Master
Pre-Calculus

We need two conditions: (1) \(\sqrt{x-3}\) must be defined → \(x - 3 \geq 0\), and (2) the denominator can't be zero → \(\sqrt{x-3} \neq 0\) → \(x - 3 \neq 0\). Combining: \(x - 3 > 0\), so \(x > 3\). Option A (\(x \geq 3\)) is the trap — it includes \(x=3\) where the denominator becomes 0.

\((f \circ g)(2) = f(g(2))\). First: \(g(2) = 2(2)-3 = 1\). Then: \(f(1) = 1^2 + 1 = 2\). The common mistake is computing \(g(f(2))\) instead — that gives \(g(5) = 7\), which isn't even an option here! Always go INSIDE first: evaluate g, then feed result into f.

Test \(x=1\): \(1^3 - 2(1)^2 - 5(1) + 6 = 1 - 2 - 5 + 6 = 0\). ✓ So \(x=1\) is a zero. The polynomial factors as \((x-1)(x-3)(x+2)\), giving zeros at \(x=1, 3, -2\). Many students skip testing small integers and go straight to complicated methods — always try ±1 first!

Focus only on the leading term: \(-2x^4\). Even degree (4) → both ends go same direction. Negative leading coefficient (−2) → both ends go DOWN to \(-\infty\). Trap: even degree makes students think "both go up," but the negative sign flips it. Rule: negative × even = both ends DOWN.

\(2^{3x} = 32 = 2^5\). Since the bases are equal, set exponents equal: \(3x = 5 \Rightarrow x = \frac{5}{3}\). Classic mistake: students write \(3x = 32\) without converting 32 to a power of 2 first. Always look for matching bases!

\(\log_2 8 + \log_2 4 = \log_2(8 \times 4) = \log_2 32 = 5\) since \(2^5 = 32\). The #1 mistake: students ADD the arguments: \(\log_2(8+4) = \log_2 12\). Wrong! The product rule says you MULTIPLY arguments when adding logs, not add them.

Convert to exponential form: \(\log_3(x+1) = 4 \Rightarrow 3^4 = x+1 \Rightarrow 81 = x+1 \Rightarrow x = 80\). Trap: Option D (\(x=81\)) looks right because \(3^4 = 81\), but students forget to subtract 1! Always isolate the log argument completely before concluding.

\(\frac{5\pi}{6}\) lies in Quadrant II (between \(\frac{\pi}{2}\) and \(\pi\)). Its reference angle is \(\pi - \frac{5\pi}{6} = \frac{\pi}{6}\). Since sine is POSITIVE in Q2: \(\sin\!\left(\frac{5\pi}{6}\right) = \sin\!\left(\frac{\pi}{6}\right) = \frac{1}{2}\). Many students confuse the reference angle with \(\frac{5\pi}{6}\) directly and compute wrong values.

Use \(\sin^2\theta + \cos^2\theta = 1\): \(\left(\frac{3}{5}\right)^2 + \cos^2\theta = 1 \Rightarrow \frac{9}{25} + \cos^2\theta = 1 \Rightarrow \cos^2\theta = \frac{16}{25} \Rightarrow \cos\theta = \pm\frac{4}{5}\). Since θ is in Q1, cosine is POSITIVE: \(\cos\theta = \frac{4}{5}\). Option C (negative) is the trap for students who forget to check the quadrant!

In \(y = 3\sin(4x - \pi)\), we have \(B = 4\). Period \(= \frac{2\pi}{B} = \frac{2\pi}{4} = \frac{\pi}{2}\). The amplitude is 3 (not the period), and \(-\pi\) is the phase shift offset, not related to the period. Many students accidentally use 3 or \(\pi\) in the period formula.

\(a_5 = a_1 + (5-1)(4) = 17 \Rightarrow a_1 + 16 = 17 \Rightarrow a_1 = 1\). Watch out: \((n-1)\), not \(n\). The 5th term uses \((5-1)=4\) steps, not 5. Using 5 instead of 4 gives \(a_1 = -3\), a very common error!

\(S_4 = \frac{2(1-3^4)}{1-3} = \frac{2(1-81)}{-2} = \frac{2(-80)}{-2} = 80\). Or simply list: \(2+6+18+54=80\). Option A (62) = \(\frac{2(1-3^4)}{1-3}\) with arithmetic error. Option C (78) = using wrong formula \(S = a(r^n-1)/(r-1)\) with an off-by-one. Always double-check by listing the first few terms!

In vertex form \(y = a(x-h)^2 + k\), the vertex is \((h, k)\). Here \(h = 3\) and \(k = 5\) → vertex = \((3, 5)\). The most common trap: students see \((x-3)\) and write \(h = -3\). The minus sign is ALREADY BUILT INTO the formula — \((x-h)\) with \(h=3\) gives \((x-3)\). Don't negate h again!

\(r^2 = 25 \Rightarrow r = \sqrt{25} = 5\). The equation equals \(r^2\), so you must take the square root. Option B (r=25) is the #1 trap — forgetting to take the square root. The center is \((2, -3)\) — note that \(+3\) inside the formula gives \(k = -3\).

Both numerator and denominator have degree 2 (same degree). Rule: divide leading coefficients → \(\frac{3}{1} = 3\). So H.A. is \(y = 3\). If top degree > bottom, no H.A. (oblique instead). If top degree < bottom, H.A. is \(y = 0\). Here degrees match, so use the ratio of leading coefficients.

Swap x and y: \(x = \frac{y+2}{3}\). Solve for y: multiply both sides by 3 → \(3x = y+2\) → \(y = 3x - 2\). So \(f^{-1}(x) = 3x - 2\). Verify: \(f(f^{-1}(x)) = \frac{(3x-2)+2}{3} = \frac{3x}{3} = x\) ✓. Always verify by composing f and its inverse!

FOIL: \((2+3i)(1-i) = 2(1) + 2(-i) + 3i(1) + 3i(-i) = 2 - 2i + 3i - 3i^2\). Since \(i^2 = -1\): \(= 2 - 2i + 3i - 3(-1) = 2 + 3 + i = 5 + i\). The trap: students leave \(i^2\) as is, getting \(2 - 3i^2\) instead of substituting \(i^2 = -1\). ALWAYS replace \(i^2\) with \(-1\) at the end!

Direct substitution gives \(\frac{0}{0}\) — indeterminate form. Factor: \(\frac{x^2-4}{x-2} = \frac{(x+2)(x-2)}{x-2} = x+2\) for \(x \neq 2\). Now substitute: \(\lim_{x\to2}(x+2) = 4\). Trap: Many say "undefined" because \(\frac{0}{0}\) looks undefined. But limits describe the behavior as x APPROACHES 2, not AT 2. Factor and cancel first!

The term with \(x^2\) has \(k=2\): \(\binom{4}{2}(x)^{4-2}(3)^2 = 6 \cdot x^2 \cdot 9 = 54x^2\). So the coefficient is 54. Common mistakes: (A) using only \(\binom{4}{2} = 6\) and forgetting to multiply by \(3^2\), or (B) computing \(3^2 = 9\) alone. The coefficient includes ALL numerical factors: \(\binom{n}{k} \times b^k\).

Right 3: replace \(x\) with \((x-3)\) → note it's MINUS for right shift (counterintuitive!). Down 2: subtract 2 from the output. Result: \(y = f(x-3) - 2\). The #1 error: thinking "right = plus" → writing \(f(x+3)\). Remember: horizontal shifts are OPPOSITE to the sign. LEFT = + inside, RIGHT = − inside. Vertical shifts match: UP = +, DOWN = −.