School of Mathematics and Statistics
Carleton University
Math. 69.104
TEST 2 SOLUTIONS

This test is out of a Total of 40.

PART I: Multiple Choice Questions
(Choose and CIRCLE only ONE answer)

1. [3 marks] Let $f(x)=\ln \sqrt{\sin x}$. Which of the following expressions represents the value of $f^{\prime}(\frac{\pi}{4})$?

   
   (a) 0, $\frac{1}{2}$, (c) $1$, (d) This derivative does not exist.

   
   

2. [3 marks] Let $$f(x)=\sqrt{{e^{x^2}} + 1}$$. Then $f^{\prime}(0)$ is equal to:

   
   0, (b) $${1}$$, (c) $${e^{-2}}$$, (d) $${ - e^{\frac{1}{2}}}$$

   
   

3. [3 marks] Let $f(x)=1+x^x$. Then the value of the derivative $f^{\prime}(1)$ is given by:

   
   (a) 0, $1$, (c) Undefined, (d) $2$.

   
   

4. [3 marks] Let $f(x) = (1.2)^x$. Then $f^{\prime}(x)$ is:

   
   (a) 16, (b) $(1.2)^x$, $(1.2)^x \ln (1.2)$, (d) $(1.2)^x (\ln x + 1)$.

   
   

5. [3 marks] Answer TRUE or FALSE: For any value of $x$ it is true that $$\ln \left ( \ln \left ( e^{e^x} \right) \right ) - \ln (e^x) = 0.$$

   
   TRUE, (b) FALSE

   
   

PART II: Show all work here.
No additional pages will be accepted

1. [6+7 marks] Evaluate the following limits using any method: a) $$\lim_{x \to \infty}{\frac{x^2}{x+2e^x}}$$. Solution Use L'Hospital's Rule since this is of the form $\frac{\infty}{\infty}$. This gives
   $$\lim_{x \to \infty}{\frac{x^2}{x+2e^x}} = \lim_{x \to \infty}{\fr... ...= \lim_{x \to \infty}{\frac{2}{2e^x}} = \lim_{x \to \infty}{\frac{1}{e^x}} = 0.$$
   Hence, the original limit is also 0. b) Evaluate $$\lim_{x \to \infty}{\sqrt{x^2+1} - x}$$ using any method. Justify your answer mathematically, not just by approximation. Solution This limit problem is of the form $\infty - \infty$ so we can't use the Rule immediately. But we can rationalize the numerator and simplify...
   $$\sqrt{x^2+1} - x = \frac{{(\sqrt{x^2+1} - x)}{(\sqrt{x^2+1} + x)}}{\sqrt{x^2+1} + x} = \frac{1}{\sqrt{x^2+1} + x}.$$
   It follows that
   $$\lim_{x \to \infty}{\sqrt{x^2+1} + x} = \lim_{x \to \infty}{\frac{1}{\sqrt{x^2+1} + x}} = 0.$$

2. [6+6 marks] Let $N(t)$ denote the amount of $Pu^{240}$ at time $t$ and assume its half-life is approximately $6500$ years. Using a Law of Decay, a) Determine the general formula for $N(t)$ given an initial sample of $N(0)$ grams. Solution
   $$N(t) = N(0)e^{kt}$$
   where $$k = - \frac{\ln 2 }{6500}$$. That is,
   $$N(t) = \frac{N(0)}{2^{t/6500}}.$$
   b) How long would it take for a $1$ gram sample of this radioactive compound to decay to $2^{-6}$ grams?? Solution $N(0)=1$ and we want $N(t) = 2^{-6}$ so we just solve for $t$ in the equation
   $$2^{-6} = \frac{1}{2^{t/6500}} = 2^{-t/6500}$$
   and so $t = 6500 \times 6 = 39,000$ years.