JEE Mathematics Inverse Circular Functions MCQs Set 05

Answer: (b) \( \frac\pi4 \)

Explanation:
1. The n-th term is \( \tan^{-1}\frac{2^{n-1}}{1+2^{2n-1}}=\tan^{-1}2^n-\tan^{-1}2^{n-1} \) (check: n = 1 gives 1/3, n = 2 gives 2/9, n = 3 gives 4/33).
2. The series telescopes: \( S_n=\tan^{-1}2^n-\tan^{-1}1 \).
3. As n → ∞: \( S=\frac\pi2-\frac\pi4=\frac\pi4 \).

Teacher's Note:
1. Use tan⁻¹A − tan⁻¹B = tan⁻¹((A − B)/(1 + AB)) to spot the telescoping pattern.
2. The terms 1/3, 2/9, 4/33 have numerators 2^(n−1).

Answer: (d) none of these

Explanation:
1. \( \tan1\approx1.557,\ \tan^{-1}1=\frac\pi4\approx0.785,\ \sin^{-1}1=\frac\pi2\approx1.571,\ \sin1\approx0.841,\ \cos1\approx0.540 \).
2. The greatest is \( \sin^{-1}1 \), which is not among (a), (b) or (c).

Teacher's Note:
1. Convert each to a decimal (1 radian ≈ 57.3°).
2. tan 1 is close to π/2 but slightly smaller.

Answer: (a) \( [\tan\sin\cos1,\ \tan\sin\cos\sin1] \)

Explanation:
1. The outer sin⁻¹ takes values up to π/2 ≈ 1.57, so the condition is \( 1\le\sin^{-1}(\cdot)\le\frac\pi2 \), i.e. \( \cos^{-1}\sin^{-1}\tan^{-1}x\in[\sin1,1] \).
2. Apply cos (decreasing): \( \sin^{-1}\tan^{-1}x\in[\cos1,\cos\sin1] \). Apply sin: \( \tan^{-1}x\in[\sin\cos1,\sin\cos\sin1] \).
3. Apply tan: \( x\in[\tan\sin\cos1,\tan\sin\cos\sin1] \).

Teacher's Note:
1. Work from the outside inwards, reversing the direction of the inequality when applying a decreasing function (cos).
2. All end values are included because the range is closed.

Answer: (c) n = 6

Explanation:
1. Square both sides: \( 1+\sin\frac\pi n=\frac n4 \).
2. Try n = 6: \( 1+\sin30^\circ=1.5=\frac64 \) ✓.
3. n = 4 gives 1 + sin 45° ≈ 1.71 ≠ 1 and n = 5 gives 1 + sin 36° ≈ 1.59 ≠ 1.25, so n = 6.

Teacher's Note:
1. Squaring removes the square root and gives 1 + sin(π/n).
2. Test the given options.

Answer: (a) 0

Explanation:
1. Both inverse cosines are non-negative, so the left side is ≥ 0, while \( \frac{n\pi}2\le0 \).
2. Equality would need \( \cos^{-1}(1-x)=0 \) (x = 0) and \( \cos^{-1}x=0 \) (x = 1) simultaneously, which is impossible.
3. So there is no solution.

Teacher's Note:
1. Inverse cosine values lie in [0, π].
2. Both terms would have to vanish at the same x.

Answer: (c) \( -\frac7{17} \)

Explanation:
1. \( \tan\left(2\tan^{-1}\frac15\right)=\frac{2/5}{1-1/25}=\frac5{12} \).
2. \( \tan\left(\theta-\frac\pi4\right)=\frac{5/12-1}{1+5/12}=\frac{-7/12}{17/12}=-\frac7{17} \).

Teacher's Note:
1. Use tan 2A = 2tan A/(1 − tan²A) first.
2. Then use tan(A − B) with B = π/4.

Answer: (a) 0

Explanation:
1. \( \sin^2 15^\circ=\frac{1-\cos30^\circ}2=\frac{2-\sqrt3}4 \), so the first term is 15°.
2. The second term: \( \cos^{-1}\frac{\sqrt{12}}4=\cos^{-1}\frac{\sqrt3}2=30^\circ \). The third: \( \sec^{-1}\sqrt2=45^\circ \).
3. The sum is 90°, \( \cot90^\circ=0 \) and \( \sin^{-1}0=0 \).

Teacher's Note:
1. Identify each inverse trigonometric value as a standard angle.
2. cot 90° = 0.

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

Explanation:
1. \( \tan^{-1}1=\frac\pi4,\ \cos^{-1}\left(-\frac12\right)=\frac{2\pi}3,\ \sin^{-1}\left(-\frac12\right)=-\frac\pi6 \).
2. \( \frac{3\pi}{12}+\frac{8\pi}{12}-\frac{2\pi}{12}=\frac{9\pi}{12}=\frac{3\pi}4 \).

Teacher's Note:
1. cos⁻¹ of a negative number lies in (π/2, π); sin⁻¹ of a negative number is negative.
2. Use a common denominator of 12.

Answer: (a) \( x\in[1,\infty) \)

Explanation:
1. For x ≥ 1, \( \sin^{-1}\frac{2x}{1+x^2}=\pi-2\tan^{-1}x \), so the sum is π.
2. For −1 ≤ x ≤ 1, \( \sin^{-1}\frac{2x}{1+x^2}=2\tan^{-1}x \) and the sum is \( 4\tan^{-1}x \), which varies.
3. For x ≤ −1 the sum is −π. Hence the sum equals π exactly for x ≥ 1.

Teacher's Note:
1. sin⁻¹(2x/(1 + x²)) equals 2tan⁻¹x only for |x| ≤ 1.
2. Outside that range the identity changes by π.

Answer: (a) \( \pi \)

Explanation:
1. \( \cos^{-1}\cos\frac{2\pi}3=\frac{2\pi}3 \) since 2π/3 lies in [0, π].
2. \( \sin^{-1}\sin\frac{2\pi}3=\sin^{-1}\frac{\sqrt3}2=\frac\pi3 \) since the principal range is [−π/2, π/2].
3. Sum = π.

Teacher's Note:
1. Check whether the angle lies in the principal range of each function.
2. sin⁻¹(sin θ) = π − θ for θ in [π/2, 3π/2].

Answer: (b) \( (-\infty,\cot3)\cup(\cot2,\infty) \)

Explanation:
1. Let \( t=\cot^{-1}x\in(0,\pi) \). Then \( (t-2)(t-3)>0 \), so t < 2 or t > 3 (note 3 < π).
2. cot⁻¹x is a decreasing function: t < 2 means \( x>\cot2 \) and t > 3 means \( x<\cot3 \).
3. So \( x\in(-\infty,\cot3)\cup(\cot2,\infty) \).

Teacher's Note:
1. cot⁻¹ x decreases from π to 0 as x goes from −∞ to ∞.
2. Solve the quadratic inequality in t first.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. For x in (0, 1), \( \sin^{-1}x>x>\tan^{-1}x \), and tan⁻¹ is increasing, so for x > y: \( \sin^{-1}x>\tan^{-1}x>\tan^{-1}y \). R is true.
2. Here x = 1/√e ≈ 0.607 > y = 1/√π ≈ 0.564, so R gives A directly.

Teacher's Note:
1. Compare sin⁻¹x with tan⁻¹x on (0, 1).
2. Use the fact that tan⁻¹ is increasing.