OCR A Level: Core 3 - Modulus [January 2013 Q3]

Algebra Level 3

$(\textbf{a})$ Given that $|t|=3$ , find the possible values of $|2t-1|$ .

$(\textbf{b})$ Solve the inequality $|x-\sqrt2|>|x+3\sqrt2|$ .

Let the two possible values of $|2t-1|$ be $y$ and $z$ , and let $x_{min}$ be the minimum value that $x$ does not satisfy.

Input $x_{min}^2+y+z$ as your answer.

There are 3 marks available for part (a) and 4 marks for part (b).

In total, this question is worth 9.72% of all available marks in the paper.

This is part of the set OCR A Level Problems .

The answer is 14.

2 solutions

Chew-Seong Cheong
Mar 9, 2016

Part (a):

$|t| = 3 \quad \Rightarrow t = \begin{cases} +3 \\ -3 \end{cases} \quad \Rightarrow |2t -1| = \begin{cases} |2(+3)-1| = 5 \quad \color{#3D99F6}{=y} \\ |2(-3)-1| = 7 \quad \color{#3D99F6}{=z} \end{cases}$

Part (b):

$\begin{cases} |x-\sqrt{2}| & = \begin{cases} x - \sqrt{2} & \quad \space \text{for } x \ge \sqrt{2} \\ \sqrt{2} -x & \quad \space \text{for } x < \sqrt{2} \end{cases} \\ |x+3\sqrt{2}| & = \begin{cases} x + 3\sqrt{2} & \text{for } x \ge -3\sqrt{2} \\ - x -3 \sqrt{2} & \text{for } x < -3\sqrt{2} \end{cases} \end{cases}$

There are $3$ cases:

$\begin{aligned} x < -3 \sqrt{2}: \quad \sqrt{2} -x & > - x - 3\sqrt{2} \quad \small \color{#D61F06}{\text{No solution}} \\ -3 \sqrt{2} \le x < \sqrt{2}: \quad \sqrt{2} -x & > x + 3\sqrt{2} \\ 2x & < - 2\sqrt{2} \\ \Rightarrow x & < -\sqrt{2} \quad \quad \space \space \space \color{#3D99F6}{ = x_{min}} \\ x \ge \sqrt{2}: \quad x - \sqrt{2} & > x + 3\sqrt{2} \quad \space \space \small \color{#D61F06}{\text{No solution}} \end{aligned}$

Therefore, $x_{min}^2 + y + z = 2 + 5 + 7 = \boxed{14}$

Same solution !!

Aakash Khandelwal - 5 years, 3 months ago

Michael Fuller
Mar 8, 2016

The mark scheme for this question: Large Version

Same Method Yay! :P

Mehul Arora - 5 years, 3 months ago