2025 JM01

1. The upward quadratic is nonnegative outside its roots.

   $$(x+4)(x-1)\ge0\iff x\le-4\ \text{or}\ x\ge1$$

2. Filter the four listed elements.

   $$A\cap B=\{-4,3\}$$

A: {−4,3}.

Checks and common pitfalls: The root −4 is included because the inequality is non-strict.

1. Recover the sum of α and β.

   $$\frac1\alpha+\frac1\beta=-1,\quad\frac1{\alpha\beta}=-\frac12\implies\alpha+\beta=2$$

2. Add exponents with the same base.

   $$2^{\alpha+1}2^{\beta+1}=2^{\alpha+\beta+2}=16$$

D: 16.

Checks and common pitfalls: α and β themselves are not the roots of the printed quadratic.

1. Square only the radius multiplier.

   $$\frac{V_{\rm new}}{V_{\rm old}}=(1.3)^2(0.7)=1.183$$

2. Subtract one to obtain the proportional increase.

   $$(1.183-1)\cdot100\%=18.3\%$$

A: increase 18.3%.

Checks and common pitfalls: Equal percentage changes in radius and height do not cancel.

1. Clear fractional powers.

   $$\frac{m^{3/2}-m^{-1/2}}{m^{1/2}+m^{-1/2}}=\frac{m^2-1}{m+1}$$

2. Factor the difference of squares; m+1 is nonzero on the domain.

   $$\frac{(m-1)(m+1)}{m+1}=m-1,\quad m>0$$

C: m−1, with m>0.

Checks and common pitfalls: Simplification does not enlarge the original domain.

1. Identify the known logarithms.

   $$p=\log_2 5,\quad q=\log_2 7$$

2. Split the quotient.

   $$\log_2\frac7{2\cdot5}=q-1-p$$

D: q−p−1.

Checks and common pitfalls: The extra factor 2 in the denominator contributes −1.

1. The upper bound gives an interior interval.

   $$x^2-5x-6<0\iff-1<x<6$$

2. The lower bound excludes the closed interval [2,3].

   $$x^2-5x+6>0\iff x<2\ \text{or}\ x>3\implies x\in(-1,2)\cup(3,6)$$

E: (−1,2)∪(3,6).

Checks and common pitfalls: All four boundary values are excluded.

1. Four girls create five gaps including the two ends.

   $$4!\binom53$$

2. Arrange the three different boys in the selected gaps.

   $$N=4!\binom53\,3!=24\cdot10\cdot6=1440$$

C: 1440.

Checks and common pitfalls: The two end gaps are allowed and must be counted.

1. The middle two values are x−4 and x+2.

   $$\frac{(x-4)+(x+2)}2=x-1=8\implies x=9$$

2. The sum of all six offsets is −6.

   $$\bar x=\frac{6x-6}6=x-1=8$$

C: 8.

Checks and common pitfalls: The unsorted third and fourth entries do not define the median.

1. The x term needs x³y⁵ from the binomial.

   $$\binom85=56$$

2. The other term needs x⁶y²; add the contributions.

   $$\binom82=28,\quad56+28=84$$

B: 84.

Checks and common pitfalls: The x⁻² factor changes the required binomial power.

1. Complete squares and compute the centre-to-fixed-point distance.

   $$(x-2)^2+y^2=4,\quad C=(2,0),\ r=2,\ CP=\sqrt2$$

2. The greatest possible line distance is CP, attained when the line is perpendicular to CP.

   $$\ell_{\min}=2\sqrt{r^2-CP^2}=2\sqrt2$$

E: 2√2.

Checks and common pitfalls: The fixed point lies inside the circle, so a tangent through it is impossible.

1. Identify the directrix using the standard coefficient 4a.

   $$y_{\rm directrix}=\frac p4,\quad y_A=-7$$

2. Apply the focus-directrix definition.

   $$7+\frac p4=15\implies p=32$$

E: 32.

Checks and common pitfalls: The coefficient p here is four times the focal distance parameter.

1. Calculate the required binomial probabilities for each person.

   $$P_J(1,2,3)=\left(\frac49,\frac29,\frac1{27}\right),\quad P_A(1,2,3)=\left(\frac29,\frac49,\frac8{27}\right)$$

2. Multiply within each independent pair, then add.

   $$P=\frac49\frac8{27}+\frac29\frac49+\frac1{27}\frac29=\frac{58}{243}$$

D: 58/243.

Checks and common pitfalls: The six shots do not share one success probability.

1. Write and substitute the three sums.

   $$S_2=8+d,\quad S_3=12+3d,\quad S_4=16+6d$$

2. Solve for d and use the seventh-term formula.

   $$2(12+3d)^2=3(8+d)(16+6d)\implies d=-2,\quad a_7=4+6d=-8$$

B: −8.

Checks and common pitfalls: S₃² means the square of the third partial sum.

1. Eliminate x from an upper bound for the objective.

   $$3x+y\le7-3y\le10$$

2. Both bounds attain equality at a feasible point.

   $$(x,y)=\left(\frac{11}3,-1\right),\quad x-2y=\frac{17}3\ge-1$$

D: maximum 10.

Checks and common pitfalls: An upper bound is a maximum only after a feasible equality case is found.

1. The translated formula is a V-shape centred at 1.

   $$f(x)=1+|x-1|\ (0\le x\le2),\quad f(0)=2,\ f(1)=1$$

2. Evaluate the relevant half-angle inputs.

   $$f(1/2)=f(-1/2)=\frac32,\quad f(\sqrt3/2)=2-\frac{\sqrt3}2<\frac32$$

3. Thus B reverses the correct inequality; A,C,D,E hold.

B is the incorrect inequality.

Checks and common pitfalls: The function is periodic but not increasing on each whole period.

1. Subtract consecutive partial sums; the same formula also gives a₁=3.

   $$a_n=(n^2+2n)-[(n-1)^2+2(n-1)]=2n+1$$

2. Use a₄=9 to determine the ratio.

   $$2r^2=18,\ r>0\implies r=3,\quad b_n=2\cdot3^{n-1}$$

aₙ=2n+1; bₙ=2·3ⁿ⁻¹.

Checks and common pitfalls: The negative square root is excluded by the given positive ratio.

1. Rewrite each summand as a difference.

   $$c_k=2k3^k-2(k-1)3^{k-1}$$

2. Sum from k=1 to n; all internal terms cancel.

   $$T_n=\sum_{k=1}^n[2k3^k-2(k-1)3^{k-1}]=2n3^n$$

Tₙ=2n·3ⁿ.

Checks and common pitfalls: A product sequence is not obtained by multiplying the two partial sums.

1. The extreme coefficients determine p and r.

   $$2p=8,\quad3r=9\implies p=4,\ r=3$$

2. The remainder theorem determines q.

   $$f(-1)=(4+3+3)(2-q+3)=10(5-q)=-10\implies q=6$$

p=4, q=6, r=3.

Checks and common pitfalls: Dividing by x+1 means substituting −1.

1. The first factor has no real zero.

   $$\Delta_1=9-48=-39<0$$

2. Solve the second factor by the quadratic formula.

   $$x=\frac{-6\pm\sqrt{36-24}}4=\frac{-3\pm\sqrt3}2$$

The real roots are (−3±√3)/2.

Checks and common pitfalls: A degree-four polynomial need not have four real roots.

1. Convert sin C to half-angle factors.

   $$2\sin\frac C2\cos\frac C2=6\sin^2\frac C2\implies\tan\frac C2=\frac13$$

2. Apply the tangent half-angle expression for cosine.

   $$\cos C=\frac{1-\tan^2(C/2)}{1+\tan^2(C/2)}=\frac45$$

cos C=4/5.

Checks and common pitfalls: C=0 is excluded because C is an interior angle of a nondegenerate triangle.

1. Double B=π−A−C.

   $$2B=\frac{3\pi}2-2C,\quad\sin2B=-\cos2C$$

2. Substitute the known cosine into the double-angle identity.

   $$\sin2B=1-2\cos^2C=1-2\left(\frac45\right)^2=-\frac7{25}$$

sin(2B)=−7/25.

Checks and common pitfalls: The factor 2 in 1−2cos²C must be retained.

1. E,G,B are collinear, and F,E,A lie on one ray from F.

   $$\angle DEG=\angle DEB=\angle AFD=\angle DFE$$

2. D,G,F lie on one ray from D, giving the second equal angle.

   $$\angle EDG=\angle FDE\implies\triangle DEG\sim\triangle DFE\quad(AA)$$

△DEG∼△DFE by AA.

Checks and common pitfalls: Preserve the correspondence D↔D, E↔F, G↔E when using ratios.

1. Equal base angles in ADE give equal supplementary angles.

   $$\angle ADE=\angle AED\implies\angle BDE=\angle DEF$$

2. The given angle gives the other equal pair.

   $$\angle DFE=\angle AFD=\angle DEB\implies\triangle DEF\sim\triangle BDE$$

△DEF∼△BDE by AA.

Checks and common pitfalls: The exterior angles here are supplements of the equal base angles.

1. The first similarity gives a mean-proportional relation.

   $$\frac{DE}{DF}=\frac{DG}{DE}\implies DE^2=DG\cdot DF$$

2. The second similarity expresses the same square.

   $$\frac{DE}{DB}=\frac{EF}{DE}\implies DE^2=DB\cdot EF$$

3. Both expressions equal DE².

   $$DG\cdot DF=DB\cdot EF$$

DG·DF=DB·EF.

Checks and common pitfalls: Ratios must follow the vertex correspondence, not visual side lengths.

1. Both slopes exist only away from the two endpoint abscissas.

   $$\frac y{x+2\sqrt2}\frac y{x-2\sqrt2}=-\frac12,\quad x\ne\pm2\sqrt2$$

2. Clear the nonzero denominator and rearrange.

   $$2y^2=8-x^2\implies\frac{x^2}8+\frac{y^2}4=1,\quad M\ne(\pm2\sqrt2,0)$$

Ellipse x²/8+y²/4=1 with the two vertices (±2√2,0) excluded under the literal slope condition.

Checks and common pitfalls: Clearing denominators can add points at which the original slopes were undefined.

1. Substitution gives a quadratic for the two intersection abscissas.

   $$(1+2k^2)x^2+4kbx+2b^2-8=0\implies x_D=-\frac{2kb}{1+2k^2}$$

2. Recover y_D and divide; k,b nonzero ensure x_D is nonzero.

   $$y_D=kx_D+b=\frac b{1+2k^2},\quad m_{OD}=\frac{y_D}{x_D}=-\frac1{2k}$$

Slope −1/(2k).

Checks and common pitfalls: The midpoint need not lie on the ellipse; it lies on the secant.