JEE Mathematics Inverse Circular Functions MCQs Set 07

Answer: (a) (B), (C) and (D) only

Explanation:
1. Take cosine of both sides: cos(θ₁+θ₂+θ₃+θ₄) = 0 with tan θ = a/x etc. This gives \( 1-e_2+e_4=0 \), where e₂ is the sum of the pairwise products of a/x, b/x, c/x, d/x and e₄ is their product.
2. Multiplying by x⁴: \( x^4-(ab+ac+ad+bc+bd+cd)x^2+abcd=0 \). The four numbers xᵢ are the roots of this biquadratic.
3. The roots are of the form ±p, ±q, so their sum is 0, not a + b + c + d: (A) is false. Then Σ1/xᵢ = 0: (B) true. Product of roots = abcd: (C) true.
4. Each bracket in (D) equals minus the missing root, so the product is x₁x₂x₃x₄ = abcd: (D) true.

Teacher's Note:
1. Reduce the sum of arctangents to a polynomial equation in x.
2. The equation has only even powers of x, so the roots come in ± pairs.

Answer: (a) (B) and (C) only

Explanation:
1. \( \tan(2\tan^{-1}a)=\frac{2a}{1-a^2} \) and \( \tan(\tan^{-1}a+\tan^{-1}a^3)=\frac{a+a^3}{1-a^4}=\frac{a}{1-a^2} \).
2. So \( 2x=\frac{2a}{1-a^2}+\frac{2a}{1-a^2}=\frac{4a}{1-a^2} \), i.e. \( x=\frac{2a}{1-a^2} \) with a ≠ ±1.
3. (A) is x(1 − a²) = 2a rearranged: valid. (D) is the required condition: valid. (B) is not equivalent (for a = 0 it gives 1 = 0), and (C) is not needed since a = 0, x = 0 works. So (B) and (C) are invalid.

Teacher's Note:
1. Simplify both tangent expressions with the addition formula.
2. Check each statement against x = 2a/(1 − a²).

Answer: (a) (A), (C) and (D)

Explanation:
1. cos(cos 0) = cos 1 ≈ 0.540. π − 1 ≈ 2.14 is outside [−π/2, π/2], so sin⁻¹(sin(π − 1)) = 1 and the value is sin 1 ≈ 0.841. cos⁻¹(cos(2π − 2)) = 2, giving sin 2 ≈ 0.909.
2. (A): 0.540 < 0.841 < 0.909 is true. (B) needs sin 2 < sin 1, which is false.
3. (C): each term on the left equals cos⁻¹(cos 1) = 1, so the sum is 5000; each term on the right equals cot⁻¹(cot 2) = 2, so the sum is 5000. True.
4. (D): sin⁻¹(sin 4) = π − 4, cos⁻¹cos(π − 4) = 4 − π, and the remaining inverse functions all return 4 − π since it lies in each principal range. True.

Teacher's Note:
1. Reduce each inverse composition to its principal value.
2. Compare numerical values for the inequalities.

Answer: (a) (A), (B) and (C)

Explanation:
1. \( \tan(\tan^{-1}x+\tan^{-1}x^3)=\frac{x+x^3}{1-x^4}=\frac{x}{1-x^2} \), so the expression is \( \frac{2x}{1-x^2} \): (A).
2. (B) tan(2 tan⁻¹x) = 2x/(1 − x²): equal. (C) cot⁻¹(−x) = π − cot⁻¹x, so the angle is π − 2cot⁻¹x and its tangent is −tan(2 cot⁻¹x) = 2x/(1 − x²): equal.
3. (D) tan(2 cot⁻¹x) = 2x/(x² − 1), the negative of the required value: not equal.

Teacher's Note:
1. Use the addition formula for the tangent first.
2. Compare each option with 2x/(1 − x²).

Answer: (a) −1, 1 or 2

Explanation:
1. Since cos⁻¹t = π/2 − sin⁻¹t, the equation becomes x² + x + 1 = ax + 1 with |x² + x + 1| ≤ 1, which forces x ∈ [−1, 0].
2. x² + (1 − a)x = 0 gives x = 0 or x = a − 1. A second distinct solution in [−1, 0] needs a − 1 ∈ [−1, 0), i.e. a ∈ [0, 1).
3. Hence a = 0 is allowed (the solutions are 0 and −1), while a = −1, 1 or 2 are not.

Teacher's Note:
1. Convert to sin⁻¹ on both sides and keep the domain restriction.
2. Find the values of a giving a second root inside the domain.

Answer: (a) \( \sin^{-1}\left(\frac{x-2}2\right)+\frac\pi6 \)

Explanation:
1. \( f(x)=2\sin\left(x-\frac\pi6\right)+2 \). For x ∈ [−π/3, 2π/3] the angle x − π/6 runs over [−π/2, π/2], where sine is one-one with values [−1, 1], so f maps onto [0, 4].
2. Solving y = 2 sin(x − π/6) + 2: \( x-\frac\pi6=\sin^{-1}\frac{y-2}2 \).
3. Hence \( f^{-1}(x)=\sin^{-1}\left(\frac{x-2}2\right)+\frac\pi6 \).

Teacher's Note:
1. Write the expression as a single sine with a phase shift.
2. Because the angle lies in [−π/2, π/2], sin⁻¹ inverts it directly.

Answer: (b) 127°

Explanation:
1. \( \tan(-37^\circ)=-\tan37^\circ=\cot127^\circ \) (since cot(90° + 37°) = −tan 37°).
2. cot⁻¹ has range (0°, 180°) and 127° lies in it, so the value is 127°.

Teacher's Note:
1. Write the tangent as a cotangent of an angle in (0°, 180°).
2. The range of cot⁻¹ is (0, π).

Answer: (b) 127°

Explanation:
1. cos(−233°) = cos 233° = cos(360° − 127°) = cos 127°.
2. 127° lies in [0°, 180°], the range of cos⁻¹, so the value is 127°.

Teacher's Note:
1. Reduce the angle using the period of cosine.
2. cos⁻¹ returns an angle in [0°, 180°].

Answer: (d) \( \frac23 \)

Explanation:
1. Let θ = cos⁻¹(1/9), an acute angle with cos θ = 1/9.
2. \( \sin\frac\theta2=\sqrt{\frac{1-\cos\theta}2}=\sqrt{\frac{8/9}2}=\sqrt{\frac49}=\frac23 \).

Teacher's Note:
1. Use the half-angle formula.
2. θ/2 is acute, so take the positive root.

Answer: (c) \( \frac34 \)

Explanation:
1. Let θ = arccos(1/8), acute, so cos θ = 1/8.
2. \( \cos\frac\theta2=\sqrt{\frac{1+\cos\theta}2}=\sqrt{\frac{9/8}2}=\sqrt{\frac9{16}}=\frac34 \).

Teacher's Note:
1. Use the half-angle formula for cosine.
2. θ/2 is acute, so the root is positive.

Answer: (b) 1

Explanation:
1. Taking tangents: \( \frac{8x}{1-15x^2}=\frac{9x}{1-14x^2} \).
2. x = 0 is a solution. Otherwise 8(1 − 14x²) = 9(1 − 15x²), so 23x² = 1 and x = ±1/√23, which are not integers.
3. The only integral solution is x = 0, so the number is 1.

Teacher's Note:
1. Apply the tangent to both sides and cross-multiply.
2. Check which roots are integers.

Answer: (b) 2

Explanation:
1. The fundamental period of sin⁻¹(sin kx) is 2π/|k|.
2. 2π/|k| = π/2 gives |k| = 4, i.e. k = 4 or k = −4: 2 values.

Teacher's Note:
1. sin⁻¹(sin t) has fundamental period 2π.
2. The sign of k does not change the period.

Answer: (b) 2

Explanation:
1. sin⁻¹ x needs |x| ≤ 1 and sec⁻¹ x needs |x| ≥ 1, so x = ±1.
2. tan⁻¹ is defined everywhere, so the domain is {−1, 1}: 2 elements.

Teacher's Note:
1. Intersect the domains of the three functions.
2. Only the end points survive.

Answer: (c) 3

Explanation:
1. cos⁻¹ is defined only when [5x] ∈ {−1, 0, 1}, giving the values π, π/2 and 0.
2. a + b + c = π + π/2 + 0 = 3π/2 = λπ/2, so λ = 3.

Teacher's Note:
1. Only integers in [−1, 1] are allowed as the argument.
2. Add the three possible values.

Answer: (b) \( \left[\frac14,\frac34\right] \)

Explanation:
1. sin⁻¹x + cos⁻¹x = π/2 on the common domain [−1, 1].
2. So f(x) = (π/2 + tan⁻¹x)/π with tan⁻¹x ∈ [−π/4, π/4], giving f ∈ [1/4, 3/4].

Teacher's Note:
1. Use sin⁻¹x + cos⁻¹x = π/2.
2. The domain is [−1, 1], where tan⁻¹ lies between ±π/4.

Answer: (d) \( \left[0,\frac\pi2\sin1\right] \)

Explanation:
1. The domain is [−1, 1], and f is even since both sin⁻¹x and sin x are odd.
2. For 0 ≤ x ≤ 1 both factors are non-negative and increasing, so f increases from f(0) = 0 to f(1) = (π/2) sin 1.
3. The range is [0, (π/2) sin 1].

Teacher's Note:
1. Check the parity of f to reduce the domain to [0, 1].
2. A product of non-negative increasing functions is increasing.

Answer: (a) P, Q and R only

Explanation:
1. The range is [−π/2, π/2] ≈ [−1.57, 1.57], which contains exactly one natural number, 1: (P) holds. g is periodic with period 2π (Q) and defined for all real x (R).
2. g is odd, not even, so its graph is not symmetric about the y-axis: (S) fails.

Teacher's Note:
1. Recall the graph of sin⁻¹(sin x): a zig-zag wave.
2. An odd function that is not zero is not symmetric about the y-axis.

Answer: (a) P only

Explanation:
1. [x] + [−x] is 0 for integers and −1 otherwise, so the square root needs x to be an integer. Also 2 − [x] ≥ 0 and x ≠ 0, so the domain is {x integer: x ≤ 2, x ≠ 0}.
2. On this domain k(x) = √(2 − x) + 1/x². The value is a natural number only for x = 1 (giving 2); the other values (e.g. 0.25, 2.73, 2.25) are not. So (P) holds.
3. The domain is not R (R fails), k is not periodic (Q fails), and the domain is not symmetric about the y-axis since it contains −5 but not 5 (S fails). So only P holds.

Teacher's Note:
1. Find the domain first: it is a set of integers.
2. Evaluate k at the first few points of the domain.

Answer: (a) −2, −1 and 0

Explanation:
1. g∘f is defined when −4 ≤ f(x) ≤ 4. On [−1, 1] the function f(x) = x² − 3x + b is decreasing (vertex at 3/2), so its largest value is f(−1) = 4 + b and its smallest f(1) = b − 2.
2. Need 4 + b ≤ 4 and b − 2 ≥ −4, so −2 ≤ b ≤ 0: the values −2, −1 and 0.

Teacher's Note:
1. Find the range of f on [−1, 1] using its monotonicity.
2. Require it to lie inside [−4, 4], the domain of sin⁻¹(x/4).

Answer: (a) 0 and 1

Explanation:
1. Put t = sin⁻¹(x/4) ∈ [−π/2, π/2]. Then (f∘g)(x) = t² + 3t + 2.
2. This has its minimum at t = −3/2 (inside the interval) equal to −1/4, and its maximum at t = π/2 equal to about 9.18.
3. The range is [−0.25, 9.18], which contains 0 and 1 but not −1 or −2.

Teacher's Note:
1. Substitute t = sin⁻¹(x/4) to get a quadratic in t.
2. Locate the vertex relative to the interval of t.

Answer: (a) 0 and 1

Explanation:
1. Domain: |∛(x² + 4x + 3)| ≤ 1 gives (x + 2)² ≤ 2, and then cos⁻¹ of the cube root takes values in (0, π] (the value 0 is excluded since ln 0 is undefined).
2. So ln(cos⁻¹ ·) takes every value in (−∞, ln π], where ln π ≈ 1.14, and the greatest integer takes all integer values ≤ 1.
3. Among 0, 1, 2, 4 the values 0 and 1 occur.

Teacher's Note:
1. Find the range of cos⁻¹ of the cube root first.
2. ln π ≈ 1.14 is the upper limit before taking the greatest integer.

Answer: (c) 4

Explanation:
1. f(x) = sinh⁻¹x, so f⁻¹(x) = sinh x. We solve sinh x = g(x).
2. For x ≤ 1: x = 0 is a solution. On (0, 1), sinh x − x^{1/3} is negative near 0 and equals sinh 1 − 1 ≈ 0.18 > 0 at x = 1, so there is a root in (0, 1); by symmetry (both sides are odd) there is a matching root in (−1, 0). That is 3 solutions with x ≤ 1.
3. For x > 1: sinh x increases from 1.18 while 2e^{1−x} decreases from 2, so they cross exactly once.
4. Total = 4.

Teacher's Note:
1. Identify f⁻¹ as sinh x.
2. Analyse the intersections separately on x ≤ 1 and x > 1.