(MGS Prelim 2016 AMath Paper 2 Q 1b)

This question is in line with the 4047 Syllabus where students have to know

“Conditions for\(ax^2+bx+c\) to be always positive (or always negative)”

*Above extracted from SEAB.*

**Question**: Determine the conditions for \(p\) and \(q\) such that the curve \(y=px^2-2x+3q\) lies **entirely above** the\(x\)axis, where \(p\) and \(q\) are constants. [3 marks]

**Solution**: The keyword is “entirely above”, implying that the graph is above the axis. We expect a Happy Face that is not intersecting the x-axis.

\(\Rightarrow\) Happy Face is ** ABOVE** axis

\(\Rightarrow\) ** No** Real Solutions

\(\Rightarrow\) \(b^2-4ac < 0\)

Observe that \(a=p\), \(b=-2\) and \(c=3q\),

\(b^2-4ac < 0 \)

\((-2)^2 – 4(p)(3q) < 0 \)

\(4-12pq<0\)

\(1-3pq<0\)

\(\frac{1}{3} < pq \)

Since the graph is a happy face, we know that the coefficient of \(x^2\) is positive and \(p>0\).

Therefore, \( p>0\) and \(pq > \frac{1}{3} \)

]]>Let’s take a look at this A maths question on **Equations and Inequalities.** A handful of students will be confused with the keywords or how to effectively determine the correct sign of the discriminant.

The example below is very similar to **St Margaret 2016** A Math Prelim Paper 1 Q2.

Original Question: Find the range of values of \(p\) for which \((p+2)x^2 – 12x + 2(p-1)\) is always negative.

__Modified Example__: Find the range of values of \(p\) for which \((p+2)x^2 – 12x > -2(p-1)\) for all real values of \(x\).

**Misconception**:

Students will mistaken REAL as the keyword and likely think that \(b^2-4ac \geq 0\).

The emphasis is that the word REAL is crucial ONLY if coupled with a friend called ROOTS. We need to see the Keyword – **REAL ROOTS** for \(b^2-4ac \geq 0\) to hold true.

**Step 1:**

Ensure that the right hand side of the inequality is** zero**.

**Step 2:**

Having done so, recognise the keyword as “\(>0\)“. This implies that the graph will be above the axis and the shape of this quadratic expression will be a **Happy Face**. Since the U-shape is **not intersecting** with the axis, the solutions will be **imaginary** and \(b^2-4ac < 0\).

**Keyword **\(>0\) implies

\(\Rightarrow\) Happy Face is ** ABOVE** axis

\(\Rightarrow\) ** No** Real Solutions

\(\Rightarrow\) \(b^2-4ac < 0\)

**Step 3:**

Observe \(a=p+2, b=-12\) and \(c=2p-2\)

For imaginary solutions, we have :

\(b^2-4ac<0\)

\((-12)^2-4(p+2)(2p-2)<0\)

\(144-4(2p^2-2p+4p-4)<0\)

\(-8p^2-8p+160<0\)

\(-p^2-p+20<0\)

\((-p+4)(p+5)<0\)

\(p<-5\) or \( p>4\)…..(1)

At this stage, most students are contented with getting the solutions and will think that the question is completed.

Well, we are almost there. This is a **typical example** of having a **variable coefficient** (\(p+2\)) in front of \(x^2\).

__Extra Step is mandatory__

Remember we mentioned earlier that the graph is a Happy Face, thus we expect the coefficient of \(x^2\) to be positive and we have

\(p+2>0\)

\(p>-2\)….. (2)

Visualise the overlapping region of answers on a **number line **from (1) and (2),

Therefore \(p>4\) (final solution)

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Increasingly asked in many different prelim questions, this type of question, which usually contains the keywords

The technique to solve this type of question is generic, as we will discuss in detail below.

Q4 (ii). The complex number \(z\) is given by $$z=-1+\text{i}c$$, where \(c\) is a non-zero real number.

Given that \(\frac{z^n}{z^*}\) is purely real, find the three smallest positive integer values of \(n\) when \(c=\sqrt{3}\).

**Solution with explanation:**

* Step 1: *To solve this type of question, we first find the

\(|z|=2\), \(\arg(z)=\frac{2\pi}{3}\)

* Step 2: *Convert to exponential form, then

Notice that by doing this, the power \(n\) is now “stuck” inside the trigo function, instead of being in the power. This is exactly what we want.

** Step 3: **For purely real, we set

Since\(\frac{z^n}{z^*}\) is purely real, Im(\(\frac{z^n}{z^*}\))\(=0\),

\(\sin\left[\left(n+1\right)\frac{2\pi}{3}\right]=0\)

**\((n+1)\frac{2\pi}{3}=k\pi, k \in \mathbb{Z}\)

\(n+1 = \frac{3k}{2}\)

We want the smallest 3 positive integers of \(n\), which corresponds to \(k=2,4,6\)

\(n+1=3, 6, 9\)

\(n = 2, 5, 8\)

**Note: We need to learn the generic solution for \(\sin \theta = 0\) and \(\cos \theta = 0\). To remember the following results, you just have to recall the graph of sine or cosine, and note down the positions of the zeros!

\(\sin \theta = 0\)

\(\theta =k\pi, k \in \mathbb{Z}\)

\(\cos \theta = 0\)

\(\theta =\frac{\pi}{2} + k\pi, k \in \mathbb{Z}\)

The following tables shows the cut off points for Junior College JC 2018 for the respective arts and science stream.

Junior College | Arts | Science/IB |
---|---|---|

Raffles Institution, RI | 5 | 5 |

Hwa Chong Institution, HCI | 5 | 5 |

Victoria Junior College, VJC | 8 | 6 |

Anglo-Chinese School (Independent), ACSi | 6 | 6 |

National Junior College, NJC | 9 | 8 |

Nanyang Junior College, NYJC | 8 | 7 |

Anglo-Chinese Junior College, ACJC | 10 | 9 |

St. Joseph’s Institution, SJI | 8 | 8 |

Eunoia Junior College, EJC | 11 | 10 |

Temasek Junior College, TJC | 11 | 10 |

St. Andrew’s Junior College, SAJC | 12 | 11 |

Anderson Serangoon Junior College | 12 | 12 |

Tampines Meridian Junior College | 13 | 14 |

Jurong Pioneer Junior College | 17 | 14 |

Catholic Junior College, CJC | 14 | 15 |

Yishun Innova Junior College | 20 | 20 |

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