Evaluation G - 12 | 1.1 Solutions (Algebraic Manipulation)
1.1.1 Simplify: $\frac{3x^2 - 12}{x^2 - x - 6}$ (A) $\frac{3(x+2)}{x+2}$ (B) $\frac{3(x-2)}{x-3}$ (C) $\frac{3(x+2)}{x-3}$ (D) $\frac{3(x-2)}{x+2}$ (E) $3$.
Factor: $3x^2-12 = 3(x^2-4) = 3(x-2)(x+2)$; $x^2-x-6 = (x-3)(x+2)$. Cancel $(x+2)$: $\frac{3(x-2)(x+2)}{(x-3)(x+2)} = \frac{3(x-2)}{x-3}$ → Answer: B. $\frac{3(x-2)}{x-3}$
1.1.2 If $x + \frac{1}{x} = 5$, find the value of $x^3 + \frac{1}{x^3}$. (A) 110 (B) 115 (C) 125 (D) 130 (E) 140.
Use identity: $x^3 + \frac{1}{x^3} = (x+\frac{1}{x})^3 - 3(x+\frac{1}{x}) = 5^3 - 3(5) = 125 - 15 = 110$ → Answer: A. 110
1.1.3 Simplify: $\frac{x+2}{x^2 - 4} - \frac{x-3}{x^2 - x - 6}$ (A) $\frac{2x-1}{(x-2)(x+2)}$ (B) $\frac{2x+1}{(x-2)(x+2)}$ (C) $\frac{5}{(x-2)(x+2)}$ (D) $\frac{4x-5}{(x-2)(x+2)}$ (E) $\frac{4}{(x-2)(x+2)}$.
Factor: $x^2-4 = (x-2)(x+2)$; $x^2-x-6 = (x-3)(x+2)$. First term: $\frac{x+2}{(x-2)(x+2)} = \frac{1}{x-2} = \frac{x-3}{(x-2)(x-3)}$. Second term: $\frac{x-3}{(x-3)(x+2)} = \frac{1}{x+2} = \frac{x-2}{(x-2)(x+2)}$. Subtract: $\frac{1}{x-2} - \frac{1}{x+2} = \frac{(x+2)-(x-2)}{(x-2)(x+2)} = \frac{4}{(x-2)(x+2)}$ → Answer: E. $\frac{4}{(x-2)(x+2)}$
1.1.4 Find the remainder when $2x^3 - 3x^2 + 4x - 5$ is divided by $x - 2$. (A) 5 (B) 7 (C) 9 (D) 11 (E) 13.
Remainder Theorem: $P(2) = 2(8) - 3(4) + 4(2) - 5 = 16 - 12 + 8 - 5 = 7$ → Answer: B. 7
1.1.5 If $x + y = 7$ and $xy = 12$, find $x^2 + y^2$. (A) 24 (B) 25 (C) 49 (D) 36 (E) 13.
$x^2+y^2 = (x+y)^2 - 2xy = 49 - 24 = 25$ → Answer: B. 25
1.1.6 Simplify: $\frac{x^2 - 5x + 6}{x^2 - 4} \times \frac{x^2 + 4x + 4}{x^2 - 9}$ (A) $\frac{(x-2)(x+2)}{(x+3)(x-3)}$ (B) $\frac{x+2}{x+3}$ (C) $\frac{x-2}{x-3}$ (D) $\frac{x+2}{x-3}$ (E) $1$.
Factor: $x^2-5x+6 = (x-2)(x-3)$; $x^2-4 = (x-2)(x+2)$; $x^2+4x+4 = (x+2)^2$; $x^2-9 = (x-3)(x+3)$. Multiply: $\frac{(x-2)(x-3)}{(x-2)(x+2)} \times \frac{(x+2)^2}{(x-3)(x+3)} = \frac{x+2}{x+3}$ → Answer: B. $\frac{x+2}{x+3}$
1.1.7 Given that $x - \frac{1}{x} = 3$, find the value of $x^4 + \frac{1}{x^4}$. (A) 119 (B) 121 (C) 123 (D) 127 (E) 129.
Square: $(x - \frac{1}{x})^2 = x^2 + \frac{1}{x^2} - 2 = 9$ → $x^2 + \frac{1}{x^2} = 11$. Square again: $(x^2 + \frac{1}{x^2})^2 = x^4 + \frac{1}{x^4} + 2 = 121$ → $x^4 + \frac{1}{x^4} = 119$ → Answer: A. 119
1.1.8
Solve for $\frac{4}{x+1} + \frac{1}{x-2} = 1$ (A) $x = 0, x = 5$ (B) $x = 0, x = 4$ (C) $x = 1, x = 5$ (D) $x
= 2, x = 3$ (E) $x = -1, x = 6$.
$\frac{4}{x+1} + \frac{1}{x-2} = 1$ → Multiply by $(x+1)(x-2)$: $4(x-2)+(x+1)=(x+1)(x-2)$ → $4x-8+x+1=x^2-2x+x-2$
→ $5x-7=x^2-x-2$ → $0=x^2-6x+5$ → $(x-1)(x-5)=0$ → $x=1$ or $x=5$ → Answer:
C. $x=1, x=5$
1.1.9 If $ax^3 + bx^2 - 5x + 6$ has $x+2$ and $x-1$ as factors, find $a$ and $b$. (A) $a=1,b=3$ (B) $a=2,b=-1$ (C) $a=-2,b=1$ (D) $a=0,b=2$ (E) $a=1,b=-2$.
Let $P(x)=ax^3+bx^2-5x+6$. $P(-2)=0$: $-8a+4b+10+6=0$ → $-8a+4b+16=0$ → $-2a+b=-4$ (1). $P(1)=0$: $a+b-5+6=0$ → $a+b=-1$ (2). Subtract (2) from (1): $(-2a+b)-(a+b) = -4 - (-1)$ → $-3a = -3$ → $a=1$. Then $1+b=-1$ → $b=-2$ → Answer: E. $a=1,b=-2$
1.1.10 Simplify: $\frac{1}{x-2} - \frac{2}{x^2-4} + \frac{1}{x+2}$ (A) $\frac{2x}{x^2-4}$ (B) $\frac{4}{x^2-4}$ (C) $\frac{2}{x+2}$ (D) $\frac{2(x-1)}{x^2-4}$ (E) $\frac{2}{x-2}$.
$x^2-4=(x-2)(x+2)$. Write over common denominator: $\frac{1}{x-2} = \frac{x+2}{(x-2)(x+2)}$; $\frac{1}{x+2} = \frac{x-2}{(x-2)(x+2)}$; $-\frac{2}{x^2-4} = -\frac{2}{(x-2)(x+2)}$. Sum numerator: $(x+2)+(x-2)-2 = 2x-2 = 2(x-1)$. Result: $\frac{2(x-1)}{x^2-4}$ → Answer: D. $\frac{2(x-1)}{x^2-4}$
1.1.11 Find the value of $k$ for which $x-3$ is a factor of $x^3 - 2x^2 - 5x + k$. (A) -6 (B) 6 (C) -12 (D) 12 (E) 0.
$P(3)=0$: $27 - 18 -15 + k = 0$ → $-6 + k = 0$ → $k=6$ → Answer: B. 6
1.1.12 If $p = 2 - \sqrt{3}$, find $p^2 + \frac{1}{p^2}$. (A) 14 (B) 12 (C) 10 (D) 8 (E) 6.
$p = 2-\sqrt{3}$, then $\frac{1}{p} = \frac{1}{2-\sqrt{3}} = \frac{2+\sqrt{3}}{4-3} = 2+\sqrt{3}$. So $p + \frac{1}{p} = 4$. Then $p^2 + \frac{1}{p^2} = (p+\frac{1}{p})^2 - 2 = 16 - 2 = 14$ → Answer: A. 14
1.1.13 Express in partial fractions: $\frac{3x+1}{(x-1)(x+2)}$ (A) $\frac{4}{3(x-1)} + \frac{5}{3(x+2)}$ (B) $\frac{5}{3(x-1)} + \frac{4}{3(x+2)}$ (C) $\frac{4}{3(x-1)} - \frac{5}{3(x+2)}$ (D) $\frac{5}{3(x-1)} - \frac{4}{3(x+2)}$ (E) $\frac{4}{3(x-1)} + \frac{1}{3(x+2)}$.
Let $\frac{3x+1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}$. Multiply: $3x+1 = A(x+2)+B(x-1)$. $x=1$: $4 = A(3)$ → $A=\frac{4}{3}$. $x=-2$: $-5 = B(-3)$ → $B=\frac{5}{3}$. So = $\frac{4}{3(x-1)} + \frac{5}{3(x+2)}$ → Answer: A
1.1.14 A rectangular field has length $(x+5)$ metres and width $(x-2)$ metres. If the area is $84m^2$, find $x$. (A) 7 (B) 8 (C) 8.31 (D) 10 (E) 11.
$(x+5)(x-2)=84$ → $x^2+3x-10=84$ → $x^2+3x-94=0$ → $x = \frac{-3 \pm \sqrt{9+376}}{2} = \frac{-3 \pm \sqrt{385}}{2} \approx \frac{-3 \pm 19.62}{2}$. Positive root $\approx 8.31$ → Answer: C. 8.31
1.1.15 Find the coefficient of $x^3$ in the expansion of $(1-2x+3x^2)(2+ x - x^2)$. (A) -3 (B) -1 (C) 1 (D) 3 (E) 5.
Collect $x^3$ terms: $(-2x)(-x^2)=2x^3$; $(3x^2)(x)=3x^3$; total $5x^3$ → coefficient 5 → Answer: E. 5
1.1.16 If $x + y = 5$ and $x^2 + y^2 = 13$, find $xy$. (A) 3 (B) 4 (C) 5 (D) 6 (E) 7.
$x^2+y^2 = (x+y)^2 - 2xy$ → $13 = 25 - 2xy$ → $2xy = 12$ → $xy=6$ → Answer: D. 6
1.1.17 Simplify: $\left(\frac{x^2-1}{x^2+2x+1}\right) \div \left(\frac{x-1}{x+1}\right)$ (A) 1 (B) $\frac{x-1}{x+1}$ (C) $\frac{x+1}{x-1}$ (D) $x-1$ (E) $x+1$.
$\frac{x^2-1}{x^2+2x+1} = \frac{(x-1)(x+1)}{(x+1)^2} = \frac{x-1}{x+1}$. Dividing by $\frac{x-1}{x+1}$ gives $\frac{x-1}{x+1} \times \frac{x+1}{x-1} = 1$ → Answer: A. 1
1.1.18 For what values of $x$ is the rational expression $\frac{2x^2-5x-3}{x^2-7x+12}$ undefined? (A) $x=3$ only (B) $x=4$ only (C) $x=3$ and $x=4$ (D) $x=-3$ and $x=-4$ (E) $x=-3$ and $x=4$.
Undefined when denominator $x^2-7x+12=0$ → $(x-3)(x-4)=0$ → $x=3$ or $x=4$ → Answer: C. $x=3$ and $x=4$
1.1.19 A businessman made a profit of $\$ (2x^2 + 5x - 3)$ from selling $x$ units of a product. If the profit per unit was $\$ (2x - 1)$, how many units did he sell? (A) $x+3$ (B) $x-3$ (C) $2x+3$ (D) $2x-3$ (E) $x-1$.
Number of units = $\frac{2x^2+5x-3}{2x-1}$. Divide: $(2x-1)(x+3)=2x^2+6x-x-3=2x^2+5x-3$ → quotient $x+3$ → Answer: A. $x+3$
1.1.20 Find the polynomial $P(x)$ of degree 3 that has zeros at $x = -2, 1, 3$ and satisfies $P(0) = 12$. (A) $2x^3-4x^2-10x+12$ (B) $x^3-2x^2-5x+6$ (C) $2x^3-2x^2-10x+12$ (D) $x^3-4x^2+x+6$ (E) $2x^3+4x^2-10x-12$.
General: $P(x)=k(x+2)(x-1)(x-3)=k[(x+2)(x^2-4x+3)]=k(x^3-4x^2+3x+2x^2-8x+6)=k(x^3-2x^2-5x+6)$. $P(0)=k(6)=12$ → $k=2$. So $P(x)=2x^3-4x^2-10x+12$ → Answer: A
Evaluation G - 12 | 1.2 Solutions (Equations)
1.2.1 Solve: $2x^2 - 5x - 3 = 0$ (A) $x = -3, \frac{1}{2}$ (B) $x = 3, -\frac{1}{2}$ (C) $x = -3, -\frac{1}{2}$ (D) $x = 3, \frac{1}{2}$ (E) $x = 6, -1$.
Factor: $(2x+1)(x-3)=0$ → $2x+1=0$ → $x=-\frac{1}{2}$; $x-3=0$ → $x=3$ → Answer: B. $x=3, -\frac{1}{2}$
1.2.2 Find roots of $x^2 + 4x + 5 = 0$. (A) $-2 \pm i$ (B) $2 \pm i$ (C) $-4 \pm i$ (D) $4 \pm i$ (E) $-2 \pm 2i$.
Quadratic formula: $x = \frac{-4 \pm \sqrt{16-20}}{2} = \frac{-4 \pm \sqrt{-4}}{2} = \frac{-4 \pm 2i}{2} = -2 \pm i$ → Answer: A. $-2 \pm i$
1.2.3 Solve: $2x + 3y = 8$ and $3x - y = 5$ (A) $x=2,y=1$ (B) $x=1,y=2$ (C) $x=-2,y=4$ (D) $x=4,y=0$ (E) $x=\frac{23}{11},y=\frac{14}{11}$.
From second: $y=3x-5$. Substitute: $2x+3(3x-5)=8$ → $2x+9x-15=8$ → $11x=23$ → $x=\frac{23}{11}$, $y=3(\frac{23}{11})-5=\frac{69}{11}-\frac{55}{11}=\frac{14}{11}$ → Answer: E. $x=\frac{23}{11}, y=\frac{14}{11}$
1.2.4 For what $k$ does $x^2 + kx + 9 = 0$ have equal roots? (A) $k=3$ only (B) $k=-3$ only (C) $k=\pm 3$ (D) $k=\pm 6$ (E) $k=6$ only.
Equal roots when discriminant $\Delta = k^2 - 4(1)(9) = k^2 - 36 = 0$ → $k^2=36$ → $k=\pm 6$ → Answer: D. $k=\pm 6$
1.2.5 Solve: $3x^2 - 2x - 5 = 0$ (A) $x=-1,\frac{5}{3}$ (B) $x=1,-\frac{5}{3}$ (C) $x=-1,-\frac{5}{3}$ (D) $x=1,\frac{5}{3}$ (E) $x=5,-3$.
Factor: $(3x-5)(x+1)=0$ → $3x-5=0$ → $x=\frac{5}{3}$; $x+1=0$ → $x=-1$ → Answer: A. $x=-1, \frac{5}{3}$
1.2.6 If one root of $2x^2 - 5x + c = 0$ is $3$, find $c$. (A) $-3$ (B) $3$ (C) $-6$ (D) $6$ (E) $12$.
Substitute $x=3$: $2(9) - 5(3) + c = 0$ → $18 - 15 + c = 0$ → $3 + c = 0$ → $c = -3$ → Answer: A. $-3$
1.2.7 Solve: $y = x^2 - 2$ and $y = 3x + 2$ (A) $x=-1,y=-1$ and $x=4,y=14$ (B) $x=1,y=-1$ and $x=4,y=14$ (C) $x=-1,y=-1$ and $x=-4,y=-14$ (D) $x=1,y=5$ and $x=4,y=14$ (E) $x=-1,y=-5$ and $x=4,y=14$.
Set equal: $x^2 - 2 = 3x + 2$ → $x^2 - 3x - 4 = 0$ → $(x-4)(x+1)=0$ → $x=4$ or $x=-1$. $x=4$ → $y=3(4)+2=14$; $x=-1$ → $y=3(-1)+2=-1$ → Answer: A. $x=-1,y=-1$ and $x=4,y=14$
1.2.8 Find complex roots of $x^2 + 2x + 3 = 0$. (A) $-1 \pm i\sqrt{2}$ (B) $1 \pm i\sqrt{2}$ (C) $-2 \pm i$ (D) $2 \pm i$ (E) $-1 \pm 2i$.
$x = \frac{-2 \pm \sqrt{4-12}}{2} = \frac{-2 \pm \sqrt{-8}}{2} = \frac{-2 \pm 2i\sqrt{2}}{2} = -1 \pm i\sqrt{2}$ → Answer: A. $-1 \pm i\sqrt{2}$
1.2.9 Solve: $2x + y = 7$ and $x^2 + y^2 = 13$ (A) $x=2,y=3$ and $x=\frac{14}{5},y=\frac{7}{5}$ (B) $x=2,y=3$ and $x=3,y=1$ (C) $x=2,y=3$ and $x=1,y=5$ (D) $x=2,y=3$ and $x=\frac{18}{5},y=-\frac{1}{5}$ (E) $x=2,y=3$ only.
$y=7-2x$. Substitute: $x^2+(7-2x)^2=13$ → $x^2+49-28x+4x^2=13$ → $5x^2-28x+36=0$ → $5x^2-28x+36=0$. Discriminant: $784-720=64$, so $x=\frac{28\pm8}{10}$ → $x=\frac{36}{10}=\frac{18}{5}$ or $x=\frac{20}{10}=2$. $x=2$ → $y=3$; $x=\frac{18}{5}=3.6$ → $y=7-7.2=-0.2=-\frac{1}{5}$ → Answer: D. $x=2,y=3$ and $x=\frac{18}{5},y=-\frac{1}{5}$
1.2.10 Sum of roots $=4$, product $=3$. Find equation. (A) $x^2-4x+3=0$ (B) $x^2+4x+3=0$ (C) $x^2-4x-3=0$ (D) $x^2+4x-3=0$ (E) $x^2-3x+4=0$.
$x^2 - (\text{sum})x + (\text{product}) = 0$ → $x^2 - 4x + 3 = 0$ → Answer: A. $x^2-4x+3=0$
1.2.11 Solve: $x^4 - 5x^2 + 4 = 0$ (A) $x=\pm 1, \pm 2$ (B) $x=\pm 1, \pm 4$ (C) $x=\pm 2, \pm 3$ (D) $x=\pm 1$ only (E) $x=\pm 2$ only.
Let $u=x^2$: $u^2-5u+4=0$ → $(u-1)(u-4)=0$ → $u=1$ or $u=4$ → $x^2=1$ → $x=\pm1$; $x^2=4$ → $x=\pm2$ → Answer: A. $x=\pm 1, \pm 2$
1.2.12 Range of $k$ for $x^2 + 2kx + 9 = 0$ to have real roots. (A) $k \leq -3$ or $k \geq 3$ (B) $-3 \leq k \leq 3$ (C) $k \leq -9$ or $k \geq 9$ (D) $-9 \leq k \leq 9$ (E) $k \geq 3$ only.
Real roots when $\Delta = (2k)^2 - 4(1)(9) = 4k^2 - 36 \geq 0$ → $4k^2 \geq 36$ → $k^2 \geq 9$ → $k \leq -3$ or $k \geq 3$ → Answer: A. $k \leq -3$ or $k \geq 3$
1.2.13 Solve: $3x - 2y = 7$ and $x + 3y = 6$ (A) $x=3,y=1$ (B) $x=1,y=-2$ (C) $x=4,y=2$ (D) $x=2,y=-1$ (E) $x=5,y=4$.
From second: $x = 6-3y$. Substitute: $3(6-3y)-2y=7$ → $18-9y-2y=7$ → $18-11y=7$ → $-11y=-11$ → $y=1$. Then $x=6-3=3$ → Answer: A. $x=3,y=1$
1.2.14 Roots of $2x^2 - 3x - 2 = 0$ are $\alpha,\beta$. Find $\frac{1}{\alpha}+\frac{1}{\beta}$. (A) $-\frac{3}{2}$ (B) $\frac{3}{2}$ (C) $-\frac{3}{4}$ (D) $\frac{3}{4}$ (E) $-\frac{2}{3}$.
$\alpha+\beta = \frac{3}{2}$, $\alpha\beta = \frac{-2}{2} = -1$. Then $\frac{1}{\alpha}+\frac{1}{\beta} = \frac{\alpha+\beta}{\alpha\beta} = \frac{3/2}{-1} = -\frac{3}{2}$ → Answer: A. $-\frac{3}{2}$
1.2.15 Quadratic with roots $2+i$ and $2-i$. (A) $x^2-4x+5=0$ (B) $x^2+4x+5=0$ (C) $x^2-4x-5=0$ (D) $x^2-4x+3=0$ (E) $x^2+4x-5=0$.
Sum $=(2+i)+(2-i)=4$, Product $=(2+i)(2-i)=4-(-1)=5$ → $x^2-4x+5=0$ → Answer: A. $x^2-4x+5=0$
1.2.16 A man buys $x$ shirts for €240. If each cost €4 less, he'd buy 5 more for same amount. Find original price per shirt. (A) €12 (B) €16 (C) €20 (D) €24 (E) €30.
Original price $p = \frac{240}{x}$. New price $p-4 = \frac{240}{x+5}$. So $\frac{240}{x} - 4 = \frac{240}{x+5}$. Multiply by $x(x+5)$: $240(x+5) - 4x(x+5) = 240x$ → $240x+1200-4x^2-20x=240x$ → $1200-4x^2-20x=0$ → divide -4: $x^2+5x-300=0$ → $(x+20)(x-15)=0$ → $x=15$. Original price $p=240/15=16$ → Answer: B. €16
1.2.17 Solve: $\sqrt{2x+3} - x = 0$ (A) $x=3$ (B) $x=-1$ (C) $x=3$ or $x=-1$ (D) $x=-3$ (E) $x=1$.
$\sqrt{2x+3}=x$ → square: $2x+3=x^2$ → $x^2-2x-3=0$ → $(x-3)(x+1)=0$ → $x=3$ or $x=-1$. Check: $x=3$ → $\sqrt{9}=3$ ✓; $x=-1$ → $\sqrt{1}=1$, LHS $=1-(-1)=2\neq0$ ✗ (extraneous). So only $x=3$ → Answer: A. $x=3$
1.2.18 If $x+y=5$ and $xy=6$, find $x^2+y^2$. (A) 11 (B) 13 (C) 15 (D) 17 (E) 19.
$x^2+y^2 = (x+y)^2 - 2xy = 25 - 12 = 13$ → Answer: B. 13
1.2.19 Find $k$ so line $y=2x+k$ is tangent to $y=x^2-2x+3$. (A) $k=-1$ (B) $k=1$ (C) $k=3$ (D) $k=5$ (E) $k=7$.
Set equal: $x^2-2x+3 = 2x+k$ → $x^2-4x+(3-k)=0$. Tangent → discriminant $=0$: $(-4)^2 - 4(1)(3-k)=0$ → $16 - 12 + 4k = 0$ → $4 + 4k = 0$ → $k=-1$ → Answer: A. $k=-1$
1.2.20 Solve: $2^{2x} - 5 \times 2^x + 4 = 0$ (A) $x=0,2$ (B) $x=1,4$ (C) $x=2,3$ (D) $x=0,4$ (E) $x=1,2$.
Let $u=2^x$: $u^2 - 5u + 4 = 0$ → $(u-1)(u-4)=0$ → $u=1$ or $u=4$. $2^x=1$ → $x=0$; $2^x=4$ → $x=2$ → Answer: A. $x=0,2$
Evaluation G - 12 | 1.3 Solutions (Binomial Expansion)
1.3.1 Coefficient of $x^3$ in $(1 + 2x)^5$. (A) 40 (B) 60 (C) 80 (D) 100 (E) 120.
$T_{r+1} = \binom{5}{r} (1)^{5-r} (2x)^r = \binom{5}{r} 2^r x^r$. For $x^3$, $r=3$: $\binom{5}{3} \cdot 2^3 = 10 \cdot 8 = 80$ → Answer: C. 80
1.3.2 Term independent of $x$ in $(2x - \frac{1}{x})^6$. (A) -120 (B) -160 (C) 120 (D) 160 (E) 240.
$T_{r+1} = \binom{6}{r} (2x)^{6-r} (-\frac{1}{x})^r = \binom{6}{r} 2^{6-r} (-1)^r x^{6-2r}$. Independent when $6-2r=0$ → $r=3$: $\binom{6}{3} 2^{3} (-1)^3 = 20 \cdot 8 \cdot (-1) = -160$ → Answer: B. -160
1.3.3 Coefficient of $x^5$ in $(1 + 2x)^8$. (A) 448 (B) 896 (C) 1120 (D) 1344 (E) 1792.
$T_{r+1} = \binom{8}{r} 2^r x^r$. For $x^5$, $r=5$: $\binom{8}{5} \cdot 2^5 = 56 \cdot 32 = 1792$ → Answer: E. 1792
1.3.4 First three terms of $(1 + 3x)^7$ ascending powers. (A) $1+21x+189x^2$ (B) $1+7x+21x^2$ (C) $1+21x+126x^2$ (D) $1+7x+147x^2$ (E) $1+3x+9x^2$.
$T_1 = \binom{7}{0}1^7=1$; $T_2 = \binom{7}{1}1^6(3x)=7\cdot3x=21x$; $T_3 = \binom{7}{2}1^5(3x)^2=21\cdot9x^2=189x^2$ → Answer: A. $1+21x+189x^2$
1.3.5 Coefficient of $x^2y^4$ in $(2x + 3y)^6$. (A) 2160 (B) 3240 (C) 4320 (D) 4860 (E) 6480.
$T_{r+1} = \binom{6}{r} (2x)^{6-r} (3y)^r = \binom{6}{r} 2^{6-r} 3^r x^{6-r} y^r$. Need $6-r=2$ → $r=4$, $y^4$: $\binom{6}{4} 2^{2} 3^4 = 15 \cdot 4 \cdot 81 = 15 \cdot 324 = 4860$ → Answer: D. 4860
1.3.6 Term containing $x^4$ in $(x - \frac{2}{x})^8$. (A) $112x^4$ (B) 112 (C) $-112x^4$ (D) $-112$ (E) $1792x^4$.
$T_{r+1} = \binom{8}{r} x^{8-r} (-\frac{2}{x})^r = \binom{8}{r} (-2)^r x^{8-2r}$. $8-2r=4$ → $2r=4$ → $r=2$: $\binom{8}{2} (-2)^2 x^4 = 28 \cdot 4 x^4 = 112x^4$ → Answer: A. $112x^4$
1.3.7 In $(1 + ax)^6$, coefficient of $x^2$ is 135. Find $a$. (A) 1.5 (B) 2 (C) 2.5 (D) 3 (E) 3.5.
$T_3 = \binom{6}{2} a^2 x^2 = 15 a^2 x^2$. So $15a^2 = 135$ → $a^2 = 9$ → $a=3$ (positive) → Answer: D. 3
1.3.8 Sum of coefficients in $(2x - 3y)^5$. (A) -1 (B) 0 (C) 1 (D) 32 (E) -32.
Sum of coefficients = substitute $x=1, y=1$: $(2(1) - 3(1))^5 = (-1)^5 = -1$ → Answer: A. -1
1.3.9 Coefficient of $x^3$ in $(1 - x)(1 + 2x)^5$. (A) 40 (B) 50 (C) 60 (D) 70 (E) 80.
$(1 + 2x)^5 = \sum_{r=0}^5 \binom{5}{r} 2^r x^r$. Multiply by $(1-x)$: coefficient of $x^3$ comes from $(1)(\binom{5}{3}2^3 x^3) + (-x)(\binom{5}{2}2^2 x^2) = \binom{5}{3}8 - \binom{5}{2}4 = 10\cdot8 - 10\cdot4 = 80 - 40 = 40$ → Answer: A. 40
1.3.10 Middle term in $(x + 2y)^8$. (A) $1120x^4y^4$ (B) $1120x^5y^3$ (C) $1120x^3y^5$ (D) $1792x^4y^4$ (E) $1792x^5y^3$.
$n=8$ even → middle term $r=4$: $T_{5} = \binom{8}{4} x^{4} (2y)^4 = 70 \cdot 16 x^4 y^4 = 1120 x^4 y^4$ → Answer: A. $1120x^4y^4$
1.3.11 Coefficient of $x^4$ in $(2 - x + 3x^2)(1 + x)^6$. (A) 45 (B) 50 (C) 55 (D) 60 (E) 65.
$(1+x)^6 = \sum_{r=0}^6 \binom{6}{r} x^r$. Multiply: $2$ gives $2\binom{6}{4}x^4 = 2\cdot15=30$; $-x$ gives $-\binom{6}{3}x^4 = -20$; $3x^2$ gives $3\binom{6}{2}x^4 = 3\cdot15=45$. Sum: $30-20+45=55$ → Answer: C. 55
1.3.12 In $(1 + kx)^8$, coeff of $x^3 = 7\times$ coeff of $x^2$. Find $k$. (A) 1.5 (B) 2 (C) 2.5 (D) 3 (E) 3.5.
Coeff of $x^2$: $\binom{8}{2}k^2 = 28k^2$; coeff of $x^3$: $\binom{8}{3}k^3 = 56k^3$. Given $56k^3 = 7 \cdot 28k^2$ → $56k^3 = 196k^2$ → $56k = 196$ → $k = 196/56 = 3.5$ → Answer: E. 3.5
1.3.13 Term independent of $x$ in $(x^2 + \frac{2}{x})^9$. (A) 5376 (B) 6720 (C) 8064 (D) 10752 (E) 13440.
$T_{r+1} = \binom{9}{r} (x^2)^{9-r} (\frac{2}{x})^r = \binom{9}{r} 2^r x^{18-3r}$. Independent when $18-3r=0$ → $r=6$: $\binom{9}{6} 2^6 = 84 \cdot 64 = 5376$ → Answer: A. 5376
1.3.14 Coefficient of $x^2y^3$ in $(x - 2y)^5$. (A) -80 (B) -40 (C) 40 (D) 80 (E) 160.
$T_{r+1} = \binom{5}{r} x^{5-r} (-2y)^r = \binom{5}{r} (-2)^r x^{5-r} y^r$. Need $5-r=2$ → $r=3$: $\binom{5}{3} (-2)^3 = 10 \cdot (-8) = -80$ → Answer: A. -80
1.3.15 First four terms of $(2 - x)^6$ ascending powers. (A) $64 - 192x + 240x^2 - 160x^3$ (B) $64 - 192x + 240x^2 - 120x^3$ (C) $64 - 192x + 160x^2 - 240x^3$ (D) $64 - 96x + 240x^2 - 160x^3$ (E) $64 - 192x + 480x^2 - 160x^3$.
$T_1 = \binom{6}{0}2^6=64$; $T_2 = \binom{6}{1}2^5(-x)=6\cdot32\cdot(-x)=-192x$; $T_3 = \binom{6}{2}2^4(-x)^2=15\cdot16\cdot x^2=240x^2$; $T_4 = \binom{6}{3}2^3(-x)^3=20\cdot8\cdot(-x^3)=-160x^3$ → Answer: A. $64 - 192x + 240x^2 - 160x^3$
1.3.16 Coefficient of $x^5$ in $(1 - 2x + x^2)^4$. (A) -16 (B) -8 (C) 8 (D) 16 (E) -56.
Note $1 - 2x + x^2 = (1-x)^2$, so expression = $(1-x)^8$. Coefficient of $x^5$: $\binom{8}{5}(-1)^5 = 56 \cdot (-1) = -56$ → Answer: E. -56
1.3.17 In $(1 + 3x)^n$, coeff of $x^2$ is 54. Find $n$. (A) 3 (B) 4 (C) 5 (D) 6 (E) 7.
$\binom{n}{2} \cdot 3^2 = 54$ → $\frac{n(n-1)}{2} \cdot 9 = 54$ → $\frac{n(n-1)}{2} = 6$ → $n(n-1)=12$ → $n^2-n-12=0$ → $(n-4)(n+3)=0$ → $n=4$ → Answer: B. 4
1.3.18 Term containing $x^3$ in $(x^2 - \frac{2}{x})^6$. (A) $120x^3$ (B) $-120x^3$ (C) $160x^3$ (D) $-160x^3$ (E) $240x^3$.
$T_{r+1} = \binom{6}{r} (x^2)^{6-r} (-\frac{2}{x})^r = \binom{6}{r} (-2)^r x^{12-3r}$. $12-3r=3$ → $3r=9$ → $r=3$: $\binom{6}{3} (-2)^3 x^3 = 20 \cdot (-8) x^3 = -160x^3$ → Answer: D. $-160x^3$
1.3.19 First three terms of $(1 + ax)^n$ are $1 + 12x + 54x^2$. Find $n,a$. (A) $n=6,a=2$ (B) $n=8,a=1.5$ (C) $n=9,a=\frac{4}{3}$ (D) $n=12,a=1$ (E) $n=4,a=3$.
$\binom{n}{1}a = na = 12$; $\binom{n}{2}a^2 = \frac{n(n-1)}{2}a^2 = 54$. From $na=12$ → $a=12/n$. Substitute: $\frac{n(n-1)}{2} \cdot \frac{144}{n^2} = 54$ → $\frac{(n-1)144}{2n} = 54$ → $\frac{72(n-1)}{n} = 54$ → $72n-72=54n$ → $18n=72$ → $n=4$, then $a=12/4=3$ → Answer: E. $n=4, a=3$
1.3.20 Coefficient of $x^4$ in $(1 + x + x^2)^5$. (A) 25 (B) 30 (C) 35 (D) 40 (E) 45.
We need combinations of exponents from five factors summing to 4. Possibilities: (4 ones, 1 zero): choose 4 factors to give $x^1$ → $\binom{5}{4}=5$; (2 ones, 1 two, 2 zeros): choose 2 for $x$, 1 for $x^2$ → $\binom{5}{2}\binom{3}{1}=10\cdot3=30$; (2 twos, 3 zeros): choose 2 for $x^2$ → $\binom{5}{2}=10$; (1 one, 1 three? impossible since max exponent 2). Total $5+30+10=45$ → Answer: E. 45
Evaluation G - 12 | 1.4 Solutions (Function Transformations)
1.4.1 The function $f(x) = x^2$ is translated 3 units right and 2 units up. What is the new equation? (A) $f(x) = (x-3)^2 + 2$ (B) $f(x) = (x+3)^2 + 2$ (C) $f(x) = (x-3)^2 - 2$ (D) $f(x) = (x+3)^2 - 2$ (E) $f(x) = x^2 + 5$.
Right 3: replace $x$ with $(x-3)$; up 2: add 2 → $f(x) = (x-3)^2 + 2$ → Answer: A
1.4.2 If $f(x) = \sin x$, what is the amplitude of $g(x) = -3\sin(2x)$? (A) -3 (B) 2 (C) 3 (D) 1 (E) 6.
Amplitude is the absolute value of the coefficient of sine: $|-3| = 3$ → Answer: C. 3
1.4.3 The graph of $y = x^3$ is reflected in the x-axis and then shifted left by 2. What is the new equation? (A) $y = -(x+2)^3$ (B) $y = -(x-2)^3$ (C) $y = (-x+2)^3$ (D) $y = -x^3 + 2$ (E) $y = (-x-2)^3$.
Reflect in x-axis: multiply by -1 → $y = -x^3$; shift left 2: replace $x$ with $(x+2)$ → $y = -(x+2)^3$ → Answer: A
1.4.4 What is the period of $y = 4\cos(3x - \pi)$? (A) $\frac{2\pi}{3}$ (B) $\frac{\pi}{3}$ (C) $2\pi$ (D) $\frac{3\pi}{2}$ (E) $4\pi$.
Period of $\cos(kx)$ is $\frac{2\pi}{|k|} = \frac{2\pi}{3}$ → Answer: A
1.4.5 If $f(x) = x^2$ and $g(x) = f(x+1) - 4$, what is the vertex of $g(x)$? (A) $(-1, -4)$ (B) $(1, -4)$ (C) $(-1, 4)$ (D) $(1, 4)$ (E) $(0, -4)$.
$g(x) = (x+1)^2 - 4$; vertex at $(-1, -4)$ → Answer: A
1.4.6 Which transformation changes $y = \sin x$ to $y = \sin(2x - \frac{\pi}{3})$? (A) Horizontal compression by $\frac{1}{2}$ and shift right $\frac{\pi}{6}$ (B) Horizontal stretch by 2 and shift right $\frac{\pi}{3}$ (C) Horizontal compression by $\frac{1}{2}$ and shift left $\frac{\pi}{6}$ (D) Horizontal stretch by 2 and shift left $\frac{\pi}{3}$ (E) Horizontal compression by $\frac{1}{2}$ and shift right $\frac{\pi}{3}$.
$y = \sin(2x - \frac{\pi}{3}) = \sin[2(x - \frac{\pi}{6})]$ → compression factor $\frac{1}{2}$, shift right $\frac{\pi}{6}$ → Answer: A
1.4.7 The graph of $y = \tan x$ has vertical asymptotes at $x = \frac{\pi}{2} + n\pi$. What happens to the asymptotes in $y = 2\tan(3x)$? (A) $x = \frac{\pi}{6} + \frac{n\pi}{3}$ (B) $x = \frac{\pi}{2} + \frac{n\pi}{3}$ (C) $x = \frac{\pi}{6} + n\pi$ (D) $x = \frac{\pi}{2} + n\pi$ (E) $x = 3(\frac{\pi}{2} + n\pi)$.
For $\tan(kx)$, asymptotes occur when $kx = \frac{\pi}{2} + n\pi$ → $3x = \frac{\pi}{2} + n\pi$ → $x = \frac{\pi}{6} + \frac{n\pi}{3}$ → Answer: A
1.4.8 If $f(x) = x^3 - 3x$, which of the following is equal to $f(-x)$? (A) $-x^3 - 3x$ (B) $x^3 - 3x$ (C) $-x^3 + 3x$ (D) $x^3 + 3x$ (E) $-x^3 + 3x$.
$f(-x) = (-x)^3 - 3(-x) = -x^3 + 3x$ → Answer: C
1.4.9 What is the range of $y = 2\sin(x) - 1$? (A) $[-3, 1]$ (B) $[-1, 3]$ (C) $[-2, 2]$ (D) $[-2, 0]$ (E) $[0, 2]$.
$\sin x \in [-1, 1]$ → $2\sin x \in [-2, 2]$ → $2\sin x - 1 \in [-3, 1]$ → Answer: A
1.4.10 The function $f(x) = \sqrt{x}$ is stretched vertically by factor 3 and reflected in the y-axis. What is the new function? (A) $f(x) = 3\sqrt{-x}$ (B) $f(x) = -3\sqrt{x}$ (C) $f(x) = 3\sqrt{x}$ (D) $f(x) = \sqrt{-3x}$ (E) $f(x) = -3\sqrt{-x}$.
Vertical stretch by 3: multiply by 3 → $3\sqrt{x}$; reflect in y-axis: replace $x$ with $-x$ → $3\sqrt{-x}$ → Answer: A
1.4.11 Given $f(x) = (x-2)^2 + 1$, which transformation gives $g(x) = (x+1)^2 - 3$? (A) Left 3, down 4 (B) Right 3, up 4 (C) Left 3, up 4 (D) Right 3, down 4 (E) Left 1, down 3.
$f(x)$ vertex at $(2,1)$; $g(x)$ vertex at $(-1,-3)$ → shift left 3 and down 4 → Answer: A
1.4.12 What is the equation of the image of $y = \cos x$ after a vertical stretch by factor 2, a horizontal compression by factor $\frac{1}{3}$, and a shift up by 4? (A) $y = 2\cos(3x) + 4$ (B) $y = 2\cos(\frac{x}{3}) + 4$ (C) $y = 2\cos(3x + 4)$ (D) $y = 3\cos(2x) + 4$ (E) $y = \frac{1}{2}\cos(3x) + 4$.
Vertical stretch ×2 → multiply by 2; horizontal compression by $\frac{1}{3}$ → multiply argument by 3; shift up 4 → add 4 → $y = 2\cos(3x) + 4$ → Answer: A
1.4.13 If $f(x) = x^2 - 4x + 3$, find the vertex after a reflection in the x-axis. (A) $(2, 1)$ (B) $(2, -1)$ (C) $(-2, 1)$ (D) $(-2, -1)$ (E) $(2, 3)$.
$f(x) = (x-2)^2 - 1$ → vertex $(2, -1)$. Reflect in x-axis: multiply by -1 → $y = -(x-2)^2 + 1$ → vertex $(2, 1)$ → Answer: A
1.4.14 The graph of $y = 3\sin(2x + \pi)$ has a phase shift of: (A) $\frac{\pi}{2}$ left (B) $\frac{\pi}{2}$ right (C) $\pi$ left (D) $\pi$ right (E) no phase shift.
$y = 3\sin(2x + \pi) = 3\sin[2(x + \frac{\pi}{2})]$ → shift left $\frac{\pi}{2}$ → Answer: A
1.4.15 Which function is even? (A) $y = x^3 - x$ (B) $y = \sin x$ (C) $y = \cos x$ (D) $y = 2\sin x \cos x$ (E) $y = x^3 + x$.
Even: $f(-x) = f(x)$. $\cos(-x) = \cos x$ → Answer: C
1.4.16 If $f(x) = 2x^2 - 3$ and $g(x) = f(x-1) + 2$, find $g(2)$. (A) 1 (B) 3 (C) 5 (D) 7 (E) 9.
$g(2) = f(1) + 2 = [2(1)^2 - 3] + 2 = (2 - 3) + 2 = -1 + 2 = 1$ → Answer: A
1.4.17 What is the maximum value of $y = -4\cos(3x) + 1$? (A) 5 (B) 3 (C) 1 (D) -3 (E) -4.
$\cos(3x) \in [-1, 1]$ → $-4\cos(3x) \in [-4, 4]$? Actually $-4 \times 1 = -4$, $-4 \times (-1) = 4$ → range of $-4\cos(3x)$ is $[-4, 4]$? Wait: when $\cos=1$, $-4$; when $\cos=-1$, $4$; so $[-4, 4]$. Then add 1 → $[-3, 5]$. Maximum = $5$ → Answer: A
1.4.18 The graph of $y = \sqrt{x}$ is stretched horizontally by factor 4. What is the new equation? (A) $y = \sqrt{4x}$ (B) $y = 4\sqrt{x}$ (C) $y = \sqrt{\frac{x}{4}}$ (D) $y = \frac{1}{4}\sqrt{x}$ (E) $y = \sqrt{x} + 4$.
Horizontal stretch by factor 4 → replace $x$ with $\frac{x}{4}$ → $y = \sqrt{\frac{x}{4}}$ → Answer: C
1.4.19 Given $f(x) = \tan x$, which transformation produces $y = -\tan(x - \frac{\pi}{4})$? (A) Right $\frac{\pi}{4}$ and reflection in x-axis (B) Left $\frac{\pi}{4}$ and reflection in x-axis (C) Right $\frac{\pi}{4}$ and reflection in y-axis (D) Left $\frac{\pi}{4}$ and reflection in y-axis (E) Right $\frac{\pi}{4}$ only.
$-\tan(x - \frac{\pi}{4})$ = reflect in x-axis ($-\tan$) and shift right $\frac{\pi}{4}$ → Answer: A
1.4.20 If the point $(4, 3)$ lies on $y = f(x)$, what point lies on $y = 2f(x-1) + 1$? (A) $(5, 7)$ (B) $(3, 7)$ (C) $(5, 4)$ (D) $(3, 4)$ (E) $(4, 7)$.
$x-1 = 4$ → $x = 5$; $y = 2f(4) + 1 = 2(3) + 1 = 7$ → point $(5, 7)$ → Answer: A
Evaluation G - 12 | 1.5 Solutions (Trigonometry)
1.5.1 Simplify: $\frac{\sin^2 x}{1 - \cos x}$ (A) $1 + \cos x$ (B) $1 - \cos x$ (C) $\sin x$ (D) $\cos x$ (E) $\tan x$.
$\sin^2 x = 1 - \cos^2 x = (1-\cos x)(1+\cos x)$ → $\frac{(1-\cos x)(1+\cos x)}{1-\cos x} = 1 + \cos x$ → Answer: A
1.5.2 What is the period of $y = 3\sin(4x) + 2$? (A) $90^\circ$ (B) $180^\circ$ (C) $360^\circ$ (D) $720^\circ$ (E) $45^\circ$.
Period of $\sin(kx)$ is $\frac{360^\circ}{k} = \frac{360^\circ}{4} = 90^\circ$ → Answer: A
1.5.3 Solve $2\cos x + 1 = 0$ on $[0^\circ, 360^\circ]$. (A) $60^\circ,300^\circ$ (B) $120^\circ,240^\circ$ (C) $30^\circ,330^\circ$ (D) $150^\circ,210^\circ$ (E) $45^\circ,315^\circ$.
$2\cos x = -1$ → $\cos x = -\frac{1}{2}$ → $x = 120^\circ, 240^\circ$ → Answer: B
1.5.4 If $\sin \theta = \frac{3}{5}$ and $\theta$ in quadrant II, find $\cos 2\theta$. (A) $\frac{7}{25}$ (B) $-\frac{7}{25}$ (C) $\frac{24}{25}$ (D) $-\frac{24}{25}$ (E) $\frac{18}{25}$.
$\cos 2\theta = 1 - 2\sin^2 \theta = 1 - 2(\frac{9}{25}) = 1 - \frac{18}{25} = \frac{7}{25}$ (cos2θ does not depend on quadrant) → Answer: A
1.5.5 Triangle ABC: $a=8$ cm, $b=10$ cm, angle $C=60^\circ$. Find $c$ using cosine rule. (A) $\sqrt{89}$ cm (B) $\sqrt{124}$ cm (C) $\sqrt{164}$ cm (D) $\sqrt{84}$ cm (E) $12$ cm.
$c^2 = a^2 + b^2 - 2ab\cos C = 64 + 100 - 2(8)(10)(\frac{1}{2}) = 164 - 80 = 84$ → $c = \sqrt{84}$ → Answer: D
1.5.6 Prove $\tan x + \cot x = \sec x \csc x$. Correct step? (A) $\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} = \frac{1}{\sin x \cos x} = \sec x \csc x$ (B) $\frac{\sin^2 x - \cos^2 x}{\sin x \cos x}$ (C) $=1$ (D) $=\sin x + \cos x$ (E) $=\frac{1}{\sin x} + \frac{1}{\cos x}$.
Option A is the correct proof → Answer: A
1.5.7 Amplitude and phase shift of $y = 4\cos(2x - 90^\circ)$. (A) Amplitude 4, phase shift $45^\circ$ right (B) Amplitude 4, phase shift $90^\circ$ left (C) Amplitude 2, phase shift $45^\circ$ right (D) Amplitude 4, phase shift $45^\circ$ left (E) Amplitude 2, phase shift $90^\circ$ right.
$y = 4\cos[2(x - 45^\circ)]$ → amplitude $4$, phase shift $45^\circ$ right → Answer: A
1.5.8 Solve $2\sin^2 x - \sin x - 1 = 0$ on $[0, 2\pi]$. (A) $x = \frac{\pi}{2}, \frac{7\pi}{6}, \frac{11\pi}{6}$ (B) $x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2}$ (C) $x = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}$ (D) $x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}$ (E) $x = 0, \frac{\pi}{2}, \pi$.
Let $u = \sin x$: $2u^2 - u - 1 = 0$ → $(2u+1)(u-1)=0$ → $u=1$ or $u=-\frac{1}{2}$. $\sin x=1$ → $x=\frac{\pi}{2}$; $\sin x=-\frac{1}{2}$ → $x=\frac{7\pi}{6}, \frac{11\pi}{6}$ → Answer: A
1.5.9 Triangle PQR: $p=12$ cm, $q=15$ cm, angle $R=75^\circ$. Find area. (A) $90\sin 75^\circ$ (B) $180\sin 75^\circ$ (C) $90\cos 75^\circ$ (D) $45\sin 75^\circ$ (E) $180\cos 75^\circ$.
Area $= \frac{1}{2}pq\sin R = \frac{1}{2}(12)(15)\sin 75^\circ = 90\sin 75^\circ$ → Answer: A
1.5.10 Equivalent to $\sin 3x$? (A) $3\sin x - 7\sin^3 x$ (B) $4\sin^3 x - 3\sin x$ (C) $3\cos x - 4\cos^3 x$ (D) $2\sin x\cos x$ (E) $\sin x(3 - 4\sin^2 x)$.
$\sin 3x = 3\sin x - 4\sin^3 x = \sin x(3 - 4\sin^2 x)$ → Answer: E
1.5.11 Exact value of $\sin 75^\circ$. (A) $\frac{\sqrt{6}-\sqrt{2}}{4}$ (B) $\frac{\sqrt{6}+\sqrt{2}}{4}$ (C) $\frac{\sqrt{3}+1}{2}$ (D) $\frac{\sqrt{3}-1}{2\sqrt{2}}$ (E) $\frac{\sqrt{2}+\sqrt{6}}{2}$.
$\sin 75^\circ = \sin(45^\circ+30^\circ) = \sin45\cos30 + \cos45\sin30 = \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2}\cdot\frac{1}{2} = \frac{\sqrt{6}+\sqrt{2}}{4}$ → Answer: B
1.5.12 Transformation $y = \sin x$ to $y = 2\sin(3x + 90^\circ) - 1$. (A) Vertical stretch 2, horizontal compression $\frac{1}{3}$, phase shift $30^\circ$ left, down 1 (B) phase shift $90^\circ$ left (C) horizontal stretch 3 (D) phase shift $30^\circ$ right (E) up 1.
$y = 2\sin[3(x + 30^\circ)] - 1$ → vertical stretch 2, horizontal compression $\frac{1}{3}$, shift left $30^\circ$, down 1 → Answer: A
1.5.13 Triangle XYZ: angle $X=40^\circ$, angle $Y=65^\circ$, side $x=10$ cm. Find $y$ using sine rule. (A) $10\sin65^\circ/\sin40^\circ$ (B) $10\sin40^\circ/\sin65^\circ$ (C) $10\sin75^\circ/\sin40^\circ$ (D) $10\sin65^\circ/\sin75^\circ$ (E) $10\sin40^\circ/\sin75^\circ$.
Angle $Z = 180-40-65=75^\circ$. Sine rule: $\frac{x}{\sin X} = \frac{y}{\sin Y}$ → $y = \frac{x\sin Y}{\sin X} = \frac{10\sin65^\circ}{\sin40^\circ}$ → Answer: A
1.5.14 Simplify $\frac{\sin 2x}{1 + \cos 2x}$. (A) $\tan x$ (B) $\cot x$ (C) $\tan 2x$ (D) $\cot 2x$ (E) $\sin x$.
$\sin 2x = 2\sin x\cos x$, $1 + \cos 2x = 2\cos^2 x$ → $\frac{2\sin x\cos x}{2\cos^2 x} = \frac{\sin x}{\cos x} = \tan x$ → Answer: A
1.5.15 Solve $\tan 2x = 1$ on $[0^\circ, 180^\circ]$. (A) $22.5^\circ, 112.5^\circ$ (B) $45^\circ,225^\circ$ (C) $22.5^\circ,202.5^\circ$ (D) $45^\circ,135^\circ$ (E) $22.5^\circ,67.5^\circ$.
$2x = 45^\circ + 180^\circ n$ → $x = 22.5^\circ + 90^\circ n$. In $[0^\circ,180^\circ]$: $n=0$ → $22.5^\circ$; $n=1$ → $112.5^\circ$ → Answer: A
1.5.16 If $\sin \theta = \frac{12}{13}$ ($\theta$ acute), find $\tan 2\theta$. (A) $\frac{120}{119}$ (B) $-\frac{120}{119}$ (C) $\frac{119}{120}$ (D) $\frac{60}{61}$ (E) $-\frac{60}{61}$.
$\cos \theta = \frac{5}{13}$, $\tan \theta = \frac{12}{5}$. $\tan 2\theta = \frac{2\tan\theta}{1-\tan^2\theta} = \frac{2(12/5)}{1 - (144/25)} = \frac{24/5}{(25-144)/25} = \frac{24/5}{-119/25} = \frac{24}{5} \cdot \frac{-25}{119} = -\frac{120}{119}$ → Answer: B
1.5.17 Graph of $y = 2\cos x - 1$? (A) Cosine curve amplitude 2, shifted down 1 (B) amplitude 1, down 2 (C) amplitude 2, up 1 (D) amplitude 3, down 1 (E) amplitude 2, period $180^\circ$.
Amplitude $|2|=2$, vertical shift $-1$ (down 1) → Answer: A
1.5.18 Triangle ABC: $a=15$ cm, $b=20$ cm, angle $C=120^\circ$. Find $c$. (A) $\sqrt{325}$ (B) $25$ (C) $\sqrt{925}$ (D) $35$ (E) $\sqrt{1225}$.
$c^2 = a^2 + b^2 - 2ab\cos C = 225 + 400 - 2(15)(20)(-\frac{1}{2}) = 625 + 300 = 925$ → $c = \sqrt{925}$ → Answer: C
1.5.19 Prove $\frac{\sin 2x}{2\sin x} = \cos x$. Correct step? (A) $\frac{2\sin x\cos x}{2\sin x} = \cos x$ (B) $\frac{2\sin x}{2\sin x}=1$ (C) $\frac{\sin x\cos x}{\sin x}=\cos x$ (D) $=\sin x$ (E) $\frac{2\cos x}{2}=\cos x$.
Option A is the correct proof → Answer: A
1.5.20 Solve $\cos 2x = \cos x$ on $[0, 2\pi]$. (A) $0,\frac{2\pi}{3},\frac{4\pi}{3}$ (B) $0,\frac{\pi}{3},\frac{5\pi}{3}$ (C) $0,\frac{2\pi}{3},\frac{4\pi}{3},2\pi$ (D) $\frac{\pi}{3},\pi,\frac{5\pi}{3}$ (E) $0,\frac{\pi}{2},\pi,\frac{3\pi}{2}$.
$\cos 2x - \cos x = 0$ → $2\cos^2 x - 1 - \cos x = 0$ → $2\cos^2 x - \cos x - 1 = 0$ → $(2\cos x+1)(\cos x-1)=0$ → $\cos x=1$ or $\cos x=-\frac{1}{2}$. $\cos x=1$ → $x=0,2\pi$; $\cos x=-\frac{1}{2}$ → $x=\frac{2\pi}{3},\frac{4\pi}{3}$. Solutions: $0,\frac{2\pi}{3},\frac{4\pi}{3},2\pi$ → Answer: C
Evaluation G - 12 | 1.6 Solutions (Vectors)
1.6.1 Given $\mathbf{a} = 3\mathbf{i} - 2\mathbf{j}$ and $\mathbf{b} = -\mathbf{i} + 4\mathbf{j}$, find $2\mathbf{a} - 3\mathbf{b}$. (A) $9\mathbf{i} - 16\mathbf{j}$ (B) $3\mathbf{i} - 16\mathbf{j}$ (C) $9\mathbf{i} + 8\mathbf{j}$ (D) $3\mathbf{i} + 8\mathbf{j}$ (E) $9\mathbf{i} - 8\mathbf{j}$.
$2\mathbf{a} = 6\mathbf{i} - 4\mathbf{j}$, $3\mathbf{b} = -3\mathbf{i} + 12\mathbf{j}$, so $2\mathbf{a} - 3\mathbf{b} = (6 - (-3))\mathbf{i} + (-4 - 12)\mathbf{j} = 9\mathbf{i} - 16\mathbf{j}$ → Answer: A
1.6.2 Find magnitude of $\mathbf{v} = 5\mathbf{i} - 12\mathbf{j}$. (A) 7 (B) 13 (C) 17 (D) 10 (E) 8.
$|\mathbf{v}| = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13$ → Answer: B
1.6.3 If $\mathbf{p} = \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix}$ and $\mathbf{q} = \begin{pmatrix} 4 \\ 0 \\ -2 \end{pmatrix}$, find $\mathbf{p} \cdot \mathbf{q}$. (A) 2 (B) 6 (C) 8 (D) 10 (E) 14.
$\mathbf{p} \cdot \mathbf{q} = 2(4) + (-3)(0) + 1(-2) = 8 + 0 - 2 = 6$ → Answer: B
1.6.4 Find unit vector in direction of $\mathbf{u} = 3\mathbf{i} + 4\mathbf{j}$. (A) $\frac{1}{5}(3\mathbf{i}+4\mathbf{j})$ (B) $\frac{1}{7}(3\mathbf{i}+4\mathbf{j})$ (C) $\frac{1}{25}(3\mathbf{i}+4\mathbf{j})$ (D) $\frac{1}{5}(4\mathbf{i}+3\mathbf{j})$ (E) $\frac{1}{12}(3\mathbf{i}+4\mathbf{j})$.
$|\mathbf{u}| = \sqrt{9+16}=5$, unit vector $= \frac{1}{5}(3\mathbf{i}+4\mathbf{j})$ → Answer: A
1.6.5 $\mathbf{a}=2\mathbf{i}-3\mathbf{j}$ and $\mathbf{b}=4\mathbf{i}+k\mathbf{j}$ are perpendicular. Find $k$. (A) -6 (B) $-\frac{8}{3}$ (C) $\frac{8}{3}$ (D) 6 (E) -2.
Perpendicular $\Rightarrow \mathbf{a}\cdot\mathbf{b}=0$: $2(4) + (-3)(k) = 8 - 3k = 0$ → $3k=8$ → $k=\frac{8}{3}$ → Answer: C
1.6.6 If $\mathbf{r}=2\mathbf{i}-\mathbf{j}+3\mathbf{k}$ and $\mathbf{s}=\mathbf{i}+2\mathbf{j}-\mathbf{k}$, find $|\mathbf{r}-2\mathbf{s}|$. (A) $5\sqrt{2}$ (B) 5 (C) $\sqrt{30}$ (D) $\sqrt{26}$ (E) $\sqrt{34}$.
$2\mathbf{s}=2\mathbf{i}+4\mathbf{j}-2\mathbf{k}$, $\mathbf{r}-2\mathbf{s} = (2-2)\mathbf{i} + (-1-4)\mathbf{j} + (3-(-2))\mathbf{k} = 0\mathbf{i} -5\mathbf{j} +5\mathbf{k}$. $|\mathbf{r}-2\mathbf{s}| = \sqrt{0+25+25}=\sqrt{50}=5\sqrt{2}$ → Answer: A
1.6.7 Position vectors $P(2,3)$, $Q(5,-1)$. Find $\overrightarrow{PQ}$. (A) $3\mathbf{i}-4\mathbf{j}$ (B) $-3\mathbf{i}+4\mathbf{j}$ (C) $7\mathbf{i}+2\mathbf{j}$ (D) $-7\mathbf{i}-2\mathbf{j}$ (E) $3\mathbf{i}+4\mathbf{j}$.
$\overrightarrow{PQ} = \mathbf{q} - \mathbf{p} = (5-2)\mathbf{i} + (-1-3)\mathbf{j} = 3\mathbf{i} - 4\mathbf{j}$ → Answer: A
1.6.8 Find angle between $\mathbf{a}=\mathbf{i}+2\mathbf{j}$ and $\mathbf{b}=2\mathbf{i}-\mathbf{j}$. (A) $0^\circ$ (B) $45^\circ$ (C) $60^\circ$ (D) $90^\circ$ (E) $120^\circ$.
$\mathbf{a}\cdot\mathbf{b}=1(2)+2(-1)=2-2=0$ → dot product zero → vectors perpendicular → angle $90^\circ$ → Answer: D
1.6.9 $\mathbf{u}=\begin{pmatrix}2\\3\end{pmatrix}$, $\mathbf{v}=\begin{pmatrix}x\\-1\end{pmatrix}$, $|\mathbf{u}+\mathbf{v}|=\sqrt{13}$, find $x$. (A) $-5$ or $1$ (B) $5$ or $-1$ (C) $2$ or $-4$ (D) $-2$ or $4$ (E) $3$ or $-3$.
$\mathbf{u}+\mathbf{v} = (2+x, 2)$. $|\mathbf{u}+\mathbf{v}|^2 = (2+x)^2 + 4 = 13$ → $(2+x)^2 = 9$ → $2+x = \pm 3$ → $x=1$ or $x=-5$ → Answer: A
1.6.10 $\mathbf{a}=2\mathbf{i}-3\mathbf{j}+\mathbf{k}$, $\mathbf{b}=\mathbf{i}+4\mathbf{j}-2\mathbf{k}$, find $\mathbf{a}\times\mathbf{b}$. (A) $2\mathbf{i}+5\mathbf{j}+11\mathbf{k}$ (B) $2\mathbf{i}-5\mathbf{j}-11\mathbf{k}$ (C) $-2\mathbf{i}+5\mathbf{j}+11\mathbf{k}$ (D) $-2\mathbf{i}-5\mathbf{j}+11\mathbf{k}$ (E) $2\mathbf{i}+5\mathbf{j}-11\mathbf{k}$.
$\mathbf{a}\times\mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -3 & 1 \\ 1 & 4 & -2 \end{vmatrix} = \mathbf{i}((-3)(-2)-1(4)) - \mathbf{j}(2(-2)-1(1)) + \mathbf{k}(2(4)-(-3)(1)) = \mathbf{i}(6-4) - \mathbf{j}(-4-1) + \mathbf{k}(8+3) = 2\mathbf{i} +5\mathbf{j} +11\mathbf{k}$ → Answer: A
1.6.11 Points A,B,C: $\mathbf{a}=\mathbf{i}+2\mathbf{j}$, $\mathbf{b}=3\mathbf{i}+5\mathbf{j}$, $\mathbf{c}=2\mathbf{i}-\mathbf{j}$. Find area of triangle ABC. (A) 2 (B) 4.5 (C) 3 (D) 5 (E) 6.
$\overrightarrow{AB} = \mathbf{b}-\mathbf{a} = 2\mathbf{i}+3\mathbf{j}$, $\overrightarrow{AC} = \mathbf{c}-\mathbf{a} = \mathbf{i}-3\mathbf{j}$. Area $= \frac{1}{2}|\overrightarrow{AB} \times \overrightarrow{AC}| = \frac{1}{2}|(2)(-3) - (3)(1)| = \frac{1}{2}|-6-3| = \frac{1}{2} \times 9 = 4.5$ → Answer: B
1.6.12 $|\mathbf{p}|=3$, $|\mathbf{q}|=4$, $\mathbf{p}\cdot\mathbf{q}=6$, find $|\mathbf{p}+\mathbf{q}|$. (A) 5 (B) 6 (C) $\sqrt{37}$ (D) 7 (E) 8.
$|\mathbf{p}+\mathbf{q}|^2 = |\mathbf{p}|^2 + |\mathbf{q}|^2 + 2\mathbf{p}\cdot\mathbf{q} = 9 + 16 + 12 = 37$ → $|\mathbf{p}+\mathbf{q}| = \sqrt{37}$ → Answer: C
1.6.13 Find $t$ so $\mathbf{a}=t\mathbf{i}+2\mathbf{j}-3\mathbf{k}$ and $\mathbf{b}=2\mathbf{i}+t\mathbf{j}+\mathbf{k}$ are perpendicular. (A) $t=0$ (B) $t=\frac{3}{4}$ (C) $t=2$ (D) $t=3$ (E) $t=-1$.
Perpendicular $\Rightarrow \mathbf{a}\cdot\mathbf{b}=0$: $t(2) + 2(t) + (-3)(1) = 2t + 2t - 3 = 4t - 3 = 0$ → $4t = 3$ → $t = \frac{3}{4}$ → Answer: B
1.6.14 Force $\mathbf{F}=4\mathbf{i}-3\mathbf{j}$ N moves particle from $A(1,2)$ to $B(5,5)$. Find work done. (A) 7 J (B) 13 J (C) 15 J (D) 17 J (E) 19 J.
Displacement $\mathbf{d} = (5-1)\mathbf{i} + (5-2)\mathbf{j} = 4\mathbf{i}+3\mathbf{j}$. Work $= \mathbf{F}\cdot\mathbf{d} = 4(4) + (-3)(3) = 16 - 9 = 7$ J → Answer: A
1.6.15 $\mathbf{a}=2\mathbf{i}-\mathbf{j}+3\mathbf{k}$, scalar projection onto $\mathbf{b}=\mathbf{i}+2\mathbf{j}-2\mathbf{k}$. (A) -2 (B) -1 (C) 0 (D) 1 (E) 2.
Scalar projection $= \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|} = \frac{2(1)+(-1)(2)+3(-2)}{\sqrt{1+4+4}} = \frac{2-2-6}{3} = \frac{-6}{3} = -2$ → Answer: A
1.6.16 Shortest distance from $P(2,3)$ to line through $A(1,1)$ with direction $\mathbf{d}=3\mathbf{i}+4\mathbf{j}$. (A) 1 (B) 1.5 (C) 2 (D) 0.4 (E) 3.
$\overrightarrow{AP} = \mathbf{i}+2\mathbf{j}$. Distance $= \frac{|\overrightarrow{AP} \times \mathbf{d}|}{|\mathbf{d}|} = \frac{|(1)(4) - (2)(3)|}{\sqrt{9+16}} = \frac{|4-6|}{5} = \frac{2}{5} = 0.4$ → Answer: D
1.6.17 $\mathbf{a}=2\mathbf{i}-2\mathbf{j}+\mathbf{k}$, $\mathbf{b}=4\mathbf{i}+2\mathbf{j}-2\mathbf{k}$, find $|\mathbf{a}\times\mathbf{b}|$. (A) $\sqrt{6}$ (B) 6 (C) $6\sqrt{6}$ (D) $6\sqrt{2}$ (E) $2\sqrt{53}$.
$\mathbf{a}\times\mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -2 & 1 \\ 4 & 2 & -2 \end{vmatrix} = \mathbf{i}((-2)(-2)-1(2)) - \mathbf{j}(2(-2)-1(4)) + \mathbf{k}(2(2)-(-2)(4)) = \mathbf{i}(4-2) - \mathbf{j}(-4-4) + \mathbf{k}(4+8) = 2\mathbf{i} +8\mathbf{j} +12\mathbf{k}$. $|\mathbf{a}\times\mathbf{b}| = \sqrt{4+64+144} = \sqrt{212} = \sqrt{4\times53} = 2\sqrt{53}$ → Answer: E
1.6.18 Points $A(1,0,2)$, $B(3,1,4)$, $C(2,2,1)$. Find $\cos$ of angle ABC. (A) $\frac{1}{3}$ (B) $\frac{1}{2}$ (C) $\frac{2}{3}$ (D) $\frac{7}{3\sqrt{11}}$ (E) $\frac{4}{5}$.
$\overrightarrow{BA} = A-B = (-2,-1,-2)$, $\overrightarrow{BC} = C-B = (-1,1,-3)$. $\cos\angle ABC = \frac{\overrightarrow{BA}\cdot\overrightarrow{BC}}{|\overrightarrow{BA}||\overrightarrow{BC}|} = \frac{(-2)(-1)+(-1)(1)+(-2)(-3)}{\sqrt{4+1+4}\sqrt{1+1+9}} = \frac{2-1+6}{\sqrt{9}\sqrt{11}} = \frac{7}{3\sqrt{11}}$ → Answer: D
1.6.19 $\mathbf{a}=\begin{pmatrix}1\\2\\3\end{pmatrix}$, $\mathbf{b}=\begin{pmatrix}4\\5\\6\end{pmatrix}$, area of parallelogram. (A) $\sqrt{3}$ (B) $3\sqrt{6}$ (C) 6 (D) $6\sqrt{6}$ (E) $2\sqrt{6}$.
$\mathbf{a}\times\mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 3 \\ 4 & 5 & 6 \end{vmatrix} = \mathbf{i}(12-15) - \mathbf{j}(6-12) + \mathbf{k}(5-8) = -3\mathbf{i} +6\mathbf{j} -3\mathbf{k}$. Area $= |\mathbf{a}\times\mathbf{b}| = \sqrt{9+36+9} = \sqrt{54} = 3\sqrt{6}$ → Answer: B
1.6.20 $\mathbf{v}(t)=2t\mathbf{i}+3t^2\mathbf{j}$, displacement from $t=0$ to $t=2$. (A) $4\mathbf{i}+8\mathbf{j}$ (B) $2\mathbf{i}+4\mathbf{j}$ (C) $8\mathbf{i}+4\mathbf{j}$ (D) $4\mathbf{i}+6\mathbf{j}$ (E) $6\mathbf{i}+8\mathbf{j}$.
Displacement $\int_0^2 \mathbf{v}(t) dt = \left[ t^2\mathbf{i} + t^3\mathbf{j} \right]_0^2 = (4\mathbf{i} + 8\mathbf{j}) - (0) = 4\mathbf{i}+8\mathbf{j}$ → Answer: A
Evaluation G - 12 | 1.7 Solutions (Differentiation)
1.7.1 Differentiate $y = 3x^5 - 2x^3 + 7x - 9$. (A) $15x^4 - 6x^2 + 7$ (B) $15x^5 - 6x^3 + 7x$ (C) $15x^4 - 6x^2 + 7x$ (D) $3x^4 - 2x^2 + 7$ (E) $15x^4 - 6x^3 + 7$.
$\frac{dy}{dx} = 3 \cdot 5x^4 - 2 \cdot 3x^2 + 7 = 15x^4 - 6x^2 + 7$ → Answer: A
1.7.2 Find gradient of $y = x^3 - 4x^2 + 2x - 1$ at $x=2$. (A) $-2$ (B) $0$ (C) $2$ (D) $4$ (E) $6$.
$\frac{dy}{dx} = 3x^2 - 8x + 2$. At $x=2$: $3(4) - 16 + 2 = 12 - 16 + 2 = -2$ → Answer: A
1.7.3 Find equation of tangent to $y = x^2 - 3x + 2$ at $x=1$. (A) $y = -x + 1$ (B) $y = -x - 1$ (C) $y = x - 1$ (D) $y = x + 1$ (E) $y = -2x + 2$.
At $x=1$, $y = 1 - 3 + 2 = 0$. $\frac{dy}{dx} = 2x - 3$, at $x=1$: $2 - 3 = -1$. Tangent: $y - 0 = -1(x - 1)$ → $y = -x + 1$ → Answer: A
1.7.4 Find equation of normal to $y = 2x^3 - 5x + 1$ at $x=1$. (A) $y = -x - 1$ (B) $y = -\frac{1}{5}x + \frac{13}{5}$ (C) $y = 5x - 7$ (D) $y = -5x + 7$ (E) $y = \frac{1}{5}x + \frac{13}{5}$.
At $x=1$, $y = 2 - 5 + 1 = -2$. $\frac{dy}{dx} = 6x^2 - 5$, at $x=1$: $6 - 5 = 1$. Normal gradient $= -1$. Equation: $y + 2 = -1(x - 1)$ → $y = -x - 1$ → Answer: A
1.7.5 Differentiate $y = (2x^3 - 5x)^4$. (A) $4(2x^3 - 5x)^3(6x^2 - 5)$ (B) $4(2x^3 - 5x)^3$ (C) $4(6x^2 - 5)^3$ (D) $(2x^3 - 5x)^3(6x^2 - 5)$ (E) $4(2x^3 - 5x)^3(6x^2)$.
Chain rule: $\frac{dy}{dx} = 4(2x^3 - 5x)^3 \cdot (6x^2 - 5)$ → Answer: A
1.7.6 Find derivative of $y = \frac{3x - 2}{x + 1}$. (A) $\frac{5}{(x+1)^2}$ (B) $\frac{3x - 5}{(x+1)^2}$ (C) $\frac{3}{(x+1)^2}$ (D) $\frac{1}{(x+1)^2}$ (E) $\frac{5x}{(x+1)^2}$.
Quotient rule: $\frac{dy}{dx} = \frac{3(x+1) - (3x-2)(1)}{(x+1)^2} = \frac{3x+3 - 3x + 2}{(x+1)^2} = \frac{5}{(x+1)^2}$ → Answer: A
1.7.7 Radius of circle increasing at $2$ cm/s. Find rate of increase of area when $r=5$ cm. (A) $10\pi$ cm²/s (B) $20\pi$ cm²/s (C) $30\pi$ cm²/s (D) $40\pi$ cm²/s (E) $50\pi$ cm²/s.
$A = \pi r^2$, $\frac{dA}{dt} = 2\pi r \frac{dr}{dt} = 2\pi(5)(2) = 20\pi$ → Answer: B
1.7.8 Find stationary points of $y = x^3 - 6x^2 + 9x + 2$. (A) $(1,6)$ and $(3,2)$ (B) $(1,6)$ and $(3,-2)$ (C) $(-1,-14)$ and $(3,2)$ (D) $(1,6)$ and $(-3,-52)$ (E) $(-1,-14)$ and $(-3,-52)$.
$\frac{dy}{dx} = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x-1)(x-3)=0$ → $x=1$ or $x=3$. $x=1$: $y=1-6+9+2=6$; $x=3$: $y=27-54+27+2=2$ → $(1,6)$ and $(3,2)$ → Answer: A
1.7.9 If $y = x\ln x$, find $\frac{dy}{dx}$. (A) $1 + \ln x$ (B) $\ln x$ (C) $1 - \ln x$ (D) $\frac{1}{x}$ (E) $x + \ln x$.
Product rule: $\frac{dy}{dx} = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1$ → Answer: A
1.7.10 Find second derivative of $y = e^{2x} \sin x$. (A) $e^{2x}(4\cos x - 3\sin x)$ (B) $e^{2x}(3\cos x + 4\sin x)$ (C) $e^{2x}(4\cos x + 3\sin x)$ (D) $e^{2x}(3\cos x - 4\sin x)$ (E) $e^{2x}(2\cos x - \sin x)$.
$\frac{dy}{dx} = 2e^{2x}\sin x + e^{2x}\cos x = e^{2x}(2\sin x + \cos x)$. $\frac{d^2y}{dx^2} = 2e^{2x}(2\sin x + \cos x) + e^{2x}(2\cos x - \sin x) = e^{2x}(4\sin x + 2\cos x + 2\cos x - \sin x) = e^{2x}(3\sin x + 4\cos x)$ → Answer: C
1.7.11 Particle $s = t^3 - 6t^2 + 9t + 2$. When is it at rest? (A) $t=1$ and $t=3$ (B) $t=1$ and $t=2$ (C) $t=2$ and $t=3$ (D) $t=1$ only (E) $t=3$ only.
$v = \frac{ds}{dt} = 3t^2 - 12t + 9 = 3(t^2 - 4t + 3) = 3(t-1)(t-3)=0$ → $t=1$ and $t=3$ → Answer: A
1.7.12 Differentiate $y = \tan(3x^2 + 1)$. (A) $6x \sec^2(3x^2 + 1)$ (B) $3x \sec^2(3x^2 + 1)$ (C) $6x \sec(3x^2 + 1)\tan(3x^2 + 1)$ (D) $6x \cos^2(3x^2 + 1)$ (E) $\sec^2(3x^2 + 1)$.
$\frac{dy}{dx} = \sec^2(3x^2 + 1) \cdot 6x = 6x \sec^2(3x^2 + 1)$ → Answer: A
1.7.13 Find gradient of tangent to $y = \ln(2x + 1)$ at $x=2$. (A) $\frac{1}{5}$ (B) $\frac{2}{5}$ (C) $\frac{1}{2}$ (D) $\frac{2}{3}$ (E) $\frac{3}{5}$.
$\frac{dy}{dx} = \frac{2}{2x+1}$. At $x=2$: $\frac{2}{4+1} = \frac{2}{5}$ → Answer: B
1.7.14 $V = \frac{4}{3}\pi r^3$, find $\frac{dV}{dr}$ when $r=3$. (A) $12\pi$ (B) $24\pi$ (C) $36\pi$ (D) $48\pi$ (E) $72\pi$.
$\frac{dV}{dr} = 4\pi r^2$. At $r=3$: $4\pi(9) = 36\pi$ → Answer: C
1.7.15 Find x-coordinate where gradient of $y = x\sqrt{x}$ is $3$. (A) $1$ (B) $2$ (C) $3$ (D) $4$ (E) $5$.
$y = x^{3/2}$, $\frac{dy}{dx} = \frac{3}{2}x^{1/2} = 3$ → $x^{1/2} = 2$ → $x = 4$ → Answer: D
1.7.16 Conical tank height $10$ m, radius $4$ m, water poured at $2$ m³/min. Find rate of rise when depth $5$ m. (A) $\frac{1}{2\pi}$ m/min (B) $\frac{1}{4\pi}$ m/min (C) $\frac{1}{8\pi}$ m/min (D) $\frac{1}{\pi}$ m/min (E) $\frac{2}{\pi}$ m/min.
By similar triangles: $\frac{r}{h} = \frac{4}{10} = \frac{2}{5}$ → $r = \frac{2}{5}h$. $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (\frac{4}{25}h^2)h = \frac{4\pi}{75}h^3$. $\frac{dV}{dt} = \frac{4\pi}{25}h^2 \frac{dh}{dt}$. $2 = \frac{4\pi}{25}(25)\frac{dh}{dt} = 4\pi \frac{dh}{dt}$ → $\frac{dh}{dt} = \frac{2}{4\pi} = \frac{1}{2\pi}$ → Answer: A
1.7.17 Differentiate $y = e^{\sin x}$. (A) $e^{\sin x} \cos x$ (B) $e^{\sin x}$ (C) $e^{\cos x} \sin x$ (D) $e^{\sin x} \sin x$ (E) $e^{\cos x}$.
Chain rule: $\frac{dy}{dx} = e^{\sin x} \cdot \cos x$ → Answer: A
1.7.18 Find point on $y = x^3 - 6x^2 + 12x - 8$ where tangent is horizontal. (A) $(2,0)$ (B) $(2,8)$ (C) $(-2,-48)$ (D) $(1,-1)$ (E) $(3,1)$.
$\frac{dy}{dx} = 3x^2 - 12x + 12 = 3(x^2 - 4x + 4) = 3(x-2)^2 = 0$ → $x=2$. $y = 8 - 24 + 24 - 8 = 0$ → $(2,0)$ → Answer: A
1.7.19 If $y = \frac{1}{x^2 + 1}$, find $\frac{dy}{dx}$ at $x=1$. (A) $-\frac{1}{2}$ (B) $-\frac{1}{4}$ (C) $0$ (D) $\frac{1}{4}$ (E) $\frac{1}{2}$.
$\frac{dy}{dx} = -\frac{2x}{(x^2+1)^2}$. At $x=1$: $-\frac{2}{(2)^2} = -\frac{2}{4} = -\frac{1}{2}$ → Answer: A
1.7.20 Particle $s(t) = t^3 - 9t^2 + 24t + 5$. Find total distance in first $5$ seconds. (A) $25$ m (B) $26$ m (C) $27$ m (D) $28$ m (E) $29$ m.
$v = 3t^2 - 18t + 24 = 3(t^2 - 6t + 8) = 3(t-2)(t-4)=0$ at $t=2,4$. $s(0)=5$, $s(2)=8-36+48+5=25$, $s(4)=64-144+96+5=21$, $s(5)=125-225+120+5=25$. Distance = $|25-5| + |21-25| + |25-21| = 20 + 4 + 4 = 28$ → Answer: D
