Inverse Trigonometric Functions: 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
The value of sin−1(sin10)+cos−1(cos10)+tan−1(tan10)+cot−1(cot10) is Convention: all inverse trigonometric functions denote principal values, sin−1x∈[−2π,2π], cos−1x∈[0,π], tan−1x∈(−2π,2π), cot−1x∈(0,π), sec−1x∈[0,π]∖{2π}, cosec−1x∈[−2π,2π]∖{0}; angles are in radians. Exactly one option is correct.
A4π−10
B20−6π
Cπ
D0
Show answer and solution
Answer: C
Note 3π≈9.425<10<27π≈10.996.
sin−1(sin10): since sin(3π−t)=sint and 3π−10≈−0.575∈[−2π,2π], it equals 3π−10.
cos−1(cos10): cos(4π−10)=cos10 and 4π−10≈2.566∈[0,π], so it equals 4π−10.
tan−1(tan10): period π; 10−3π≈0.575∈(−2π,2π), so 10−3π.
cot−1(cot10): period π; 10−3π∈(0,π), so 10−3π.
Sum =(3π−10)+(4π−10)+(10−3π)+(10−3π)=π.
Difficulty: Hard
Question 2Level 2 · ExamSingle correct
The set of all real x satisfying sin−1x≥cos−1(2x) is Convention: all inverse trigonometric functions denote principal values, sin−1x∈[−2π,2π], cos−1x∈[0,π], tan−1x∈(−2π,2π), cot−1x∈(0,π), sec−1x∈[0,π]∖{2π}, cosec−1x∈[−2π,2π]∖{0}; angles are in radians. Exactly one option is correct.
A[51,21]
B[51,1]
C[−21,51]
D[51,52]
Show answer and solution
Answer: A
Both sides are defined only for x∈[−1,1] and 2x∈[−1,1], i.e. x∈[−21,21].
On this interval sin−1x is strictly increasing and cos−1(2x) strictly decreasing, so φ(x)=sin−1x−cos−1(2x) is strictly increasing and the solution set is [x0,21] where φ(x0)=0.
At x0 the common value θ lies in [−2π,2π]∩[0,π]=[0,2π], with x0=sinθ and 2x0=cosθ. Hence tanθ=21 and x0=51≈0.447≤21.
Check: φ(21)=6π−0>0.
Answer: [51,21].
Trap: forgetting the domain of cos−1(2x) gives [51,1].
Difficulty: Hard
Question 3Level 3 · AdvancedSingle correct
Let f(x)=sin−1(1+x22x)+cos−1(1+x21−x2) for x∈R. Which of the following is correct? Convention: all inverse trigonometric functions denote principal values, sin−1x∈[−2π,2π], cos−1x∈[0,π], tan−1x∈(−2π,2π), cot−1x∈(0,π), sec−1x∈[0,π]∖{2π}, cosec−1x∈[−2π,2π]∖{0}; angles are in radians. Exactly one option is correct.
Af(x)=4tan−1x for all x≥0
Bf(x)=π for all x>1 and f(x)=−π for all x<−1
Cf(x)=0 for all x≤0
Df(x)=π for all x≥1 and f(x)=0 for all x∈[−1,0]
Show answer and solution
Answer: D
Put x=tanθ, θ∈(−2π,2π). Then 1+x22x=sin2θ, 1+x21−x2=cos2θ, 2θ∈(−π,π).
sin−1(sin2θ)=2θ if ∣x∣≤1; π−2θ if x>1; −π−2θ if x<−1.
cos−1(cos2θ)=2θ if x≥0; −2θ if x<0.
Hence f(x)=4tan−1x on [0,1]; f(x)=π on (1,∞); f(x)=0 on [−1,0); f(x)=−π−4tan−1x on (−∞,−1).
Since f(1)=π and f(0)=0: f=π on [1,∞) and f=0 on [−1,0], which is option D.
A fails at x=2 (f=π=4tan−12). B fails at x=−2 (f=−π+4tan−12≈1.29). C fails for every x<−1.
Difficulty: Very hard
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