IJMB Mathematics Past Questions for 2025

Welcome to today’s tutorial on IJMB Mathematics Past Questions or JUPEB candidates. In this session, we’ll solve some important past questions focusing on trigonometry and quadratic equations. Grab your calculators and writing materials – let’s dive into two likely exam questions that could easily appear in your next paper.

You can watch the full class in the video below:

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Question 1: Solve When Given sec θ = 25/24

Problem Statement:
Given that: sec⁡θ=2524\sec \theta = \frac{25}{24}secθ=2425​

Find the value of: cot⁡θ+cos⁡θ\cot \theta + \cos \thetacotθ+cosθ

Step-by-Step Breakdown:

1. Recall trigonometric identities:

• sec⁡θ=1cos⁡θ\sec \theta = \frac{1}{\cos \theta}secθ=cosθ1​
• cot⁡θ=1tan⁡θ\cot \theta = \frac{1}{\tan \theta}cotθ=tanθ1​

2. Since sec⁡θ=2524\sec \theta = \frac{25}{24}secθ=2425​, then:

cos⁡θ=1sec⁡θ=2425\cos \theta = \frac{1}{\sec \theta} = \frac{24}{25}cosθ=secθ1​=2524​

Using the identity: cos⁡θ=adjacenthypotenuse⇒adjacent=24,hypotenuse=25\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} \Rightarrow \text{adjacent} = 24, \text{hypotenuse} = 25cosθ=hypotenuseadjacent​⇒adjacent=24,hypotenuse=25

3. Use Pythagoras’ Theorem to find the opposite side:

hypotenuse2=opposite2+adjacent2⇒252=x2+242⇒625=x2+576⇒x2=49⇒x=7\text{hypotenuse}^2 = \text{opposite}^2 + \text{adjacent}^2 \Rightarrow 25^2 = x^2 + 24^2 \Rightarrow 625 = x^2 + 576 \Rightarrow x^2 = 49 \Rightarrow x = 7hypotenuse2=opposite2+adjacent2⇒252=x2+242⇒625=x2+576⇒x2=49⇒x=7

So, the opposite side is 7.

4. Find tan⁡θ\tan \thetatanθ:

tan⁡θ=oppositeadjacent=724⇒cot⁡θ=1tan⁡θ=247\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{7}{24} \Rightarrow \cot \theta = \frac{1}{\tan \theta} = \frac{24}{7}tanθ=adjacentopposite​=247​⇒cotθ=tanθ1​=724​

5. Recall earlier:

cos⁡θ=2425\cos \theta = \frac{24}{25}cosθ=2524​

Add both values:

cot⁡θ+cos⁡θ=247+2425\cot \theta + \cos \theta = \frac{24}{7} + \frac{24}{25}cotθ+cosθ=724​+2524​

6. Find LCM of 7 and 25 = 175

24×25+24×7175=600+168175=768175≈4.39⇒4.4\frac{24 \times 25 + 24 \times 7}{175} = \frac{600 + 168}{175} = \frac{768}{175} \approx 4.39 \Rightarrow \boxed{4.4}17524×25+24×7​=175600+168​=175768​≈4.39⇒4.4​

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Question 2: Find α−β\alpha – \betaα−β from a Quadratic Equation

Given Equation: x2−5x+6=0x^2 – 5x + 6 = 0x2−5x+6=0

Find: α−β\alpha – \betaα−β

Step-by-Step Breakdown:

1. Factor the quadratic:

x2−5x+6=0⇒(x−2)(x−3)=0⇒x=2 or x=3x^2 – 5x + 6 = 0 \Rightarrow (x – 2)(x – 3) = 0 \Rightarrow x = 2 \text{ or } x = 3×2−5x+6=0⇒(x−2)(x−3)=0⇒x=2 or x=3

So:

• α=3\alpha = 3α=3
• β=2\beta = 2β=2

2. Subtract:

α−β=3−2=1\alpha – \beta = 3 – 2 = \boxed{1}α−β=3−2=1​

Even if reversed (i.e., β−α=2−3\beta – \alpha = 2 – 3β−α=2−3), the answer would be −1-1−1, but the magnitude remains 1 if only difference is asked.

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Watch more educational videos and past questions: https://youtube.com/@allcbts

Summary & Key Tips

• Use trigonometric identities to rewrite unknowns in solvable form.
• Always draw a triangle to relate adjacent, opposite, and hypotenuse.
• Pythagoras’ Theorem is critical for most trigonometry questions.
• Factorization is the fastest way to solve most quadratic expressions.
• Always double-check your roots before using them in further calculations.

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