Intermediate Mathematics Quiz(with Detailed Solution)

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

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?

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)?

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?

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

Answers & Detailed Explanations

Question 1: Algebra

Answer: x=5x=5

Explanation:

Use the logarithm product rule: log⁡b(m)+log⁡b(n)=log⁡b(m⋅n)logb(m)+logb(n)=logb(m⋅n).log⁡2((x−3)(x−1))=3log2((x−3)(x−1))=3
Rewrite the logarithmic equation in exponential form: log⁡2(A)=Blog2(A)=B means A=2BA=2B.(x−3)(x−1)=23(x−3)(x−1)=23(x−3)(x−1)=8(x−3)(x−1)=8
Expand and simplify into a standard quadratic equation.x2−4x+3=8x2−4x+3=8×2−4x−5=0x2−4x−5=0
Factor the quadratic equation.(x−5)(x+1)=0(x−5)(x+1)=0So, x=5x=5 or x=−1x=−1.
Check the domain. The original equation has log⁡2(x−3)log2(x−3) and log⁡2(x−1)log2(x−1). The arguments 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.
  Therefore, the only valid solution is x=5x=5.

Question 2: Geometry

Answer: 80% increase.

Explanation:

Recall the formula for the volume of a cylinder: V=πr2hV=πr2h.
Let the original radius be rr and the original height be h=10h=10. The original volume is:Voriginal=πr2(10)=10πr2Voriginal=πr2(10)=10πr2
Apply the changes:
- New radius: rnew=r+0.5r=1.5rrnew=r+0.5r=1.5r
- New height: hnew=10−(0.2×10)=10−2=8hnew=10−(0.2×10)=10−2=8 cm
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
Find the ratio of the new volume to the original volume:VnewVoriginal=18πr210πr2=1810=1.8VoriginalVnew=10πr218πr2=1018=1.8This means the new volume is 1.8 times the original volume.
Calculate the percentage increase:
- A multiplier of 1.8 corresponds to an 80% increase because (1.8−1)×100%=80%(1.8−1)×100%=80%.

Question 3: Number Theory & Algebra

Answer: The remainder is −x+7−x+7.

Explanation:

Use the Remainder Theorem. It states that when a polynomial P(x)P(x) is divided by (x−c)(x−c), the remainder is P(c)P(c).
- P(2)=5P(2)=5
- P(3)=4P(3)=4
We are dividing P(x)P(x) by (x−2)(x−3)(x−2)(x−3), which is a quadratic. This means the remainder must be of a degree less than 2—so it is linear. Let the remainder be R(x)=ax+bR(x)=ax+b.
We can write the polynomial as:P(x)=(x−2)(x−3)⋅Q(x)+(ax+b)P(x)=(x−2)(x−3)⋅Q(x)+(ax+b)where Q(x)Q(x) is the quotient.
Now, use the values we know from the Remainder Theorem.
- For x=2x=2: P(2)=(0)⋅Q(2)+(a⋅2+b)=2a+bP(2)=(0)⋅Q(2)+(a⋅2+b)=2a+b. We know P(2)=5P(2)=5.2a+b=5(Equation 1)2a+b=5(Equation 1)
- For x=3x=3: P(3)=(0)⋅Q(3)+(a⋅3+b)=3a+bP(3)=(0)⋅Q(3)+(a⋅3+b)=3a+b. We know P(3)=4P(3)=4.3a+b=4(Equation 2)3a+b=4(Equation 2)
Solve the system of equations.
Subtract Equation 1 from Equation 2:(3a+b)−(2a+b)=4−5(3a+b)−(2a+b)=4−5a=−1a=−1Substitute a=−1a=−1 into Equation 1:2(−1)+b=52(−1)+b=5−2+b=5−2+b=5b=7b=7
Therefore, the remainder is R(x)=(−1)x+7=−x+7R(x)=(−1)x+7=−x+7.

Question 4: Coordinate Geometry

Answer: m+b=13m+b=31

Explanation:

The point of tangency (3, 4) lies on both the circle and the line.
- Since it’s on the line y=mx+by=mx+b, we can write: 4=m(3)+b4=m(3)+b, or 3m+b=43m+b=4. (Equation A)
A key property of a tangent to a circle is that the radius to the point of tangency is perpendicular to the tangent line.
- The center of the circle x2+y2=25x2+y2=25 is (0, 0).
- The radius to the point (3, 4) has a slope of 4−03−0=433−04−0=34.
For perpendicular lines, the product of their slopes is -1. Let the slope of the tangent line be mm, and the slope of the radius be 4334.m⋅43=−1m⋅34=−1m=−34m=−43
Now 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
Finally, calculate m+bm+b:m+b=−34+254=224=112m+b=−43+425=422=211

Correction: Let’s re-check the perpendicularity calculation.m⋅43=−1 ⟹ m=−34m⋅34=−1⟹m=−43

Substitute into 3m+b=43m+b=4:3(−34)+b=43(−43)+b=4−94+b=4−49+b=4b=4+94=164+94=254b=4+49=416+49=425m+b=−34+254=224=112m+b=−43+425=422=211

The calculation is correct. The initial answer of 1331 was an error. The correct final answer is 112211.

Question 5: Functions

Answer: g(x)=x2+3x−5g(x)=x2+3x−5

Explanation:

We are given f(x)=2x−1f(x)=2x−1 and g(f(x))=4×2−2x−1g(f(x))=4x2−2x−1.
The notation g(f(x))g(f(x)) means we plug f(x)f(x) into the function gg. Let’s set u=f(x)u=f(x).u=2x−1u=2x−1
The goal is to find g(u)g(u), which is the same as g(x)g(x). We need to express g(f(x))g(f(x)) entirely in terms of uu.
From u=2x−1u=2x−1, we can solve for xx:u+1=2xu+1=2xx=u+12x=2u+1
Now, substitute this into the expression for g(f(x))g(f(x)):g(f(x))=4×2−2x−1g(f(x))=4x2−2x−1g(u)=4(u+12)2−2(u+12)−1g(u)=4(2u+1)2−2(2u+1)−1
Simplify the expression step-by-step:g(u)=4⋅(u+1)24−(u+1)−1g(u)=4⋅4(u+1)2−(u+1)−1g(u)=(u+1)2−u−1−1g(u)=(u+1)2−u−1−1g(u)=(u2+2u+1)−u−2g(u)=(u2+2u+1)−u−2g(u)=u2+2u+1−u−2g(u)=u2+2u+1−u−2g(u)=u2+u−1g(u)=u2+u−1
Since uu is just a placeholder, we can write the function gg as:g(x)=x2+x−1g(x)=x2+x−1

Correction: Let’s double-check the simplification.
g(u)=4∗(u+1)24−(u+1)−1g(u)=4∗4(u+1)2−(u+1)−1
g(u)=(u2+2u+1)−u−1−1g(u)=(u2+2u+1)−u−1−1
g(u)=u2+2u+1−u−2g(u)=u2+2u+1−u−2
g(u)=u2+u−1g(u)=u2+u−1
This is correct. The initial answer of x2+3x−5x2+3x−5 was an error. The correct final answer is g(x)=x2+x−1g(x)=x2+x−1.

Final Corrected Answers:

x=5x=5
80%
−x+7−x+7
112211
g(x)=x2+x−1g(x)=x2+x−1

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