2023 JM01

1. Solve the quadratic inequality.

   $$(x-4)(x+2)\ge0\iff x\le-2\text{ or }x\ge4$$

2. Intersect with the open interval.

   $$M\cap N=[4,6)$$

E: [4,6).

Checks and common pitfalls: The endpoint 4 belongs; 6 does not.

1. Introduce the polynomial quotient q.

   $$f(x)=(x^2-x-6)q(x)+3x-2$$

2. Substitute the root of the divisor.

   $$f(3)=0q(3)+9-2=7$$

D: 7.

Checks and common pitfalls: The remainder is evaluated at 3; it is not itself f(3).

1. The middle reciprocal pair multiplies to one.

   $$\log_{12}17\log_{17}12=1$$

2. Convert the remaining pair to natural logarithms.

   $$\frac{3\ln5}{2\ln3}\frac{\ln3}{2\ln5}=\frac34$$

C: 3/4.

Checks and common pitfalls: Apply powers to both bases and arguments.

1. Factor the equation in t.

   $$t^2+4t-12=(t-2)(t+6)=0\implies t=2$$

2. Solve the remaining quadratic and check its radicand.

   $$x^2-3x=4\iff(x-4)(x+1)=0$$

C: {−1,4}.

Checks and common pitfalls: Reject t=−6 because t is a principal square root.

1. Set the discriminant equal to zero.

   $$4(a+3)^2-144a^2=0\iff(a+3)^2=36a^2$$

2. Solve the two linear equations; both satisfy a≠0.

   $$a+3=\pm6a\implies a=3/5\text{ or }-3/7$$

D: a=−3/7 or 3/5.

Checks and common pitfalls: a=0 gives a linear equation and violates the stated quadratic condition.

1. Write the general term.

   $$\binom6k2^{6-k}(-1)^kx^{3-k}$$

2. Choose k=3 and evaluate.

   $$\binom632^3(-1)^3=-160$$

C: −160.

Checks and common pitfalls: The odd number of negative factors makes the term negative.

1. Differentiate.

   $$f'(x)=2ax+4$$

2. In the negative-a case require 8a+4≥0; equality is allowed because 4 is excluded.

   $$a\ge-1/2$$

D: a≥−1/2.

Checks and common pitfalls: The endpoint case a=−1/2 is still strictly increasing on the open interval.

1. Use the logarithmic branch with its domain.

   $$f(5)=2,\quad\log_2(1/2-3|x|)>-2$$

2. Since the base is greater than one, preserve the inequality direction.

   $$1/2-3|x|>1/4\iff|x|<1/12$$

A: −1/12<x<1/12.

Checks and common pitfalls: The positive logarithm argument is automatically enforced by the stronger final bound.

1. Find the displaced volume.

   $$V=\frac43\pi(2)^3=\frac{32\pi}3$$

2. Convert volume to height and check no overflow.

   $$\Delta h=\frac{32\pi/3}{9\pi}=\frac{32}{27},\quad5+32/27<8$$

E: 32/27 m.

Checks and common pitfalls: Use the tank area, not the sphere’s great-circle area.

1. Find d.

   $$d=(26-80)/(16-7)=-6$$

2. Evaluate the required term.

   $$a_{34}=26+18(-6)=-82$$

B: −82.

Checks and common pitfalls: Use index differences, not the indices themselves.

1. The midpoint is (−2,−2); the required slope is −4/3.

   $$m_0=3/4,\quad m=-4/3$$

2. Use point-slope form.

   $$y+2=-\frac43(x+2)\iff4x+3y+14=0$$

A.

Checks and common pitfalls: The perpendicular slope changes sign as well as taking a reciprocal.

1. Use c²=a²+b² and c/a=3.

   $$b^2=8a^2,\quad\frac{b^2+2}{a}=8a+\frac2a$$

2. The product is constant and equality is attainable.

   $$8a+2/a\ge2\sqrt{16}=8,\quad8a=2/a\iff a=1/2$$

E: 8.

Checks and common pitfalls: The numerator is b²+2 as a whole.

1. Determine the cosines with the quadrant signs.

   $$\cos A=-\sqrt{21}/5,\quad\cos B=-3/5$$

2. Substitute into the addition formula.

   $$\sin(A+B)=\frac25\left(-\frac35\right)+\left(-\frac{\sqrt{21}}5\right)\frac45=\frac{-6-4\sqrt{21}}{25}$$

A.

Checks and common pitfalls: Positive sine does not imply positive cosine in quadrant II.

1. Recover the midline and amplitude.

   $$b=(4+0)/2=2,\quad|a|=(4-0)/2=2$$

2. Use the maximum point to determine the sign.

   $$4=a\sin(3\pi/2)+2=-a+2\implies a=-2$$

B: a=−2, b=2.

Checks and common pitfalls: Amplitude gives |a|, not its sign.

1. Rotate and reflect.

   $$(-2,3)\to(3,2)\to(3,-2)$$

2. Subtract 3 from the final y-coordinate.

   $$(3,-2)\to(3,-5)$$

E: (3,−5).

Checks and common pitfalls: Changing the transformation order changes the result.

1. Write the two disjoint cases.

   $$P=(3/4)^{10}+\binom{10}1(1/4)(3/4)^9$$

2. Factor the common term.

   $$P=\frac{13}4(3/4)^9$$

Probability (13/4)(3/4)⁹.

Checks and common pitfalls: “At most” includes zero, unlike “exactly one”.

1. Specify the unique outcome pattern.

   $$T^9H$$

2. Multiply the independent probabilities.

   $$P=(3/4)^9(1/4)=3^9/4^{10}$$

3⁹/4¹⁰.

Checks and common pitfalls: No binomial coefficient is needed because the head position is fixed.

1. Count the choices for the earlier two heads.

   $$P=\binom92(1/4)^2(3/4)^7(1/4)$$

2. Simplify the powers.

   $$P=\frac{36\cdot3^7}{4^{10}}=(3/4)^9$$

(3/4)⁹.

Checks and common pitfalls: Choosing three heads anywhere in ten does not enforce the last-head time.

1. Identify the focal parameter.

   $$4p=4\implies p=1$$

2. Place the focus p units above the vertex.

   $$F=(0,p)=(0,1)$$

F=(0,1).

Checks and common pitfalls: The coefficient 4 is four times the focal distance.

1. Find the two points.

   $$x^2=3x+4\implies x=-1,4;\quad A=(-1,1/4),\ B=(4,4)$$

2. Use the distance formula.

   $$AB=\sqrt{5^2+(15/4)^2}=25/4$$

AB=25/4.

Checks and common pitfalls: The horizontal difference alone is not the chord length.

1. Vieta gives the sum, and the diagram gives the signed ratio.

   $$x^2-4x-4t=0,\quad x_C+x_D=4,\quad x_D=-3x_C$$

2. Solve and convert the horizontal span to chord length.

   $$x_C=-2,\ x_D=6,\quad CD=\sqrt{1+1^2}(6+2)=8\sqrt2$$

CD=8√2.

Checks and common pitfalls: The diagram’s betweenness condition is essential; unsigned lengths alone allow another configuration.

1. The first three terms are 9−2k,18,54.

   $$18^2=54(9-2k)\implies k=3/2$$

2. Identify first term 6 and ratio 3.

   $$a_n=6\cdot3^{n-1}=2\cdot3^n$$

k=3/2; aₙ=2·3ⁿ.

Checks and common pitfalls: The difference formula for n≥2 cannot determine a₁ without checking S₁.

1. Expand the logarithm.

   $$b_n=\frac1{2\cdot3^n}+1+n\log_2 3$$

2. Sum each component.

   $$T_n=\frac14(1-3^{-n})+n+\frac{n(n+1)}2\log_2 3$$

Tₙ=(1−3⁻ⁿ)/4+n+n(n+1)log₂3/2.

Checks and common pitfalls: The first reciprocal term is 1/6, not 1/2.

1. The continuous maximum is at c=1/10.

   $$-5c^2+c=1/20-5(c-1/10)^2$$

2. Of the decreasing values 1/3,1/9,1/27,…, 1/9 is closest to 1/10.

   $$|1/9-1/10|=1/90<1/10-1/27,\quad n=2,\ f(2)=4/81$$

n=2; the maximum is 4/81.

Checks and common pitfalls: The continuous optimum need not be an available sequence value.

1. Combine the sine and cosine.

   $$f(x)=\sqrt3\sin(2wx)-\cos(2wx)-1=2\sin(2wx-\pi/6)-1$$

2. The period determines two possible signs.

   $$\frac{2\pi}{|2w|}=3\pi\implies w=\pm1/3,\quad f(x)=2\sin(\pm2x/3-\pi/6)-1$$

w=±1/3; the official answer selects the positive branch.

Checks and common pitfalls: A period alone determines |w|, not its sign.

1. For w<0 the sine argument is in (−5π/6,−π/6), so f(C)<0. Hence w=1/3.

2. Solve the positive-branch equation in its allowed argument range.

   $$\sin(2C/3-\pi/6)=1/2,\quad-\pi/6<2C/3-\pi/6<\pi/2\implies C=\pi/2$$

3. Reduce to a quadratic in sin A and keep its positive root.

   $$\sin^2B=\sin A\implies1-\sin^2A=\sin A\implies\sin A=\frac{\sqrt5-1}2$$

sin A=(√5−1)/2.

Checks and common pitfalls: The square in sin²B denotes a power, not sin(2B).

1. Solve for the three vertices.

   $$A=(2,7/2),\quad B=(5,13/2),\quad C=(5,-1)$$

2. The region is the closed triangle above y=(13−3x)/2, below y=x+3/2 and left of x=5.

The closed triangle with vertices (2,7/2), (5,13/2), (5,−1).

Checks and common pitfalls: All three inequalities are non-strict, so all edges belong.

1. For each x, bound the ratio.

   $$\frac{13}{2x}-\frac32\le\frac yx\le1+\frac{3}{2x},\quad2\le x\le5$$

2. Both bounds decrease with x; extrema occur at C and A.

   $$z_{\min}=\frac{13}{10}-\frac32=-\frac15,\quad z_{\max}=1+\frac34=\frac74$$

−1/5≤z≤7/4.

Checks and common pitfalls: A ratio is not a linear objective; justify the range using x>0 or slopes.

1. Bound the squared distance.

   $$13(x^2+y^2)\ge(3x+2y)^2\ge169\implies t\ge13$$

2. Equality requires (x,y) proportional to (3,2), with 3x+2y=13; the resulting point is feasible.

   $$(x,y)=(3,2),\quad x\le5,\quad2x-2y+3=5\ge0$$

Minimum 13 at (3,2).

Checks and common pitfalls: The nearest point may lie on an edge rather than at a vertex.