2026 JM01

1. The upward quadratic is nonpositive between its roots.

   $$(x+2)(x-1)\le0\iff -2\le x\le1$$

2. Keep only points in both sets.

   $$B=(-\infty,0]\cup[2,\infty),\quad A\cap B=[-2,0]$$

A: [−2,0].

Checks and common pitfalls: The symbol ≥ includes both boundary points of B.

1. The original equation requires a nonzero denominator.

   $$2y-x\ne0,\quad3x+5y=10y-5x$$

2. If y were zero, x would also be zero, contradicting the denominator condition.

   $$8x=5y\implies \frac xy=\frac58$$

B: 5/8.

Checks and common pitfalls: Do not invert x/y when rearranging the ratio.

1. Rearrange to a linear polynomial that vanishes for every real x.

   $$(3+b)x=a+5$$

2. Both sides must have zero constant discrepancy and zero x coefficient.

   $$b=-3,\ a=-5,\quad a-b=-2$$

C: −2.

Checks and common pitfalls: One solution and infinitely many solutions impose different conditions.

1. Compute the discriminant.

   $$\Delta=(-6)^2-4k=36-4k$$

2. Require positivity and solve.

   $$36-4k>0\iff k<9$$

E: k<9.

Checks and common pitfalls: At k=9 the two roots coincide.

1. Substitute −3, not 3.

   $$R=P(-3)=\sum_{j=0}^{1000}(-3)^j$$

2. Apply the finite geometric-sum formula.

   $$R=\frac{1-(-3)^{1001}}{1-(-3)}=\frac{1+3^{1001}}4$$

D: (1+3¹⁰⁰¹)/4.

Checks and common pitfalls: Degree 1000 gives 1001 terms because the constant term is included.

1. Vieta gives both quantities.

   $$\alpha+\beta=\frac52,\quad\alpha\beta=\frac12$$

2. Subtract twice the product.

   $$\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=\frac{25}4-1=\frac{21}4$$

C: 21/4.

Checks and common pitfalls: The middle term is 2αβ, not αβ.

1. The logarithms require positive bases different from one.

   $$a,b>0,\ a,b\ne1,\quad\log_a b=x-1\ne0$$

2. Use change of base.

   $$\log_b(ab)=\frac{\log_a(ab)}{\log_a b}=\frac{x}{x-1}$$

B: x/(x−1).

Checks and common pitfalls: x=1 would force b=1, which is not a logarithm base.

1. The centre-to-point vector is normal to the tangent.

   $$(x+1)^2+(y-3)^2=25,\ C=(-1,3),\ \overrightarrow{CT}=(-3,-4)$$

2. Use the point-normal form.

   $$3(x+4)+4(y+1)=0\iff3x+4y+16=0$$

D: 3x+4y+16=0.

Checks and common pitfalls: Use the centre, rather than the origin, to obtain the radius direction.

1. Only the even-power and constant terms remain.

   $$f(x)+f(-x)=-16x^2-8$$

2. Evaluate the identity at x=2.

   $$-6+f(-2)=-72\implies f(-2)=-66$$

E: −66.

Checks and common pitfalls: Do not assume a polynomial with mixed powers is odd.

1. Choose the members independently within each case.

   $$N_{3,2}=\binom63\binom42=120,\quad N_{4,1}=\binom64\binom41=60$$

2. The two cases are disjoint, so add.

   $$N=120+60=180$$

A: 180 committees.

Checks and common pitfalls: A committee has no ordering; permutations would overcount.

1. Isolate and square, retaining the nonnegative-root condition.

   $$\sqrt{3x^2-2x+4}=\sqrt{3x^2-4x+1}+1\implies\sqrt{3x^2-4x+1}=x+1$$

2. Square again and test both candidates in the original equation.

   $$3x^2-4x+1=(x+1)^2\implies2x(x-3)=0;\quad x=0:1-2=-1,\ x=3:4-5=-1$$

D: {0,3}.

Checks and common pitfalls: Squaring gives candidates; substitution is still required.

1. The supplied angle differs from x+π/4 by two full tangent periods.

   $$\frac{\sin x+\cos x}{\cos x-\sin x}=-7$$

2. Use the reciprocal quotient; the supplied value excludes the cancelled zero factors.

   $$\frac{1-\sin2x}{\cos2x}=\frac{(\cos x-\sin x)^2}{(\cos x-\sin x)(\cos x+\sin x)}=-\frac17$$

A: −1/7.

Checks and common pitfalls: The requested quotient is the reciprocal, not the same tangent.

1. The nonzero sum ensures a is nonzero.

   $$2a(1+r+r^2)=a(r^3-1)\implies(r-3)(r^2+r+1)=0$$

2. For real r, the quadratic factor is positive; shifting both terms by two multiplies the sum by r².

   $$r=3,\quad a_5+a_7=r^2(a_3+a_5)=9\cdot7=63$$

D: 63.

Checks and common pitfalls: There is no real root of r²+r+1=0.

1. Write the vertex coordinate.

   $$x_v=\frac{a-1}3$$

2. To be decreasing throughout the open interval, the vertex may equal its left endpoint.

   $$\frac{a-1}3\le-1\iff a\le-2$$

B: a≤−2.

Checks and common pitfalls: Equality is allowed: the interval excludes −1.

1. The range of g must include 0 but cannot include any t in (−4,0). A downward half-line through 0 is therefore impossible.

2. The only upward half-line satisfying these conditions starts at 0; its image under f is the desired range.

   $$g(\mathbb R)=[0,\infty),\quad f([0,\infty))=[0,4)\cup[2,\infty)=[0,\infty)$$

C: [0,∞).

Checks and common pitfalls: A disconnected set cannot be the range of a real quadratic.

1. For each shooter, use miss probability plus hit probability times t.

   $$G(t)=\left(\frac14+\frac34t\right)\left(\frac13+\frac23t\right)\left(\frac12+\frac12t\right)$$

2. Expand; the coefficient of t to power k is the probability of score k.

   $$G(t)=\frac{1+6t+11t^2+6t^3}{24}$$

Scores 0,1,2,3 have probabilities 1/24, 1/4, 11/24, 1/4.

Checks and common pitfalls: The three success probabilities differ, so one binomial distribution does not apply to team A.

1. Find B’s probabilities with the binomial formula.

   $$P(B=3)=\frac{27}{64},\quad P(B=2)=\binom32\left(\frac34\right)^2\frac14=\frac{27}{64}$$

2. Independence allows multiplication; the two disjoint cases are then added.

   $$P=\frac1{24}\frac{27}{64}+\frac14\frac{27}{64}=\frac{63}{512}$$

63/512.

Checks and common pitfalls: Total score three alone would also include pairs where A wins.

1. At x=0 the ordinate is 5.

   $$P=(0,5)$$

2. The square vanishes at the vertex.

   $$y=9-(x-2)^2,\quad Q=(2,9)$$

P=(0,5), Q=(2,9).

Checks and common pitfalls: The vertex ordinate is 9, not the constant term 5.

1. Reflect P to P′; distances to points on the x-axis are unchanged.

   $$P'=(0,-5),\quad PM=P'M,\quad P'M+MQ\ge P'Q$$

2. Equality holds where the segment crosses the axis.

   $$P'Q:\ y=7x-5,\quad M=\left(\frac57,0\right)$$

3. Compute the straight-line distance.

   $$\min(PM+QM)=\sqrt{2^2+14^2}=10\sqrt2$$

Minimum 10√2, attained at M=(5/7,0).

Checks and common pitfalls: Connecting P directly to Q does not cross the x-axis and cannot realise the constrained path.

1. Solve the two linear constraints.

   $$a_1+4d=44,\quad\frac52(a_1+44)=140\implies a_1=12,\ d=8$$

2. Substitute into the nth-term formula.

   $$a_n=12+8(n-1)=8n+4$$

aₙ=8n+4 for n≥1.

Checks and common pitfalls: The fifth term is four differences after the first.

1. Find and decompose the reciprocal partial sum.

   $$S_n=\frac n2(12+8n+4)=4n(n+2),\quad b_n=\frac18\left(\frac1n-\frac1{n+2}\right)$$

2. All interior terms cancel, leaving two initial and two terminal terms.

   $$T_n=\frac18\left(1+\frac12-\frac1{n+1}-\frac1{n+2}\right)=\frac3{16}-\frac18\left(\frac1{n+1}+\frac1{n+2}\right)$$

Tₙ=3/16−(1/8)(1/(n+1)+1/(n+2)).

Checks and common pitfalls: A gap of two in the denominators leaves two terminal terms.

1. Positive summands make the partial sums increasing, with minimum at n=1.

   $$T_n\ge T_1=b_1=\frac1{4\cdot1\cdot3}=\frac1{12}$$

2. For every finite positive n the remainder is strictly positive.

   $$\frac18\left(\frac1{n+1}+\frac1{n+2}\right)>0\implies T_n<\frac3{16}$$

The lower bound is attained only at n=1; the upper bound is never attained.

Checks and common pitfalls: A limit need not be an attained maximum of the sequence.

1. Square the eccentricity relation.

   $$\frac{c^2}{a^2}=\frac32,\quad b^2=c^2-a^2=\frac{a^2}2$$

2. Substitute the point and solve for positive semiaxes.

   $$\frac{12}{a^2}-\frac4{a^2/2}=1\implies a^2=4,\ b^2=2$$

x²/4−y²/2=1.

Checks and common pitfalls: For a hyperbola c² is a²+b², unlike the corresponding ellipse relation.

1. Substitution gives two real roots and their symmetric sums.

   $$x^2-2(x-6)^2=4\implies x^2-24x+76=0,\quad x_1+x_2=24,\ x_1x_2=76$$

2. Neither root equals 2, so both slopes exist.

   $$m_1m_2=\frac{(x_1-6)(x_2-6)}{(x_1-2)(x_2-2)}=\frac{76-6(24)+36}{76-2(24)+4}=-1$$

3. A slope product of −1 proves perpendicularity.

AM⊥BM.

Checks and common pitfalls: The two slope denominators use x₁−2 and x₂−2 respectively.

1. Evaluate the first substitution.

   $$x_2=\frac{3(3/2)}{3/2+3}=1$$

2. Apply the same function once more.

   $$x_3=\frac{3\cdot1}{1+3}=\frac34$$

x₂=1, x₃=3/4.

Checks and common pitfalls: The recurrence iterates f; it does not substitute the index n into f.

1. The base case n=1 gives 3/2, as required.

   $$x_1=\frac3{1+1}$$

2. Assume the formula for k, then apply the recurrence.

   $$x_{k+1}=\frac{3\cdot3/(k+1)}{3/(k+1)+3}=\frac3{k+2}$$

3. Thus induction establishes xₙ; applying f increases the index by one.

   $$x_n=\frac3{n+1},\quad f(x_n)=x_{n+1}=\frac3{n+2}\quad(n\ge1)$$

xₙ=3/(n+1); therefore f(xₙ)=3/(n+2).

Checks and common pitfalls: The printed question asks for f(xₙ), whereas the suggested answer proves xₙ. Both are shown to avoid an off-by-one error.