Trigonometric Functions and Equations: questions with solutions
3 free questions from this chapter (the full set has 30). Marking: Multi-correct: +4 all correct, +1 per correct option if none wrong, −2 if any wrong · Numerical: +4 / 0 · Matching: +3 / −1.
Question 1Level 2 · ExamSingle correct
Let f(x)=sin6x+cos6x+∣sin3x∣ for x∈R. The fundamental (least positive) period of f is ______. Exactly one option is correct.
A6π
B2π
C32π
Dπ
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Answer: D
sin6x+cos6x=1−3sin2xcos2x=1−43sin22x=85+83cos4x, a smooth function with fundamental period 2π. The term ∣sin3x∣ has fundamental period 3π, so π is a common period and f(x+π)=f(x).
No smaller period: ∣sin3x∣ is not differentiable exactly at x=3kπ, while 85+83cos4x is differentiable everywhere. So f is non-differentiable exactly on {3kπ}, and any period T must map this set to itself, so T is a multiple of 3π. If T=3π or 32π, then ∣sin3x∣ is already T-periodic, so cos(4x+4T)=cos4x for all x, forcing 4T∈2πZ, i.e. T a multiple of 2π, which is false. Hence the fundamental period is π.
Difficulty: Hard
Question 2Level 3 · AdvancedSingle correct
Let x,y,z be positive real numbers with xy+yz+zx=1. The maximum value of E=1+x21−x2+1+y21−y2+1+z22z is ______. Exactly one option is correct.
A23
B233
C3
D2
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Answer: B
Put x=tan2A,y=tan2B,z=tan2C with A,B,C∈(0,π). Since cos(2A+2B+2C)=cos2Acos2Bcos2C(1−xy−yz−zx)=0 and the half-angle sum lies in (0,23π), we get 2A+B+C=2π, i.e. A+B+C=π. Conversely every triangle gives admissible x,y,z.
Then E=cosA+cosB+sinC. Now cosA+cosB=2sin2Ccos2A−B≤2sin2C, so with u=2C∈(0,2π): E≤φ(u)=2sinu+sin2u.
φ′(u)=2cosu+2cos2u=2(2cosu−1)(cosu+1), positive for u<3π and negative after, so maxφ=φ(3π)=3+23=233.
Equality: A=B=6π, C=32π, i.e. x=y=2−3, z=3 (check: (2−3)2+23(2−3)=1). Maximum =233.
Difficulty: Peak challenge
Question 3Level 3 · AdvancedSingle correct
Consider the equation tanx+tan2x+tan3x=0, where x ranges over all real numbers for which tanx, tan2x and tan3x are all defined. Here n∈Z and cos−1 denotes the principal value in [0,π]. The complete solution set is ______. Exactly one option is correct.
Ax=3nπ only
Bx=6nπ or x=nπ±21cos−131
Cx=3nπ or x=nπ±21cos−131
Dx=3nπ or x=2nπ±cos−132
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Answer: C
tanx+tan2x=cosxcos2xsin3x, so the equation is sin3x(cosxcos2x1+cos3x1)=0.
Case 1: sin3x=0, x=3nπ. Then cos3x=±1, cos2x=cos32nπ=0, cosx=cos3nπ=0: all valid.
Case 2: cos3x+cosxcos2x=0. Using cosxcos2x=21(cos3x+cosx): 3cos3x+cosx=0, i.e. 3(4c3−3c)+c=0, so 4c(3c2−2)=0 with c=cosx. c=0 is excluded (tanx undefined), so cos2x=32, i.e. cos2x=31. Check: cos2x=31=0 and cos3x=c(4c2−3)=−3c=0, so all are valid. cos2x=31⟺x=nπ±21cos−131.
Option (B) contains x=6π, where tan3x is undefined; (D) gives only cosx=+2/3 and misses cosx=−2/3; (A) misses Case 2. Answer (C).
Difficulty: Very hard
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