Q: The functions $$g$$ and $$h$$ are given by$$g(x)=3\log a-2\log b+\frac{1}{2}\log a+\log_{10}c$$$$h(x)=\sec^2 x\tan x-\tan x$$(A) i. Rewrite $$g(x)$$ as a single logarithm with base $$10$$ without negative exponents. Your answer should be in the form $$\log(\text{expression})$$ii. Rewrite $$h(x)$$ as an expression in which $$\tan x$$ appears once and no other trigonometric functions are involved.(B) The functions $$j$$ and $$k$$ are given by$$j(x)=2\csc^2 x-3$$$$k(x)=5e^{4-2x}+1$$i. Solve $$j(x)=1$$ for all values of $$x$$ in the interval $$[0,2\pi)$$.ii. Solve $$k(x)=6$$ for all real values of $$x$$. Express your answer exactly.(C) Two functions $$a$$ and $$b$$ are given by$$a(x)=4-2\sec(2x)$$ $$b(x)=1+\sec(2x)$$Find all values of $$x$$ in the interval $$[0,2\pi)$$ for which $$a(x)=b(x)$$.

(A)(i) Combine like logarithms.$$3\log a+\frac{1}{2}\log a=\frac{7}{2}\log a.$$So:$$g(x)=\frac{7}{2}\log a-2\log b+\log c.$$Use the power rule.$$\frac{7}{2}\log a=\log\left(a^{7/2}\right)$$$$2\log b=\log\left(b^2\right).$$So:$$g(x)=\log\left(a^{7/2}\right)-\log\left(b^2\right)+\log(c).$$Use the product and quotient rules.$$g(x)=\log\left(\frac{a^{7/2}c}{b^2}\right).$$Final answer: $$\log\left(\frac{a^{7/2}c}{b^2}\right)$$(A)(ii) Start with:$$h(x)=\sec^2 x\tan x-\tan x.$$Factor out $$\tan x.$$$$h(x)=\tan x(\sec^2 x-1).$$Use the identity:$$\sec^2 x-1=\tan^2 x.$$Substitute.$$h(x)=\tan x\cdot\tan^2 x$$$$h(x)=\tan^3 x.$$Final answer: $$\tan^3 x$$(B)(i) Solve:$$2\csc^2 x-3=1.$$Add $$3.$$$$2\csc^2 x=4.$$Divide by $$2.$$$$\csc^2 x=2.$$Take reciprocals.$$\sin^2 x=\frac{1}{2}.$$So:$$\sin x=\pm\frac{\sqrt{2}}{2}.$$On $$[0,2\pi),$$ the solutions are:$$x=\frac{\pi}{4}$$$$x=\frac{3\pi}{4}$$$$x=\frac{5\pi}{4}$$$$x=\frac{7\pi}{4}.$$Final answer: $$\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}$$(B)(ii) Solve:$$5e^{4-2x}+1=6.$$Subtract $$1.$$$$5e^{4-2x}=5.$$Divide by $$5.$$$$e^{4-2x}=1.$$Take the natural logarithm.$$4-2x=0.$$$$-2x=-4.$$$$x=2.$$Final answer: $$2$$(C) Set the functions equal.$$4-2\sec(2x)=1+\sec(2x).$$Subtract $$1.$$$$3-2\sec(2x)=\sec(2x).$$Add $$2\sec(2x).$$$$3=3\sec(2x).$$Divide by $$3.$$$$1=\sec(2x).$$So:$$\cos(2x)=1.$$This happens when:$$2x=2\pi n.$$$$x=\pi n.$$On $$[0,2\pi),$$ the solutions are:$$x=0$$$$x=\pi.$$Final answer: $$0,\pi$$

Question 1

A rotating spotlight is mounted on a tower and sweeps around at a constant speed. The light completes one full rotation every $$40$$ seconds. The light is mounted $$60$$ feet above the ground, and the beam has a length of $$20$$ feet. The height of the tip of the beam above the ground varies periodically. At time $$t=0$$, the beam is pointing straight upward.(A) This situation can be modeled by a sinusoidal function of the form $$h(t)=a\cos(bt)+d,$$ where $$t$$ is time in seconds and $$h(t)$$ is the height of the tip of the beam in feet. Find the function $$h(t).$$(B) On the interval $$0$$<$$t$$<$$40,$$ when is the height of the tip equal to $$60+10\sqrt{2}$$ feet? Give your answers in exact form.(C) i. Determine all times on the interval $$0\le t$$<$$40$$ at which the graph of $$h(t)$$ has a point of inflection.ii. Explain the meaning of these points in the context of the problem, in terms of how the height is changing and how the rate of change is changing.(D) Find the average rate of change of the height from $$t=10$$ to $$t=30,$$ and interpret its meaning in context.

Accepted Answer

(A) The spotlight is mounted $$60$$ feet above the ground, so the midline is $$d=60.$$The beam has length $$20$$ feet, so the amplitude is $$a=20.$$The light completes one full rotation every $$40$$ seconds, so the period is $$T=40.$$For a cosine function,$$\frac{2\pi}{b}=40.$$Solve for $$b$$:$$b=\frac{2\pi}{40}=\frac{\pi}{20}.$$At $$t=0$$ the beam points straight upward, so the height is at its maximum. A positive cosine model fits directly.Therefore,$$h(t)=20\cos\left(\frac{\pi}{20}t\right)+60.$$Final answer: $$h(t)=20\cos\left(\frac{\pi}{20}t\right)+60$$(B) Set the height equal to $$60+10\sqrt{2}.$$$$20\cos\left(\frac{\pi}{20}t\right)+60=60+10\sqrt{2}.$$Subtract $$60$$:$$20\cos\left(\frac{\pi}{20}t\right)=10\sqrt{2}.$$Divide by $$20$$:$$\cos\left(\frac{\pi}{20}t\right)=\frac{\sqrt{2}}{2}.$$On one full cycle, cosine equals $$\frac{\sqrt{2}}{2}$$ at $$\frac{\pi}{4}$$ and $$\frac{7\pi}{4}.$$So,$$\frac{\pi}{20}t=\frac{\pi}{4}$$ or $$\frac{\pi}{20}t=\frac{7\pi}{4}.$$Solve for $$t$$:$$t=5$$ or $$t=35.$$Final answer: $$t=5,35$$(C)(i) Points of inflection occur where the concavity changes.Start with$$h(t)=20\cos\left(\frac{\pi}{20}t\right)+60.$$Differentiate:$$h'(t)=-20\sin\left(\frac{\pi}{20}t\right)\cdot\frac{\pi}{20}.$$$$h'(t)=-\pi\sin\left(\frac{\pi}{20}t\right).$$Differentiate again:$$h''(t)=-\pi\cos\left(\frac{\pi}{20}t\right)\cdot\frac{\pi}{20}.$$$$h''(t)=-\frac{\pi^2}{20}\cos\left(\frac{\pi}{20}t\right).$$Set the second derivative equal to zero:$$-\frac{\pi^2}{20}\cos\left(\frac{\pi}{20}t\right)=0.$$So,$$\cos\left(\frac{\pi}{20}t\right)=0.$$This happens when $$\frac{\pi}{20}t=\frac{\pi}{2}$$ or $$\frac{\pi}{20}t=\frac{3\pi}{2}.$$Solve for $$t$$:$$t=10$$ or $$t=30.$$Final answer: $$t=10,30$$(C)(ii) These points mean the tip of the beam is at the midline height of $$60$$ feet.At these times, the height is changing most rapidly.Also, the graph changes concavity there, which means the rate of change stops increasing and starts decreasing or stops decreasing and starts increasing.(D) The average rate of change from $$t=10$$ to $$t=30$$ is:$$\frac{h(30)-h(10)}{30-10}.$$Compute the function values:$$h(10)=20\cos\left(\frac{\pi}{20}\cdot10\right)+60.$$$$h(10)=20\cos\left(\frac{\pi}{2}\right)+60=60.$$$$h(30)=20\cos\left(\frac{\pi}{20}\cdot30\right)+60.$$$$h(30)=20\cos\left(\frac{3\pi}{2}\right)+60=60.$$So,$$\frac{h(30)-h(10)}{20}=\frac{60-60}{20}=0.$$Interpretation:From $$t=10$$ to $$t=30,$$ the average change in height is $$0$$ feet per second. Over that interval, the tip ends at the same height where it started.Final answer: $$0$$

Question 2

The functions $$g$$ and $$h$$ are given by$$g(x)=3\log a-2\log b+\frac{1}{2}\log a+\log_{10}c$$$$h(x)=\sec^2 x	an x-	an x$$(A) i. Rewrite $$g(x)$$ as a single logarithm with base $$10$$ without negative exponents. Your answer should be in the form $$\log(	ext{expression})$$ii. Rewrite $$h(x)$$ as an expression in which $$	an x$$ appears once and no other trigonometric functions are involved.(B) The functions $$j$$ and $$k$$ are given by$$j(x)=2\csc^2 x-3$$$$k(x)=5e^{4-2x}+1$$i. Solve $$j(x)=1$$ for all values of $$x$$ in the interval $$[0,2\pi)$$.ii. Solve $$k(x)=6$$ for all real values of $$x$$. Express your answer exactly.(C) Two functions $$a$$ and $$b$$ are given by$$a(x)=4-2\sec(2x)$$ $$b(x)=1+\sec(2x)$$Find all values of $$x$$ in the interval $$[0,2\pi)$$ for which $$a(x)=b(x)$$.

Accepted Answer

(A)(i) Combine like logarithms.$$3\log a+\frac{1}{2}\log a=\frac{7}{2}\log a.$$So:$$g(x)=\frac{7}{2}\log a-2\log b+\log c.$$Use the power rule.$$\frac{7}{2}\log a=\log\left(a^{7/2}ight)$$$$2\log b=\log\left(b^2ight).$$So:$$g(x)=\log\left(a^{7/2}ight)-\log\left(b^2ight)+\log(c).$$Use the product and quotient rules.$$g(x)=\log\left(\frac{a^{7/2}c}{b^2}ight).$$Final answer: $$\log\left(\frac{a^{7/2}c}{b^2}ight)$$(A)(ii) Start with:$$h(x)=\sec^2 x	an x-	an x.$$Factor out $$	an x.$$$$h(x)=	an x(\sec^2 x-1).$$Use the identity:$$\sec^2 x-1=	an^2 x.$$Substitute.$$h(x)=	an x\cdot	an^2 x$$$$h(x)=	an^3 x.$$Final answer: $$	an^3 x$$(B)(i) Solve:$$2\csc^2 x-3=1.$$Add $$3.$$$$2\csc^2 x=4.$$Divide by $$2.$$$$\csc^2 x=2.$$Take reciprocals.$$\sin^2 x=\frac{1}{2}.$$So:$$\sin x=\pm\frac{\sqrt{2}}{2}.$$On $$[0,2\pi),$$ the solutions are:$$x=\frac{\pi}{4}$$$$x=\frac{3\pi}{4}$$$$x=\frac{5\pi}{4}$$$$x=\frac{7\pi}{4}.$$Final answer: $$\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}$$(B)(ii) Solve:$$5e^{4-2x}+1=6.$$Subtract $$1.$$$$5e^{4-2x}=5.$$Divide by $$5.$$$$e^{4-2x}=1.$$Take the natural logarithm.$$4-2x=0.$$$$-2x=-4.$$$$x=2.$$Final answer: $$2$$(C) Set the functions equal.$$4-2\sec(2x)=1+\sec(2x).$$Subtract $$1.$$$$3-2\sec(2x)=\sec(2x).$$Add $$2\sec(2x).$$$$3=3\sec(2x).$$Divide by $$3.$$$$1=\sec(2x).$$So:$$\cos(2x)=1.$$This happens when:$$2x=2\pi n.$$$$x=\pi n.$$On $$[0,2\pi),$$ the solutions are:$$x=0$$$$x=\pi.$$Final answer: $$0,\pi$$

Section II Part B

Frequently Asked Questions