High School Math · Self-Study

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20 carefully crafted problems on Logarithms, Quadratic Functions & Discriminants — with instant feedback and explanations.

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Section 01
Logarithm Operations
Key concept: logs undo exponentials. Master the change-of-base and log laws — these appear on every exam.
⚡ Ultra-Quick Memory Points
PRODUCT → ADDlog(ab) = log a + log b
QUOTIENT → SUBTRACTlog(a/b) = log a − log b
POWER → MULTIPLYlog(aⁿ) = n·log a
BASE = ANSWERlog_a(a) = 1, log_a(1) = 0
CHANGE BASElog_a(b) = log b / log a
Q 01 Logarithm
Simplify: \(\log_2 8 + \log_2 4\)
💡 Hint: Use the Product Rule — adding logs means multiplying their arguments.
Explanation
\(\log_2 8 + \log_2 4 = \log_2(8 \times 4) = \log_2 32 = \log_2 2^5 = 5\)
Key: Product Rule → logs add, arguments multiply.
Q 02 Logarithm
Evaluate: \(\log_3 81 - \log_3 9\)
💡 Hint: Subtraction of logs = division of arguments.
Explanation
\(\log_3 81 - \log_3 9 = \log_3\!\left(\dfrac{81}{9}\right) = \log_3 9 = \log_3 3^2 = 2\)
Key: Quotient Rule → subtract logs = divide arguments.
Q 03 Logarithm
What is \(\log_5 125^{2/3}\,\)?
💡 Hint: Power Rule first — bring the exponent down as a multiplier.
Explanation
\(\log_5 125^{2/3} = \dfrac{2}{3}\log_5 125 = \dfrac{2}{3} \times 3 = 2\)
Since \(125 = 5^3\), so \(\log_5 125 = 3\).
Q 04 Logarithm
If \(\log x = 2\) and \(\log y = 3\), find \(\log\!\left(\dfrac{x^2}{y}\right)\).
💡 Hint: Apply Power Rule then Quotient Rule step by step.
Explanation
\(\log\!\left(\dfrac{x^2}{y}\right) = 2\log x - \log y = 2(2) - 3 = 4 - 3 = 1\)
Q 05 Logarithm
Solve for \(x\): \(\log_2(x+1) = 4\)
💡 Hint: Convert log → exponential form. "log_b(N) = k means b^k = N"
Explanation
\(\log_2(x+1) = 4 \Rightarrow x+1 = 2^4 = 16 \Rightarrow x = 15\)
Common mistake: Forgetting to subtract 1 at the end.
Q 06 Logarithm
Using the change-of-base formula, \(\log_4 64 = \,?\)
Explanation
\(64 = 4^3\), so \(\log_4 64 = 3\).
Or: \(\log_4 64 = \dfrac{\log 64}{\log 4} = \dfrac{3\log 2}{2\log 2} \times \dfrac{2}{2} = \dfrac{6}{2} = 3\). Wait — directly: \(\dfrac{\log 64}{\log 4} = \dfrac{\log 2^6}{\log 2^2} = \dfrac{6}{2} = 3\).
Q 07 Logarithm
Which expression equals \(\log_a\!\left(\sqrt{a^3}\right)\)?
💡 Hint: \(\sqrt{a^3} = a^{3/2}\). Then use the Power Rule.
Explanation
\(\log_a\!\left(\sqrt{a^3}\right) = \log_a a^{3/2} = \dfrac{3}{2}\log_a a = \dfrac{3}{2}(1) = \dfrac{3}{2}\)
Section 02
Quadratic Functions — Maximum & Minimum
Vertex form is your best friend. Know when you're looking for max vs. min, and how the domain affects the answer.
⚡ Ultra-Quick Memory Points
a > 0 → MINIMUMOpens up, vertex is the lowest point
a < 0 → MAXIMUMOpens down, vertex is the highest point
VERTEX Xx = −b/(2a) from standard form
COMPLETE THE SQUAREConvert ax²+bx+c → a(x−h)²+k
RESTRICTED DOMAINCheck endpoints! Max/Min may shift.
Q 08 Quadratic
Find the minimum value of \(f(x) = x^2 - 6x + 11\).
💡 Hint: Complete the square, or use x = −b/(2a) to find the vertex.
Explanation
\(f(x) = (x-3)^2 + 2\). Vertex at \(x=3\), minimum value = \(2\).
Key: Since \(a=1 > 0\), the parabola opens up → minimum at vertex.
Q 09 Quadratic
What is the maximum value of \(f(x) = -2x^2 + 8x - 3\)?
Explanation
Vertex x: \(x = -\dfrac{8}{2(-2)} = 2\).
\(f(2) = -2(4)+8(2)-3 = -8+16-3 = 5\). Maximum = \(5\).
Key: \(a = -2 < 0\) → opens downward → vertex is the max.
Q 10 Quadratic
The function \(f(x) = x^2 - 4x + 1\) is defined on \(0 \le x \le 5\). Find its maximum value.
💡 Hint: Restricted domain! Evaluate both endpoints AND the vertex. The max may be at a boundary.
Explanation
Vertex at \(x=2\): \(f(2) = 4-8+1=-3\) (minimum on this domain).
Check endpoints: \(f(0)=1\), \(f(5)=25-20+1=6\).
Maximum = \(6\) at \(x=5\).
Common trap: Students only check the vertex and miss endpoint maxima!
Q 11 Quadratic
Which vertex form represents \(f(x) = x^2 + 4x + 7\)?
Explanation
\(x^2+4x+7 = (x^2+4x+4)+3 = (x+2)^2+3\).
Trick to complete the square: Take half of the middle coefficient → \(4/2=2\), square it → \(4\). Add and subtract.
Q 12 Quadratic
A ball is thrown and its height is \(h(t) = -t^2 + 6t + 2\) (in meters). What is the maximum height?
Explanation
Vertex at \(t = -\dfrac{6}{2(-1)} = 3\).
\(h(3) = -9+18+2 = 11\) meters.
Q 13 Quadratic
For \(f(x) = 2(x-3)^2 - 5\), the minimum value is \(-5\).
On the domain \(-1 \le x \le 2\), what is the minimum value?
💡 Hint: The vertex \(x=3\) is outside the domain! The closest point to the vertex within the domain gives the min.
Explanation
The vertex \(x=3\) is outside \([-1, 2]\). The function decreases as \(x\) moves toward 3, so on this domain it's smallest at \(x=2\).
\(f(2) = 2(2-3)^2-5 = 2(1)-5 = -3\).
Rule: If vertex is outside the domain, min/max is at the nearest endpoint.
Section 03
The Discriminant
The discriminant \(D = b^2 - 4ac\) tells you everything about a quadratic's roots — without actually solving it.
⚡ Ultra-Quick Memory Points
D > 0 → TWO REALTwo distinct real roots (cuts x-axis twice)
D = 0 → ONE REALRepeated (double) root — tangent to x-axis
D < 0 → NO REALTwo complex (imaginary) roots — no x-intercept
FORMULAD = b² − 4ac (from ax² + bx + c = 0)
TANGENT CONDITIONLine tangent to curve ⟺ D = 0
Q 14 Discriminant
For \(2x^2 - 5x + 3 = 0\), how many real roots does it have?
Explanation
\(D = (-5)^2 - 4(2)(3) = 25 - 24 = 1 > 0\)
Since \(D > 0\) → two distinct real roots.
Q 15 Discriminant
For what value of \(k\) does \(x^2 - kx + 9 = 0\) have exactly one real root?
💡 Hint: Set D = 0 for a repeated (double) root.
Explanation
Set \(D = 0\): \((-k)^2 - 4(1)(9) = 0 \Rightarrow k^2 = 36 \Rightarrow k = \pm 6\).
Don't forget: \(k^2 = 36\) gives both \(+6\) and \(-6\)!
Q 16 Discriminant
The equation \(x^2 + 4x + m = 0\) has no real roots. Which condition on \(m\) is correct?
Explanation
\(D = 4^2 - 4(1)(m) = 16 - 4m\). For no real roots: \(D < 0 \Rightarrow 16 - 4m < 0 \Rightarrow m > 4\).
Q 17 Discriminant
Find the value of \(k\) so that the line \(y = kx + 1\) is tangent to the parabola \(y = x^2 + 3x + 2\).
💡 Hint: Tangent means exactly one intersection → set equations equal and use D = 0.
Explanation
Set equal: \(kx+1 = x^2+3x+2 \Rightarrow x^2+(3-k)x+1=0\).
For tangency, \(D=0\): \((3-k)^2 - 4 = 0 \Rightarrow (3-k)^2 = 4 \Rightarrow 3-k = \pm 2\).
\(k = 1\) or \(k = 5\).
Q 18 Discriminant
For \(3x^2 + 6x + 3 = 0\), the discriminant equals:
Explanation
\(D = 6^2 - 4(3)(3) = 36 - 36 = 0\).
This means the equation has a repeated root: \(x = -1\) (double root).
Q 19 Discriminant
If the sum of roots of \(x^2 + bx + 12 = 0\) is \(-7\), what is the discriminant?
💡 Hint: Vieta's formula — sum of roots = −b/a, product = c/a.
Explanation
Sum of roots = \(-b = -7 \Rightarrow b = 7\).
\(D = 7^2 - 4(1)(12) = 49 - 48 = 1\).
Since \(D = 1 > 0\), there are two distinct real roots.
Q 20 Discriminant
Which of the following is a necessary condition for \(ax^2+bx+c=0\) to have two distinct positive real roots?
💡 This is tricky! You need D > 0, sum of roots > 0, AND product of roots > 0.
Explanation
For two distinct positive real roots, you need all three:
1. \(D > 0\) — two real roots exist
2. Sum \(= -b/a > 0\) — both roots are positive (their sum is positive)
3. Product \(= c/a > 0\) — both roots have the same sign (both positive)
Common mistake: Only checking D > 0 doesn't ensure both roots are positive.