JUPEB Mathematics Questions and Answers.

Welcome to AllCBTs Jupeb Mathematics Questions and Answers! Today, we’ll be solving some high-yield math problems, particularly focusing on limits and logarithmic equations. These types of questions commonly appear in NECO, WAEC, and JUPEB exams. Mastering them will help boost your confidence and accuracy in exams.

You can watch the full class in the video below:

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Question 1: Evaluate

Evaluate: lim⁡x→0tan⁡(3x)sin⁡(2x)\lim_{x \to 0} \frac{\tan(3x)}{\sin(2x)}x→0lim​sin(2x)tan(3x)​

Step-by-Step Solution:

To solve this limit problem:

1. Express tangent as a ratio of sine and cosine: tan⁡(3x)=sin⁡(3x)cos⁡(3x)\tan(3x) = \frac{\sin(3x)}{\cos(3x)}tan(3x)=cos(3x)sin(3x)​
2. Plug this back into the expression: sin⁡(3x)cos⁡(3x)÷sin⁡(2x)\frac{\sin(3x)}{\cos(3x)} \div \sin(2x)cos(3x)sin(3x)​÷sin(2x)
3. Convert division into multiplication: sin⁡(3x)cos⁡(3x)×1sin⁡(2x)=sin⁡(3x)sin⁡(2x)×1cos⁡(3x)\frac{\sin(3x)}{\cos(3x)} \times \frac{1}{\sin(2x)} = \frac{\sin(3x)}{\sin(2x)} \times \frac{1}{\cos(3x)}cos(3x)sin(3x)​×sin(2x)1​=sin(2x)sin(3x)​×cos(3x)1​
4. Using the limit identity: lim⁡x→0sin⁡(ax)x=a\lim_{x \to 0} \frac{\sin(ax)}{x} = ax→0lim​xsin(ax)​=a This means: sin⁡(3x)x÷sin⁡(2x)x=32\frac{\sin(3x)}{x} \div \frac{\sin(2x)}{x} = \frac{3}{2}xsin(3x)​÷xsin(2x)​=23​
5. Also, cos⁡(3x)→1\cos(3x) \to 1cos(3x)→1 as x→0x \to 0x→0

Final Answer:

32\frac{3}{2}23​

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Question 2: Solve Logarithmic Equation

Solve for xxx: log⁡2(x+3)+log⁡2(x−1)=3\log_2(x + 3) + \log_2(x – 1) = 3log2​(x+3)+log2​(x−1)=3

Step-by-Step Solution:

1. Apply Log Law:

log⁡b(m)+log⁡b(n)=log⁡b(mn)\log_b(m) + \log_b(n) = \log_b(mn)logb​(m)+logb​(n)=logb​(mn) log⁡2[(x+3)(x−1)]=3\log_2[(x + 3)(x – 1)] = 3log2​[(x+3)(x−1)]=3

2. Convert logarithmic form to exponential form:

(x+3)(x−1)=23=8(x + 3)(x – 1) = 2^3 = 8(x+3)(x−1)=23=8

3. Expand the quadratic:

x2+2x−3=8x^2 + 2x – 3 = 8×2+2x−3=8

4. Rearrange the equation:

x2+2x−11=0x^2 + 2x – 11 = 0x2+2x−11=0

5. Use the quadratic formula:

x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}x=2a−b±b2−4ac​​

Where:

• a=1a = 1a=1
• b=2b = 2b=2
• c=−11c = -11c=−11

x=−2±(2)2−4(1)(−11)2(1)=−2±4+442=−2±482x = \frac{-2 \pm \sqrt{(2)^2 – 4(1)(-11)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 44}}{2} = \frac{-2 \pm \sqrt{48}}{2}x=2(1)−2±(2)2−4(1)(−11)​​=2−2±4+44​​=2−2±48​​ x=−2±432=−1±23x = \frac{-2 \pm 4\sqrt{3}}{2} = -1 \pm 2\sqrt{3}x=2−2±43​​=−1±23​

6. Check for valid solutions:
   Since logarithms are only defined for positive arguments, we must ensure:

• x+3>0x + 3 > 0x+3>0
• x−1>0x – 1 > 0x−1>0

The value x=−1−23x = -1 – 2\sqrt{3}x=−1−23​ is negative → Not valid.
The value x=−1+23x = -1 + 2\sqrt{3}x=−1+23​ is approximately 2.46 → Valid.

Final Answer:

x=−1+23x = -1 + 2\sqrt{3}x=−1+23​

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

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Takeaway Tips for NECO/WAEC Math Exams

• Master log laws and basic identities—they always come up.
• Always simplify trigonometric limits using fundamental identities.
• Be careful with domain restrictions, especially in logarithmic equations.
• Don’t forget to check the validity of roots after solving equations.

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