Complete Calculus Exam Solutions

Question 1

Calculate without a calculator: \[e^{\tanh^{-1}(1/2)}\]

  1. Recall that \(\tanh^{-1}(x) = \frac{1}{2}\ln(\frac{1+x}{1-x})\) for |x| < 1
  2. With x = 1/2: \[\tanh^{-1}(1/2) = \frac{1}{2}\ln(\frac{1+1/2}{1-1/2}) = \frac{1}{2}\ln(\frac{3/2}{1/2}) = \frac{1}{2}\ln(3)\]
  3. Therefore: \[e^{\tanh^{-1}(1/2)} = e^{\frac{1}{2}\ln(3)} = (e^{\ln(3)})^{1/2} = \sqrt{3}\]

Answer: a) \(\sqrt{3}\)

Question 2

Calculate: \[\lim_{x \to \infty} \sin\left(\frac{\sin x}{x}\right)\left(\frac{\sin x}{x}\right)^{-1}\]

  1. Note that as x → ∞, \(\frac{\sin x}{x} \to 0\)
  2. For small θ, we have \(\sin θ \approx θ\)
  3. Therefore: \[\lim_{x \to \infty} \frac{\sin(\sin x/x)}{(\sin x/x)} = 1\]
  4. This is because when y = \(\frac{\sin x}{x}\) → 0, \(\frac{\sin y}{y}\) → 1

Answer: a) 1

Question 3

Solutions to: \(x^{\ln(x^2)} = e^3 x\)

  1. Simplify the exponent: \[x^{\ln(x^2)} = x^{2\ln(x)}\] Equation becomes: \[x^{2\ln(x)} = e^3 x\]
  2. Take \(\ln\) of both sides: \[\ln\left(x^{2\ln(x)}\right) = \ln\left(e^3 x\right)\] Using logarithmic properties: \[2(\ln(x))^2 = 3 + \ln(x)\]
  3. Let \(y = \ln(x)\): \[2y^2 = 3 + y\] Rearrange into a quadratic equation: \[2y^2 - y - 3 = 0\]
  4. Solve the quadratic equation: \[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] Substituting \(a = 2\), \(b = -1\), \(c = -3\): \[y = \frac{1 \pm \sqrt{25}}{4}\] \[y = \frac{6}{4} = \frac{3}{2} \quad \text{and} \quad y = -1\]
  5. Back-substitute \(y = \ln(x)\): \[x = e^{\frac{3}{2}} \quad \text{and} \quad x = e^{-1}\]

Answer: a) \(x_{1,2} = e^{-1}, e^{\frac{3}{2}}\)

Question 4

Domain of \[f(x) = \ln\sqrt{1-e^{-x}}\]

  1. For ln to be defined, we need: \[\sqrt{1-e^{-x}} > 0\]
  2. This means: \[1-e^{-x} > 0\] \[-e^{-x} > -1\] \[e^{-x} < 1\] \[-x < 0\] \[x> 0\]

Answer: a) {x ∈ ℝ: x > 0}

Question 5

Range of \[f(x) = \ln\sqrt{1-e^{-x}}\]

  1. Let y = f(x) \[e^y = \sqrt{1-e^{-x}}\]
  2. \[e^{2y} = 1-e^{-x}\]
  3. Since x > 0, we have 0 < e^{-x} < 1
  4. Therefore: \[0 < 1-e^{-x} < 1\] \[0 < e^{2y} < 1\] \[-\infty < y < 0\]

Answer: a) {y = f(x) ∈ ℝ: -∞ < y ≤ 0}

Question 6

Inverse function of \[f(x) = e^{\sinh^{-1} x}\]

  1. Let y = f(x). Then: \[y = e^{\sinh^{-1} x}\]
  2. Take ln of both sides: \[\ln y = \sinh^{-1} x\]
  3. Apply sinh to both sides: \[x = \sinh(\ln y)\]
  4. Using the definition of sinh: \[x = \frac{e^{\ln y} - e^{-\ln y}}{2} = \frac{y - \frac{1}{y}}{2}\]
  5. Therefore for the inverse function, replace y with x: \[f^{-1}(x) = \frac{x^2 - 1}{2x}\]
  6. We can verify this works by composing with f: \[f(f^{-1}(x)) = e^{\sinh^{-1}(\frac{x^2 - 1}{2x})} = x\]

Answer: a) \(f^{-1}(x) = \frac{x^2 - 1}{2x}\)

Question 7

The functional relationship \[y(x) = \tanh^2(\cosh^{-1} x)\] is equivalent to:

  1. For \(\cosh^{-1} x\), we know x ≥ 1
  2. Let t = \(\cosh^{-1} x\), then x = cosh(t)
  3. Recall that \(\tanh^2 t = \frac{\sinh^2 t}{\cosh^2 t}\)
  4. And \(\cosh^2 t - \sinh^2 t = 1\)
  5. Therefore: \[y = \tanh^2 t = \frac{\sinh^2 t}{\cosh^2 t} = \frac{\cosh^2 t - 1}{\cosh^2 t} = 1 - \frac{1}{\cosh^2 t} = \frac{x^2-1}{x^2}\]

Answer: a) \(y = \frac{x^2-1}{x^2}\)

Question 8

The function \[f(x) = \sin\left(\frac{\sinh x}{\sin x}\right)\] for x ≠ kπ, k ∈ ℤ is:

  1. Let's check if f(-x) = f(x) (even) or f(-x) = -f(x) (odd)
  2. Note that sinh(-x) = -sinh(x) and sin(-x) = -sin(x)
  3. Therefore: \[f(-x) = \sin\left(\frac{\sinh(-x)}{\sin(-x)}\right) = \sin\left(\frac{-\sinh(x)}{-\sin(x)}\right) = \sin\left(\frac{\sinh(x)}{\sin(x)}\right) = f(x)\]

Answer: a) Even function

Question 9

For \[f(x) = a^{\sin^{2m+1}(x)}\] where a > 0, a ≠ 1, m ∈ ℕ:

  1. Consider f(-x): \[f(-x) = a^{\sin^{2m+1}(-x)} = a^{-\sin^{2m+1}(x)} = \frac{1}{a^{\sin^{2m+1}(x)}} = \frac{1}{f(x)}\]

Answer: a) \(f(-x) = \frac{1}{f(x)}\)

Question 10

For \[f(x) = \frac{x^3\sin(1/x)}{\sin^2 x}\], which statement is most correct?

  1. As x → 0: - sin(1/x) oscillates between -1 and 1 - x³ → 0 - sin²x ≈ x² for small x
  2. Therefore: \[\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{x^3\sin(1/x)}{x^2} = 0\]
  3. This limit exists and equals 0

Answer: a) The function has a limit at x = 0 and its value is 0

Question 11

Given that f, f⁻¹ are inverse functions with limit at x₀ ≠ 0, find: \[\lim_{x \to x_0} f\left(f^{-1}\left(\frac{1}{f^{-1}(f(x))}\right)\right)\]

  1. Note that f⁻¹(f(x)) = x
  2. Therefore: \[\lim_{x \to x_0} f\left(f^{-1}\left(\frac{1}{x}\right)\right)\]
  3. As x → x₀, 1/x → 1/x₀

Answer: a) 1/x₀

Question 12

Calculate: \[\lim_{x \to 1} \frac{x^2-1}{2\ln x}\]

  1. This is a 0/0 indeterminate form
  2. Using L'Hôpital's rule: \[\lim_{x \to 1} \frac{2x}{2/x} = \lim_{x \to 1} x^2 = 1\]

Answer: a) 1

Question 13

Calculate: \[\lim_{x \to 0} \frac{5-\sqrt{x+25}}{x}\]

  1. Multiply by conjugate: \[\lim_{x \to 0} \frac{(5-\sqrt{x+25})(5+\sqrt{x+25})}{x(5+\sqrt{x+25})}\]
  2. \[\lim_{x \to 0} \frac{25-(x+25)}{x(5+\sqrt{x+25})} = \lim_{x \to 0} \frac{-x}{x(5+\sqrt{x+25})} = -\frac{1}{10}\]

Answer: a) -1/10

Question 14

For which \( k \neq 0 \) is the function \[ f(x) = \begin{cases} (1 + k\sin x)^{\cot x} & x > 0, \\ e^{k+x} & x \leq 0 \end{cases} \] continuous at \( x = 0 \)?

  1. Continuity condition: For \( f(x) \) to be continuous at \( x = 0 \), the left-hand limit (\( \lim_{x \to 0^-} f(x) \)), the right-hand limit (\( \lim_{x \to 0^+} f(x) \)), and the function value \( f(0) \) must all be equal.
  2. Left-hand limit (\( x \to 0^- \)): \[ f(x) = e^{k+x} \quad \text{for } x \leq 0. \] Taking the limit as \( x \to 0^- \): \[ \lim_{x \to 0^-} f(x) = e^k. \]
  3. Right-hand limit (\( x \to 0^+ \)): \[ f(x) = (1 + k \sin x)^{\cot x} \quad \text{for } x > 0. \] As \( x \to 0^+ \):
    • \( \sin x \to 0 \), so \( 1 + k \sin x \to 1 \).
    • \( \cot x = \frac{\cos x}{\sin x} \to \infty \) as \( \sin x \to 0^+ \).
    For small \( \sin x \), we approximate: \[ \ln(1 + k \sin x) \approx k \sin x \quad \text{(valid for small \( k \sin x \))}. \] Thus: \[ (1 + k \sin x)^{\cot x} \approx e^{k \sin x \cdot \cot x}. \] Now \( \sin x \cdot \cot x = \sin x \cdot \frac{\cos x}{\sin x} = \cos x \), so: \[ \lim_{x \to 0^+} (1 + k \sin x)^{\cot x} = e^{k \cos 0} = e^k. \]
  4. Function value at \( x = 0 \): \[ f(0) = e^k. \]
  5. Conclusion: The left-hand limit, right-hand limit, and \( f(0) \) all equal \( e^k \). Thus, \( f(x) \) is continuous at \( x = 0 \) for any \( k \neq 0 \).

Answer: a) Any \( k \neq 0 \)

Question 15

Find the value of the product: \[(e^x-1)(1 + e^x + e^{2x} + e^{3x} + ... + e^{(m-1)x}), m \in \mathbb{N}\]

  1. Let's denote the second factor as S: \[S = 1 + e^x + e^{2x} + e^{3x} + ... + e^{(m-1)x}\]
  2. This is a geometric series with first term a = 1 and ratio r = e^x \[S = \frac{1-e^{mx}}{1-e^x}\]
  3. Therefore: \[(e^x-1)\cdot\frac{1-e^{mx}}{1-e^x} = e^{mx}-1\]

Answer: a) e^{mx}-1

Summary of Final Answers:

  1. e^{tanh⁻¹(1/2)} = √3
  2. lim[sin(sin x/x)/(sin x/x)] = 1
  3. x₁,₂ = e⁻¹, e^{3/2}
  4. Domain: {x ∈ ℝ: x > 0}
  5. Range: {y = f(x) ∈ ℝ: -∞ < y ≤ 0}
  6. f⁻¹(x) = (x²-1)/2x
  7. y = (x²-1)/x²
  8. Even function
  9. f(-x) = 1/f(x)
  10. Function has limit at x = 0 equal to 0
  11. 1/x₀
  12. 1
  13. -1/10
  14. Any k ≠ 0
  15. e^{mx}-1

All questions required careful analysis of limits, continuity, and properties of exponential and logarithmic functions. The key to solving many of these problems was recognizing patterns and using standard calculus techniques like L'Hôpital's rule, analysis of even/odd functions, and properties of inverse functions.