Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 3.8 AMC 10/12 Find the answer

A. In the rectangular coordinate system, the parametric equation of curve C is given by $\begin{cases} x = 2\cos\theta \\ y = \sqrt{3}\sin\theta \end{cases}$ where $\theta$ is the parameter. Establish a polar coordinate system with the coordinate origin as the pole and the positive half of the x-axis as the polar axis. The line $l$ passes through two points $A(\sqrt{2}, \frac{\pi}{4})$ and $B(3, \frac{\pi}{2})$ in the polar coordinate system.

(I) Write the ordinary equation of curve C and find the slope of line $l$ .

(II) Suppose line $l$ intersects curve C at points $P$ and $Q$ . Find $|BP| \cdot |BQ|$ .

B. Given the function $f(x) = |x - 1| + |2x - a|$ .

(I) When $a = 1$ , find the solution set of $f(x) \geq 1$ .

(II) When $x \in [-1, 1]$ , $f(x) \geq 1$ always holds true. Find the range of real number values for $a$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

A. (I) From the given information, the ordinary equation of curve C is $\frac{x^2}{4} + \frac{y^2}{3} = 1$ .

Given points $A(1,1)$ and $B(0, 3)$ , the slope of line $l$ is $-2$ .

(II) The parametric equation of line $l$ is $\begin{cases} x = -\frac{1}{\sqrt{5}}t \\ y = 3 + \frac{2}{\sqrt{5}}t \end{cases}$ where $t$ is the parameter.

Substitute this equation into the equation of curve C, we get $\frac{19}{5}t^2 + \frac{48}{\sqrt{5}}t + 24 = 0$ .

Let the two roots of the equation be ${t}_1$ and ${t}_2$ . Then $|BP| \cdot |BQ| = |{t}_1{t}_2| = \boxed{\frac{120}{19}}$ .

B. (I) When $a = 1$ , we have the inequality $|x - 1| + |2x - 1| \geq 1$ .

This leads to three cases:

1. When $x 1$ , solve $3x - 2 \geq 1$ .

The solutions for the three cases are $x \leq \frac{1}{3}$ , $x = 1$ , and $x > 1$ , respectively.

Combining all three cases, the solution set for the inequality is $\boxed{(-\infty, \frac{1}{3}] \cup [1, +\infty)}$ .

(II) When $x \in [-1, 1]$ , we have $|2x - a| \geq 1 - |x - 1| = x$ .

When $x \in [-1, 0)$ , $a \in \mathbb{R}$ .

When $x \in [0, 1]$ , either $2x - a \geq x$ or $2x - a \leq -x$ holds true.

Thus, $a \leq x$ or $a \geq 3x$ always holds true.

Hence, $a \leq 0$ or $a \geq 3$ .

In conclusion, the range of values for $a$ is $\boxed{(-\infty, 0] \cup [3, +\infty)}$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.