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Quiz 1 Solutions

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Question 1

Which of the following quantities is undefined?

  1. sin⁡−1(−12)\sin^{-1}\p{-\frac{1}{2}}
  2. cos⁡−1(−2)\cos^{-1}\p{-2}
  3. csc⁡−1(12)\csc^{-1}\p{\frac{1}{2}}
  4. csc⁡−1(2)\csc^{-1}\p{2}
Solution.

Only (2) and (3) are undefined.

The range of sin⁡x\sin{x} and cos⁡x\cos{x} is [−1,1]\br{-1, 1}, so (1) is defined and (2) is undefined. Similarly, the range of csc⁡x\csc{x} is (−∞,−1)∪(1,∞)\p{-\infty, -1} \cup \p{1, \infty}, so (3) is undefined and (4) is defined.

Question 2

Which of the following statements is true?

  1. log⁡b(a)=ln⁡bln⁡a\log_b\p{a} = \frac{\ln{b}}{\ln{a}}
  2. log⁡b(ba)=a\log_b\p{b^a} = a
  3. ln⁡(a+b)=ln⁡(a)+ln⁡b\ln\p{a + b} = \ln\p{a} + \ln{b}
Solution.

Only (2) is correct.

(1) looks like the change-of-base formula, but the fraction is flipped, so it's false. (2) is true because log⁡b(x)\log_b\p{x} is the inverse of bxb^x. (3) is false in general; for example, if a=b=1a = b = 1, then

ln⁡(1+1)=ln⁡(2)≠0=ln⁡1+ln⁡1.\ln\p{1 + 1} = \ln\p{2} \neq 0 = \ln{1} + \ln{1}.

So the only true statement is (2).

Question 3

Which of the following equals the slope of the tangent to f(x)=ln⁡(1x)f\p{x} = \ln\p{\frac{1}{x}} at the point P(e,−1)P\p{e, -1}?

Solution.

The slope of the tangent line is just the derivative of f(x)f\p{x} at x=ex = e:

f′(e)=lim⁡h→0f(e+h)−f(e)h=lim⁡h→0ln⁡(1e+h)+1h.f'\p{e} = \lim_{h\to0} \frac{f\p{e + h} - f\p{e}}{h} = \boxed{\lim_{h\to0} \frac{\ln\p{\frac{1}{e + h}} + 1}{h}}.

Question 4

Compute ddx4sin⁡(πx)\displaystyle \deriv{}{x} 4^{\sin\p{\pi x}}.

Solution.

By the chain rule,

ddx4sin⁡(πx)=ln⁡(4)4sin⁡(πx)⋅(ddxsin⁡(πx))=ln⁡(4)4sin⁡(πx)⋅(πcos⁡(πx))=4sin⁡(πx)πln⁡(4)cos⁡(πx).\begin{aligned} \deriv{}{x} 4^{\sin\p{\pi x}} &= \ln\p{4} 4^{\sin\p{\pi x}} \cdot \p{\deriv{}{x} \sin\p{\pi x}} \\ &= \ln\p{4} 4^{\sin\p{\pi x}} \cdot \p{\pi \cos\p{\pi x}} \\ &= \boxed{4^{\sin\p{\pi x}} \pi\ln\p{4} \cos\p{\pi x}}. \end{aligned}

Question 5

Compute lim⁡x→1−x1/(1−x)\displaystyle \lim_{x\to1^-} x^{1/\p{1-x}}.

Solution.

This has the indeterminate form 1∞1^\infty, so we can try to apply L'Hôpital's. Set y=x1/(1−x)y = x^{1/\p{1-x}} so that

lim⁡x→1−ln⁡y=lim⁡x→1−ln⁡x1−x=Hlim⁡x→1−1x−1=−1.\lim_{x\to1^-} \ln{y} = \lim_{x\to1^-} \frac{\ln{x}}{1 - x} \overset{H}{=} \lim_{x\to1^-} \frac{\frac{1}{x}}{-1} = -1.

Thus,

lim⁡x→1−x1/(1−x)=lim⁡x→1−eln⁡y=elim⁡x→1−ln⁡y=e−1=1e.\lim_{x\to1^-} x^{1/\p{1-x}} = \lim_{x\to1^-} e^{\ln{y}} = e^{\lim_{x\to1^-} \ln{y}} = e^{-1} = \boxed{\frac{1}{e}}.