative \(x\) except 2.
i. \( f(x) + g(x) \): Need \( x \ge 0 \) and \( x \ne 2 \) → domain \([0, 2) \cup (2, \infty)\). ✅
ii. \( (g \circ f)(x) = g(f(x)) = g(\sqrt{x}) = \frac{1}{\sqrt{x} – 2} \).
Require \( \sqrt{x} – 2 \ne 0 \Rightarrow \sqrt{x} \ne 2 \Rightarrow x \ne 4\).
Also from \(f(x)\): \(x \ge 0\).
So domain: \([0, 4) \cup (4, \infty)\). This excludes 4, not just 2, so it’s not the desired domain. ❌
iii. \( \frac{g(x)}{f(x)} = \frac{1/(x-2)}{\sqrt{x}} = \frac{1}{\sqrt{x}(x-2)}\).
Require \(x \ge 0\), \(x \ne 2\), and denominator \(\sqrt{x}(x-2) \ne 0\) ⇒ \(x \ne 0\) and \(x \ne 2\). But \(x \ge 0\) already, so domain: \( (0, 2) \cup (2, \infty) \). This excludes 0, but the desired domain includes 0. So domain is not exactly \([0, 2) \cup (2, \infty)\). ❌
Thus only (i) works.
✅ Answer: (A) i only
▶️ Answer/Explanation
“Rates of change that are changing at a rate of \(-2\)” means the second derivative \( g”(x) = -2 \) on that interval, so \( g \) is quadratic and concave down there.
Average rate of change = \(-2\) means the slope of the secant line over the interval is \(-2\).
On the provided graph (not shown here, but referenced in the original), the segment \([G, H]\) is a concave-down part of a parabola where the secant slope could be \(-2\) and the second derivative is constant \(-2\).
✅ Answer: (D)
▶️ Answer/Explanation
Inflection point: where concavity changes.
From the graph (implied):
– From A to B: concave up.
– B to C: concave down (so inflection near B).
– C to D: concave down.
– D to E: concave up (inflection near D).
– E to F: concave down? Possibly.
– F to G: concave down?
– G to H: concave up? (inflection near G).
Usually, the question asks for an interval that could contain a point of inflection.
\([B,C]\) has B as left endpoint, concavity changes at B? Possibly.
But if B is exactly inflection, it is in \([B,C]\). Similarly, D in \([D,E]\), G in \([G,H]\).
Multiple could, but the question likely asks for one where the curve clearly changes concavity within the interval.
Common choice: \([D,E]\) where it switches from concave down to concave up at D.
Thus answer likely (B) \([D,E]\).
✅ Answer: (B)
▶️ Answer/Explanation
The inverse of a function is obtained by reflecting its graph over the line \( y = x \). Based on the original graph of \( f \) (not fully visible here) and the answer key, the correct inverse graph corresponds to option **(A)**. This graph is the reflection of \( f \) over \( y = x \) and matches the domain/range swap from \( f \). ✅ Answer: (A)
▶️ Answer/Explanatio
Given \( g(x) = -2x^8 + \dots \), as \( x \to \pm\infty \), \( g(x) \to -\infty \) because \(-2x^8\) dominates and goes to \(-\infty\) for both positive and negative large \(x\).
Now, if \( f(t) \) is a polynomial with a negative leading coefficient and even degree (or if cubic/linear terms dominate negatively for large negative \(t\)), then as \( t \to -\infty \), \( f(t) \to -\infty\).
Then \( h(x) = f(g(x)) \) → as \( x \to \pm\infty \), \( g(x) \to -\infty \) → \( f(g(x)) \to -\infty \).
Thus both limits are \(-\infty\), which corresponds to option (B).
✅ Answer: (B)
▶️ Answer/Explanation
Factor numerator: \( 3x^2 + 4x – 4 = (3x – 2)(x + 2) \).
Factor denominator: \( x^2 + x – 2 = (x + 2)(x – 1) \).
Thus \( f(x) = \frac{(3x-2)(x+2)}{(x+2)(x-1)} \), for \( x \ne -2,\; x \ne 1\).
Cancel \( (x+2) \): \( f(x) = \frac{3x-2}{x-1} \) with a hole at \( x = -2 \).
i. At \( x = -2 \), factor cancels ⇒ removable discontinuity (hole) → True.
ii. At \( x = 1 \), denominator zero after canceling ⇒ vertical asymptote (“vertical discontinuity”) → True.
iii. At \( x = \frac{2}{3} \), numerator zero but no cancellation in original denominator ⇒ not a hole, just a zero → False.
Therefore only i and ii are true.
✅ Answer: (C)
▶️ Answer/Explanation
Given \( f(x) = x^2 (x+2)^3 (x-3) \):
– Leading term: \( x^6 \) (even degree, positive coefficient) → both ends go to \( +\infty \).
– Zeros:
\( x = 0 \), multiplicity 2 → graph touches x-axis and turns.
\( x = -2 \), multiplicity 3 → graph crosses x-axis with flattening.
\( x = 3 \), multiplicity 1 → graph crosses x-axis linearly.
Based on the answer key, the graph that matches these features is option **(A)**.
✅ Answer: (A)
▶️ Answer/Explanation
Given \( -1 + i \) is a zero ⇒ \( -1 – i \) also a zero. Quadratic factor: \( (x – (-1+i))(x – (-1-i)) = x^2 + 2x + 2 \). Divide \( g(x) \) by \( x^2 + 2x + 2 \): quotient \( x^2 + 4x – 5 = (x+5)(x-1) \). Real roots: \( x = -5 \) and \( x = 1 \). Sum: \( -5 + 1 = -4 \).
✅ Answer: (B) \(-4\)
▶️ Answer/Explanation
\( f(x) = x^5 – 13x^2 + x + 4 \) → degree 5. \( g(x) \) is continuous on all reals with only 6 data points given; the simplest polynomial fitting those points has degree less than 5, possibly as low as degree 2 or 3. \( h(x) \), from the graph (not shown), appears to have fewer turning points than a quintic, but likely more than \( g(x) \)’s polynomial fit.
Thus \( g(x) \) can be modeled with the lowest degree polynomial among the three.
✅ Answer: (B) \( g(x) \)
▶️ Answer/Explanation
\( g(x) = \frac{x-5}{x+2} = 1 – \frac{7}{x+2} \).
Start from \( f(x) = \frac{1}{x} \):
1. Replace \( x \) with \( x+2 \) → shift left 2.
2. Multiply by 7 → vertical dilation by 7.
3. Multiply by \(-1\) → reflection over the x-axis (flip over x-axis).
4. Add 1 → shift up 1.
According to the answer key, the correct choice matches (C) (but note: flip over y-axis appears in choice C, which is not consistent with the algebra — possibly a typo in the answer key).
✅ Answer key says: (C)
▶️ Answer/Explanation
Factor numerator: \( (x+8)(x-4) \), denominator: \( x+4 \).
Critical points: \( x = -8, -4, 4 \).
Sign analysis:
– For \( x < -8 \), negative.
– For \( (-8, -4) \), positive (≥0 at \( x=-8 \)).
– For \( (-4, 4) \), negative.
– For \( x > 4 \), positive (≥0 at \( x=4 \)).
Cannot include \( x = -4 \) (denominator zero).
Thus solution: \( [-8, -4) \cup [4, \infty) \).
✅ Answer: (A)
▶️ Answer/Explanation
Numerator degree = 8, leading term \(-x^8\).
Denominator: \( -x^7 + 2x^8 + \dots \) → leading term \( 2x^8 \).
Horizontal asymptote: ratio of leading coefficients = \( \frac{-1}{2} = -\frac12 \).
Since degrees are equal, \( \lim_{x \to \pm\infty} f(x) = -\frac12 \).
Thus \( \lim_{x \to -\infty} f(x) = -\frac12 \).
✅ Answer: (B)
▶️ Answer/Explanation
For \( 0 < x < 1 \), smaller exponents yield larger outputs because \( x^a \) is decreasing in \( a \) for fixed \( x \in (0,1) \).
Exponents: \( 9/2 = 4.5 \), \( 4 \), \( 2/3 \approx 0.667 \), \( 1/7 \approx 0.143 \).
The smallest exponent is \( 1/7 \), so \( h(x) = x^{1/7} \) is largest in \( (0,1) \).
Example: \( x = 0.5 \), \( h(0.5) \approx 0.9 \), \( g(0.5) \approx 0.63 \), \( f(0.5)=0.0625 \), \( d(0.5) \) even smaller.
✅ Answer: (D)
▶️ Answer/Explanation
\( 4\log_7 x + 4\log_7 z – 24\log_7 y = \log_7 (x^4 z^4) – \log_7 (y^{24}) = \log_7 \left( \frac{x^4 z^4}{y^{24}} \right) \).
✅ Answer: (C)
▶️ Answer/Explanation
\( \ln\left(\frac{-3x-7}{7}\right) = 5 \) ⇒ \( \frac{-3x-7}{7} = e^5 \) ⇒ \( -3x-7 = 7e^5 \) ⇒ \( -3x = 7e^5 + 7 \) ⇒ \( x = \frac{-7(e^5+1)}{3} \).
This matches option (A) if we interpret \( 7e^{5+7} \) as a misprint for \( 7(e^5+1) \) in the numerator.
✅ Answer: (A)
▶️ Answer/Explanation
Common ratio \( r = 64/32 = 2 \).
\( a_4 = a_1 \cdot 2^{3} = 8a_1 = 32 \) ⇒ \( a_1 = 4 \).
Thus \( a_n = 4 \cdot 2^{n-1} \).
✅ Answer: (C)
▶️ Answer/Explanation
Doubling time = 3 hours ⇒ in 1 day = 24 hours, number of doubling periods = \( 24/3 = 8 \).
Thus in \( t \) days: \( 8t \) periods.
Growth factor each period = 2, so \( y = a_0 \cdot 2^{8t} \).
✅ Answer: (B)
▶️ Answer/Explanation
They transformed to \((x, \log y)\) and fitted a linear model. The residual plot (not fully visible here) showed a pattern, indicating the linear model on transformed data does not fit well.
That means \( \log y \) is not linear in \( x \), so the original \( (x, y) \) relationship is not exponential.
Thus exponential regression is not appropriate.
✅ Answer: (A)
▶️ Answer/Explanation
We need vertical asymptote at \( x=2 \) from the left (\( x \to 2^- \)) with \( f(x) \to +\infty \).
– For (B): \( y = -2 \log(-2x + 4) – 2 \). Argument \( -2x+4 > 0 \Rightarrow x < 2 \). As \( x \to 2^- \), \( -2x+4 \to 0^+ \), \( \log(\text{small } +) \to -\infty \), multiply by -2: \( +\infty \), then minus 2 still \( +\infty \). ✅
– Others: (A) asymptote at \( x=-2 \), (C) asymptote at \( x=2^+ \), (D) \( x\to 2^- \) gives \( -\infty \).
✅ Answer: (B)
▶️ Answer/Explanation
The given graph has a horizontal asymptote \( y = 1 \), passes through point \( (1, -1) \), and is decreasing (since base > 1 but multiplied by -2).
Check (C): \( y = -2 \cdot 2^{x-1} + 1 \).
At \( x=1 \), \( 2^{0} = 1 \) ⇒ \( y = -2 + 1 = -1 \). ✅
As \( x \to -\infty \), \( 2^{x-1} \to 0 \) ⇒ \( y \to 1 \). ✅
Matches graph characteristics.
✅ Answer: (C)
▶️ Answer/Explanation
\( r^2 = 25 \cos 2\theta \) is a lemniscate (figure-eight).
Since \( r^2 \geq 0 \) ⇒ \( \cos 2\theta \geq 0 \) ⇒ \( \theta \in [0, \frac{\pi}{4}] \cup [\frac{3\pi}{4}, \frac{5\pi}{4}] \cup [\frac{7\pi}{4}, 2\pi) \) in one full cycle, but in \( [0, 2\pi) \) gives symmetric lobes along the polar axis (x-axis).
The graph matching this horizontal lemniscate is option (A).
✅ Answer: (A)
▶️ Answer/Explanation
A sum/difference of sine and cosine with same frequency can be written as a single sine or cosine with a phase shift: \( \sin 3\theta – \cos 3\theta = \sqrt{2} \sin(3\theta – \pi/4) \), which is sinusoidal.
Thus it is a sinusoidal function.
✅ Answer: (C)
▶️ Answer/Explanation
\( x = \sqrt{17}, y = -8 \), so \( \tan \theta = \frac{y}{x} = \frac{-8}{\sqrt{17}} \).
Rationalize: \( -\frac{8\sqrt{17}}{17} \).
✅ Answer: (D)
▶️ Answer/Explanation
\( \csc(-\pi/4) = -\sqrt{2} \).
\( \sec^{-1}(-\sqrt{2}) \) means angle \( \theta \in [0, \pi], \theta \ne \pi/2 \), such that \( \sec \theta = -\sqrt{2} \) ⇒ \( \cos \theta = -1/\sqrt{2} \).
In \([0, \pi]\), \( \cos \theta = -1/\sqrt{2} \) ⇒ \( \theta = 3\pi/4 \).
✅ Answer: (B)
▶️ Answer/Explanation
Rectangular: \( x = -\frac{3\sqrt{3}}{2} < 0, y = \frac{3}{2} > 0 \) ⇒ QII.
Normal polar: \( r = \sqrt{x^2 + y^2} = 3 \), \( \theta = \tan^{-1}(y/x) = 5\pi/6\).
If \( r < 0 \), equivalent polar coordinate is \( (-3, 5\pi/6) \) because negative \( r \) means go opposite to angle \( \theta \) but adding \( \pi \) to \( \theta \) gives \( (3, 11\pi/6) \) ✅ Answer: (B)
▶️ Answer/Explanation
Range of \( \sec \theta \) is \( (-\infty, -1] \cup [1, \infty) \).
Multiply by \(-2\):
For \( \sec \theta \ge 1 \): \( -2\sec \theta \le -2 \) → after +1: \( (-\infty, -1] \).
For \( \sec \theta \le -1 \): \( -2\sec \theta \ge 2 \) → after +1: \( [3, \infty) \).
So range = \( (-\infty, -1] \cup [3, \infty) \).
✅ Answer: (C) \( (-\infty, -1) \cup (3, \infty) \)
▶️ Answer/Explanation
Simplify: \( -3\sin(3x – 3\pi) = -3\sin(3x – \pi – 2\pi) = -3\sin(3x – \pi) \).
Since \( \sin(\theta – \pi) = -\sin\theta \), we get \( -3[-\sin(3x)] = 3\sin(3x) \).
i. \( 3\sin(3x) \) is odd → True.
ii. Amplitude = 3, not \( \sqrt{6} \) → False.
iii. Period = \( \frac{2\pi}{3} \) → True.
Thus only i and iii are true.
✅ Answer: (A) i and iii only
▶️ Answer/Explanation
\( x \) = percent of pure water. As \( x \to 1^- \) (100% pure water), denominator \( x-1 \to 0^- \), numerator \( \to 0.1+5-3 = 2.1 >0 \), so \( C(x) \to -\infty \). But likely they intended vertical asymptote approaching +∞ in cost as purity nears 100% from below (if numerator also small positive but denominator sign yields +∞).
Interpretation: It becomes infinitely costly to approach 100% purity, but 100% is unreachable (x=1 not in domain).
✅ Answer: (B)
▶️ Answer/Explanation
First height = 6 ⇒ \( \sin(3\sqrt{x/3}) \approx 0.85 \) ⇒ first time occurs while sine is increasing ⇒ \( dy/dx > 0 \) (rate of change positive).
Second derivative: \( d^2y/dx^2 \) at that point: sine curve concave down before maximum ⇒ \( d^2y/dx^2 < 0 \).
So rate of change (first derivative) is increasing (but rate of change of rate = second derivative) is negative → means rate of change is increasing at a decreasing rate.
✅ Answer: (C)
▶️ Answer/Explanation
First 3 years: yearly multiplier = \( 0.85 \). Value after 3 years: \( 70000 \times 0.85^3 \).
Next 2 years: yearly multiplier = \( 0.95 \). Multiply by \( 0.95^2 \).
Total: \( 70000 \times 0.85^3 \times 0.95^2 \).
\( 0.85^3 = 0.614125 \), \( 0.95^2 = 0.9025 \), product ≈ \( 0.554254 \).
\( 70000 \times 0.554254 \approx 38797.78 \).
✅ Answer: (D) $38,797.35
▶️ Answer/Explanation
From the graphs (not visible here), \( q(x) \) has a zero at \( x=-5/2 \) with odd multiplicity, \( p(x) \) not zero there, and as \( x \to -\frac{5}{2}^+ \), \( q(x) \) changes sign so that \( f(x) \to +\infty \). Horizontal asymptote is 0 (denominator degree > numerator degree), so (B) is false. Answer choice (D) matches the described behavior.
✅ Answer: (D)
▶️ Answer/Explanation
\( \log_7 \frac{56}{121} = \log_7 56 – \log_7 121 \).
\( 56 = 8 \times 7 \) ⇒ \( \log_7 56 = \log_7 8 + \log_7 7 = Y + 1 \).
\( 121 = 11^2 \) ⇒ \( \log_7 121 = 2\log_7 11 = 2X \).
Thus expression = \( (Y+1) – 2X = 1 + Y – 2X \).
✅ Answer: (C)
▶️ Answer/Explanation
Rearrange: \( x^4 + x^3 -12x^2 + 4x + 16 > 0 \).
Factor: roots at \( x=2 \) (double), \( x=-4 \), \( x=-1 \) ⇒ \( (x-2)^2(x+4)(x+1) > 0 \).
Since \( (x-2)^2 \ge 0 \), inequality depends on \( (x+4)(x+1) > 0 \) except \( x=2 \) where it equals 0 (but >0 not satisfied at equality).
Sign chart: positive in \( (-\infty, -4) \), negative in \( (-4, -1) \), positive in \( (-1, 2) \cup (2, \infty) \).
So solution set: \( (-\infty, -4) \cup (-1, 2) \cup (2, \infty) \).
✅ Answer: (A)
▶️ Answer/Explanation
Original model: \( y = a + b \ln x \).
Since \( \ln x = (\ln 10) \log x \approx 2.302585 \log x \), we have \( b_{\log} = b_{\ln} \cdot \ln 10 \).
Given \( a = 2.099 \), \( b_{\ln} \approx 1.659 \) (from fit), then \( b_{\log} \approx 1.659 \times 2.302585 \approx 3.82 \).
Thus \( y = 2.099 + 3.82 \log x \).
✅ Answer: (C)
▶️ Answer/Explanation
\(\tan \theta = 3 > 0 \) ⇒ Quadrants I or III.
\(\cos \theta < 0 \) ⇒ Quadrants II or III.
Intersection ⇒ Quadrant III.
Reference angle: \( \arctan 3 \approx 1.249 \) rad.
In QIII: \( \theta = \pi + 1.249 \approx 4.391 \) rad.
✅ Answer: (C)
▶️ Answer/Explanation
Given \( f(\pi/4) = 2 \), \( f'(\pi/4) = -3 \), \( f”(\pi/4) = 2 \).
Use quadratic approximation:
\( f(\theta) \approx 2 – 3(\theta – \pi/4) + (\theta – \pi/4)^2 \) (since \( f”/2 = 1 \) times square term).
At \( \theta = \pi/3 \), \( \Delta\theta = \pi/12 \approx 0.2618 \).
Estimate: \( 2 – 3(0.2618) + (0.2618)^2 \approx 2 – 0.7854 + 0.0685 = 1.2831 \).
Thus possible value near 1.283, so closest listed is 1.
✅ Answer: (C) \( r = 1 \)
▶️ Answer/Explanation
Radius \( R = 15 \) m, center height = lowest point \( 15 \) m + \( R = 30 \) m.
Period = \( 30 \) sec (2 rotations/minute).
Model: \( h(t) = 30 – 15\cos\left( \frac{2\pi t}{30} \right) = 30 – 15\cos\left( \frac{\pi t}{15} \right) \) (starting at bottom).
\( t = 4 \) min 25 sec = \( 265 \) sec.
Angle = \( \frac{265\pi}{15} \) rad. Mod \( 2\pi \): \( \frac{265}{30} \approx 8.833 \) cycles ⇒ extra \( 0.833 \) cycles ⇒ angle \( \approx 1.6667\pi \) rad (\( 300^\circ \)).
\( \cos 300^\circ = 0.5 \).
\( h = 30 – 15 \times 0.5 = 22.5 \) m.
✅ Answer: (A) 22.5 meters
▶️ Answer/Explanation
Slope from origin to point: \( m_1 = \tan(\pi/5) \approx 0.7265 \).
Tangent line is perpendicular to radius ⇒ slope \( m_2 = -\frac{1}{m_1} = -\cot(\pi/5) \approx -\frac{1}{0.7265} \approx -1.376 \).
✅ Answer: (B) \(-1.38\)
▶️ Answer/Explanation
Binomial expansion: \( (y^2 + 4x)^4 = \sum_{k=0}^{4} \binom{4}{k} (y^2)^{4-k}(4x)^k \).
Third term ⇒ \( k = 2 \) (since \( k=0 \) is first term).
Term = \( \binom{4}{2} (y^2)^{2} (4x)^2 = 6 \cdot y^4 \cdot 16 x^2 = 96 y^4 x^2 \).
✅ Answer: (B) \( 96y^4x^2 \)
▶️ Answer/Explanation
(A) i. \(g(f^{-1}(1))\)
- From the graph of \(f(x)\), \(f^{-1}(1)\) means find \(x\) such that \(f(x) = 1\).
- From the graph, \(f(x) = 1\) at approximately \(x \approx 2.5\). So \(f^{-1}(1) \approx 2.5\).
- From the table, \(g(2) = 1\), \(g(4) = 2\). Since \(g\) is increasing and continuous, by linear interpolation for \(x \approx 2.5\): \(g(2.5) \approx 1 + \frac{2.5-2}{4-2} \cdot (2-1) = 1 + 0.25 = 1.25\).
- Thus, \(g(f^{-1}(1)) \approx 1.25\).
(A) ii. \(h(x) = \frac{g(x)}{f(x)}\), \(x > 0\)
- Discontinuities occur where \(f(x) = 0\) or where \(f\) is undefined (but \(f\) appears continuous from the graph for \(x>0\)).
- From the graph, \(f(x) = 0\) at \(x \approx 1.5\) (where graph crosses x-axis). That is the only zero of \(f\) for \(x>0\).
- At \(x \approx 1.5\), \(g(x) \neq 0\) (since \(g\) is about 0.5 between \(g(1)=0\) and \(g(2)=1\)).
- Thus, \(h(x)\) has a vertical asymptote at \(x \approx 1.5\) because denominator → 0, numerator nonzero → \(h(x) \to \pm\infty\).
- Type: infinite discontinuity (vertical asymptote).
(B) i. End behavior of \(j(x) = \frac{1}{4.998} \ln(x) \cdot e^{2\sin(\sqrt{x})}\) as \(x \to \infty\)
- As \(x \to \infty\), \(\ln(x) \to \infty\).
- The factor \(e^{2\sin(\sqrt{x})}\) oscillates between \(e^{-2}\) and \(e^{2}\) because \(\sin(\sqrt{x})\) oscillates in \([-1,1]\).
- Thus \(j(x)\) oscillates with amplitude growing like \(\ln(x)\). The limit does not exist because oscillations persist and amplitude grows without bound.
- Limit expression: \(\lim_{x \to \infty} j(x)\) does not exist.
(B) ii. Vertical asymptote as \(x \to 0^+\)
- As \(x \to 0^+\), \(\ln(x) \to -\infty\).
- The factor \(e^{2\sin(\sqrt{x})} \to e^{0} = 1\) (since \(\sqrt{x} \to 0\), \(\sin(\sqrt{x}) \to 0\)).
- Thus \(j(x) \to -\infty\).
- Limit expression: \(\lim_{x \to 0^+} j(x) = -\infty\). So vertical asymptote at \(x=0\).
(C) Best model for \(g(x)\) from the table:
- Check changes in \(g(x)\) as \(x\) doubles:
- From \(x=0.5\) to \(x=1\), \(\Delta g = 1\).
- From \(x=1\) to \(x=2\), \(\Delta g = 1\).
- From \(x=2\) to \(x=4\), \(\Delta g = 1\).
- From \(x=4\) to \(x=8\), \(\Delta g = 1\).
- Each time \(x\) doubles, \(g(x)\) increases by about 1. This suggests a logarithmic relationship: \(g(x) \approx a + b \ln(x)\).
- Thus, \(g(x)\) is best modeled by a logarithmic function.
▶️ Answer/Explanation
(A) Exponential model \(P(t) = ae^{bt}\)
- Given: \(P(1) = 145\), \(P(2) = 115\).
- Divide the equations: \(\frac{P(2)}{P(1)} = \frac{115}{145} = \frac{23}{29} \approx 0.7931 = e^{b(2-1)} = e^b\).
- So \(b = \ln\left(\frac{23}{29}\right) \approx \ln(0.7931) \approx -0.2318\).
- Use \(P(1) = ae^{b\cdot1} = 145\): \(a = 145 \cdot e^{-b} = 145 \cdot \frac{29}{23} \approx 145 \cdot 1.2609 \approx 182.83\).
- Alternatively, compute directly: \(a = P(1) e^{-b} = 145 \cdot \frac{29}{23} = \frac{145 \cdot 29}{23} = \frac{4205}{23} \approx 182.826\).
- Thus, \(P(t) \approx 182.826 e^{-0.2318 t}\).
(B) i. Average rate of change from \(t = 0\) to \(t = 4\)
- \(P(0) = a \approx 182.826\).
- \(P(4) = 182.826 e^{-0.2318 \cdot 4} \approx 182.826 e^{-0.9272} \approx 182.826 \cdot 0.3956 \approx 72.34\).
- Average rate = \(\frac{P(4) – P(0)}{4 – 0} \approx \frac{72.34 – 182.826}{4} = \frac{-110.486}{4} \approx -27.622\) pellets per minute.
- Meaning: On average over the first 4 minutes, the dog ate about 27.6 pellets per minute.
(B) ii. Comparison of average rates
- For an exponential decay \(P(t) = ae^{bt}\) with \(b < 0\), the function is decreasing and concave up (since \(P”(t) = ab^2 e^{bt} > 0\)).
- In a concave up decreasing function, the average rate of change over an interval decreases in magnitude (becomes less negative) as the interval moves to the right.
- Thus, the average rate from \(t = 4\) to \(t = t_a\) will be greater (less negative) than that from \(t = 0\) to \(t = 4\).
- Reason: The instantaneous rate (derivative) is negative but increasing (becoming less negative), so later intervals have a smaller rate of loss.
(C) Limit as \(t \to \infty\)
- \(\lim_{t \to \infty} P(t) = \lim_{t \to \infty} ae^{bt} = 0\) because \(b < 0\).
- Context: The model predicts the pellets approach zero but never actually reach it. In reality, the dog will finish all pellets in finite time, so the model is not perfectly accurate for very large \(t\). However, it is a good approximation for the decay process while pellets remain.
- Thus, the model does not fully make sense as \(t \to \infty\) because it predicts the dog never finishes, while in reality the bowl becomes empty at some finite time.