Inverse Trigonometric Functions — Chapter Test | Class 12 IMO

Q1

The minimum value of (sin⁻¹(x))² + (cos⁻¹(x))² is:

From the quadratic 2a² − πa + π²/4, the minimum occurs at the vertex a = -(-π)/4 = π/4. The value is 2(π/4)² − π(π/4) + π²/4 = π²/8.

Q2

The solution of sin⁻¹(1−x) − 2sin⁻¹(x) = π/2 is:

Substitute x = 0: sin⁻¹(1) − 0 = π/2, which is true. If x = 1/2: sin⁻¹(1/2) − 2sin⁻¹(1/2) = π/6 − 2(π/6) = −π/6 ≠ π/2. Thus x=0 is the only solution.

Q3

The value of sin⁻¹[sin(2π/3)] + cos⁻¹[cos(4π/3)] is:

sin⁻¹(sin 2π/3) = π/3 (principal value). cos⁻¹(cos 4π/3) = cos⁻¹(−1/2) = 2π/3. Sum = π/3 + 2π/3 = π.

Q4

The value of sin(2 tan⁻¹(1/3)) + cos(tan⁻¹(2√2)) is:

Let θ = tan⁻¹(1/3). sin 2θ = 2tanθ/(1+tan²θ) = 2(1/3)/(1+1/9) = (2/3)/(10/9) = 3/5. Let φ = tan⁻¹(2√2). cos φ = 1/√(1+8) = 1/3. Sum = 3/5 + 1/3 = 14/15.

Q5

The number of real solutions of the equation √(1+cos 2x) = √2 cos⁻¹(cos x) in [π/2, π] is:

LHS = √(2cos²x) = √2|cos x| = −√2 cos x (since cos x ≤ 0 in [π/2, π]). RHS = √2 cos⁻¹(cos x). For x ∈ [π/2, π], cos⁻¹(cos x) = 2π−x? No, in [0,π], cos⁻¹(cos x)=x. So RHS = √2 x. Equation: −√2 cos x = √2 x → cos x = −x. Drawing graphs of y=cos x and y=−x in [π/2, π] shows 1 intersection.

Q6

The value of tan⁻¹(tan(3π/4)) is:

3π/4 is outside (−π/2, π/2). tan(3π/4) = tan(π − π/4) = −tan(π/4) = tan(−π/4). Hence the value is −π/4.

Q7

The principal value of sec⁻¹(−2) is:

sec⁻¹(−2) = π − sec⁻¹(2) = π − π/3 = 2π/3, which lies in [0, π] − {π/2}, the principal branch of sec⁻¹.

Q8

The value of sin⁻¹(√3/2) + cos⁻¹(1/2) − tan⁻¹(1) is:

sin⁻¹(√3/2) = π/3. cos⁻¹(1/2) = π/3. tan⁻¹(1) = π/4. Sum = π/3+π/3−π/4 = 2π/3−π/4 = 5π/12? Not in options. We compute: 2π/3 = 8π/12, π/4=3π/12, difference = 5π/12. Options: π/2=6π/12, 2π/3=8π/12, π/3=4π/12, 5π/6=10π/12. None matches 5π/12. Fix question: sin⁻¹(√3/2) + cos⁻¹(√3/2) − tan⁻¹(√3). sin⁻¹(√3/2)=π/3, cos⁻¹(√3/2)=π/6, sum=π/2. tan⁻¹(√3)=π/3. π/2−π/3=π/6. Not in options. Better: sin⁻¹(1/2)+cos⁻¹(1/2)−tan⁻¹(1) = π/6+π/3−π/4 = π/2−π/4=π/4. Not in options. We use sin⁻¹(√3/2)+cos⁻¹(1/2)+tan⁻¹(0) = π/3+π/3+0=2π/3. That's option 1. We'll change question accordingly.

Q9

The trajectory of a water fountain jet at India Gate makes an angle tan⁻¹(4/3) with the horizontal. The sine of this angle is:

If tan(θ) = 4/3, opp = 4, adj = 3, hyp = 5. Therefore, sin(θ) = 4/5.

Q10

The domain of f(x) = sin⁻¹(x² − 3x + 2) is:

For sin⁻¹(y), −1 ≤ y ≤ 1. Solve −1 ≤ x² − 3x + 2 ≤ 1. x² − 3x + 2 ≤ 1 → x² − 3x + 1 ≤ 0 → x ∈ [(3−√5)/2, (3+√5)/2]. x² − 3x + 2 ≥ −1 → x² − 3x + 3 ≥ 0, discriminant negative so always true. Domain is [(3−√5)/2, (3+√5)/2].

Q11

A wheelchair ramp at a Mumbai hospital rises 1m for every 5m distance along the ground. The angle of inclination is:

The rise (opposite) is 1m and run (adjacent) is 5m. tan(θ) = 1/5. Hence, θ = tan⁻¹(1/5).

Q12

A CCTV camera is on a 10m pole at a shop. It targets a till 10m away from the pole's base. The angle of depression is:

tan(θ) = opposite / adjacent = 10 / 10 = 1. Angle of depression equals angle of elevation, so tan⁻¹(1) = 45°.

Q13

If tan⁻¹x = sin⁻¹(1/√2), then x equals:

sin⁻¹(1/√2) = π/4. So tan⁻¹x = π/4 → x = tan(π/4) = 1.

Q14

The maximum value of (sin⁻¹(x))² + (cos⁻¹(x))² is:

Let a = sin⁻¹(x). Then (cos⁻¹(x))² = (π/2 − a)². The sum is 2a² − πa + π²/4. The maximum occurs at the domain boundary a = −π/2 (when x = −1), yielding 5π²/4.

Q15

If sin⁻¹x + sin⁻¹y = π, then the value of x + y is:

The maximum value of sin⁻¹(x) is π/2. For the sum to be π, both terms must be π/2. Thus x = 1 and y = 1. The sum x + y is 2.