2022 JM01

1. The trigonometric sum cannot attain 3.

   $$|\sin x+\cos x|=|\sqrt2\sin(x+\pi/4)|\le\sqrt2<3$$

2. The other sets contain 0, ±1, ∅, or (0,0), respectively.

B.

Checks and common pitfalls: Distinguish ∅ from {∅}.

1. Put the difference over a common denominator.

   $$\frac{b+m}{a+m}-\frac ba=\frac{m(a-b)}{a(a+m)}$$

2. Every factor has the required positive sign.

   $$m(a-b)>0,\quad a(a+m)>0\implies\frac ba<\frac{b+m}{a+m}$$

C.

Checks and common pitfalls: Expressions involving b−m can change sign or be undefined.

1. Evaluate the two remainders.

   $$f(2)=20+2a+b,\quad f(-1)=2-a+b$$

2. Solve the resulting linear equation.

   $$20+2a=2-a\implies a=-6$$

E: −6.

Checks and common pitfalls: The root of x+1 is −1, not 1.

1. Identify the perfect square.

   $$7-4\sqrt3=(2-\sqrt3)^2$$

2. Since 2>√3, the principal root is positive.

   $$\sqrt{(2-\sqrt3)^2}=|2-\sqrt3|=2-\sqrt3$$

B: 2−√3.

Checks and common pitfalls: √(u²)=|u|, not automatically u.

1. The quadratic discriminant simplifies to a linear expression.

   $$\Delta=4(p+3)^2-4p(p-1)=4(7p+9)\ge0\iff p\ge-9/7$$

2. At p=0, the linear equation −6x−1=0 also has a real solution, already inside this range.

D: p≥−9/7.

Checks and common pitfalls: The question says equation, so the linear case must not be discarded.

1. Express the reciprocals.

   $$\frac1a=\frac{\ln2}{\ln\sqrt{10}},\quad\frac1b=\frac{\ln5}{\ln\sqrt{10}}$$

2. Combine using ln2+ln5=ln10.

   $$\frac1a+\frac1b=\frac{\ln10}{(1/2)\ln10}=2$$

A: 2.

Checks and common pitfalls: The common value is √10, so its logarithm is half ln10.

1. Derive and add the two horizontal gaps.

   $$d^2=(R+r)^2-(R-r)^2=4Rr,\quad2\sqrt{3r}+2\sqrt{4r}=2\sqrt{12}$$

2. Solve for r and rationalise.

   $$r=\frac{12}{(\sqrt3+2)^2}=\frac{12}{7+4\sqrt3}=84-48\sqrt3$$

A: 84−48√3.

Checks and common pitfalls: The bounded-gap condition selects the sum of the two horizontal separations.

1. Multiply the three loss probabilities.

   $$P(\text{no wins})=(3/4)^3=27/64$$

2. Subtract from one.

   $$P(\text{at least one})=1-27/64=37/64$$

D: 37/64.

Checks and common pitfalls: Adding three win probabilities double-counts outcomes with multiple wins.

1. Convert the system to vertical comparisons.

   $$y\le2x+4,\quad y\ge5-x,\quad y\ge2$$

2. This is the region above both lower boundaries and below the rising line: IV.

D: region IV.

Checks and common pitfalls: All three conditions must hold simultaneously.

1. Form the line equation.

   $$y-2=-\frac32(x-1)$$

2. Set x=0.

   $$y=2+3/2=7/2$$

A: (0,7/2).

Checks and common pitfalls: A y-intercept has x=0.

1. The first 30 seats contain six disjoint five-seat blocks, hence at least six vacancies per row.

   $$\text{occupied per row}\le32-6=26$$

2. Repeat four occupied then one empty six times, then occupy the last two.

   $$\text{rows}=6400/32=200,\quad\max=26(200)=5200$$

B: 5200.

Checks and common pitfalls: The final partial block need not contain another empty seat.

1. Apply Vieta and the mean condition.

   $$\frac1{x_1}+\frac1{x_2}=\frac{8/3}{m/3}=\frac8m=4\implies m=2$$

2. Check reality and nonzero roots.

   $$\Delta=64-12(2)=40>0,\quad x_1x_2=2/3\ne0$$

E: m=2.

Checks and common pitfalls: Mean 2 means sum 4, not sum 2.

1. Count all ordered pairs.

   $$N=5\cdot4=20$$

2. Count favourable pairs.

   $$N_f=3\cdot4=12,\quad P=12/20=3/5$$

C: 3/5.

Checks and common pitfalls: The digits cannot repeat, but the order matters.

1. Collect the two exponential bases.

   $$6^x-2^x=6\cdot2^x-6\cdot6^x\implies7\cdot6^x=7\cdot2^x$$

2. Use injectivity of the exponential.

   $$3^x=1\implies x=0$$

C: x=0.

Checks and common pitfalls: 6ˣ⁺¹ is 6·6ˣ, not 6ˣ+1.

1. Form the distance equation.

   $$(n-3\sqrt3/2)^2+(-3/2)^2=3$$

2. Factor the resulting quadratic.

   $$n^2-3\sqrt3n+6=(n-\sqrt3)(n-2\sqrt3)=0$$

E: n=√3 or 2√3.

Checks and common pitfalls: The turning angle is 150°; the triangle’s interior angle is its supplement 30°.

1. Write f(x)=a(x−2)²−9 with a>0.

   $$0=9a-9\implies a=1$$

2. Expand the result.

   $$f(x)=(x-2)^2-9=x^2-4x-5$$

a=1, b=−4, c=−5.

Checks and common pitfalls: The minimum specifies the vertex ordinate, not the constant term c.

1. Apply the two transformations.

   $$y=f(x+3)+3$$

2. Simplify.

   $$y=(x+1)^2-6=x^2+2x-5$$

y=(x+1)²−6.

Checks and common pitfalls: Horizontal and vertical changes act in different places in the formula.

1. The vertex t=2 belongs to the inner range.

   $$g_{\min}=f(2)=-9$$

2. The farthest endpoint from 2 is −3.

   $$g_{\max}=f(-3)=25-9=16$$

Maximum 16; minimum −9.

Checks and common pitfalls: Optimise f over the inner range, not over all real t.

1. Solve the repeated-root quadratic for q.

   $$8q=1+16q^2\iff(4q-1)^2=0\implies q=1/4$$

2. Insert q into the two definitions.

   $$a_n=4^{1-n},\quad b_n=\frac n4\,4^{1-n}=\frac n{4^n}$$

aₙ=4¹⁻ⁿ; bₙ=n/4ⁿ.

Checks and common pitfalls: The factors 4 and 16 apply to the sequence terms before testing arithmeticity.

1. Compute the geometric sum.

   $$S_n=\frac{1-4^{-n}}{1-1/4}=\frac43(1-4^{-n})$$

2. Shift and subtract the arithmetic-geometric sum.

   $$3T_n=1+\frac14+\cdots+\frac1{4^{n-1}}-\frac n{4^n}=\frac43(1-4^{-n})-n4^{-n}$$

3. Divide by three.

   $$T_n=\frac49-\frac{3n+4}{9\cdot4^n}$$

Sₙ=4(1−4⁻ⁿ)/3; Tₙ=4/9−(3n+4)/(9·4ⁿ).

Checks and common pitfalls: The shifted subtraction leaves a final −n/4ⁿ term.

1. The squared semiaxes satisfy b²=a²/2.

   $$1-\frac{b^2}{a^2}=1/2$$

2. Use the point condition to find both.

   $$\frac8{a^2/2}+\frac{16}{a^2}=1\implies a^2=32,\quad b^2=16$$

x²/16+y²/32=1.

Checks and common pitfalls: The larger denominator lies under y² for a vertical major axis.

1. Obtain the midpoint abscissa.

   $$(2+k_1^2)x^2+2k_1dx+d^2-32=0\implies x_M=-\frac{k_1d}{2+k_1^2}$$

2. Compute its ordinate and ratio; xM≠0 follows from k₁d≠0.

   $$y_M=\frac{2d}{2+k_1^2},\quad k_2=\frac{y_M}{x_M}=-\frac2{k_1}\implies k_1k_2=-2$$

k₁k₂=−2.

Checks and common pitfalls: The nonzero slope and intercept conditions ensure OM has a defined finite slope.

1. Choose r=2 and β=π/6.

   $$2\cos(\theta+\pi/6)=\sqrt3\cos\theta-\sin\theta$$

2. Use the full cosine range.

   $$-2\le\sqrt3\cos\theta-\sin\theta\le2$$

2cos(θ+π/6), range [−2,2].

Checks and common pitfalls: The minus sine coefficient corresponds to a positive phase inside cosine.

1. The shifted angle is in [π/6,13π/6].

   $$\theta+\pi/6=2\pi/3\text{ or }4\pi/3$$

2. Subtract the phase.

   $$\theta=\pi/2\text{ or }7\pi/6$$

θ=π/2 or 7π/6.

Checks and common pitfalls: Both cosine solutions must be retained.

1. The auxiliary-angle expression gives cos2u=−1/4.

   $$2\cos(\theta+\pi/6)=-1/2\implies\sin^2u=\frac{1+1/4}2=5/8$$

2. u lies in [π/12,7π/12], where sine is positive.

   $$\sin u=\sqrt{5/8}=\frac{\sqrt{10}}4$$

√10/4.

Checks and common pitfalls: This subpart has right side −1/2, different from part (b).

1. The base case is a multiple of 14.

   $$3^6+5^3=729+125=854=14\cdot61$$

2. Assume 14 divides Aₖ=3⁴ᵏ⁺²+5²ᵏ⁺¹.

3. Express the next term as a sum of two multiples of 14.

   $$A_{k+1}=81\cdot3^{4k+2}+25\cdot5^{2k+1}=25A_k+56\cdot3^{4k+2}$$

4. The induction hypothesis and 56=4·14 complete the inductive step, hence the claim holds for all positive n.

Divisible by 14 for every positive integer n.

Checks and common pitfalls: State both the induction hypothesis and how it proves the next case.