Fundamental OF Mathematics flash cards

1. 1.If $\alpha$ is a real number and $|x-\alpha| < a$ (with $a>0$), what is the equivalent interval form? Also state the solution of $|x-\alpha|>a$.

   $|x-\alpha|a \Rightarrow x<\alpha-a$ or $x>\alpha+a$.

   Hint: Distance from $\alpha$ on number line

2. 2.State the condition for the quadratic $ax^2+bx+c=0$ (with $a,b,c$ rational, $a\neq0$) to have rational roots.

   Discriminant $D=b^2-4ac$ must be a perfect square (and $D\ge0$), with $a,b,c\in\mathbb{Q}$.

   Hint: D = perfect square

3. 3.For $x\in\mathbb{R}$, how do you solve an inequality of the type $\dfrac{P(x)}{Q(x)} \ge 0$ (rational inequality)?

   Use the wavy curve (sign-scheme) method: mark zeros of $P(x)$ and $Q(x)$ on number line, exclude zeros of $Q(x)$, alternate signs across simple roots (repeat sign at even-multiplicity roots), then read off intervals where expression is $\ge 0$.

   Hint: Wavy curve method

4. 4.What is the key property used to solve equations/inequalities involving $\log_a x$ regarding domain and base restrictions?

   Require $x>0$ (argument positive), base $a>0,\; a\neq1$; if $0decreasing, so inequality signs reverse when taking logs/removing log.

   Hint: Base < 1 flips inequality

5. 5.How many real roots can a cubic equation $ax^3+bx^2+cx+d=0$ (real coefficients, $a\neq0$) have at minimum, and why?

   At least 1 real root, since complex (non-real) roots of a real-coefficient polynomial occur in conjugate pairs, and an odd-degree polynomial cannot have all roots paired.

   Hint: Odd degree ⇒ complex roots pair up

6. 6.For positive reals $a,b,c$ with $a+b+c=9$, find the maximum possible value of $abc$.

   By AM-GM, $\dfrac{a+b+c}{3}\ge (abc)^{1/3}$, so $3\ge(abc)^{1/3}$, giving $abc\le 27$. Equality holds when $a=b=c=3$, so the maximum value is $27$.

   Hint: Apply AM-GM to $a,b,c$ and use the equality condition.

7. 7.If $a,b,c>0$ and $abc=1$, prove that $a+b+c\ge 3$ and state when equality holds.

   By AM-GM, $\dfrac{a+b+c}{3}\ge (abc)^{1/3}=1^{1/3}=1$, so $a+b+c\ge 3$. Equality holds iff $a=b=c=1$.

   Hint: Use AM-GM on three positive numbers whose product is fixed at $1$.

8. 8.For $x>0$, find the minimum value of $f(x)=x^2+\dfrac{2}{x}$.

   Split $\dfrac2x$ into two equal parts and apply AM-GM to three terms: $x^2+\dfrac1x+\dfrac1x \ge 3\left(x^2\cdot\dfrac1x\cdot\dfrac1x\right)^{1/3}=3$. Equality needs $x^2=\dfrac1x\Rightarrow x^3=1\Rightarrow x=1$, giving minimum value $3$.

   Hint: Split the $2/x$ term into two equal pieces before applying AM-GM to three terms.

9. 9.For two positive reals $a,b$, prove the chain $\dfrac{2ab}{a+b}\le\sqrt{ab}\le\dfrac{a+b}{2}$ (HM $\le$ GM $\le$ AM), stating the equality condition.

   Since $(\sqrt a-\sqrt b)^2\ge 0$, expanding gives $a+b\ge 2\sqrt{ab}$, i.e. AM $\ge$ GM. Also $\mathrm{GM}^2=ab=\dfrac{a+b}{2}\cdot\dfrac{2ab}{a+b}=\mathrm{AM}\cdot\mathrm{HM}$; combined with AM $\ge$ GM this forces $\mathrm{GM}\ge\mathrm{HM}$. Hence HM $\le$ GM $\le$ AM, with equality throughout iff $a=b$.

   Hint: Use $(\sqrt a-\sqrt b)^2\ge0$ and the identity $\mathrm{GM}^2=\mathrm{AM}\cdot\mathrm{HM}$.

10. 10.For positive reals $a,b,c$, prove $(a+b+c)\left(\dfrac1a+\dfrac1b+\dfrac1c\right)\ge 9$.

    By AM-GM, $a+b+c\ge 3(abc)^{1/3}$ and $\dfrac1a+\dfrac1b+\dfrac1c\ge \dfrac{3}{(abc)^{1/3}}$. Multiplying the two inequalities gives $(a+b+c)\left(\dfrac1a+\dfrac1b+\dfrac1c\right)\ge 9$, with equality iff $a=b=c$.

    Hint: Apply AM-GM separately to the sum and to the sum of reciprocals, then multiply.

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