2021 JM02

1. The given perpendicularities allow Pythagoras.

   $$PD^2=PA^2+AD^2=5,\quad BD^2=AB^2+AD^2=5,\quad PB^2=PA^2+AB^2=8$$

2. The altitude to PB bisects this base.

   $$h=\sqrt{5-(\sqrt8/2)^2}=\sqrt3$$

3. Multiply half the base by its corresponding height.

   $$A_{PBD}=\tfrac12\sqrt8\sqrt3=\sqrt6$$

Area √6.

Checks and common pitfalls: PA is the pyramid height, not the altitude to PB inside triangle PBD.

1. Compute the base area and the original perpendicular height.

   $$A_{ABD}=\tfrac12(2)(1)=1,\quad V=\tfrac13(1)(2)=\tfrac23$$

2. Changing the base does not change the volume.

   $$\tfrac13 A_{PBD}d=\tfrac23\implies \tfrac{\sqrt6}{3}d=\tfrac23$$

3. Use a positive perpendicular distance.

   $$d=2/\sqrt6=\sqrt{2/3}=\sqrt6/3$$

Volume 2/3; distance √(2/3).

Checks and common pitfalls: The official √(2/3) has the entire fraction under the radical.

1. AD lies in the base plane, so the perpendicular PA is perpendicular to AD.

   $$AD\perp PA$$

2. The given angle gives a second intersecting line in plane PAB.

   $$AD\perp AB,\quad PA\cap AB=\{A\}$$

3. A line perpendicular to both intersecting lines is perpendicular to their plane; transfer this direction to BC.

   $$AD\perp\operatorname{plane}PAB,\quad BC\parallel AD\implies CB\perp\operatorname{plane}PAB$$

CB⊥plane PAB.

Checks and common pitfalls: Perpendicularity to only AB does not prove perpendicularity to the plane.

1. The midpoint segment has the required direction and length.

   $$NM\parallel BC\parallel AD,\quad NM=BC/2=1=AD$$

2. One pair of equal parallel opposite sides makes ADMN a parallelogram, and AD⊥AN gives a right angle.

   $$AN\subset\operatorname{plane}PAB,\quad AD\perp AN$$

3. Thus ADMN is a rectangle and DM is parallel to AN.

   $$DM\parallel AN$$

4. D is outside plane PAB, while AN lies in it. A line through D parallel to AN therefore has no intersection with the plane.

   $$DM\parallel\operatorname{plane}PAB$$

ADMN is a rectangle; DM∥plane PAB.

Checks and common pitfalls: A line parallel to a line in a plane also needs to be outside the plane to establish line–plane parallelism.

1. Use the power rule.

   $$f'(x)=6x^2-18x+12$$

2. Constants vanish when differentiated.

   $$f''(x)=12x-18$$

f′(x)=6x²−18x+12; f″(x)=12x−18.

Checks and common pitfalls: Do not retain the constant −5 in f′.

1. The stationary abscissae are 1 and 2.

   $$f'(x)=6(x-1)(x-2)=0\iff x=1,2$$

2. The signs are positive, negative, positive.

   $$x<1:f'>0;\quad1<x<2:f'<0;\quad x>2:f'>0$$

3. Evaluate f at the two points and classify the changes.

   $$f(1)=0\text{ (local maximum)},\quad f(2)=-1\text{ (local minimum)}$$

Local maximum 0 at x=1; local minimum −1 at x=2.

Checks and common pitfalls: These local values are not the global bounds of the cubic on the real line.

1. The second derivative vanishes at 3/2.

   $$12x-18=0\implies x=\tfrac32$$

2. It changes from negative to positive; evaluate the ordinate.

   $$x<\tfrac32:f''<0;\quad x>\tfrac32:f''>0;\quad f(\tfrac32)=-\tfrac12$$

Inflection point (3/2,−1/2).

Checks and common pitfalls: An inflection requires a concavity change; f″=0 alone is insufficient.

1. Factor f to locate both x-intercepts, including the double root.

   $$f(x)=(x-1)^2(2x-5)$$

2. The double zero touches the axis; the simple zero crosses it.

   $$(-1,-28),\ (0,-5),\ (1,0),\ (3/2,-1/2),\ (2,-1),\ (5/2,0),\ (3,4)$$

3. Increase to x=1, decrease to x=2, then increase; concavity changes at 3/2. Keep the two domain endpoints.

The plotted cubic is restricted to [−1,3], with the listed key points.

Checks and common pitfalls: The double root x=1 touches the axis rather than crossing it.

1. For negative x substitute −x; for nonnegative x retain x, then subtract one.

   $$g(x)=\begin{cases}-2x^3-9x^2-12x-6&-1\le x<0\\2x^3-9x^2+12x-6&0\le x\le3\end{cases}$$

2. Mark the corner at zero and transformed turning points/endpoints.

   $$(-1,-1),\ (0,-6),\ (1,-1),\ (2,-2),\ (3,3)$$

3. Only the reflected portion with −1≤x≤0 remains in the requested domain.

The plotted reflected-and-shifted curve has a corner at (0,−6).

Checks and common pitfalls: Do not reflect the original negative-x branch; f(|x|) uses the original nonnegative-x branch.

1. Set the two ordinates equal and factor.

   $$2x^3-8x=2x(x-2)(x+2)=0\implies x=-2,0,2$$

2. On (−2,0) the cubic is above; on (0,2) the quadratic is above.

   $$A=\int_{-2}^0(2x^3-8x)\,dx+\int_0^2(8x-2x^3)\,dx$$

3. Evaluate both nonnegative contributions.

   $$A=[x^4/2-4x^2]_{-2}^0+[4x^2-x^4/2]_0^2=8+8=16$$

Total area 16.

Checks and common pitfalls: A single signed integral over [−2,2] would cancel the two regions.

1. The leading quadratic coefficient remains one for every m.

   $$x^2-8mx-8c=0$$

2. Set its discriminant to zero and rearrange.

   $$\Delta=64m^2+32c=0\implies c=-2m^2$$

c=−2m².

Checks and common pitfalls: Here the elimination remains quadratic, so the double-root criterion is valid even when m=0.

1. The two lines have different slopes, since a slope specifies one tangent of this parabola.

   $$k=m_1h-2m_1^2=m_2h-2m_2^2$$

2. Cancel the nonzero slope difference.

   $$(m_1-m_2)h=2(m_1-m_2)(m_1+m_2)\implies h=2(m_1+m_2)$$

3. Substitute back to obtain the ordinate.

   $$k=2m_1(m_1+m_2)-2m_1^2=2m_1m_2$$

A=(2(m₁+m₂),2m₁m₂).

Checks and common pitfalls: Cancellation requires the stated distinctness of the tangents.

1. The ordinate is fixed by the slope product.

   $$k=2m_1m_2=-2$$

2. For any real h, solve the tangent-slope equation.

   $$2m^2-hm-2=0,\quad\Delta=h^2+16>0$$

3. Its two distinct real roots have product −1, so their tangents meet at (h,−2) and are perpendicular.

   $$m_1+m_2=h/2,\quad m_1m_2=-1$$

The full straight line y=−2.

Checks and common pitfalls: A necessary equation for a locus also needs a sufficiency argument.

1. Perpendicular slopes are excluded by the specified π/4 angle.

   $$1=\left|\frac{2-m_2}{1+2m_2}\right|,\quad m_2\ne-1/2$$

2. Squaring and factoring gives two valid slopes.

   $$(2-m_2)^2=(1+2m_2)^2\iff3m_2^2+8m_2-3=0\implies m_2=-3,1/3$$

3. Use the sum and product formulas from part (i).

   $$A=(-2,-12)\quad\text{or}\quad A=(14/3,4/3)$$

A=(−2,−12) or (14/3,4/3).

Checks and common pitfalls: Using a signed angle formula without the absolute value would lose one of the two unoriented-line configurations.

1. Define the half-sum and half-difference.

   $$u=(A+B)/2,\quad v=(A-B)/2$$

2. Expand and add.

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

3. Substitute the definitions back.

   $$\sin A+\sin B=2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$$

The sum-to-product identity follows.

Checks and common pitfalls: The two expansions have opposite signs on cos u sin v.

1. Combine the first two terms and the double-angle form of the third.

   $$\sin A+\sin B+\sin C=2\cos(C/2)\left[\cos\frac{A-B}{2}+\sin(C/2)\right]$$

2. The angle-sum condition converts sine into cosine.

   $$\sin(C/2)=\cos\frac{A+B}{2}$$

3. Expand the two cosines; the sine-product terms cancel.

   $$\cos\frac{A-B}{2}+\cos\frac{A+B}{2}=2\cos(A/2)\cos(B/2)$$

4. Multiply the remaining factors, without dividing by a potentially zero cosine.

   $$\sin A+\sin B+\sin C=4\cos(A/2)\cos(B/2)\cos(C/2)$$

The identity is proved whenever A+B+C=π.

Checks and common pitfalls: The proof uses no division by cos(C/2), so zero-cosine cases remain included.

1. The base case holds for every real x.

   $$n=1:\quad2\sin x\cos x=\sin2x$$

2. Assume the statement for a fixed positive integer k.

   $$2\sin x\sum_{j=1}^k\cos(2j-1)x=\sin2kx$$

3. The addition formulas give a telescoping increment.

   $$2\sin x\cos(2k+1)x=\sin(2k+2)x-\sin2kx$$

4. Add the new term to the induction hypothesis; the old right side cancels.

   $$2\sin x\sum_{j=1}^{k+1}\cos(2j-1)x=\sin2(k+1)x$$

5. The base and implication establish the identity for all positive integers n.

The identity holds for all n≥1 and every real x.

Checks and common pitfalls: No division by sin x is needed in the induction proof.

1. The product identity reduces the equation where sin x is nonzero.

   $$2\sin x(\cos x+\cos3x+\cos5x)=\sin6x$$

2. At 0,π,2π the original left side is 3,−3,3, so none is a solution.

   $$\sin x=0\implies x=0,\pi,2\pi$$

3. Solve sin6x=0 and exclude those three points.

   $$x=j\pi/6,\quad j\in\{1,2,3,4,5,7,8,9,10,11\}$$

x=jπ/6 for j=1,2,3,4,5,7,8,9,10,11.

Checks and common pitfalls: Multiplying by sin x introduces candidate zeros at 0,π,2π; the original equation rejects them.

1. Row subtraction produces two polynomial factors.

   $$D=(b-a)(c-a)\begin{vmatrix}a&a^2+1&bc\\1&a+b&-c\\1&a+c&-b\end{vmatrix}$$

2. Subtract the new row 2 from row 3, then extract c−b.

   $$D=(b-a)(c-a)(c-b)\begin{vmatrix}a&a^2+1&bc\\1&a+b&-c\\0&1&1\end{vmatrix}$$

3. Subtract column 3 from column 2 and expand along the last row.

   $$D=(b-a)(c-a)(c-b)[a(a+b+c)-(a^2+1-bc)]$$

4. Simplify the final factor and account for the two sign reversals.

   $$D=(a-b)(b-c)(c-a)(ab+bc+ca-1)$$

D=(a−b)(b−c)(c−a)(ab+bc+ca−1).

Checks and common pitfalls: Factoring a polynomial identity does not require a,b,c to be distinct; equal-variable cases give determinant zero.

1. Expand along the first row.

   $$\det A=k(k^2-1)-(k+1)-(k+1)=k^3-3k-2$$

2. A square system is uniquely solvable for any right side exactly when this determinant is nonzero.

   $$\det A=(k+1)^2(k-2)\ne0\iff k\ne-1,2$$

All real k except −1 and 2.

Checks and common pitfalls: A zero determinant does not distinguish no solution from infinitely many solutions without checking consistency.

1. The coefficient dependence forces the same relation on the right side.

   $$R_2=R_1+R_3\implies q=p+r\iff p-q+r=0$$

2. The first two rows are independent, so this condition is also sufficient and leaves one free variable.

   $$\begin{vmatrix}2&1\\1&2\end{vmatrix}=3\ne0$$

3. The specified right side is consistent; set z=t and solve the first two equations.

   $$5-1-4=0,\quad2x+y-t=5,\quad x+2y+t=1$$

4. Substitution also satisfies the third equation.

   $$(x,y,z)=(3+t,-1-t,t),\quad t\in\mathbb R$$

p−q+r=0; for the specified values, (x,y,z)=(3+t,−1−t,t), t∈ℝ.

Checks and common pitfalls: State the free parameter domain and retain the compatibility condition; one particular solution is not the full answer.