Calculus Midterm Exam Solutions
Question 1
Calculate without L'Hôpital's rule: \[\lim_{x \to 0} \frac{\sin x^3}{\sin^3 x}\]
- We start by rewriting the fraction: \[ \frac{\sin x^3}{\sin^3 x} = \frac{\sin x^3}{x^3} \cdot \frac{x^3}{\sin^3 x}. \]
- First, evaluate \(\lim_{x \to 0} \frac{\sin x^3}{x^3}\):
Since \(\lim_{u \to 0} \frac{\sin u}{u} = 1\), substituting \(u = x^3\) gives: \[ \lim_{x \to 0} \frac{\sin x^3}{x^3} = 1. \] - Next, evaluate \(\lim_{x \to 0} \frac{x^3}{\sin^3 x}\):
Rewrite as \(\left(\frac{x}{\sin x}\right)^3\), and since \(\lim_{x \to 0} \frac{x}{\sin x} = 1\), we have: \[ \lim_{x \to 0} \frac{x^3}{\sin^3 x} = 1^3 = 1. \] - Combine the results: \[ \lim_{x \to 0} \frac{\sin x^3}{\sin^3 x} = \left(\lim_{x \to 0} \frac{\sin x^3}{x^3}\right) \cdot \left(\lim_{x \to 0} \frac{x^3}{\sin^3 x}\right) = 1 \cdot 1 = 1. \]
Answer: a) 1
Note: we can also assert that the limit is 1 by using the fact that \(\sin x \approx x\) for small \(x\).
So the limit is approximately \(\frac{x^3}{x^3} = 1\) as \(x \to 0\).
Question 2
Given function \[f(x) = \frac{1}{1-\sin x}, \, x \neq \frac{\pi}{2} + 2k\pi, \, k \in \mathbb{Z}\] Find \[f(f(\sin^{-1} x))\]
- Let \(y = \sin^{-1}x\), so \(\sin y = x\).
- Substitute into \(f(y)\): \[ f(y) = \frac{1}{1-\sin y} = \frac{1}{1-x}. \]
- Now substitute \(f(y) = \frac{1}{1-x}\) into \(f(x)\): \[ f(f(y)) = f\left(\frac{1}{1-x}\right). \]
- Evaluate \(f\left(\frac{1}{1-x}\right)\): \[ f\left(\frac{1}{1-x}\right) = \frac{1}{1 - \sin\left(\frac{1}{1-x}\right)}. \] There is no further simplification possible for \(\sin\left(\frac{1}{1-x}\right)\).
Answer: a) \[\frac{1}{1 - \sin\left(\frac{1}{1-x}\right)}\]
Question 3
Find domain of: \[f(x) = \sqrt{\sin 2x}\]
- For the square root to be defined, we need: \[\sin 2x \geq 0\]
- This means: \[2x \in [2k\pi, \pi + 2k\pi]\] \[x \in [k\pi, \frac{\pi}{2} + k\pi], k \in \mathbb{Z}\]
Answer: a) \[x \in [k\pi, \frac{\pi}{2} + k\pi], k \in \mathbb{Z}\]
Question 4
Calculate: \[\lim_{x \to 0} \frac{\sin^{-1}(\cosh x)}{\cos^{-1} x}\]
- As x → 0: - cosh x → 1 - cos⁻¹ x → \pi/2 - sin⁻¹(1) = \pi/2
- Therefore: \[\lim_{x \to 0} \frac{\sin^{-1}(\cosh x)}{\cos^{-1} x} = 1\]
Answer: a) 1
Question 5
For the function: \[f(x) = \frac{|x+1|}{x^2 + x}\]
Choose the correct statement:
- a) Has a non-removable discontinuity at \(x = -1\)
- b) Has a removable discontinuity at \(x = -1\)
- c) Has a removable discontinuity at \(x = 0\)
- d) Is continuous in the range \((-∞, 0)\)
- Factorizing the denominator: \[ x^2 + x = x(x + 1) \] The denominator is zero at \(x = -1\) and \(x = 0\), making \(f(x)\) undefined at these points.
- **At \(x = -1\):** - Both the numerator \(|x + 1|\) and the denominator \(x^2 + x = x(x+1)\) are zero, giving the indeterminate form \(0/0\). - To analyze the behavior, simplify \(f(x)\) for \(x \neq -1\): \[ f(x) = \frac{|x+1|}{x(x+1)} = \frac{\text{sgn}(x+1)}{x}, \quad x \neq -1 \] - Here, \(\text{sgn}(x+1)\) is the sign function, which equals \(1\) for \(x > -1\) and \(-1\) for \(x < -1\). - As \(x\) approaches \(-1\) from the left (\(x \to -1^-\)): \[ f(x) \to \frac{-1}{-1} = 1 \] - As \(x\) approaches \(-1\) from the right (\(x \to -1^+\)): \[ f(x) \to \frac{1}{-1} = -1 \] - Since the left-hand limit (\(1\)) does not equal the right-hand limit (\(-1\)), the discontinuity at \(x = -1\) is **non-removable**.
- **At \(x = 0\):** - The denominator \(x^2 + x = x(x+1)\) is zero, but the numerator \(|x + 1| = 1\). - Since the numerator is nonzero while the denominator is zero, \(f(x)\) approaches \(\pm \infty\) as \(x \to 0\). This is a **non-removable discontinuity.**
- **On the range \((-∞, 0)\):** - The function is undefined at both \(x = -1\) and \(x = 0\), so it is not continuous on the range \((-∞, 0)\).
Answer: a) Has a non-removable discontinuity at \(x = -1\)
Question 6
Find smallest period of: \[f(x) = \sin(2x)\cos(2x)\]
- Start by simplifying the function using the trigonometric identity: \[ \sin(2x)\cos(2x) = \frac{1}{2}\sin(4x). \] This means \(f(x)\) can be rewritten as: \[ f(x) = \frac{1}{2}\sin(4x). \] Scaling the sine function by \(\frac{1}{2}\) does not affect its periodicity.
- The period of \(\sin(kx)\) is given by \(\frac{2\pi}{k}\). Here, \(k = 4\), so the period of \(\sin(4x)\) is: \[ \frac{2\pi}{4} = \frac{\pi}{2}. \]
- Since \(f(x)\) is directly proportional to \(\sin(4x)\), the smallest period of \(f(x)\) is the same as the period of \(\sin(4x)\).
Answer: a) \(\frac{\pi}{2}\)
Question 7
For \(x > 0\), given: \[f(\ln x) = \frac{\ln^2 x}{x}\]
We can rewrite the function as:
- a) \[f(x) = \frac{x^{2}}{e^{x}}\]
- b) \[f(\frac{1}{x}) = \frac{x^{2}}{e^x}\]
- c) \[f(x) = \frac{e^{2x}}{x}\]
- d) \[f(\frac{1}{x}) = \frac{e^{2x}}{x}\]
- Start by letting \(y = \ln x\), so \(x = e^y\).
- Substitute \(x = e^y\) into \(f(\ln x)\): \[ f(\ln x) = \frac{\ln^2 x}{x} = \frac{(\ln e^y)^2}{e^y} = \frac{y^2}{e^y}. \] Thus, \(f(y) = \frac{y^2}{e^y}\).
- Rewrite \(f(y)\) as \(f(x)\) by substituting \(y = x\): \[ f(x) = \frac{x^2}{e^x}. \]
- None of the other transformations (such as \(f(\frac{1}{x})\)) lead to valid simplifications that match the function \(f(\ln x)\).
Answer: a) \[f(x) = \frac{x^2}{e^x}\]
Question 8
Find the domain and range of: \[y(x) = \tan^{-1}(\sin x)\]
- a) \(-\infty < x < +\infty, -\pi/4 < y < \pi/4\)
- b) \(-\infty < x < +\infty, -\infty < y < +\infty\)
- c) \(-\pi/2 \leq x \leq \pi/2, -\infty < y < +\infty\)
- d) \(-\pi/2 \leq x \leq \pi/2, -\pi/4 < y < \pi/4\)
- **Domain of \(y(x)\):** - The function \( \sin x \) is defined for all real numbers (\(-\infty < x < +\infty\)). - Since \( \tan^{-1}(z) \) is also defined for all real \( z \), there are no restrictions on \(x\). - Therefore, the **domain** of \(y(x)\) is: \[ -\infty < x < +\infty \]
- **Range of \(\sin x\):** - The sine function maps any real number \(x\) to the interval \([-1, 1]\), regardless of the value of \(x\).
- Behavior of \(\tan^{-1}(z)\):
- The inverse tangent function \(\tan^{-1}(z)\) maps any real number \(z\) to the interval \((- \pi / 2, \pi / 2)\). However, when restricted to the input range \([-1, 1]\), it maps to the interval \([- \pi / 4, \pi / 4]\).
- To verify:
- When \(z = -1\): \(\tan^{-1}(-1) = -\pi / 4\).
- When \(z = 1\): \(\tan^{-1}(1) = \pi / 4\).
- Since \(\sin(x)\) produces values only within \([-1, 1]\), the output of \(\tan^{-1}(\sin x)\) is restricted to: \[ - \pi / 4 \leq y \leq \pi / 4 \]
- **Answer Verification:** - Combining the domain (\(-\infty < x < +\infty\)) and range (\(-\pi/4 \leq y \leq \pi/4\)), the correct option is: \[ \text{a) } -\infty < x < +\infty, -\pi/4 \leq y \leq \pi/4 \]
Answer: a) \(-\infty < x < +\infty, -\pi/4 \leq y \leq \pi/4\)
Question 9
For the function: \[f(x) = \frac{e^{\tan x} - e^{-\tan x}}{e^{\tan x} + e^{-\tan x}}\]
Choose the correct option:
- a) Odd function
- b) Even function
- c) Neither odd nor even
- d) Fulfills \[f(-x) = f^{-1}(x)\]
- To determine the nature of \(f(x)\), we check \(f(-x)\): \[ f(-x) = \frac{e^{\tan(-x)} - e^{-\tan(-x)}}{e^{\tan(-x)} + e^{-\tan(-x)}} \]
- Using the property of the tangent function, \(\tan(-x) = -\tan(x)\), we rewrite the equation: \[ f(-x) = \frac{e^{-\tan x} - e^{\tan x}}{e^{-\tan x} + e^{\tan x}} \]
- Simplify the numerator and denominator: \[ f(-x) = \frac{-(e^{\tan x} - e^{-\tan x})}{e^{\tan x} + e^{-\tan x}} \] \[ f(-x) = -\frac{e^{\tan x} - e^{-\tan x}}{e^{\tan x} + e^{-\tan x}} \]
- From this, we observe that: \[ f(-x) = -f(x) \]
- By definition, a function \(f(x)\) is odd if \(f(-x) = -f(x)\). Since this condition is satisfied, \(f(x)\) is an odd function.
Answer: a) Odd function
Question 10
For a,b ≠ 0 and function \[f(x) = \frac{1}{x}\], which is true:
- Let's verify each option
- \[f(a) - f(b) = \frac{1}{a} - \frac{1}{b} = \frac{b-a}{ab}\]
- \[f(\frac{ab}{b-a}) = \frac{b-a}{ab}\]
Answer: a) \[f(a) - f(b) = f(\frac{ab}{b-a})\]