Q1. Solve the equation: $\ln x + \ln(x-3) = 0$

Background

Topic: Exponential and Logarithmic Equations

This question tests your understanding of logarithmic properties and how to solve equations involving logarithms.

Key Terms and Formulas

• Logarithm properties: $\ln a + \ln b = \ln(ab)$

• Exponential and logarithmic relationship: $\ln a = 0 \implies a = 1$

Step-by-Step Guidance

1. Combine the two logarithms using the property $\ln a + \ln b = \ln(ab)$.

2. Set the resulting logarithmic expression equal to 0 and rewrite it in exponential form.

3. Solve the resulting equation for $x$.

4. Check for any extraneous solutions by ensuring the arguments of all logarithms are positive.

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Q2. Find the exact value: $\log_{15} 9 + \log_{15} 25$

Background

Topic: Properties of Logarithms

This question tests your ability to use logarithm addition properties and evaluate logarithms.

Key Terms and Formulas

• $\log_b a + \log_b c = \log_b (ac)$

• Change of base and evaluating logarithms

Step-by-Step Guidance

1. Use the property $\log_b a + \log_b c = \log_b (ac)$ to combine the two logarithms.

2. Multiply the arguments: $9 \times 25$.

3. Express the result as a single logarithm: $\log_{15}(225)$.

4. Rewrite $225$ in terms of $15$ if possible, to simplify the logarithm.

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Q3. Compute the limit: $\displaystyle \lim_{x \to 0} \frac{\sin(5x)}{2x}$

Background

Topic: Limits and Trigonometric Limits

This question tests your understanding of limits involving trigonometric functions, especially the standard limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$.

Key Terms and Formulas

• Standard limit: $\lim_{x \to 0} \frac{\sin x}{x} = 1$

• Algebraic manipulation to match the standard form

Step-by-Step Guidance

1. Rewrite the limit to resemble the standard form $\frac{\sin x}{x}$.

2. Factor or multiply numerator and denominator to get $\sin(5x)$ over $5x$.

3. Express the limit as $\frac{5}{2} \cdot \lim_{x \to 0} \frac{\sin(5x)}{5x}$.

4. Recall the value of $\lim_{x \to 0} \frac{\sin(5x)}{5x}$.

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Q4. Find the values of $a$ and $b$ so that the function $f(x) = \begin{cases} x^2 - a & x \leq 1 \\ \frac{3x^2 + 12x - b}{x^2 + 2x - 3} & x > 1 \end{cases}$ is continuous on $(-\infty, \infty)$.

Background

Topic: Continuity of Piecewise Functions

This question tests your understanding of how to ensure a piecewise function is continuous everywhere by matching function values at the point where the definition changes.

Key Terms and Formulas

• Continuity at a point: $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$

• For piecewise functions, set the left and right limits equal at the transition point.

Step-by-Step Guidance

1. Identify the point where the function changes definition ($x = 1$).

2. Compute $\lim_{x \to 1^-} f(x)$ using the first piece.

3. Compute $\lim_{x \to 1^+} f(x)$ using the second piece. Simplify the denominator and numerator if possible.

4. Set the two limits equal and solve for $a$ and $b$.

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Q5. Using the Intermediate Value Theorem, determine on which of the intervals (0,1), (1,2), (2,3), (3,4) the equation $x^3 - 3x = 5$ has a root.

Background

Topic: Intermediate Value Theorem (IVT)

This question tests your understanding of the IVT, which states that if a function is continuous on $[a, b]$ and takes values of opposite sign at $a$ and $b$, then it must cross zero somewhere in $(a, b)$.

Key Terms and Formulas

• Intermediate Value Theorem: If $f$ is continuous on $[a, b]$ and $f(a)$ and $f(b)$ have opposite signs, then $f$ has a root in $(a, b)$.

• Check $f(x)$ at the endpoints of each interval.

Step-by-Step Guidance

1. Define $f(x) = x^3 - 3x - 5$.

2. Evaluate $f(x)$ at the endpoints of each interval: $x = 0, 1, 2, 3, 4$.

3. Check the sign of $f(x)$ at each endpoint.

4. Identify intervals where $f(x)$ changes sign between endpoints.

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