Master Grade 12 Math Practice Test: 17 Problems with Solutions

1. Move all terms to one side:

\[ \frac{x+1}{x+2} - 1 - \frac{1}{x-1} \geq 0 \]

2. Find a common denominator, which is $(x+2)(x-1)$:

\[ \frac{(x+1)(x-1) - (x+2)(x-1) - (x+2)}{(x+2)(x-1)} \geq 0 \] \[ \frac{(x^2-1) - (x^2+x-2) - (x+2)}{(x+2)(x-1)} \geq 0 \] \[ \frac{x^2 - 1 - x^2 - x + 2 - x - 2}{(x+2)(x-1)} \geq 0 \] \[ \frac{-2x - 1}{(x+2)(x-1)} \geq 0 \]

3. Identify critical points: $x = -0.5$ (numerator zero), $x = -2$ and $x = 1$ (denominator undefined).

4. Test intervals:

• $(-\infty, -2)$: Test $x = -3 \implies \frac{5}{(-1)(-4)} > 0$ (True)
• $(-2, -0.5]$: Test $x = -1 \implies \frac{1}{(1)(-2)} < 0$ (False)
• $[-0.5, 1)$: Test $x = 0 \implies \frac{-1}{(2)(-1)} > 0$ (True)
• $(1, \infty)$: Test $x = 2 \implies \frac{-5}{(4)(1)} < 0$ (False)

Solution Set:

Inequality: $x < -2$ or $-0.5 \leq x < 1$

Interval: $(-\infty, -2) \cup [-0.5, 1)$

1. Rewrite $1/\sec x$ as $\cos x$:

\[ \cos(2x) - 2\cos x = 2 \]

2. Use the double angle identity $\cos(2x) = 2\cos^2 x - 1$:

\[ 2\cos^2 x - 1 - 2\cos x = 2 \] \[ 2\cos^2 x - 2\cos x - 3 = 0 \]

3. Apply the quadratic formula for $\cos x$:

\[ \cos x = \frac{2 \pm \sqrt{(-2)^2 - 4(2)(-3)}}{2(2)} = \frac{2 \pm \sqrt{28}}{4} = \frac{1 \pm \sqrt{7}}{2} \]

4. Evaluate the values:

• $\frac{1 + \sqrt{7}}{2} \approx 1.82$ (Impossible, as $|\cos x| \leq 1$)
• $\frac{1 - \sqrt{7}}{2} \approx -0.823$ (Valid)

5. Find $x$: $x = \arccos\left(\frac{1-\sqrt{7}}{2}\right)$. This occurs in Quadrants II and III.

$x \approx 2.54 \text{ rad}$ and $x \approx 3.74 \text{ rad}$.

1. Combine logarithms using the quotient rule:

\[ \ln\left(\frac{x^2 + 2}{x - 1}\right) = 2 \]

2. Convert to exponential form:

\[ \frac{x^2 + 2}{x - 1} = e^2 \]

3. Rearrange into a quadratic equation:

\[ x^2 + 2 = e^2(x - 1) \implies x^2 - e^2x + (e^2 + 2) = 0 \]

4. Solve using the quadratic formula:

\[ x = \frac{e^2 \pm \sqrt{(-e^2)^2 - 4(1)(e^2 + 2)}}{2} \]

Note: $x$ must be greater than 1 for the original $\ln(x-1)$ to be defined. Use a calculator to find the decimal values and verify the domain.

1. Let $u = 2^x$. Then $4^x = (2^2)^x = (2^x)^2 = u^2$.

2. Substitute into the equation:

\[ 3u^2 + u - 30 = 0 \]

3. Factor the quadratic:

\[ (3u + 10)(u - 3) = 0 \]

4. Solve for $u$:

• $u = -10/3$ (Rejected, as $2^x$ must be positive)
• $u = 3$ (Valid)

5. Solve for $x$:

\[ 2^x = 3 \implies x = \frac{\ln 3}{\ln 2} \approx 1.58 \]

1. Work with the Left-Hand Side (LHS). Use $\tan(2x) = \frac{\sin(2x)}{\cos(2x)}$:

\[ \text{LHS} = \frac{1}{2 \sin x} \cdot \frac{\sin(2x)}{\cos(2x)} \]

2. Use the double angle identity $\sin(2x) = 2 \sin x \cos x$:

\[ \text{LHS} = \frac{1}{2 \sin x} \cdot \frac{2 \sin x \cos x}{\cos(2x)} = \frac{\cos x}{\cos(2x)} \]

3. Work with the Right-Hand Side (RHS). Recall the double angle identity $\cos(2x) = 1 - 2 \sin^2 x$:

\[ \text{RHS} = \frac{\cos x}{1 - 2 \sin^2 x} = \frac{\cos x}{\cos(2x)} \]

4. Conclusion: LHS = RHS. The identity is verified.

1. Simplify the angle: $\frac{13\pi}{12} = \pi + \frac{\pi}{12}$. Since $\tan(\pi + \theta) = \tan \theta$:

\[ \tan\left(\frac{13\pi}{12}\right) = \tan\left(\frac{\pi}{12}\right) \]

2. Use the difference identity $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$ with $A = \frac{\pi}{4}$ and $B = \frac{\pi}{6}$:

\[ \tan\left(\frac{\pi}{12}\right) = \tan\left(\frac{\pi}{4} - \frac{\pi}{6}\right) = \frac{1 - 1/\sqrt{3}}{1 + (1)(1/\sqrt{3})} \]

3. Rationalize and simplify:

\[ = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = \frac{(\sqrt{3} - 1)^2}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3} \]

1. From the Factor Theorem, since $x - 1$ is a factor, $P(1) = 0$.

2. From the Remainder Theorem, $P(-1) = 4$ and $P(2) = 4$.

3. Let $P(x) = (x - 1)(x^2 + bx + c)$ since the degree is 3 and leading coefficient is 1.

4. Use $P(-1) = 4$:

\[ (-1 - 1)((-1)^2 - b + c) = 4 \implies -2(1 - b + c) = 4 \implies 1 - b + c = -2 \implies c - b = -3 \]

5. Use $P(2) = 4$:

\[ (2 - 1)((2)^2 + 2b + c) = 4 \implies 1(4 + 2b + c) = 4 \implies 2b + c = 0 \implies c = -2b \]

6. Solve for $b, c$: Substitute $c = -2b$ into $c - b = -3$:

\[ -2b - b = -3 \implies -3b = -3 \implies b = 1, \quad c = -2 \]

7. Final polynomial: $P(x) = (x-1)(x^2 + x - 2) = x^3 + x^2 - 2x - x^2 - x + 2 = x^3 - 3x + 2$.

1. Factor out $-x$: $f(x) = -x(x^3 + 5x^2 + 3x - 9)$.

2. Test for roots of the cubic term. $P(1) = 1 + 5 + 3 - 9 = 0$. So $(x - 1)$ is a factor.

3. Use division to find the quadratic factor: $x^3 + 5x^2 + 3x - 9 = (x - 1)(x^2 + 6x + 9)$.

4. Factor the quadratic: $(x^2 + 6x + 9) = (x + 3)^2$.

Complete Factored Form: $f(x) = -x(x - 1)(x + 3)^2$.

5. Zeros: $x = 0$ (mult 1), $x = 1$ (mult 1), $x = -3$ (mult 2).

1. Identify zeros and multiplicities from the graph:

• At $x = -2$: Graph crosses (multiplicity 1).
• At $x = -1$: Graph touches (multiplicity 2).
• At $x = 2$: Graph crosses (multiplicity 1).

2. Write the general form: $g(x) = a(x + 2)(x + 1)^2(x - 2)$.

3. Use the y-intercept $(0, 2)$ to solve for $a$:

\[ 2 = a(0 + 2)(0 + 1)^2(0 - 2) \] \[ 2 = a(2)(1)(-2) \] \[ 2 = -4a \implies a = -0.5 \]

Final Equation:

\[ g(x) = -0.5(x + 2)(x + 1)^2(x - 2) \]

1. Identify key parameters:

• Amplitude: 0.5 (Reflected over the midline)
• Period: $2\pi/4 = \pi/2$
• Phase Shift: $-\pi/16$ (left)
• Vertical Shift (Midline): $y = 2.5$

2. Key points (start at phase shift, increment by Period/4 = $\pi/8$):

• $x = -\pi/16 \implies y = 2.5$
• $x = \pi/16 \implies y = 2.0$ (min due to negative sine)
• $x = 3\pi/16 \implies y = 2.5$
• $x = 5\pi/16 \implies y = 3.0$ (max)
• $x = 7\pi/16 \implies y = 2.5$

3. Graph of $y = - 0.5 \sin \left( 4(x+\dfrac{\pi}{16}) \right) + 2.5$:

1. Find the amplitude ($A$): Max = 1, Min = 0.

\[ A = \frac{\text{Max} - \text{Min}}{2} = \frac{1 - 0}{2} = 0.5 \]

2. Find the vertical shift ($d$):

\[ d = \frac{\text{Max} + \text{Min}}{2} = \frac{1 + 0}{2} = 0.5 \]

3. Find the period ($P$): Given as 3 seconds.

4. Calculate $b$:

\[ b = \frac{2\pi}{P} = \frac{2\pi}{3} \]

5. Find the phase shift ($c$): For a cosine function, the shift corresponds to the horizontal position of the maximum. The problem states the maximum is at $t = 2.25$.

\[ c = 2.25 \]

Final Equation:

\[ V(t) = 0.5 \cos\left(\frac{2\pi}{3}(t - 2.25)\right) + 0.5 \]

a) Domain: All real numbers except where the denominator is zero. $x \neq -2$.

b) Intercepts:

• x-intercept (set $y=0$): $2x - 4 = 0 \implies x = 2$.
• y-intercept (set $x=0$): $y = -4/2 = -2$.

c) Asymptotes:

• Vertical: $x = -2$.
• Horizontal (limit as $x \to \infty$): $y = 2/1 = 2$.

Graph of $y = \dfrac{2 x - 4}{x+2}$:

a) Domain: $x \in \mathbb{R}, x \neq -2$.

b) Intercepts: x-int at $x = \pm 3$; y-int at $y = -4.5$.

c) Asymptotes:

• Vertical: $x = -2$.
• Horizontal: None (degree of numerator > denominator).
• Slant (oblique): Use long division. $(x^2-9) \div (x+2) = (x-2)$ with remainder $-5$. Asymptote is $y = x - 2$.
• Table of signs \[ \begin{array}{|c|c|c|c|c|c|c|c|} \hline x & (-\infty, -3) & -3 & (-3, -2) & -2 & (-2, 3) & 3 & (3, \infty) \\ \hline (x+3) & - & 0 & + & + & + & + & + \\ \hline (x-3) & - & - & - & - & - & 0 & + \\ \hline (x+2) & - & - & - & \text{und} & + & + & + \\ \hline y = \frac{(x+3)(x-3)}{x+2} & \mathbf{-} & 0 & \mathbf{+} & \text{und} & \mathbf{-} & 0 & \mathbf{+} \\ \hline \text{Position} & \text{Below} & x\text{-int} & \text{Above} & \text{VA} & \text{Below} & x\text{-int} & \text{Above} \\ \hline \end{array} \]
• $ y = \dfrac{x^2-9}{x+2}$:

1. Identify the hole: The hole is at $x = 2$. This means $(x - 2)$ must be a factor in both the numerator and denominator.

2. Identify the Vertical Asymptote: VA at $x = 1$. This means $(x - 1)$ is a factor in the denominator.

3. Identify the Horizontal Asymptote (HA): The HA is at $y = 0$. This confirms that the degree of the denominator (degree 2) is greater than the degree of the numerator (which must be degree 1).

4. Construct the function:

\[ h(x) = a \cdot \frac{x - 2}{(x - 1)(x - 2)} = a \cdot \frac{1}{(x - 1)} \]

5. Determine the constant \( a \) using the hole at $(2, 2)$ :

\[ 2 = a \cdot \frac{1}{(2 - 1)} \implies a = 2 \]

6. Final Equation:

\[ h(x) = \frac{2(x - 2)}{(x - 1)(x - 2)} \]

Note: The simplified form for sketching (excluding the hole) is \( y = \dfrac{2}{x - 1} \).

a) Domain: Argument must be positive. $x^2 - 1 > 0 \implies x^2 > 1 \implies x > 1$ or $x < -1$.

b) Intercepts: No y-intercept (0 is not in domain). For x-intercept, set $y=0$:

\[ -0.5 \log_2(x^2 - 1) = 1 \implies \log_2(x^2 - 1) = -2 \implies x^2 - 1 = 2^{-2} = 0.25 \] \[ x^2 = 1.25 \implies x \approx \pm 1.12 \]

c) Asymptotes: Vertical asymptotes where argument approaches zero: $x = 1$ and $x = -1$.

d) Note that $f(-x) = f(x) $ and therefore the graph is symmetric with respect to the y-axis.

Graph of $ y = -0.5 \log_2(x^2 - 1)-1 $:

a) Domain: All real numbers $(-\infty, \infty)$.

b) Intercepts: y-intercept at $x = 0 \implies y = 2 + e^{-2} \approx 2.135$. No x-intercept because $e^{(x-2)} + 2$ is always greater than 2.

c) Asymptotes: Horizontal asymptote as $x \to -\infty$ is $y = 2$.

Graph of $ y = 2 + e^{(x-2)} $:

a) Domain: $2x - 1 > 0 \implies x > 0.5$. Range: All real numbers.

b) Finding the inverse:

\[ x = \ln(2y - 1) + 2 \] \[ x - 2 = \ln(2y - 1) \] \[ e^{x-2} = 2y - 1 \] \[ y = \frac{e^{x-2} + 1}{2} \]

Inverse Function: $h^{-1}(x) = 0.5(e^{x-2} + 1)$.

Inverse Domain: All real numbers. Inverse Range: $y > 0.5$.