2024 JM02

1. Calculate AC and AD.

   $$AC=PA/\tan(\pi/6)=\sqrt3,\quad AD=PA=1$$

2. The converse of Pythagoras gives DA⊥AC.

   $$AD^2+AC^2=1+3=4=DC^2$$

3. Also DA⊥AP since PAD is right at A; therefore DA⊥plane APC, hence DA⊥PC.

DA⊥PC.

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

1. The base triangle is right at C.

   $$\sin\angle DCA=1/2\implies\angle DCA=\pi/6,\quad\angle DCB=\pi/6+\pi/3=\pi/2$$

2. Use base area and perpendicular height.

   $$[BCD]=\frac12(2)(\sqrt3)=\sqrt3,\quad V=\frac13[BCD]\,PA=\frac{\sqrt3}3$$

Volume √3/3.

Checks and common pitfalls: The relative positions in the diagram determine that the two angles add.

1. The equal sides in each triangle make its median an altitude.

   $$PA=DA=1,\quad PC=\sqrt{1+3}=2=DC,\quad AM\perp PD,\ CM\perp PD$$

2. Compute lengths using right triangles.

   $$PD=\sqrt2,\quad AM=\frac{\sqrt2}2,\quad CM=\sqrt{4-1/2}=\frac{\sqrt{14}}2$$

3. Use the cosine rule in AMC.

   $$\cos\angle AMC=\frac{AM^2+CM^2-AC^2}{2AM\,CM}=\frac{1/2+7/2-3}{\sqrt7}=\frac{\sqrt7}7$$

Cosine √7/7.

Checks and common pitfalls: Use the plane angle perpendicular to the common edge, not an arbitrary face angle.

1. The height is the nonnegative square root.

   $$h=\sqrt{1-x^2},\quad0\le x\le1$$

2. Substitute into the volume and square.

   $$V=\frac\pi3x^2h\implies V^2=\frac{\pi^2}{9}(x^4-x^6)$$

V²=π²(x⁴−x⁶)/9, 0≤x≤1.

Checks and common pitfalls: The endpoints describe degenerate cones and are included for the closed model.

1. The only interior stationary point is √(2/3).

   $$(V^2)'=\frac{2\pi^2}{9}x^3(2-3x^2)=0\implies x=\frac{\sqrt6}3$$

2. Its derivative changes from positive to negative. There is no interior local minimum.

   $$V^2(\sqrt6/3)=\frac{\pi^2}9\left(\frac49-\frac8{27}\right)=\frac{4\pi^2}{243}$$

Local maximum 4π²/243 at x=√6/3; no local minimum on (0,1).

Checks and common pitfalls: Do not count endpoint minima in an explicitly open interval.

1. Solve the interior second-derivative equation.

   $$(V^2)''=\frac{2\pi^2}3x^2(2-5x^2),\quad x=\sqrt{2/5}=\frac{\sqrt{10}}5$$

2. Concavity changes from up to down; evaluate the ordinate.

   $$V^2(\sqrt{2/5})=\frac{\pi^2}9\left(\frac4{25}-\frac8{125}\right)=\frac{4\pi^2}{375}$$

Inflection point (√10/5,4π²/375).

Checks and common pitfalls: x=0 is a boundary and is not an inflection point of this radius-domain curve.

1. The graph starts flat at zero, rises to its maximum and falls to zero.

   $$0<\sqrt{2/5}<\sqrt{2/3}<1$$

2. It is concave up before √(2/5) and concave down afterwards.

   $$\max y=4\pi^2/243,\quad y(0)=y(1)=0$$

The plot shows the full domain, maximum, and inflection.

Checks and common pitfalls: The vertical coordinate is volume squared, not volume.

1. Both endpoint volumes vanish and the interior maximum is larger.

   $$\max V^2=4\pi^2/243$$

2. Recover volume with cubic-metre units.

   $$V_{\max}=\sqrt{\frac{4\pi^2}{243}}=\frac{2\sqrt3\pi}{27}\ \mathrm{m}^3$$

Maximum volume 2√3π/27 m³.

Checks and common pitfalls: Do not report the squared-volume maximum as a volume.

1. Split the area into two integrals.

   $$A=\int_0^k\left(\frac{x^2}k-\frac{x^2}{2k}\right)dx+\int_k^{2k}\left(x-\frac{x^2}{2k}\right)dx$$

2. Evaluate and retain the positive parameter.

   $$A=\frac{k^2}6+\frac{k^2}3=\frac{k^2}2=1\implies k=\sqrt2$$

k=√2.

Checks and common pitfalls: Integrating one fixed pair of curves over the whole range gives the wrong region.

1. Clear the denominator after substitution.

   $$4x^2-m^2(x-\sqrt5)^2=4$$

2. Collect and reverse the signs.

   $$(m^2-4)x^2-2\sqrt5m^2x+5m^2+4=0$$

(m²−4)x²−2√5m²x+5m²+4=0.

Checks and common pitfalls: Keep the cross term −2√5x in the square.

1. Calculate the discriminant.

   $$\Delta=20m^4-4(m^2-4)(5m^2+4)=64(m^2+1)>0$$

2. Only the degenerate leading-coefficient values are excluded.

   $$m\in\mathbb R\setminus\{-2,2\}$$

All real m except ±2.

Checks and common pitfalls: A positive discriminant alone is insufficient when the equation becomes linear.

1. Read Vieta’s formulae.

   $$s=x_1+x_2=\frac{2\sqrt5m^2}{m^2-4},\quad p=x_1x_2=\frac{5m^2+4}{m^2-4}$$

2. Expand the dot product and substitute.

   $$0=p+m^2(p-\sqrt5s+5)=\frac{4-11m^2}{m^2-4}$$

3. Solve and check the excluded values.

   $$m=\pm\frac2{\sqrt{11}}=\pm\frac{2\sqrt{11}}{11}\ne\pm2$$

m=±2√11/11.

Checks and common pitfalls: The denominator restriction m≠±2 remains in force.

1. Calculate the root separation.

   $$x_1+x_2=10\sqrt5,\quad x_1x_2=29,\quad|x_1-x_2|=\sqrt{500-116}=8\sqrt6$$

2. Use y=√5x−5 in the determinant.

   $$[AOB]=\frac12|x_1y_2-x_2y_1|=\frac52|x_2-x_1|=20\sqrt6$$

Area 20√6.

Checks and common pitfalls: Area uses the absolute determinant, regardless of vertex ordering.

1. Write the two numbers in polar form.

   $$\sqrt3+i=2\operatorname{cis}(\pi/6),\quad1+i=\sqrt2\operatorname{cis}(\pi/4)$$

2. The resulting argument is already principal.

   $$z=\sqrt2[\cos(-\pi/12)+i\sin(-\pi/12)]$$

Modulus √2, principal argument −π/12.

Checks and common pitfalls: Arguments subtract under division.

1. Apply De Moivre’s theorem.

   $$z^{26}=2^{13}\operatorname{cis}(-13\pi/6)=2^{13}\operatorname{cis}(-\pi/6)$$

2. Evaluate the exact trigonometric values.

   $$z^{26}=2^{12}\sqrt3-2^{12}i$$

4096√3−4096i.

Checks and common pitfalls: The modulus is (√2)²⁶=2¹³, not 2²⁶.

1. The two powers have equal real parts and opposite imaginary parts.

   $$\omega^{\pm n}=\cos n\alpha\pm i\sin n\alpha$$

2. Adding and subtracting proves both requested formulae.

   $$\omega^n+\omega^{-n}=2\cos n\alpha,\quad\omega^n-\omega^{-n}=2i\sin n\alpha$$

3. Expand the product using those formulae.

   $$\sin^2\alpha\cos^3\alpha=-\frac1{32}(\omega^2-2+\omega^{-2})(\omega^3+3\omega+3\omega^{-1}+\omega^{-3})$$

4. Collect symmetric powers and convert back to cosine.

   $$=-\frac1{32}\big[(\omega^5+\omega^{-5})+(\omega^3+\omega^{-3})-2(\omega+\omega^{-1})\big]=\frac{2\cos\alpha-\cos3\alpha-\cos5\alpha}{16}$$

Both power formulae and the product identity follow.

Checks and common pitfalls: The symmetric first-power pair is ω+ω⁻¹, which converts to 2cosα.

1. Reduce to sine and cosine zeros.

   $$16\sin^2\alpha\cos^3\alpha=0\iff\sin\alpha=0\ \text{or}\ \cos\alpha=0$$

2. Merge the two families.

   $$\alpha=k\pi\ \text{or}\ \alpha=\pi/2+k\pi\iff\alpha=n\pi/2,\ n\in\mathbb Z$$

α=nπ/2, n∈ℤ.

Checks and common pitfalls: Dividing by sine or cosine would discard a whole solution family.

1. The second row becomes constant.

   $$D=(a+b+c)\begin{vmatrix}a&b&c\\1&1&1\\1+b&1+c&1+a\end{vmatrix}$$

2. Column differences make expansion along row 2 simple.

   $$D=-(a+b+c)\begin{vmatrix}b-a&c-a\\c-b&a-b\end{vmatrix}$$

3. Expand and collect the quadratic factor.

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

(a+b+c)(a²+b²+c²−ab−bc−ca).

Checks and common pitfalls: The cofactor in position (2,1) carries a minus sign.

1. Expand the determinant.

   $$\begin{vmatrix}1&1&1\\k&5&1\\1&k&-1\end{vmatrix}=(-5-k)+(k+1)+(k^2-5)=k^2-9$$

2. Exclude its two zeros.

   $$k\in\mathbb R\setminus\{-3,3\}$$

k≠±3.

Checks and common pitfalls: A zero determinant does not automatically imply no solution; it only rules out uniqueness.

1. Subtract three times equation 1 from equation 2.

   $$2y-2z=0\implies y=z=t$$

2. Recover x and verify the third relation.

   $$x=1-2t,\quad x+3y-z=1-2t+3t-t=1$$

(x,y,z)=(1−2t,t,t), t∈ℝ.

Checks and common pitfalls: The third equation is dependent, so t remains free.

1. Reduce the final equation with its domain.

   $$t\ge0,\quad a=2(1-2t)+2\sqrt t+3t=2-t+2\sqrt t$$

2. The square has minimum zero.

   $$a=3-(\sqrt t-1)^2\le3,\quad a=3\iff t=1$$

3. Substitute the equality case.

   $$(x,y,z)=(-1,1,1)$$

Maximum a=3, attained only at (−1,1,1).

Checks and common pitfalls: The square root requires y=t≥0.