2024 JM01

1. The quadratic is nonpositive between its roots.

   $$A=[-1,4],\quad B=[-a/3,\infty)$$

2. The intersection begins at 2, so −a/3=2.

   $$a=-6$$

B: a=−6.

Checks and common pitfalls: Dividing by positive 3 does not reverse the inequality.

1. Each unit increase of the argument lowers the value by one.

   $$f(x+1)=f(x)-1$$

2. Iterate from zero.

   $$f(15)=16-15=1$$

B: 1.

Checks and common pitfalls: The recurrence decreases, rather than increases, the value.

1. Cancel the xy terms.

   $$4x+5y=2x+y-1\implies y=-\frac x2-\frac14$$

2. Multiply the slope by the change in x.

   $$\Delta y=-\frac12(4)=-2$$

C: y decreases by 2.

Checks and common pitfalls: The minus sign outside the product changes both inner signs.

1. Collect the required coefficients in both factors.

   $$[x^0](\sqrt x-2)^5=-32,\quad[x](\sqrt x-2)^5=\binom52(-2)^3=-80$$

2. The second factor has constant 1 and x coefficient −8.

   $$[x]=(-32)(-8)+(-80)(1)=176$$

C: 176.

Checks and common pitfalls: A √x term cannot combine with an integer power to produce x.

1. Square the radius.

   $$r^2=2^2+3^2=13$$

2. Expand the standard circle equation.

   $$(x-2)^2+(y-3)^2=13\iff x^2+y^2-4x-6y=0$$

D.

Checks and common pitfalls: The constant cancels because the circle passes through (0,0).

1. Simplify the numerator.

   $$3\log(1/2)+\log16=(-3+4)\log2=\log2$$

2. Simplify the denominator and divide.

   $$\log4+\log5-1=\log(20/10)=\log2\ne0$$

A: 1.

Checks and common pitfalls: The logarithm is of 1/2; it is not log 12.

1. Divide the two nonzero sums.

   $$r^5=288/9=32\implies r=2$$

2. Find a₂ and advance eighteen places.

   $$a_2(1+2^3)=9\implies a_2=1,\quad a_{20}=2^{18}=262144$$

D: 262144.

Checks and common pitfalls: From the second to the twentieth term there are eighteen ratios.

1. Solve for n.

   $$13n+8=5(6.8)=34\implies n=2$$

2. Take the central sorted entry.

   $$-1,2,4,9,20\implies\operatorname{median}=4$$

A: 4.

Checks and common pitfalls: The third listed expression need not be the median before sorting.

1. The numerator is a perfect square.

   $$1+\frac{(m^4-1)^2}{4m^4}=\frac{(m^4+1)^2}{4m^4}$$

2. Both m⁴+1 and 2m² are positive.

   $$\sqrt{\frac{(m^4+1)^2}{4m^4}}=\frac{m^4+1}{2m^2}=\frac{m^2}2+\frac1{2m^2}$$

C.

Checks and common pitfalls: Square roots return the nonnegative value; m=0 is excluded.

1. Find the cosine.

   $$\cos C=\sqrt{1-48/49}=1/7$$

2. Let a=BC>0 and solve.

   $$64=a^2+49-2(7)a(1/7)\implies(a-5)(a+3)=0\implies a=5$$

E: 5.

Checks and common pitfalls: The acute-angle condition selects the positive cosine.

1. Determine the leading coefficient.

   $$y=a(x+2)(x-6),\quad4=-12a\implies a=-1/3$$

2. Evaluate at the symmetry axis.

   $$x=2,\quad y=-\frac13(4)(-4)=\frac{16}3$$

B: 16/3.

Checks and common pitfalls: The negative leading coefficient makes the vertex a maximum.

1. Find one complete increasing interval.

   $$\pi<x+\pi/3<2\pi\iff2\pi/3<x<5\pi/3$$

2. Only option C lies wholly inside this interval.

   $$(\pi,3\pi/2)\subset(2\pi/3,5\pi/3)$$

C.

Checks and common pitfalls: An option that overlaps an increasing interval is insufficient.

1. Factor the quadratic in sine.

   $$2\sin^2\theta+\sin\theta-1=(2\sin\theta-1)(\sin\theta+1)=0$$

2. Discard sinθ=−1 and retain both acute/supplementary angles.

   $$\sin\theta=1/2\implies\theta=\pi/6,5\pi/6$$

A.

Checks and common pitfalls: Do not omit the second-quadrant solution.

1. Find x²+x⁻².

   $$x+x^{-1}=3\implies x^2+x^{-2}=9-2=7$$

2. Square again.

   $$x^4+x^{-4}=7^2-2=47$$

B: 47.

Checks and common pitfalls: The product x²·x⁻² is 1, so the cross term is 2.

1. Obtain strict size bounds.

   $$0<a<1/4<1/2<b<1,\quad2/7<1/3\implies c<-1$$

2. Order by absolute arguments.

   $$|c|>b>a>0\implies f(c)=f(|c|)>f(b)>f(a)$$

E.

Checks and common pitfalls: A negative argument can have the largest value because f is even.

1. The two cases are disjoint.

   $$N=\binom32\binom72+\binom33\binom71=63+7=70$$

2. Use the equally likely sample count.

   $$P=\frac{70}{\binom{10}4}=\frac{70}{210}=\frac13$$

Probability 1/3.

Checks and common pitfalls: Sampling is without replacement, so a binomial model is inappropriate.

1. Write the count as a sum.

   $$X=I_1+I_2+I_3+I_4,\quad E(I_j)=3/10$$

2. Linearity does not require independence.

   $$E(X)=4(3/10)=6/5$$

Expected count 6/5.

Checks and common pitfalls: An expected count need not be an integer.

1. Recover the other trigonometric ratios of β.

   $$\sin\beta=\frac{2\sqrt{13}}{13},\quad\tan\beta=2/3$$

2. The denominator is nonzero.

   $$\tan(\alpha+\beta)=\frac{1/5+2/3}{1-(1/5)(2/3)}=1$$

tan(α+β)=1.

Checks and common pitfalls: The tangent addition denominator has a minus sign.

1. Tangent equals one only at π/4 in this range.

   $$\alpha+\beta=\pi/4$$

2. Apply cosine addition.

   $$\cos(\alpha+2\beta)=\frac{\sqrt2}2\left(\frac{3\sqrt{13}}{13}-\frac{2\sqrt{13}}{13}\right)=\frac{\sqrt{26}}{26}$$

√26/26.

Checks and common pitfalls: Do not infer an angle from tangent without its range.

1. Form and factor the equation.

   $$(3+d)^2=3(3+4d)\iff d(d-6)=0$$

2. Retain both valid common differences.

   $$d=0\implies a_n=3;\quad d=6\implies a_n=6n-3$$

aₙ=3 or aₙ=6n−3.

Checks and common pitfalls: A constant sequence is also both arithmetic and geometric.

1. The constant case cannot satisfy the inequality.

   $$S_n=3n:\quad3n\ge12n+36\iff n\le-4$$

2. For the increasing case, solve the quadratic inequality.

   $$S_n=3n^2,\quad n^2-4n-12=(n-6)(n+2)\ge0\implies n\ge6$$

No positive n for aₙ=3; least n=6 for aₙ=6n−3.

Checks and common pitfalls: An answer of 6 alone omits the other valid sequence.

1. Between 3 and 7 the distance sum is 4; outside it is 2|x−5|.

   $$|x-3|+|x-7|=\max(4,2|x-5|)$$

2. Require that sum not exceed 8.

   $$|x-5|\le4\iff1\le x\le9$$

x∈[1,9].

Checks and common pitfalls: Include endpoints because the inequality is non-strict.

1. Reduce to a quadratic and locate its vertex.

   $$g(x)=2x^2+(a-10)x,\quad x_0=(10-a)/4$$

2. For 6≤a≤14 the minimum is the vertex value.

   $$-\frac{(a-10)^2}{8}=-1\implies a=10\pm2\sqrt2$$

3. Outside this range the minimum is at an endpoint; each resulting candidate violates its case.

   $$a<6:\ a-8=-1\implies a=7;\quad a>14:\ 12-a=-1\implies a=13$$

a=10±2√2.

Checks and common pitfalls: The unconstrained vertex formula requires its vertex to be in [−1,1].

1. Relate the two squared parameters.

   $$\frac{b^2}{a^2}=\frac52-1=\frac32$$

2. Solve using the point condition.

   $$\frac8{a^2}-\frac9{(3/2)a^2}=1\implies a^2=2,\ b^2=3$$

x²/2−y²/3=1.

Checks and common pitfalls: For a hyperbola, c²=a²+b².

1. Substitute the line and read the root sum/product.

   $$x^2-4mx-2m^2-6=0,\quad x_1+x_2=4m,\quad x_1x_2=-2m^2-6$$

2. Impose the dot product condition.

   $$0=x_1x_2+(x_1+m)(x_2+m)=2(-2m^2-6)+4m^2+m^2=m^2-12$$

3. Both signs give two distinct intersections.

   $$m=\pm2\sqrt3,\quad\Delta=24m^2+24>0$$

m=±2√3.

Checks and common pitfalls: A perpendicularity condition uses a dot product, not equal slopes.