Unit 4: Functions and Graphs – Review Exercise 4

Q.No	Question	A	B	C	D	Correct
1	\( f(x) = \frac{5x-6}{3} \), then \( f(3) = \)	\(-1\)	\(3\)	\(9\)	\(15\)	B
2	A function \( f \) from \( X \) to \( Y \) is represented by:	\( f:XY \)	\( f:Y \to X \)	\( f:X \to Y \)	\( f:\frac{X}{Y} \)	C
3	\( (f \circ g)(x) = \)	\( (f+g)(x) \)	\( (f-g)(x) \)	\( f(g(x)) \)	\( f(x) \div g(x) \)	C
4	If \( f(x)=2x+3, g(x)=x+1 \), then \( f(x)+g(x)= \)	\( 3x \)	\( 3x+4 \)	\( 4 \)	\( 2x^2+3 \)	B
5	If \( f(x)=5x+2, h(x)=2x-2 \), then \( f(x)-h(x)= \)	\( 3x \)	\( 5x^2-4 \)	\( 3x+4 \)	\( -3x-4 \)	C
6	If \( f(x)=3x+1, g(x)=2x \), then \( g(x) \times f(x)= \)	\( 6x+2x \)	\( 5x^2+1 \)	\( x+1 \)	\( 6x^2+2x \)	D
7	If \( f(x)=x^2-4, g(x)=x+2, x\neq -2 \), then \( \frac{f(x)}{g(x)}= \)	\( \frac{1}{x-2} \)	\( \frac{1}{x+2} \)	\( x+2 \)	\( x-2 \)	D
8	Shape of absolute value function graph?	U-shaped	V-shaped	L-shaped	M-shaped	B
9	Vertical line test for a function: every vertical line intersects at:	4 points	3 points	2 points	1 point	D
10	If \( f(x)=x^3 \), then \( f(-2)= \)	\(-8\)	\(8\)	\(4\)	\(-6\)	A

Explanation & Correct answers:
\[ \begin{aligned} &\textbf{Q1: } f(3)=\frac{5(3)-6}{3}= \frac{15-6}{3}=3 \quad \Rightarrow \text{(B)}\\[4pt] &\textbf{Q2: } \text{Function from }X\text{ to }Y: f:X \rightarrow Y \quad \Rightarrow \text{(C)}\\[4pt] &\textbf{Q3: } (f \circ g)(x)=f(g(x)) \quad \Rightarrow \text{(C)}\\[4pt] &\textbf{Q4: } f(x)+g(x)=2x+3+x+1=3x+4 \quad \Rightarrow \text{(B)}\\[4pt] &\textbf{Q5: } f(x)-h(x)=5x+2-(2x-2)=3x+4 \quad \Rightarrow \text{(C)}\\[4pt] &\textbf{Q6: } g(x)\times f(x)=2x(3x+1)=6x^2+2x \quad \Rightarrow \text{(D)}\\[4pt] &\textbf{Q7: } \frac{f(x)}{g(x)}=\frac{x^2-4}{x+2}=x-2 \;(x\neq -2) \quad \Rightarrow \text{(D)}\\[4pt] &\textbf{Q8: } |x| \text{ graph is V-shaped } \Rightarrow \text{(B)}\\[4pt] &\textbf{Q9: } \text{vertical line test: at most 1 point } \Rightarrow \text{(D)}\\[4pt] &\textbf{Q10: } f(-2)=(-2)^3=-8 \quad \Rightarrow \text{(A)} \end{aligned} \]

(i) \( f(x)+g(x) \) \[ \begin{aligned} f(x)+g(x) &= (25 - x^{2}) + (5 + x) \\ &= -x^{2} + x + 30 \end{aligned} \]

(ii) \( f(x)-g(x) \) \[ \begin{aligned} f(x)-g(x) &= (25 - x^{2}) - (5 + x) \\ &= 25 - x^{2} -5 - x \\ &= -x^{2} - x + 20 \end{aligned} \]

(iii) \( f(x)\cdot g(x) \) \[ \begin{aligned} f(x)\cdot g(x) &= (25 - x^{2})(5 + x) \\ &= 125 + 25x -5x^{2} - x^{3} \\ &= -x^{3} -5x^{2} + 25x +125 \end{aligned} \]

(iv) \( \frac{f(x)}{g(x)} \) \[ \begin{aligned} \frac{f(x)}{g(x)} &= \frac{25 - x^{2}}{5 + x} = \frac{(5-x)(5+x)}{5+x} \\ &= 5 - x, \quad x \neq -5 \end{aligned} \]

(v) \( f(7) \) \[ \begin{aligned} f(7) &= 25 - (7)^{2} = 25 - 49 = -24 \end{aligned} \]

(vi) \( g(-8) \) \[ \begin{aligned} g(-8) &= 5 + (-8) = -3 \end{aligned} \]

(i) \( (f \circ g)(x) \) \[ \begin{aligned} (f\circ g)(x) &= f(g(x)) = f(14+2x) \\ &= (14 + 2x)^{3} \end{aligned} \]

(ii) \( (g \circ f)(x) \) \[ \begin{aligned} (g\circ f)(x) &= g(f(x)) = g(x^{3}) \\ &= 14 + 2x^{3} = 2(7+x^{3}) \end{aligned} \]

(iii) \( (f \circ f)(x) \) \[ \begin{aligned} (f\circ f)(x) &= f(f(x)) = f(x^{3}) \\ &= (x^{3})^{3} = x^{9} \end{aligned} \]

(iv) \( (g \circ g)(x) \) \[ \begin{aligned} (g\circ g)(x) &= g(g(x)) = g(14+2x) \\ &= 14 + 2(14+2x) = 14+28+4x \\ &= 4x + 42 = 2(2x+21) \end{aligned} \]

(i) \( f(x)=9x-1 \) \[ \begin{aligned} y &= 9x-1 \;\Rightarrow\; y+1 = 9x \\ x &= \frac{y+1}{9} \;\Rightarrow\; f^{-1}(y)=\frac{y+1}{9} \\ \therefore\; f^{-1}(x) &= \frac{x+1}{9} \end{aligned} \]

(ii) \( f(x)=\frac{5}{x-1},\; x\neq 1 \) \[ \begin{aligned} y &= \frac{5}{x-1} \;\Rightarrow\; y(x-1)=5 \\ xy - y &=5 \;\Rightarrow\; xy = 5+y \\ x &= \frac{5+y}{y} \;\Rightarrow\; f^{-1}(y)=\frac{5+y}{y} \\ \therefore\; f^{-1}(x) &= \frac{5+x}{x} \end{aligned} \]

(iii) \( f(x)=\sqrt{x-5},\; x\ge 5 \) \[ \begin{aligned} y &= \sqrt{x-5} \;\Rightarrow\; y^{2}=x-5 \\ x &= y^{2}+5 \;\Rightarrow\; f^{-1}(y)=y^{2}+5 \\ \therefore\; f^{-1}(x) &= x^{2}+5,\; x\ge 0 \end{aligned} \]

(iv) \( f(x)=\frac{3-x}{2} \) \[ \begin{aligned} y &= \frac{3-x}{2} \;\Rightarrow\; 2y = 3-x \\ x &= 3-2y \;\Rightarrow\; f^{-1}(y)=3-2y \\ \therefore\; f^{-1}(x) &= 3-2x \end{aligned} \]

(i) \( f(x)=7|x| \) Vertex at \((0,0)\), V-shaped symmetric.

(ii) \( f(x)=|x+6|-2 \) Vertex \((-6,-2)\), opens upward.

(i) \( |3x-2| = 1 \) \[ \begin{aligned} 3x-2 = 1 &\;\text{or}\; 3x-2 = -1 \\ 3x=3 &\;\Rightarrow\; x=1 \\ 3x=1 &\;\Rightarrow\; x=\frac{1}{3} \end{aligned} \] Solution set: \( \left\{\frac{1}{3},\;1\right\} \)

(ii) \( |6x+1| = 9 \) \[ \begin{aligned} 6x+1 = 9 &\;\Rightarrow\; 6x=8 \;\Rightarrow\; x=\frac{4}{3}=1\frac{1}{3}\\ 6x+1 = -9 &\;\Rightarrow\; 6x=-10 \;\Rightarrow\; x=-\frac{5}{3}=-1\frac{2}{3} \end{aligned} \] S.S = \( \left\{ -1\frac{2}{3},\; 1\frac{1}{3} \right\} \)

(i) \( |7-2x| \le 1 \) \[ \begin{aligned} -1 \le 7-2x \le 1 &\;\Rightarrow\; -8 \le -2x \le -6 \\ \text{multiply by } -1 &: 8 \ge 2x \ge 6 \;\Rightarrow\; 3 \le x \le 4 \end{aligned} \] Interval: \([3,4]\)

(ii) \( |6x+18| \le 24 \) \[ \begin{aligned} -24 \le 6x+18 \le 24 &\;\Rightarrow\; -42 \le 6x \le 6 \\ &\;\Rightarrow\; -7 \le x \le 1 \end{aligned} \] Solution: \([-7,\;1]\)

(i) \( f^{-1}(x) \) \[ \begin{aligned} y &= 3x+7 \;\Rightarrow\; y-7=3x \\ x &= \frac{y-7}{3} \;\Rightarrow\; f^{-1}(x)=\frac{x-7}{3} \end{aligned} \]

(ii) Solve \( f(x)=f^{-1}(x) \) \[ \begin{aligned} 3x+7 &= \frac{x-7}{3} \\ 9x+21 &= x-7 \\ 8x &= -28 \\ x &= -\frac{7}{2} \end{aligned} \] Hence \( x = -\frac{7}{2} \).

\[ \begin{aligned} P(x) &= 0 \\ 100x - 10000 &= 0 \\ 100x &= 10000 \\ x &= 100 \end{aligned} \]

\[ \begin{aligned} D(2000) &= 0.85 \times 2000 \\ &= 1700 \end{aligned} \]

\[ \begin{aligned} |100 - r| &\le 6 \\ -6 &\le 100 - r \le 6 \\ -106 &\le -r \le -94 \\ 94 &\le r \le 106 \end{aligned} \]