JUPEB Mathematics 2025: Exam Questions.

Welcome to today’s JUPEB Mathematics tutorial from AllCBTs!
In this session, we’ll tackle some of the most examinable JUPEB questions — including determinants, surds, and complex number manipulation. These are topics that students often find tricky but are guaranteed to show up in the JUPEB exam.

So grab your writing materials and calculator, and let’s begin!

You can watch the full class in the video below:

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Question 1: Find the Non-Zero Negative Value of xxx

Problem:
Solve for the non-zero negative value of xxx that satisfies the determinant equation of the matrix: ∣x101x101x∣=0\begin{vmatrix} x & 1 & 0 \\ 1 & x & 1 \\ 0 & 1 & x \end{vmatrix} = 0​x10​1×1​01x​​=0

Solution:

To find the determinant, expand the 3×3 matrix: Det=x(x⋅x−1⋅1)−1(1⋅x−1⋅0)+0(1⋅1−x⋅0)\text{Det} = x(x \cdot x – 1 \cdot 1) – 1(1 \cdot x – 1 \cdot 0) + 0(1 \cdot 1 – x \cdot 0)Det=x(x⋅x−1⋅1)−1(1⋅x−1⋅0)+0(1⋅1−x⋅0) =x(x2−1)−(x)+0=x3−x−x=x3−2x= x(x^2 – 1) – (x) + 0 = x^3 – x – x = x^3 – 2x=x(x2−1)−(x)+0=x3−x−x=x3−2x

Set it to 0: x3−2x=0⇒x(x2−2)=0x^3 – 2x = 0 \Rightarrow x(x^2 – 2) = 0x3−2x=0⇒x(x2−2)=0

Solving this:

• x=0x = 0x=0
• x2=2⇒x=±2x^2 = 2 \Rightarrow x = \pm \sqrt{2}x2=2⇒x=±2​

Non-zero negative value: −2\boxed{-\sqrt{2}}−2​​

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Question 2: Find the Determinant of the Matrix ZZZ

Matrix: Z=[233410140]Z = \begin{bmatrix} 2 & 3 & 3 \\ 4 & 1 & 0 \\ 1 & 4 & 0 \end{bmatrix}Z=​241​314​300​​

Solution:

Use the cofactor expansion or Sarrus’ Rule: Det=2(1⋅0−0⋅4)−3(4⋅0−0⋅1)+3(4⋅4−1⋅1)\text{Det} = 2(1 \cdot 0 – 0 \cdot 4) – 3(4 \cdot 0 – 0 \cdot 1) + 3(4 \cdot 4 – 1 \cdot 1)Det=2(1⋅0−0⋅4)−3(4⋅0−0⋅1)+3(4⋅4−1⋅1) =2(0)−3(0)+3(16−1)=0+0+3(15)=45= 2(0) – 3(0) + 3(16 – 1) = 0 + 0 + 3(15) = \boxed{45}=2(0)−3(0)+3(16−1)=0+0+3(15)=45​

Correct Option: D. 45

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Question 3: Simplify

(1+31+i)2\left(1 + \frac{3}{1 + i}\right)^2(1+1+i3​)2

Where i=−1i = \sqrt{-1}i=−1​

Step-by-Step:

1. Multiply numerator and denominator by the conjugate of the denominator:

31+i⋅1−i1−i=3(1−i)(1+i)(1−i)=3−3i1−(−1)=3−3i2\frac{3}{1 + i} \cdot \frac{1 – i}{1 – i} = \frac{3(1 – i)}{(1 + i)(1 – i)} = \frac{3 – 3i}{1 – (-1)} = \frac{3 – 3i}{2}1+i3​⋅1−i1−i​=(1+i)(1−i)3(1−i)​=1−(−1)3−3i​=23−3i​

2. Add 1:

1+3−3i2=2+3−3i2=5−3i21 + \frac{3 – 3i}{2} = \frac{2 + 3 – 3i}{2} = \frac{5 – 3i}{2}1+23−3i​=22+3−3i​=25−3i​

3. Now square:

(5−3i2)2=(5−3i)24\left(\frac{5 – 3i}{2}\right)^2 = \frac{(5 – 3i)^2}{4}(25−3i​)2=4(5−3i)2​

4. Expand numerator:

(5−3i)2=25−30i+9i2=25−30i−9=16−30i(5 – 3i)^2 = 25 – 30i + 9i^2 = 25 – 30i – 9 = 16 – 30i(5−3i)2=25−30i+9i2=25−30i−9=16−30i

5. Final answer:

16−30i4=4−7.5i\frac{16 – 30i}{4} = \boxed{4 – 7.5i}416−30i​=4−7.5i​

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

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Summary: Key Concepts Covered

• How to solve determinant-based equations.
• Working with surds like 2\sqrt{2}2​ in algebra.
• Rationalizing complex denominators using conjugates.
• Expanding and simplifying binomial squares involving imaginary numbers.

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