2023 JM02

1. The midpoint F gives EF=BF=√2.

   $$EF=BC/2=\sqrt2$$

2. In equilateral DBC, DF is an altitude.

   $$DF^2=DB^2-BF^2=8-2=6$$

3. Apply the cosine rule at F.

   $$\cos\angle DFE=\frac{DF^2+EF^2-DE^2}{2DF\,EF}=\frac{6+2-5}{2\sqrt6\sqrt2}=\frac{\sqrt3}4$$

cos∠DFE=√3/4.

Checks and common pitfalls: The cosine formula must use the angle at F, opposite DE.

1. From part (a), compute the squared lengths.

   $$FG=EF\cos\angle DFE=\sqrt6/4,\quad EG^2=2-6/16=13/8$$

2. DF⊥BC, so BFG is right at F; EB²=4.

   $$BG^2=BF^2+FG^2=2+6/16=19/8,\quad EG^2+BG^2=4=EB^2$$

3. Thus EG⊥GB. It is also perpendicular to DF by construction, and GB,DF are intersecting base-plane lines; hence EG⊥plane ABCD.

EG⊥plane ABCD.

Checks and common pitfalls: Verify two intersecting lines in the base plane, not only DF.

1. The angle ADG equals π/3+π/6=π/2.

   $$DG=\sqrt6-\sqrt6/4=3\sqrt6/4,\quad AG^2=AD^2+DG^2=8+54/16=91/8$$

2. Use right triangle AGE.

   $$AE^2=AG^2+EG^2=91/8+13/8=13\implies AE=\sqrt{13}$$

AE=√13.

Checks and common pitfalls: Use the perpendicular height EG; EF is a sloping segment.

1. Apply the power rule.

   $$f'(x)=3x^2-12$$

2. The derivative of the constant −12 is zero.

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

f′=3x²−12, f″=6x.

Checks and common pitfalls: The original constant 6 disappears on differentiation.

1. Factor the derivative and read its signs +,−,+.

   $$f'(x)=3(x-2)(x+2)$$

2. Evaluate the two stationary values.

   $$f(-2)=22\text{ (local maximum)},\quad f(2)=-10\text{ (local minimum)}$$

Maximum 22 at x=−2; minimum −10 at x=2.

Checks and common pitfalls: These are local extrema; the cubic has no global bounds.

1. Concavity changes from down to up at zero.

   $$x<0\implies f''<0,\quad x>0\implies f''>0$$

2. Evaluate the ordinate.

   $$f(0)=6\implies(0,6)$$

Inflection point (0,6).

Checks and common pitfalls: The point need not have a horizontal tangent.

1. Increase on (−∞,−2) and (2,∞); decrease between them.

   $$(-2,22),\quad(2,-10),\quad(0,6)$$

2. Draw concave down for x<0 and concave up for x>0, with unbounded cubic tails.

   $$x\to-\infty:f(x)\to-\infty;\quad x\to\infty:f(x)\to\infty$$

The plot marks all required features; the curve continues beyond the viewing window.

Checks and common pitfalls: The inflection point differs from either extremum.

1. Find the possible tangent abscissae.

   $$3x^2+6x+1=1\implies x=0,-2$$

2. Check the curve values against the line.

   $$x=0:\ 0\ne4;\quad x=-2:\ -8+12-2=2=-2+4$$

A=(−2,2).

Checks and common pitfalls: Equal slopes alone do not guarantee tangency to this particular line.

1. Factor the difference.

   $$x^3+3x^2-4=(x-1)(x+2)^2=0\implies x=-2,1$$

2. The integrand is nonnegative on the bounded interval.

   $$A=\int_{-2}^1(4-x^3-3x^2)\,dx=\left[4x-\frac{x^4}4-x^3\right]_{-2}^1=\frac{27}4$$

Area 27/4.

Checks and common pitfalls: Tangency gives a repeated intersection root but does not eliminate the bounded region.

1. Substitute and collect the quadratic in x.

   $$(4+9m^2)x^2+18mdx+9d^2-36=0$$

2. The leading coefficient is positive, and the discriminant is zero exactly under the required condition.

   $$\Delta=144(9m^2+4-d^2)=0\iff9m^2-(k-mh)^2+4=0$$

The stated condition is necessary and sufficient.

Checks and common pitfalls: This slope form treats only nonvertical tangents; vertical cases require separate handling.

1. Form the genuine quadratic assumed in this subpart.

   $$(9-h^2)m^2+2hkm+4-k^2=0,\quad h^2\ne9$$

2. Read the two symmetric functions.

   $$m_1+m_2=\frac{2hk}{h^2-9},\quad m_1m_2=\frac{k^2-4}{h^2-9}$$

Sum 2hk/(h²−9); product (k²−4)/(h²−9).

Checks and common pitfalls: When h=±3 the slope equation is not quadratic.

1. The finite-slope case yields a circle.

   $$\frac{k^2-4}{h^2-9}=-1\implies h^2+k^2=13$$

2. The missing cases are the intersections x=±3 and y=±2; all lie on the same circle.

   $$(h,k)=(\pm3,\pm2)$$

3. Every point of this circle lies outside the ellipse, so no further exclusion is needed.

   $$h^2/9+k^2/4\ge(h^2+k^2)/9=13/9>1$$

The full circle x²+y²=13.

Checks and common pitfalls: Omitting the vertical cases would leave four unjustified holes.

1. Evaluate the symmetric quantities.

   $$m_1+m_2=5/2,\quad m_1m_2=3/4,\quad|m_1-m_2|=\sqrt{13}/2$$

2. Use the acute line-angle formula.

   $$\tan\alpha=\frac{|m_1-m_2|}{|1+m_1m_2|}=\frac{2\sqrt{13}}7\implies\alpha=\arctan\frac{2\sqrt{13}}7$$

α=arctan(2√13/7).

Checks and common pitfalls: Use the absolute ratio to obtain the angle between unoriented lines.

1. Separate the equations.

   $$4y=16\implies y=4,\quad\sqrt{x^2+16}=2x-1\ge0\implies x\ge1/2$$

2. Square and factor.

   $$x^2+16=(2x-1)^2\iff(x-3)(3x+5)=0$$

3. Only x=3 satisfies the unsquared equation.

   $$z=3+4i,\quad|z|=5$$

z=3+4i.

Checks and common pitfalls: The root x=−5/3 introduced by squaring is extraneous.

1. Equate the expansion to cis(3θ).

   $$\cos3\theta+i\sin3\theta=(\cos^3\theta-3\cos\theta\sin^2\theta)+i(3\cos^2\theta\sin\theta-\sin^3\theta)$$

2. Thus each component supplies its respective formula.

   $$\cos3\theta=\cos^3\theta-3\cos\theta\sin^2\theta,\quad\sin3\theta=3\cos^2\theta\sin\theta-\sin^3\theta$$

3. For cosθ≠0 and cos3θ≠0, divide by cos³θ.

   $$\tan3\theta=\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}$$

The sine/cosine identities hold for all θ; the tangent identity requires defined denominators.

Checks and common pitfalls: The tangent formula cannot be used when 1−3tan²θ=0.

1. The potential denominator zeros x=±1/√3 are not roots of the cubic, so division is valid.

   $$\frac{3x-x^3}{1-3x^2}=\sqrt3\iff\tan3\theta=\sqrt3$$

2. Solve and restrict θ to its selected interval.

   $$\theta=\pi/9+n\pi/3\implies\theta=-2\pi/9,\pi/9,4\pi/9$$

3. Transform back to all three real roots.

   $$x=\tan(-2\pi/9),\ \tan(\pi/9),\ \tan(4\pi/9)$$

tan(−2π/9), tan(π/9), tan(4π/9).

Checks and common pitfalls: Taking only the principal arctangent of √3 would miss two cubic roots.

1. For sine, the sin u cos v terms cancel.

   $$\sin(u+v)-\sin(u-v)=2\cos u\sin v$$

2. For cosine, the cos u cos v terms cancel. Substitute u,v back.

   $$\cos(u+v)-\cos(u-v)=-2\sin u\sin v$$

Both identities follow by addition-formula cancellation.

Checks and common pitfalls: The cosine difference identity has a leading minus sign.

1. Expand along the simplified first row.

   $$D=\begin{vmatrix}\sin4\theta-\sin2\theta&\sin8\theta-\sin2\theta\\\cos4\theta-\cos2\theta&\cos8\theta-\cos2\theta\end{vmatrix}$$

2. Extract the common sine factors from the columns.

   $$D=4\sin\theta\sin3\theta\begin{vmatrix}\cos3\theta&\cos5\theta\\-\sin3\theta&-\sin5\theta\end{vmatrix}$$

3. The remaining determinant is −sin(5θ−3θ).

   $$D=-4\sin\theta\sin3\theta\sin2\theta$$

D=−4sinθ sin2θ sin3θ.

Checks and common pitfalls: Column subtraction leaves a determinant unchanged; swapping columns would change its sign.

1. Find the candidate angles.

   $$\sin\theta\sin2\theta\sin3\theta=0\implies\theta=0,\pi/3,\pi/2$$

2. At 0 and π/2 the second equation becomes 0=√3, so both are impossible.

3. At π/3 the cosine equation is dependent; the other two equations give y=2 and x+z=4.

   $$x+y+z=6,\quad x-y+z=2\implies(x,y,z)=(4-t,2,t),\quad t\in\mathbb R$$

θ=π/3; (x,y,z)=(4−t,2,t), t∈ℝ.

Checks and common pitfalls: A zero determinant is necessary but consistency must still be checked.