2025 JM02

1. The altitude bisects AD.

   $$AM=\frac a2,\quad VA=a,\quad VM\perp AD$$

2. Take the positive square root.

   $$VM=\sqrt{a^2-\left(\frac a2\right)^2}=\frac{\sqrt3}2a$$

VM=√3a/2.

Checks and common pitfalls: Use the altitude, not the sloping side VA, for VM.

1. VM is perpendicular to the planes’ intersection AD within the face, so it is the pyramid height.

   $$h=VM=\frac{\sqrt3}2a,\quad A_{\rm base}=a^2$$

2. Apply the volume formula.

   $$V=\frac13a^2\frac{\sqrt3}2a=\frac{\sqrt3}6a^3$$

Volume √3a³/6.

Checks and common pitfalls: The volume coefficient is 1/3, not 1/2.

1. In equilateral VAD, AP⊥VD. Also VB=BD=√2a, so the median BP of isosceles VBD is perpendicular to VD.

2. Thus ∠APB measures the dihedral angle. AB is perpendicular to face VAD, so triangle APB is right at A.

   $$AP=\frac{\sqrt3}2a,\quad AB=a$$

3. Use the opposite-to-adjacent ratio at P.

   $$\tan x=\frac{AB}{AP}=\frac2{\sqrt3}=\frac{2\sqrt3}3$$

tan x=2√3/3.

Checks and common pitfalls: A dihedral angle requires two lines perpendicular to the same common edge.

1. Compute the base, equilateral face, and two right faces.

   $$A_{ABCD}=a^2,\quad A_{VAD}=\frac{\sqrt3}4a^2,\quad A_{VAB}=A_{VDC}=\frac12a^2$$

2. For VBC, its equal sides are √2a and its base is a.

   $$h_{VBC}=\sqrt{2a^2-\frac{a^2}4}=\frac{\sqrt7}2a,\quad A_{VBC}=\frac{\sqrt7}4a^2$$

3. Sum all five faces.

   $$A_{\rm total}=\left(2+\frac{\sqrt3+\sqrt7}4\right)a^2$$

Total area [2+(√3+√7)/4]a².

Checks and common pitfalls: Total surface area includes the square base.

1. Factor the cubic.

   $$f(x)=(x-1)(x^2+4x+4)=(x-1)(x+2)^2$$

2. Set the factors equal to zero.

   $$x=1\quad\text{or}\quad x=-2\text{ (double root)}$$

x=1 and x=−2; −2 is a double root.

Checks and common pitfalls: A repeated root is one distinct zero but has multiplicity two.

1. Differentiate once.

   $$f'(x)=3x^2+6x$$

2. Differentiate again.

   $$f''(x)=6x+6$$

f′=3x²+6x; f″=6x+6.

Checks and common pitfalls: The derivative of 3x² is 6x.

1. Solve the stationary equation.

   $$f'(x)=3x(x+2)=0\iff x=-2,0$$

2. The derivative signs are +,−,+. Evaluate f at the critical points.

   $$f(-2)=0\text{ (local maximum)},\quad f(0)=-4\text{ (local minimum)}$$

Local maximum 0 at x=−2; local minimum −4 at x=0.

Checks and common pitfalls: The double zero at −2 touches the axis at a local maximum.

1. The second derivative changes from negative to positive at −1.

   $$f''(x)=6(x+1)$$

2. Evaluate its ordinate.

   $$f(-1)=-1+3-4=-2$$

Inflection point (−1,−2).

Checks and common pitfalls: An inflection concerns concavity, not necessarily a stationary point.

1. Mark the interval endpoints and zeros.

   $$f(-3)=-4,\quad f(1.2)=2.048,\quad f(-2)=f(1)=0$$

2. Join smoothly, increasing to (−2,0), decreasing to (0,−4), then increasing; change concavity at (−1,−2).

The plotted cubic touches the axis at −2 and crosses at 1.

Checks and common pitfalls: A double root touches rather than crosses the axis here.

1. Find where their difference vanishes and its sign between the roots.

   $$y_2-y_1=-3(x+2)(x-1)>0\quad(-2<x<1)$$

2. Integrate the positive difference.

   $$A=\int_{-2}^1(-3x^2-3x+6)\,dx=\left[-x^3-\frac32x^2+6x\right]_{-2}^1=\frac{27}2$$

Area 27/2.

Checks and common pitfalls: Subtracting the curves before integration simplifies the calculation.

1. Calculate the slope from A and P.

   $$m=\frac{1-(-1)}{2-0}=1$$

2. Insert A in the point-slope equation.

   $$y+1=x\implies y=x-1$$

y=x−1.

Checks and common pitfalls: The focus coordinates are not needed for this first step.

1. Set y=0 in the line.

   $$0=x-1\implies F_1=(1,0)$$

2. Use the central symmetry of the ellipse.

   $$F_2=(-1,0)$$

F₁=(1,0), F₂=(−1,0).

Checks and common pitfalls: a>b establishes a horizontal major axis.

1. A lies on the ellipse, giving b²=1.

   $$\frac1{b^2}=1,\quad b>0\implies b=1$$

2. Recover a from the focal relation.

   $$a^2=b^2+c^2=2\implies a=\sqrt2$$

a=√2, b=1; ellipse x²/2+y²=1.

Checks and common pitfalls: Choose positive semiaxes as stipulated.

1. Obtain the quadratic in x.

   $$3x^2+4mx+2m^2-2=0$$

2. Set its discriminant to zero.

   $$16m^2-12(2m^2-2)=0\implies m^2=3\implies m=\pm\sqrt3$$

m=±√3.

Checks and common pitfalls: There are two parallel tangents, on opposite sides of the ellipse.

1. Use the difference-of-cubes identity.

   $$z^3-8=(z-2)(z^2+2z+4)$$

2. The root 2 violates b≠0; solve the other factor.

   $$z=-1\pm\sqrt3i$$

z=−1±√3i.

Checks and common pitfalls: The root 2 must be excluded because the imaginary part is nonzero.

1. Reduce the large exponent to a square.

   $$z^{2024}=8^{674}z^2=2^{2022}z^2$$

2. Square each conjugate and preserve the linked sign.

   $$(-1\pm\sqrt3i)^2=-2\mp2\sqrt3i\implies z^{2024}=-2^{2023}(1\pm\sqrt3i)$$

For z=−1±√3i, z²⁰²⁴=−2²⁰²³(1±√3i), using matching signs.

Checks and common pitfalls: The outer negative sign changes both real and imaginary components.

1. Since w≠1, the sum has a nonzero denominator.

   $$\sum_{k=0}^nw^k=\frac{e^{i(n+1)\theta}-1}{e^{i\theta}-1}$$

2. Factor numerator and denominator and simplify the exponential phase.

   $$\frac{2ie^{i(n+1)\theta/2}\sin((n+1)\theta/2)}{2ie^{i\theta/2}\sin(\theta/2)}=e^{in\theta/2}\frac{\sin((n+1)\theta/2)}{\sin(\theta/2)}=w^{n/2}\frac{\sin((n+1)\theta/2)}{\sin(\theta/2)}$$

The required identity holds; n/2 is an integer and sin(θ/2)≠0.

Checks and common pitfalls: Using exponential phases avoids ambiguity from choosing complex square roots of w.

1. For θ≠π, the identity gives a product, and its denominator is nonzero.

   $$\sum_{k=1}^6\sin(k\theta)=\frac{\sin3\theta\sin(7\theta/2)}{\sin(\theta/2)}$$

2. Solve each factor and retain the open interval.

   $$\theta=\frac{k\pi}3\ (k=1,2,3,4,5),\quad\text{or}\quad\theta=\frac{2m\pi}7\ (m=1,2,3,4,5,6)$$

3. At θ=π every summand is zero, so it is indeed included; the two lists have no overlap in this interval.

Eleven solutions: kπ/3 (k=1,…,5) and 2mπ/7 (m=1,…,6).

Checks and common pitfalls: Do not discard θ=π from this part just because the previous part excluded it.

1. Expansion and collection leave a symmetric cubic.

   $$|C|=2(a^3+b^3+c^3-3abc)$$

2. Factor and rewrite the quadratic factor as a sum of squares.

   $$|C|=(a+b+c)\big[(a-b)^2+(b-c)^2+(c-a)^2\big]$$

|C|=(a+b+c)[(a−b)²+(b−c)²+(c−a)²].

Checks and common pitfalls: A row interchange would reverse the determinant sign.

1. The linear factor describes one plane of solutions.

   $$a+b+c=0\implies(a,b,c)=(r,s,-r-s)$$

2. The squared-difference factor describes the diagonal line.

   $$a=b=c\implies(a,b,c)=(r,r,r),\quad r,s\in\mathbb R$$

All triples (r,s,−r−s), together with all triples (r,r,r).

Checks and common pitfalls: The two families form a union, not simultaneous conditions.

1. The row operation preserves the determinant.

   $$\begin{vmatrix}k&2&-1\\0&k&1\\k&3&0\end{vmatrix}=\begin{vmatrix}k&2&-1\\0&k&1\\0&1&1\end{vmatrix}=k(k-1)$$

2. A nonsingular coefficient matrix gives exactly one solution for every p,q.

   $$k\ne0,1$$

Unique solution exactly when k≠0 and k≠1.

Checks and common pitfalls: A zero determinant needs a separate consistency analysis, not an automatic no-solution conclusion.

1. When k=0, the last two equations fix y and z, leaving x free if the first equation agrees.

   $$k=0:\quad p+q=4,\quad(x,y,z)=(t,2,q),\ t\in\mathbb R$$

2. When k=1, add the first two equations and compare with the third.

   $$k=1:\quad p+q=6,\quad(x,y,z)=(6-3t,t,q-t),\ t\in\mathbb R$$

3. If the corresponding consistency condition fails, there is no solution; no other k can give multiple solutions.

k=0 requires p+q=4; k=1 requires p+q=6, with the parameterisations above.

Checks and common pitfalls: Both singular cases must be included.