2026 JM02

1. Choose coordinates consistent with the original vertex labels.

   $$E=(0,0,0),\ C=(1,1,1),\ B=(1,0,1),\ D=(0,1,1),\ G=(1,1,0)$$

2. Write M=(t,t,t) and impose the perpendicularity condition.

   $$(G-M)\cdot(1,1,1)=2-3t=0\implies M=\left(\frac23,\frac23,\frac23\right)$$

3. All three triangles have the same three side lengths, so SSS applies.

   $$CG=CB=CD=1,\quad CM=\frac1{\sqrt3},\quad MG=MB=MD=\sqrt{\frac23}$$

△CMG≅△CMB≅△CMD by SSS.

Checks and common pitfalls: The diagram is not to scale; equality comes from cube geometry or coordinates.

1. The three noncollinear vertices determine the plane.

   $$BDG:\ x+y+z=2,\quad\mathbf n=(1,1,1)$$

2. M satisfies the plane equation and CM is parallel to its normal.

   $$\frac23+\frac23+\frac23=2,\quad\overrightarrow{CM}=-\frac13(1,1,1)$$

M∈plane BDG and CM⊥plane BDG.

Checks and common pitfalls: Perpendicularity to one line alone does not establish perpendicularity to a plane.

1. M is in the base plane and equidistant from its three vertices, so the radius is MG.

   $$r^2=MG^2=\frac23$$

2. The body diagonal passes perpendicularly through the base centre.

   $$h=EM=\sqrt{3\left(\frac23\right)^2}=\frac{2\sqrt3}3$$

3. Apply the cone-volume formula.

   $$V=\frac13\pi r^2h=\frac13\pi\frac23\frac{2\sqrt3}3=\frac{4\sqrt3\pi}{27}$$

4√3π/27 cubic units.

Checks and common pitfalls: EC is the whole body diagonal, not the cone height EM.

1. Differentiate the cubic.

   $$f'(x)=3x^2-2x-1$$

2. Differentiate the quadratic derivative.

   $$f''(x)=6x-2$$

f′=3x²−2x−1; f″=6x−2.

Checks and common pitfalls: The derivative of −x is −1.

1. Locate the stationary points.

   $$f'(x)=(3x+1)(x-1)=0\iff x=-\frac13,1$$

2. The derivative signs are +,−,+, hence a maximum then a minimum.

   $$f\left(-\frac13\right)=\frac5{27},\quad f(1)=-1$$

Local maximum 5/27 at x=−1/3; local minimum −1 at x=1.

Checks and common pitfalls: A stationary x-coordinate is not itself the extremum value.

1. The second derivative is negative to the left and positive to the right.

   $$f''(x)=6x-2=0\iff x=\frac13$$

2. Evaluate the original function, not its derivative.

   $$f\left(\frac13\right)=\frac1{27}-\frac19-\frac13=-\frac{11}{27}$$

Inflection point (1/3,−11/27).

Checks and common pitfalls: f″=0 alone does not guarantee an inflection point without a change of concavity.

1. Factor to find all three intercepts.

   $$f(x)=x(x^2-x-1),\quad x=0,\ \frac{1-\sqrt5}2,\ \frac{1+\sqrt5}2$$

2. Plot the closed endpoints and the previously found critical points.

   $$f(-2)=-10,\ f(2)=2,\quad\left(-\frac13,\frac5{27}\right),\ (1,-1),\ \left(\frac13,-\frac{11}{27}\right)$$

3. Increase to x=−1/3, decrease to x=1, then increase; change concavity at x=1/3.

The original generated plot shows the complete graph on [−2,2].

Checks and common pitfalls: Do not extend the answer beyond the requested interval.

1. Write an even expression for the transformed curve.

   $$y=-|x|^3+x^2+|x|$$

2. Use the right half of the old curve and its mirror; there are peaks at ±1 and a corner at the origin.

   $$(0,0),\ (\pm1,1),\ (\pm2,-2),\quad x\text{-intercepts}:0,\ \pm\frac{1+\sqrt5}2$$

The graph is even, with maxima at (±1,1) and endpoints (±2,−2).

Checks and common pitfalls: f(|x|) copies only the original right half, not the original left half.

1. Find the three intersection abscissas.

   $$f(x)-x=x(x-2)(x+1)=0\implies x=-1,0,2$$

2. Integrate upper minus lower separately on the two intervals.

   $$A=\int_{-1}^0(x^3-x^2-2x)\,dx+\int_0^2(-x^3+x^2+2x)\,dx$$

3. Evaluate the two positive areas and add.

   $$A=\left[\frac{x^4}4-\frac{x^3}3-x^2\right]_{-1}^0+\left[-\frac{x^4}4+\frac{x^3}3+x^2\right]_0^2=\frac5{12}+\frac83=\frac{37}{12}$$

37/12 square units.

Checks and common pitfalls: One signed integral over [−1,2] would cancel part of the area.

1. Find the focus and the circle equation.

   $$a^2=9-4=5,\quad a=\sqrt5,\quad x^2+y^2=5$$

2. Eliminate y² and choose positive coordinates.

   $$4x^2+9y^2=36,\quad5x^2=9,\quad Q=\left(\frac{3\sqrt5}5,\frac{4\sqrt5}5\right)$$

a=√5; Q=(3√5/5,4√5/5).

Checks and common pitfalls: Here a names the focal distance in the question, not the major semiaxis.

1. Compute the line.

   $$m=-2,\quad y=-2x+2\sqrt5$$

2. Substitute into the ellipse, then select the other root.

   $$5x^2-9\sqrt5x+18=0,\quad x_Q+x_M=\frac{9\sqrt5}5\implies x_M=\frac{6\sqrt5}5$$

3. Recover the ordinate from the line.

   $$y_M=-\frac{2\sqrt5}5$$

M=(6√5/5,−2√5/5).

Checks and common pitfalls: Returning Q again does not satisfy the requested distinct intersection.

1. Find the two slopes and check the ray dot product is positive.

   $$m_{MQ}=-2,\quad m_{MF_1}=-\frac2{11},\quad\overrightarrow{MQ}\cdot\overrightarrow{MF_1}=9>0$$

2. Apply the tangent of the angle between lines.

   $$\tan\angle QMF_1=\left|\frac{-2+2/11}{1+4/11}\right|=\frac43$$

tan∠QMF₁=4/3.

Checks and common pitfalls: An absolute-value slope formula gives an acute line angle; check the requested ray angle.

1. Use slope −2/11 and point F₁.

   $$MF_1:\ 2x+11y+2\sqrt5=0$$

2. Substitute Q and normalise the normal vector.

   $$d=\frac{|2(3\sqrt5/5)+11(4\sqrt5/5)+2\sqrt5|}{\sqrt{2^2+11^2}}=\frac{12\sqrt5}{5\sqrt5}=\frac{12}5$$

Distance 12/5.

Checks and common pitfalls: The distance denominator is √(A²+B²), not A²+B².

1. The discriminant is −1.

   $$z=\frac{\sqrt3\pm i}{2}$$

2. Identify cosine and sine coordinates on the unit circle.

   $$z=\cos\left(\pm\frac\pi6\right)+i\sin\left(\pm\frac\pi6\right)$$

Arguments π/6 and −π/6, both of modulus one.

Checks and common pitfalls: The ± sign changes the sine term as well as the angle.

1. Apply the quadratic result to z².

   $$z^2=e^{i\pi/6}\ \text{or}\ e^{-i\pi/6}$$

2. Halve the arguments after allowing an extra full turn, then reduce to the principal interval.

   $$z=\cos\theta+i\sin\theta,\quad\theta\in\left\{\frac\pi{12},-\frac\pi{12},\frac{11\pi}{12},-\frac{11\pi}{12}\right\}$$

Four roots with arguments ±π/12 and ±11π/12.

Checks and common pitfalls: Halving only the two principal arguments would omit two roots.

1. For the pair with arguments ±π/12, the sum and product give the first factor.

   $$(z-e^{i\pi/12})(z-e^{-i\pi/12})=z^2-2z\cos\frac\pi{12}+1$$

2. The other pair has cosine −cos(π/12). Multiplying the two monic factors accounts for all four roots.

   $$z^4-\sqrt3z^2+1=\left(z^2-2z\cos\frac\pi{12}+1\right)\left(z^2+2z\cos\frac\pi{12}+1\right)$$

The required product follows by pairing conjugate roots.

Checks and common pitfalls: Do not assume the half-angle value that the next part asks you to prove.

1. Let c=cos(π/12); the product has this middle coefficient.

   $$(z^2+1)^2-4c^2z^2=z^4+(2-4c^2)z^2+1$$

2. Match coefficients and choose the positive square root because π/12 is acute.

   $$2-4c^2=-\sqrt3\implies c=\frac{\sqrt{2+\sqrt3}}2$$

cos(π/12)=√(2+√3)/2.

Checks and common pitfalls: The outer square root covers 2+√3 together.

1. After row subtraction, factor b−a and c−a from the last two rows.

   $$D=(b-a)(c-a)\big[(c^3+ac^2+a^2c+a^3)-(b^3+ab^2+a^2b+a^3)\big]$$

2. Factor the bracket by c−b. The polynomial identity also holds when two parameters coincide.

   $$D=(b-a)(c-a)(c-b)(a^2+b^2+c^2+ab+ac+bc)$$

(b−a)(c−a)(c−b)(a²+b²+c²+ab+ac+bc).

Checks and common pitfalls: Reversing one difference changes the sign; keep the row order consistent.

1. Elimination gives the relation between x and z.

   $$2x-2z=2\implies x=z+1$$

2. Set z=t and recover y from the first equation.

   $$(x,y,z)=(1+t,1-2t,t),\quad t\in\mathbb R$$

(x,y,z)=(1+t,1−2t,t), t∈ℝ.

Checks and common pitfalls: Two independent equations in three unknowns leave one free parameter.

1. Expand after substitution.

   $$a(1+t)^2+b(1-2t)^2+ct^2=(a+4b+c)t^2+(2a-4b)t+(a+b)$$

2. Match coefficients with the constant 3 and solve.

   $$a+4b+c=0,\quad2a-4b=0,\quad a+b=3\implies(a,b,c)=(2,1,-6)$$

a=2, b=1, c=−6.

Checks and common pitfalls: Checking one parameter value is insufficient when every solution is required.

1. Substitute and solve the scalar quadratic.

   $$(1+t)^2+(1-2t)^2+t^2=3\iff6t^2-2t-1=0\implies t=\frac{1\pm\sqrt7}6$$

2. Use matching signs in x and z and the opposite sign in y.

   $$(x,y,z)=\left(\frac{7\pm\sqrt7}6,\frac{4\mp2\sqrt7}6,\frac{1\pm\sqrt7}6\right)$$

Exactly two triples, corresponding to the consistent upper and lower signs.

Checks and common pitfalls: The ± choices are linked; they do not produce eight independent triples.