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Question 1: Algebra
Solve for xx:log⁡2(x−3)+log⁡2(x−1)=3log2​(x−3)+log2​(x−1)=3

A) x=−1x=−1
B) x=1x=1
C) x=4x=4
D) x=5x=5

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Question 2: Geometry
A right circular cylinder has a height of 10 cm. If the radius of the base is increased by 50% and the height is decreased by 20%, what is the percentage increase in the volume of the cylinder?

A) 20%
B) 50%
C) 80%
D) 100%

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Question 3: Number Theory & Algebra
When a polynomial P(x)P(x) is divided by (x−2)(x−2), the remainder is 5. When P(x)P(x) is divided by (x−3)(x−3), the remainder is 4. What is the remainder when P(x)P(x) is divided by (x−2)(x−3)(x−2)(x−3)?

A) x+3x+3
B) −x+7−x+7
C) x−7x−7
D) 2x−12x−1

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Question 4: Coordinate Geometry
The line y=mx+by=mx+b is tangent to the circle x2+y2=25x2+y2=25. If the point of tangency is (3, 4), what is the value of m+bm+b?

A) 1331​
B) 112211​
C) 44
D) 77

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Question 5: Functions
If f(x)=2x−1f(x)=2x−1 and g(f(x))=4×2−2x−1g(f(x))=4x2−2x−1, what is the function g(x)g(x)?

A) g(x)=x2+3x−5g(x)=x2+3x−5
B) g(x)=x2+x−1g(x)=x2+x−1
C) g(x)=4×2−2x−1g(x)=4x2−2x−1
D) g(x)=2×2+xg(x)=2x2+x

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Answers & Detailed Explanations

Question 1: Algebra

Correct Answer: D) x=5x=5

Explanation:

1. Combine the logarithms using the product rule: log⁡2((x−3)(x−1))=3log2​((x−3)(x−1))=3.
2. Convert to exponential form: (x−3)(x−1)=23(x−3)(x−1)=23, so (x−3)(x−1)=8(x−3)(x−1)=8.
3. Expand and simplify: x2−4x+3=8x2−4x+3=8 → x2−4x−5=0x2−4x−5=0.
4. Factor the quadratic: (x−5)(x+1)=0(x−5)(x+1)=0, giving potential solutions x=5x=5 or x=−1x=−1.
5. Check the domain: The arguments of the logarithms (x−3x−3 and x−1x−1) must be positive.
   • For x=5x=5: 5−3=2>05−3=2>0 and 5−1=4>05−1=4>0. Valid.
   • For x=−1x=−1: −1−3=−4<0−1−3=−4<0. Invalid.
     Thus, the only valid solution is x=5x=5.

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Question 2: Geometry

Correct Answer: C) 80%

Explanation:

1. The original volume is Vorig=πr2h=πr2(10)=10πr2Vorig​=πr2h=πr2(10)=10πr2.
2. Apply the changes:
   • New radius: rnew=1.5rrnew​=1.5r
   • New height: hnew=10×0.8=8hnew​=10×0.8=8 cm
3. Calculate the new volume: Vnew=π(1.5r)2×8=π(2.25r2)×8=18πr2Vnew​=π(1.5r)2×8=π(2.25r2)×8=18πr2.
4. Find the ratio: VnewVorig=18πr210πr2=1.8Vorig​Vnew​​=10πr218πr2​=1.8. This means the volume is 1.8 times the original, which is an 80% increase (1.8−1=0.8 or 80%)(1.8−1=0.8 or 80%).

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Question 3: Number Theory & Algebra

Correct Answer: B) −x+7−x+7

Explanation:

1. By the Remainder Theorem:
   • P(2)=5P(2)=5
   • P(3)=4P(3)=4
2. Since we are dividing by a quadratic (x−2)(x−3)(x−2)(x−3), the remainder must be linear. Let the remainder be R(x)=ax+bR(x)=ax+b.
3. We can write: P(x)=(x−2)(x−3)⋅Q(x)+(ax+b)P(x)=(x−2)(x−3)⋅Q(x)+(ax+b).
4. Use the known values:
   • P(2)=0+(2a+b)=5P(2)=0+(2a+b)=5 → 2a+b=52a+b=5 …(1)
   • P(3)=0+(3a+b)=4P(3)=0+(3a+b)=4 → 3a+b=43a+b=4 …(2)
5. Subtract equation (1) from equation (2): (3a+b)−(2a+b)=4−5(3a+b)−(2a+b)=4−5 → a=−1a=−1.
6. Substitute a=−1a=−1 into equation (1): 2(−1)+b=52(−1)+b=5 → −2+b=5−2+b=5 → b=7b=7.
7. Therefore, the remainder is R(x)=−x+7R(x)=−x+7.

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Question 4: Coordinate Geometry

Correct Answer: B) 112211​

Explanation:

1. The point (3, 4) lies on the line, so it must satisfy the line’s equation: 4=m(3)+b4=m(3)+b, or 3m+b=43m+b=4. (Equation A)
2. The radius drawn to the point of tangency is perpendicular to the tangent line.
   • The center of the circle is (0, 0).
   • The slope of the radius to (3, 4) is 4−03−0=433−04−0​=34​.
3. The slope of the tangent line mm must be the negative reciprocal of the radius’s slope for them to be perpendicular.m⋅43=−1  ⟹  m=−34m⋅34​=−1⟹m=−43​
4. Substitute m=−34m=−43​ into Equation A:3(−34)+b=43(−43​)+b=4−94+b=4−49​+b=4b=4+94=164+94=254b=4+49​=416​+49​=425​
5. Calculate m+bm+b:m+b=−34+254=224=112m+b=−43​+425​=422​=211​

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Question 5: Functions

Correct Answer: B) g(x)=x2+x−1g(x)=x2+x−1

Explanation:

1. We are given f(x)=2x−1f(x)=2x−1 and g(f(x))=4×2−2x−1g(f(x))=4x2−2x−1.
2. Let u=f(x)=2x−1u=f(x)=2x−1. Our goal is to find g(u)g(u).
3. Solve for xx in terms of uu: u=2x−1u=2x−1 → u+1=2xu+1=2x → x=u+12x=2u+1​.
4. Substitute this expression for xx into g(f(x))g(f(x)):g(u)=4(u+12)2−2(u+12)−1g(u)=4(2u+1​)2−2(2u+1​)−1
5. Simplify step-by-step:g(u)=4⋅(u+1)24−(u+1)−1g(u)=4⋅4(u+1)2​−(u+1)−1g(u)=(u2+2u+1)−u−1−1g(u)=(u2+2u+1)−u−1−1g(u)=u2+2u+1−u−2g(u)=u2+2u+1−u−2g(u)=u2+u−1g(u)=u2+u−1
6. Since uu is the input, the function is g(x)=x2+x−1g(x)=x2+x−1.