JEE Mathematics Inverse Circular Functions MCQs Set 06

Answer: (b) \( \tan^{-1}(2) \)

Explanation:
1. \( \cot^{-1}\left(n^2+\frac34\right)=\tan^{-1}\frac{1}{n^2+\frac34}=\tan^{-1}\left(n+\tfrac12\right)-\tan^{-1}\left(n-\tfrac12\right) \), because \( \frac{(n+\frac12)-(n-\frac12)}{1+(n^2-\frac14)}=\frac{1}{n^2+\frac34} \).
2. The series telescopes: sum to ∞ = \( \frac{\pi}{2}-\tan^{-1}\frac12=\tan^{-1}2 \).

Teacher's Note:
1. Write each term as a difference of two inverse tangents to get a telescoping sum.
2. The last term tends to tan⁻¹(∞) = π/2.

Answer: (a) \( 4\tan^{-1}(1) \)

Explanation:
1. Let t = tan A with 0 < t < 1. Then \( \frac12\tan2A=\frac{t}{1-t^2} \), cot A = 1/t and cot³A = 1/t³.
2. For the first two terms: \( \tan(\alpha+\beta)=\frac{\frac{t}{1-t^2}+\frac1t}{1-\frac1{1-t^2}}=-\frac{1}{t^3} \). Both angles are positive and the product of the tangents is greater than 1, so the sum lies in (π/2, π): it equals \( \pi-\tan^{-1}\left(\frac1{t^3}\right) \).
3. Adding the third term tan⁻¹(cot³A) cancels this, so the total is π = 4 tan⁻¹(1).

Teacher's Note:
1. When the product of the arguments exceeds 1, the sum of two inverse tangents picks up an extra π.
2. A numerical check at one value of A confirms the constant result.

Answer: (a) \( \tan\alpha=\cot\beta \)

Explanation:
1. \( \cos(\sin^{-1}x)=\sqrt{1-x^2} \) and \( \sin(\cos^{-1}x)=\sqrt{1-x^2} \).
2. So α = sin⁻¹√(1 − x²) and β = cos⁻¹√(1 − x²), both in [0, π/2].
3. α + β = π/2, hence tan α = cot β.

Teacher's Note:
1. Both inner expressions equal √(1 − x²).
2. sin⁻¹y + cos⁻¹y = π/2 for y ∈ [0, 1].

Answer: (a) \( \cos^{-1}\left(\ln\left(\frac{2e+4}{3}\right)\right) \)

Explanation:
1. (a) (2e + 4)/3 ≈ 3.15, so ln of it ≈ 1.15 > 1, which is outside [−1, 1]: cos⁻¹ is not defined.
2. (b) π/3 ≈ 1.05 ≥ 1 and (d) π ≥ 1, so cosec⁻¹ and sec⁻¹ are defined. (c) cot⁻¹ is defined for every real number.

Teacher's Note:
1. cos⁻¹ and sin⁻¹ need arguments in [−1, 1]; sec⁻¹ and cosec⁻¹ need |argument| ≥ 1.
2. Estimate the logarithm to see if it exceeds 1.

Answer: (c) 2

Explanation:
1. Let t = x² + x. For arcsin to be defined we need 0 ≤ t + 1 ≤ 1, and for the square root, t ≥ 0. So t = 0.
2. t = 0 gives x = 0 or x = −1.
3. Check: arctan 0 + arcsin 1 = π/2 for both. So there are 2 solutions.

Teacher's Note:
1. Restrict the variable using the domains of the square root and arcsin.
2. Always verify the candidate solutions in the original equation.

Answer: (a) (0, 1)

Explanation:
1. Put x = cos θ, θ ∈ (0, π). Then the argument is cos 2θ / sin 2θ = cot 2θ (as sin θ > 0).
2. cot⁻¹(cot 2θ) = 2θ only when 2θ ∈ (0, π), i.e. θ ∈ (0, π/2).
3. That means x = cos θ ∈ (0, 1).

Teacher's Note:
1. Substitute x = cos θ so that √(1 − x²) = sin θ.
2. The principal value of cot⁻¹ lies in (0, π), which restricts θ.

Answer: (d) \( \frac{17\pi}{20} \)

Explanation:
1. sin(2π/5) = cos(π/2 − 2π/5) = cos(π/10), so the bracket is cos(7π/5) − cos(π/10).
2. Evaluating: (cos 252° − sin 72°)/√2 = (−0.3090 − 0.9511)/1.4142 = −0.8910 = cos(153°) = cos(17π/20).
3. 17π/20 lies in [0, π], so the principal value is 17π/20.

Teacher's Note:
1. Compute the numerical value of the bracket and recognise it as a cosine of a standard angle.
2. The principal value of cos⁻¹ lies in [0, π].

Answer: (c) \( \frac{2\pi}{5} \)

Explanation:
1. \( \cos(\tan^{-1}x)=\frac{1}{\sqrt{1+x^2}}=x\Rightarrow x^4+x^2-1=0 \), so \( x^2=\frac{\sqrt5-1}{2} \).
2. \( \frac{x^2}{2}=\frac{\sqrt5-1}{4}=\cos72^\circ \).
3. Hence cos⁻¹(x²/2) = 72° = 2π/5.

Teacher's Note:
1. Convert the equation into a quadratic in x².
2. (√5 − 1)/4 is the exact value of cos 72°.

Answer: (b) 1

Explanation:
1. sin⁻¹a + sin⁻¹b = π/2 means a² + b² = 1 with a, b ≥ 0.
2. So 25/x² + 144/x² = 1, giving x = ±13. For x = −13 both inverse sines are negative, so the sum cannot be π/2.
3. Only x = 13 works: 1 solution.

Teacher's Note:
1. If sin⁻¹a + sin⁻¹b = π/2 then b = cos(sin⁻¹a), hence a² + b² = 1.
2. Reject roots that make the sum negative.

Answer: (a) \( (3+\sqrt{10})^{-1} \)

Explanation:
1. Let θ = cot⁻¹3, so tan θ = 1/3. Let t = tan(θ/2): \( \frac{2t}{1-t^2}=\frac13 \).
2. t² + 6t − 1 = 0, and t > 0 gives t = −3 + √10.
3. \( \sqrt{10}-3=\frac{1}{\sqrt{10}+3} \).

Teacher's Note:
1. Use tan θ = 2t/(1 − t²) with t = tan(θ/2).
2. Choose the positive root because θ/2 is acute.

Answer: (d) \( \left[\frac1{\sqrt2},1\right] \)

Explanation:
1. Put x = cos θ, θ ∈ [0, π]. LHS = 2θ and RHS = sin⁻¹(sin 2θ).
2. sin⁻¹(sin 2θ) = 2θ only when 2θ ∈ [−π/2, π/2], i.e. θ ∈ [0, π/4].
3. So x = cos θ ∈ [1/√2, 1].

Teacher's Note:
1. Substituting x = cos θ turns the square root into sin θ.
2. The principal range of sin⁻¹ limits 2θ.

Answer: (c) [−1, 0) ∪ {1}

Explanation:
1. Put x = cos θ with θ ∈ [0, π]. Then √(1 − x²) = sin θ, cos⁻¹x = θ, sin⁻¹x = π/2 − θ, and the cot⁻¹ argument is tan θ (x ≠ 0).
2. If x > 0 (θ < π/2): LHS = θ + θ = 2θ and cot⁻¹(tan θ) = π/2 − θ, so RHS = 0. Equality needs θ = 0, i.e. x = 1.
3. If x < 0 (θ > π/2): LHS = (π − θ) + θ = π and cot⁻¹(tan θ) = 3π/2 − θ, so RHS = 3π/2 − θ − (π/2 − θ) = π. Equality holds for every x in [−1, 0).
4. Solution set: [−1, 0) ∪ {1}.

Teacher's Note:
1. Test both sides in each region of x: negative, positive and the endpoints.
2. Inverse-trigonometric identities often hold only on part of the domain.

Answer: (c) 25°

Explanation:
1. tan 65° − 2 tan 40° = tan 65° − tan 40° − tan 40°.
2. tan 25° = tan(65° − 40°) = (tan 65° − tan 40°)/(1 + tan 65° tan 40°), and tan 65° = cot 25°, so the expression simplifies to tan 25° (numerically 2.1445 − 1.6782 = 0.4663 = tan 25°).
3. tan⁻¹(tan 25°) = 25°.

Teacher's Note:
1. Use tan 65° = cot 25° and the difference formula for tangent.
2. A calculator check of both sides confirms the value.

Answer: (c) \( \{1,(1+\pi)^2\} \)

Explanation:
1. sec⁻¹x needs |x| ≥ 1 and cos⁻¹x needs |x| ≤ 1, so the domain is x = ±1.
2. f(1) = (1 + 0)(1 + 0) = 1. f(−1) = (1 + π)(1 + π) = (1 + π)².
3. Range = {1, (1 + π)²}.

Teacher's Note:
1. Find the common domain first; here it contains just two points.
2. Evaluate the function at those points.

Answer: (d) \( \frac{3\pi}{4} \)

Explanation:
1. (3/2)(y + 1) = 2 gives y = 1/3.
2. cot⁻¹(1/2) + cot⁻¹(1/3) = tan⁻¹2 + tan⁻¹3. Since 2·3 > 1, the sum is π + tan⁻¹((2 + 3)/(1 − 6)) = π + tan⁻¹(−1) = π − π/4.
3. The measure is 3π/4.

Teacher's Note:
1. When the product of the tangents exceeds 1, the sum of inverse tangents is greater than π/2.
2. tan⁻¹(−1) = −π/4.

Answer: (c) 26

Explanation:
1. cos⁻¹x + cos⁻¹2x = π − cos⁻¹3x. Take cosine: \( 2x^2-\sqrt{(1-x^2)(1-4x^2)}=-3x \).
2. So \( \sqrt{(1-x^2)(1-4x^2)}=2x^2+3x \). Squaring: 4x⁴ + 12x³ + 9x² = 1 − 5x² + 4x⁴, i.e. 12x³ + 14x² − 1 = 0.
3. Comparing with ax³ + bx² + cx − 1 = 0: a = 12, b = 14, c = 0, so a + b + c = 26.

Teacher's Note:
1. Isolate one inverse cosine, take cosines, and square to remove the root.
2. Compare coefficients with the given cubic.

Answer: (a) f(−1)

Explanation:
1. Since a > 0, f is increasing: f(−1) = 0 and f(1) = 2, so a = 1, b = 1 and f(x) = x + 1.
2. tan⁻¹1 + tan⁻¹(1/2) + tan⁻¹(1/3) = π/4 + π/4 = π/2, so the cotangent is 0.
3. f(−1) = 0, so the value equals f(−1).

Teacher's Note:
1. Find f from the onto condition first.
2. cot of a right angle is 0.

Answer: (d) \( \cot(\sin^{-1}(\sin4)) \)

Explanation:
1. tan⁻¹(tan 4) = 4 − π ≈ 0.858, so (a) = cos(0.858) > 0.
2. cot⁻¹(cot 4) = 4 − π ≈ 0.858, so (b) = sin(0.858) > 0.
3. cos⁻¹(cos 5) = 2π − 5 ≈ 1.283, so (c) = tan(1.283) > 0.
4. sin⁻¹(sin 4) = π − 4 ≈ −0.858, so (d) = cot(−0.858) < 0.

Teacher's Note:
1. Reduce each inverse function to its principal value first.
2. cot is negative for angles in (−π/2, 0).

Answer: (a) 9

Explanation:
1. Let t = |x| > 0 and u = (2t² + 10t + 4)/(t² + 5t + 3) = 2 − 2/(t² + 5t + 3), which lies in (4/3, 2].
2. Since 0 < u < π, sin⁻¹cos u = π/2 − u. The right side is (2 − 18t)/(9t) + π/2.
3. So −u = 2/(9t) − 2, giving 2/(t² + 5t + 3) = 2/(9t), i.e. t² − 4t + 3 = 0, t = 1 or 3. The roots are x = ±1, ±3 and their product is (−1)(1)(−3)(3) = 9.

Teacher's Note:
1. Use sin⁻¹(cos u) = π/2 − u for u in [0, π].
2. Each positive value of |x| gives two real roots ±x.

Answer: (d) (−∞, cot 1)

Explanation:
1. Need ln(cot⁻¹x) > 0, so cot⁻¹x > 1.
2. cot⁻¹ is a decreasing function with values in (0, π), so cot⁻¹x > 1 means x < cot 1.
3. Domain: (−∞, cot 1).

Teacher's Note:
1. The reciprocal of a square root requires the quantity under the root to be strictly positive.
2. cot⁻¹ is decreasing, so the inequality reverses when you invert it.

Answer: (b) \( \frac2\pi+1 \)

Explanation:
1. Domain is [−1, 1]. Since b > 0, the range is [a, a + bπ].
2. Equating: a = −1 and a + bπ = 1, so b = 2/π.
3. b − a = 2/π + 1.

Teacher's Note:
1. For b > 0 the function is increasing in cos⁻¹x, so the range runs from a to a + bπ.
2. Match the endpoints with −1 and 1.

Answer: (a) \( \frac\pi2+\cos1 \)

Explanation:
1. Domain is [−1, 1]; arcsin x + arccos x = π/2.
2. f′(x) = cos x − sin x + sec²x + 1/(1 + x²) > 0 on [−1, 1], so f is increasing: m = f(−1), M = f(1).
3. f(1) + f(−1) = 2cos 1 + π (the odd terms cancel, and the arctan terms cancel). The mean is π/2 + cos 1.

Teacher's Note:
1. Show the function is increasing so extreme values occur at the end points.
2. Odd terms cancel in f(1) + f(−1).

Answer: (b) \( \left[-\frac{3\pi}{2},-\frac\pi2\right] \)

Explanation:
1. Let θ = tan⁻¹x ∈ (−π/2, π/2) and cot⁻¹x = π/2 − θ. The equation becomes g(θ) = θ² + (a + π)θ − π²/2 = 0.
2. g(0) = −π²/2 < 0, so the parabola has one root on each side of 0. There is no root in the open interval exactly when g(π/2) ≤ 0 and g(−π/2) ≤ 0.
3. g(π/2) = π²/4 + aπ/2 ≤ 0 gives a ≤ −π/2; g(−π/2) = −3π²/4 − aπ/2 ≤ 0 gives a ≥ −3π/2. So a ∈ [−3π/2, −π/2].

Teacher's Note:
1. Use cot⁻¹x = π/2 − tan⁻¹x to get a quadratic in θ = tan⁻¹x.
2. Both roots of the quadratic must lie outside the open interval (−π/2, π/2).

Answer: (c) \( B<\frac{7\pi}{12}<A \)

Explanation:
1. 2√2 − 1 ≈ 1.828, so tan⁻¹ of it ≈ 61.3° and A ≈ 122.6°.
2. 3 sin⁻¹(1/3) ≈ 3 × 19.47° = 58.4° and sin⁻¹(3/5) ≈ 36.9°, so B ≈ 95.3°.
3. 7π/12 = 105°. Hence B < 7π/12 < A.

Teacher's Note:
1. Convert each quantity to degrees to compare them.
2. 7π/12 radians equals 105°.

Answer: (a) Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation for Statement-1

Explanation:
1. cot⁻¹x − tan⁻¹x = π/2 − 2tan⁻¹x, which is positive iff tan⁻¹x < π/4, i.e. x < 1. So Statement-1 is true.
2. Statement-2 says the same thing graphically (cot⁻¹x above tan⁻¹x means cot⁻¹x − tan⁻¹x > 0), so it is true and explains Statement-1.

Teacher's Note:
1. Reduce the difference to π/2 − 2tan⁻¹x.
2. Statement-2 is the graphical restatement of Statement-1.

Answer: (b) (B), (C) and (D)

Explanation:
1. \( \cos\left(-\frac{14\pi}{5}\right)=\cos\frac{4\pi}{5} \), so cos⁻¹ of it is 4π/5 and half of it is 2π/5. The value is cos(2π/5).
2. (A) cos(7π/5) = −cos(2π/5): not equal.
3. (B) sin(π/10) = cos(π/2 − π/10) = cos(2π/5), (C) is the value itself, and (D) −cos(3π/5) = cos(2π/5). So (B), (C), (D).

Teacher's Note:
1. Reduce 14π/5 modulo 2π and use cos⁻¹(cos θ) = θ on [0, π].
2. Convert sines and negative cosines to cos(2π/5).