2021 JM01

1. The upward-opening quadratic is negative between its roots.

   $$(x-3)(x-12)<0\Rightarrow Q=(3,12)$$

2. Intersect with the finite set.

   $$P\cap Q=\{5,7,11\};\quad|P\cap Q|=3$$

B: 3 elements.

Checks and common pitfalls: The endpoint 3 is excluded by the strict inequality.

1. Express the two individual rates.

   $$r=1/4+1/3=7/12$$

2. Use time=amount/rate.

   $$T=5/(7/12)=60/7$$

E: 60/7 hours.

Checks and common pitfalls: Averaging 4 and 3 would not give a joint work rate.

1. Recover the general term.

   $$a_n=S_n-S_{n-1}=n^2-(n-1)^2=2n-1$$

2. Evaluate at n=10.

   $$a_{10}=20-1=19$$

A: 19.

Checks and common pitfalls: S₁₀ is a sum of ten terms, not the tenth term itself.

1. For m<0 the tails are negative; for m=0 the expression 6x changes sign. Thus m>0 is necessary.

2. Exclude real roots and tangency to zero.

   $$\Delta=36-12m^2<0\iff m^2>3$$

3. Combine the two restrictions.

   $$m>\sqrt3$$

B: m>√3.

Checks and common pitfalls: A zero discriminant permits the value zero, so it cannot give strict positivity.

1. Read the original root relations.

   $$\alpha+\beta=3/2,\quad\alpha\beta=7/2$$

2. Transform the root relations.

   $$1/\alpha+1/\beta=3/7,\quad1/(\alpha\beta)=2/7$$

3. Construct the polynomial and clear its denominator.

   $$x^2-(3/7)x+2/7=0\iff7x^2-3x+2=0$$

B: 7x²−3x+2=0.

Checks and common pitfalls: The original roots need not be real for Vieta’s identities to hold.

1. Each radius is 2 cm, so each disk area is 4π.

   $$r=4/2=2,\quad|D_1|=|D_2|=4\pi$$

2. Use inclusion–exclusion before subtracting from the rectangle.

   $$|D_1\cup D_2|=8\pi-3;\quad S=28-(8\pi-3)=31-8\pi$$

A: 31−8π cm².

Checks and common pitfalls: The disks overlap; no external tangency between the disks is assumed.

1. The exponent is nonpositive, and a positive exponential never reaches zero.

   $$u=3^{-x^2}\in(0,1]$$

2. Compare endpoints and the unattained limiting value.

   $$k=u^2-4u=(u-2)^2-4\in[-3,0)$$

3. Every u in (0,1] is attainable by a real x.

   $$x=\pm\sqrt{-\log_3u}$$

C: −3≤k<0 is the exact range.

Checks and common pitfalls: Do not include k=0: it is a limiting value, not an attained one.

1. Apply the factor theorem at the root of the linear factor.

   $$0=f(-1/2)=2+m/2-m=2-m/2$$

2. Solve the linear equation.

   $$m=4$$

D: 4.

Checks and common pitfalls: The relevant input is −1/2, not −1.

1. Reduce the base before exponentiating.

   $$103^{10}\equiv3^{10}\pmod{100}$$

2. Read the last two digits of the small power.

   $$3^{10}=59049\equiv49\pmod{100}$$

C: the tens digit is 4.

Checks and common pitfalls: The units digit 9 is not the requested tens digit.

1. The first equation gives y=u+3.

   $$2u^2=1-u\iff(2u-1)(u+1)=0$$

2. Convert both logarithm values back to x.

   $$u=1/2,-1\Rightarrow x=4^{1/2}=2,\ 4^{-1}=1/4$$

A: x=1/4 or 2.

Checks and common pitfalls: Both resulting x values are positive and satisfy the logarithm domain.

1. The graph ranges from 1 to 5, so its midline is 3 and amplitude is 2.

   $$c=(5+1)/2=3,\quad A=(5-1)/2=2$$

2. Peaks at zero and π give period π and frequency 2.

   $$T=\pi\Rightarrow\omega=2\pi/T=2$$

3. The graph begins at a maximum, agreeing with positive cosine.

   $$y=3+2\cos2x$$

B: y=3+2cos 2x.

Checks and common pitfalls: Frequency is obtained from a full cycle, not the peak-to-trough half-cycle.

1. Compute midpoint and segment slope.

   $$M=(2,-2),\quad m_{PQ}=2/6=1/3$$

2. Write the perpendicular line.

   $$y+2=-3(x-2)\iff3x+y-4=0$$

E: 3x+y−4=0.

Checks and common pitfalls: A perpendicular line through an endpoint need not bisect the segment.

1. Transform the mean.

   $$\overline{10x}=10\bar x=10$$

2. Transform the variance.

   $$\operatorname{Var}(10X)=10^2\operatorname{Var}(X)=100(0.01)=1$$

D: mean 10, variance 1.

Checks and common pitfalls: Standard deviation scales by 10, but variance scales by 100.

1. Simplify the numerator before rationalizing.

   $$\frac{2(\sqrt{35}-\sqrt{33})}{\sqrt{35}+\sqrt{33}}$$

2. The conjugate denominator is 35−33=2.

   $$\frac{2(\sqrt{35}-\sqrt{33})^2}{35-33}=68-2\sqrt{1155}$$

A: 68−2√1155.

Checks and common pitfalls: Keep the cross term −2√(35·33) when squaring.

1. Apply AM–GM to the two positive new variables.

   $$u+v=3\Rightarrow uv\le(3/2)^2=9/4$$

2. Invert the positive bound and verify equality.

   $$ab=20/(uv)\ge80/9;\quad a=10/3,\ b=8/3$$

C: 80/9.

Checks and common pitfalls: Equality requires 5/a=4/b, not a=b.

1. The total and favourable counts are combination counts.

   $$N=\binom93=84,\quad F=4\cdot2\cdot3=24$$

2. Reduce the fraction.

   $$P=24/84=2/7$$

2/7.

Checks and common pitfalls: The categories have unequal sizes; treating the three categories as equally likely draws would be wrong.

1. There are 3! block orders, and internal orders 4!,2!,3!.

   $$F=3!\,4!\,2!\,3!$$

2. Divide by all permutations of the distinct books.

   $$P=\frac{3!4!2!3!}{9!}=1/210$$

1/210.

Checks and common pitfalls: The books remain distinct even within the same category.

1. Use the point-line distance formula.

   $$r=\frac{|2(4)-4|}{\sqrt{2^2+(-1)^2}}=4/\sqrt5$$

2. Insert the centre and radius.

   $$(x-4)^2+(y-4)^2=16/5\iff5x^2+5y^2-40x-40y+144=0$$

(x−4)²+(y−4)²=16/5.

Checks and common pitfalls: The oblique distance from the origin to M is not the circle radius.

1. Equate the squared centre-to-line distances.

   $$\frac{(4m-4)^2}{m^2+1}=16/5$$

2. Solve and distinguish the two tangents.

   $$5(m-1)^2=m^2+1\iff(2m-1)(m-2)=0$$

3. The root m=2 is L₁; the distinct second line has the other root.

   $$m=1/2$$

m=1/2.

Checks and common pitfalls: Both algebraic roots are tangent slopes, but only one is the requested second line.

1. Split the y² coefficient after completing the x square.

   $$x^2-2mxy+y^2=(x-my)^2+(1-m^2)y^2$$

2. Use the original equality and rearrange.

   $$(1-m^2)y^2=a^2-(x-my)^2$$

(1−m²)y²=a²−(x−my)².

Checks and common pitfalls: The square is x−my, not y−mx in this particular rearrangement.

1. Obtain a sharp upper bound for positive y.

   $$y^2=\frac{a^2}{1-m^2}-\frac{(x-my)^2}{1-m^2}\le\frac{a^2}{1-m^2}$$

2. Equality is attained at positive coordinates because 0<m<1 and feasibility forces a≠0.

   $$y=\frac{|a|}{\sqrt{1-m^2}},\quad x=my>0$$

3. Since m is nonzero, the equality relation is equivalent to the printed form.

   $$x=my\iff y=x/m$$

The maximum occurs when x=my, equivalently y=x/m.

Checks and common pitfalls: The fraction x/m must not be misread as the product mx.

1. Positive y and the squared bound give its maximum.

   $$y_{\max}=\sqrt{a^2/(1-m^2)}=|a|/\sqrt{1-m^2}$$

2. Multiply by m using x=my.

   $$x_0=\frac{m|a|}{\sqrt{1-m^2}}$$

x=m|a|/√(1−m²).

Checks and common pitfalls: The sign of a is unspecified, so √(a²)=|a|.

1. Compute the common difference.

   $$d=(39-3)/(20-2)=2$$

2. Find the first and general terms.

   $$a_1=3-2=1;\quad a_n=1+2(n-1)=2n-1$$

aₙ=2n−1.

Checks and common pitfalls: The index difference is 18, not 19 or 20.

1. Use the difference between consecutive odd reciprocals.

   $$\frac1{(2k-1)(2k+1)}=\frac12\left(\frac1{2k-1}-\frac1{2k+1}\right)$$

2. Telescope the sum and solve the rational equation.

   $$S_n=\frac12\left(1-\frac1{2n+1}\right)=\frac n{2n+1}$$

3. The resulting n is a positive integer.

   $$n/(2n+1)=10/21\Rightarrow21n=20n+10\Rightarrow n=10$$

n=10.

Checks and common pitfalls: Retain the factor 1/2 in the partial-fraction identity.

1. The cosine determines positive sine of the interior angle B.

   $$\sin B=\sqrt{1-8/9}=1/3$$

2. Combine C−A=π/2 with A+B+C=π.

   $$2C=3\pi/2-B\Rightarrow\cos2C=-\sin B=-1/3$$

3. Use the power-reduction identity.

   $$\sin^2C=(1-\cos2C)/2=(1+1/3)/2=2/3$$

sin²C=2/3.

Checks and common pitfalls: The requested expression is sine squared, not sine of twice the angle.

1. Choose the correct cosine sign for the obtuse angle.

   $$\sin C=\sqrt{2/3},\quad\cos C=-1/\sqrt3,\quad\sin A=\sin(C-\pi/2)=1/\sqrt3$$

2. Match opposite sides in the sine rule.

   $$BC=AC\frac{\sin A}{\sin B}=5\frac{1/\sqrt3}{1/3}=5\sqrt3$$

3. The included angle between AC and BC is C.

   $$S=\frac12(5)(5\sqrt3)\sqrt{2/3}=\frac{25\sqrt2}2$$

Area =25√2/2.

Checks and common pitfalls: Taking cos C positive would contradict the obtuse-angle condition.