Calculus S.M Yusuf Chapter 1 Ex 1.1 Solved Notes & Quiz

Question 1 (Part i): Solve the absolute value inequality $|2x - 5| < 7$ and express the solution set in interval notation.

Solution:
Using the absolute value property $|u| < a \iff -a < u < a$ for $a > 0$: $$-7 < 2x - 5 < 7$$ Add $5$ to all three parts of the inequality: $$-7 + 5 < 2x < 7 + 5$$ $$-2 < 2x < 12$$ Divide all parts by $2$: $$-1 < x < 6$$ In interval notation, the solution set is $(-1, 6)$.
Answer: $$(-1, 6)$$

Question 1 (Part ii): Solve the absolute value inequality $|3x + 4| \ge 10$ and write the solution set in interval notation.

Solution:
Using the absolute value property $|u| \ge a \iff u \ge a \text{ or } u \le -a$: $$\text{Case 1: } 3x + 4 \ge 10 \implies 3x \ge 6 \implies x \ge 2$$ $$\text{Case 2: } 3x + 4 \le -10 \implies 3x \le -14 \implies x \le -\frac{14}{3}$$ Combining both cases gives $x \in \left(-\infty, -\frac{14}{3}\right] \cup [2, \infty)$.
Answer: $$\left(-\infty, -\frac{14}{3}\right] \cup [2, \infty)$$

Question 2 (Part i): Solve the quadratic inequality $x^2 - 3x - 10 \le 0$.

Solution:
First, factor the quadratic expression: $$x^2 - 3x - 10 = (x - 5)(x + 2) \le 0$$ Find the critical numbers by setting $(x - 5)(x + 2) = 0$, which gives $x = -2$ and $x = 5$.
These points divide the real number line into three test intervals: $(-\infty, -2)$, $(-2, 5)$, and $(5, \infty)$.

• Test $x = -3 \in (-\infty, -2)$: $(-3 - 5)(-3 + 2) = (-8)(-1) = 8 > 0$ (False)
• Test $x = 0 \in (-2, 5)$: $(0 - 5)(0 + 2) = (-5)(2) = -10 \le 0$ (True)
• Test $x = 6 \in (5, \infty)$: $(6 - 5)(6 + 2) = (1)(8) = 8 > 0$ (False)

Question 2 (Part ii): Solve the rational inequality $\frac{x + 2}{x - 4} > 0$.

Solution:
Find the boundary points where the numerator is zero or the denominator is zero: $$\text{Numerator } x + 2 = 0 \implies x = -2$$ $$\text{Denominator } x - 4 = 0 \implies x = 4$$ The critical numbers are $x = -2$ and $x = 4$. Test the three intervals $(-\infty, -2)$, $(-2, 4)$, and $(4, \infty)$:

• Test $x = -3 \in (-\infty, -2)$: $\frac{-3 + 2}{-3 - 4} = \frac{-1}{-7} = \frac{1}{7} > 0$ (True)
• Test $x = 0 \in (-2, 4)$: $\frac{0 + 2}{0 - 4} = \frac{2}{-4} = -\frac{1}{2} < 0$ (False)
• Test $x = 5 \in (4, \infty)$: $\frac{5 + 2}{5 - 4} = \frac{7}{1} = 7 > 0$ (True)

Question 3 (Part i): Evaluate the limit $\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$.

Solution:
Direct substitution of $x = 3$ yields the indeterminate form $\frac{3^2 - 9}{3 - 3} = \frac{0}{0}$.
Factor the numerator using the difference of squares $a^2 - b^2 = (a - b)(a + b)$: $$\lim_{x \to 3} \frac{(x - 3)(x + 3)}{x - 3}$$ Since $x \to 3$, $x \neq 3$, so we can cancel the non-zero common factor $(x - 3)$: $$\lim_{x \to 3} (x + 3) = 3 + 3 = 6$$ Answer: $$6$$

Question 3 (Part ii): Evaluate the limit $\lim_{x \to 2} \frac{x^3 - 8}{x^2 - 4}$.

Solution:
Direct substitution gives $\frac{2^3 - 8}{2^2 - 4} = \frac{0}{0}$.
Factor numerator using $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ and denominator using $a^2 - b^2 = (a - b)(a + b)$: $$\lim_{x \to 2} \frac{(x - 2)(x^2 + 2x + 4)}{(x - 2)(x + 2)}$$ Cancel the common term $(x - 2)$: $$\lim_{x \to 2} \frac{x^2 + 2x + 4}{x + 2} = \frac{(2)^2 + 2(2) + 4}{2 + 2} = \frac{4 + 4 + 4}{4} = \frac{12}{4} = 3$$ Answer: $$3$$

Question 4 (Part i): Evaluate the trigonometric limit $\lim_{x \to 0} \frac{\sin(5x)}{3x}$.

Solution:
Direct substitution yields $\frac{0}{0}$. Use the fundamental limit theorem $\lim_{\theta \to 0} \frac{\sin\theta}{\theta} = 1$.
Multiply and divide by $5$: $$\lim_{x \to 0} \frac{\sin(5x)}{3x} = \lim_{x \to 0} \left( \frac{5}{3} \cdot \frac{\sin(5x)}{5x} \right)$$ Let $\theta = 5x$. As $x \to 0$, $\theta \to 0$: $$= \frac{5}{3} \lim_{\theta \to 0} \frac{\sin\theta}{\theta} = \frac{5}{3} (1) = \frac{5}{3}$$ Answer: $$\frac{5}{3}$$

Question 4 (Part ii): Evaluate the limit $\lim_{x \to 0} \frac{1 - \cos x}{x^2}$.

Solution:
Using the trigonometric double-angle identity $1 - \cos x = 2\sin^2\left(\frac{x}{2}\right)$: $$\lim_{x \to 0} \frac{2\sin^2(x/2)}{x^2}$$ Rewrite the expression: $$= 2 \lim_{x \to 0} \left[ \frac{\sin(x/2)}{x} \right]^2 = 2 \lim_{x \to 0} \left[ \frac{1}{2} \cdot \frac{\sin(x/2)}{x/2} \right]^2$$ Since $\lim_{x \to 0} \frac{\sin(x/2)}{x/2} = 1$: $$= 2 \left( \frac{1}{2} \cdot 1 \right)^2 = 2 \cdot \frac{1}{4} = \frac{1}{2}$$ Answer: $$\frac{1}{2}$$

Question 5 (Part i): Evaluate the limit at infinity $\lim_{x \to \infty} \frac{4x^2 - 3x + 7}{2x^2 + 5}$.

Solution:
Divide both numerator and denominator by the highest power of $x$ present in the denominator, which is $x^2$: $$\lim_{x \to \infty} \frac{\frac{4x^2}{x^2} - \frac{3x}{x^2} + \frac{7}{x^2}}{\frac{2x^2}{x^2} + \frac{5}{x^2}} = \lim_{x \to \infty} \frac{4 - \frac{3}{x} + \frac{7}{x^2}}{2 + \frac{5}{x^2}}$$ Apply the limit property $\lim_{x \to \infty} \frac{1}{x^n} = 0$ for $n > 0$: $$= \frac{4 - 0 + 0}{2 + 0} = \frac{4}{2} = 2$$ Answer: $$2$$

Question 5 (Part ii): Evaluate the limit $\lim_{x \to \infty} \left(\sqrt{x^2 + 4x} - x\right)$.

Solution:
This has the indeterminate form $\infty - \infty$. Rationalize the expression by multiplying and dividing by its conjugate $(\sqrt{x^2 + 4x} + x)$: $$\lim_{x \to \infty} \frac{(\sqrt{x^2 + 4x} - x)(\sqrt{x^2 + 4x} + x)}{\sqrt{x^2 + 4x} + x}$$