JEE Mathematics Inverse Trigonometric Functions MCQs Set 09

Answer: (a) No solution

Explanation:
1. \( 2\cos^{-1}x+\sin^{-1}x=\cos^{-1}x+\left(\cos^{-1}x+\sin^{-1}x\right)=\cos^{-1}x+\frac\pi2 \).
2. \( \cos^{-1}x+\frac\pi2=\frac{11\pi}{6}\Rightarrow\cos^{-1}x=\frac{4\pi}{3} \), which exceeds \( \pi \).
3. No such \( x \) exists.

Teacher's Note:
1. Group the terms so that sin inverse x + cos inverse x = pi/2 can be used.
2. Always compare the resulting value with the range [0, pi] of cos inverse.

Answer: (b) \( n\pi \)

Explanation:
1. \( \alpha+\beta+\gamma=-m,\ \alpha\beta+\beta\gamma+\gamma\alpha=3,\ \alpha\beta\gamma=-m \).
2. \( \tan\left(\sum\tan^{-1}\right)=\frac{S_1-S_3}{1-S_2}=\frac{-m+m}{1-3}=0 \).
3. A tangent equal to 0 means the angle is a multiple of \( \pi \): \( n\pi \).

Teacher's Note:
1. Use tan(A + B + C) = (S1 - S3)/(1 - S2) with the symmetric functions of the roots.
2. The parameter m cancels, so the result does not depend on it.

Answer: (a) \( -\frac{2\sqrt6}{5} \)

Explanation:
1. \( \sin^{-1}x+2\cos^{-1}x=\left(\sin^{-1}x+\cos^{-1}x\right)+\cos^{-1}x=\frac\pi2+\cos^{-1}x \).
2. \( \cos\left(\frac\pi2+\cos^{-1}x\right)=-\sin\left(\cos^{-1}x\right)=-\sqrt{1-x^2} \).
3. At \( x=\frac15 \): \( -\sqrt{1-\frac1{25}}=-\frac{2\sqrt6}{5} \).

Teacher's Note:
1. Split 2 cos inverse x into two copies so that the pair sin inverse + cos inverse gives pi/2.
2. cos (pi/2 + A) = -sin A, which fixes the negative sign.

Answer: (d) \( \pi-\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) \)

Explanation:
1. For \( x>1 \): \( 2\tan^{-1}x=\cos^{-1}\frac{1-x^2}{1+x^2} \) (valid for \( x\ge0 \)), so option (a) is equal.
2. \( 2\tan^{-1}x=\pi-\sin^{-1}\frac{2x}{1+x^2} \) for \( x>1 \), so option (b) is equal; and \( 2\tan^{-1}x=\pi+\tan^{-1}\frac{2x}{1-x^2} \) for \( x>1 \), so option (c) is equal.
3. Option (d) equals \( \pi-2\tan^{-1}x \), which is different.

Teacher's Note:
1. Memorise all three conversions of 2 tan inverse x together with their ranges of validity.
2. For x greater than 1, 2 tan inverse x lies between pi/2 and pi, which is where the pi corrections appear.

Answer: (d) \( 8\pi-24 \)

Explanation:
1. \( \cos^{-1}(\cos12)=\cos^{-1}\left(\cos(4\pi-12)\right)=4\pi-12 \) since \( 4\pi-12\approx0.57\in[0,\pi] \).
2. \( \sin^{-1}(\sin12)=\sin^{-1}\left(\sin(12-4\pi)\right)=12-4\pi \) since \( 12-4\pi\approx-0.57 \).
3. Difference: \( (4\pi-12)-(12-4\pi)=8\pi-24 \).

Teacher's Note:
1. Use 4 pi (two full periods) to bring 12 radians into the principal interval of both functions.
2. The two answers are negatives of each other, so the difference is twice the first.

Answer: (d) \( \frac54 \)

Explanation:
1. Use \( \sin^{-1}t=\frac\pi2-\cos^{-1}t \) in the second equation: \( \cos^{-1}\frac yb-\cos^{-1}\frac xa=\frac\pi{12} \).
2. Adding to the first equation: \( 2\cos^{-1}\frac yb=\frac{6\pi}{12}\Rightarrow\cos^{-1}\frac yb=\frac\pi4\Rightarrow\frac yb=\frac1{\sqrt2} \).
3. Then \( \cos^{-1}\frac xa=\frac{5\pi}{12}-\frac\pi4=\frac\pi6\Rightarrow\frac xa=\frac{\sqrt3}{2} \).
4. \( \frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac34+\frac12=\frac54 \).

Teacher's Note:
1. Convert both equations to cos inverse so that the two unknown angles can be found by adding and subtracting.
2. Check that both angles are inside [0, pi]: pi/4 and pi/6 are.

Answer: (d) \( \frac1{\sqrt2}\le x\le1 \)

Explanation:
1. Put \( x=\cos y,\ 0\le y\le\pi \). The right side becomes \( \sin^{-1}(\sin2y) \) and the left side is \( 2y \).
2. \( \sin^{-1}(\sin2y)=2y \) only when \( -\frac\pi2\le2y\le\frac\pi2 \), i.e. \( 0\le y\le\frac\pi4 \).
3. Hence \( \cos\frac\pi4\le x\le1 \), that is \( \frac1{\sqrt2}\le x\le1 \).

Teacher's Note:
1. The doubled angle must stay inside the principal range of sin inverse, which limits the original angle to 45 degrees.
2. Compare with the identity for 2 sin inverse x, which is valid on the symmetric interval [-1/root 2, 1/root 2].

Answer: (c) \( 1 \)

Explanation:
1. A sum of squares is zero only if each term is zero, so \( x=y=z \).
2. \( 3\tan^{-1}x=\frac\pi2\Rightarrow\tan^{-1}x=\frac\pi6\Rightarrow x=\frac1{\sqrt3} \).
3. \( x^2+y^2+z^2=3\cdot\frac13=1 \).

Teacher's Note:
1. Recognise the sum of squared differences and conclude that all three variables are equal.
2. As a check, xy + yz + zx = 1 also holds, which is the condition for the sum to be pi/2.

Answer: (a) \( \tan^{-1}\left(\frac{n^2+n}{n^2+n+2}\right) \)

Explanation:
1. \( m^4+m^2+2=1+(m^2+m+1)(m^2-m+1) \) and \( (m^2+m+1)-(m^2-m+1)=2m \).
2. So the general term is \( \tan^{-1}(m^2+m+1)-\tan^{-1}(m^2-m+1) \).
3. Telescoping (note \( (m+1)^2-(m+1)+1=m^2+m+1 \)): \( \text{sum}=\tan^{-1}(n^2+n+1)-\tan^{-1}1=\tan^{-1}\frac{n^2+n}{n^2+n+2} \).

Teacher's Note:
1. Factor m^4 + m^2 + 1 as (m^2 + m + 1)(m^2 - m + 1) and add 1 to reach the denominator.
2. The upper term of one step equals the lower term of the next, which is why only the first and last terms survive.

Answer: (c) 2

Explanation:
1. Let \( \theta=\sin^{-1}\frac27 \). Then \( \cos2\theta=1-2\sin^2\theta=1-\frac{8}{49}=\frac{41}{49} \).
2. \( \cos^{-1}\frac{41}{49}=2\theta \) because \( 2\theta\in[0,\pi] \), so the ratio is 2.

Teacher's Note:
1. Look for a double-angle relation: 41/49 = 1 - 2 (2/7) squared.
2. The angle 2 theta is small and acute, so cos inverse returns exactly 2 theta.

Answer: (a) \( a\ge x\ge b\ \text{or}\ a\le x\le b \)

Explanation:
1. Both radicals must be defined with values in \( [0,1] \): \( 0\le\frac{a-x}{a-b}\le1 \) and \( 0\le\frac{x-b}{a-b}\le1 \).
2. These say that \( x \) lies between \( a \) and \( b \) (inclusive), in either order.
3. For such \( x \): \( 1-\frac{a-x}{a-b}=\frac{x-b}{a-b} \), so \( \cos^{-1}u=\sin^{-1}\sqrt{1-u^2} \) holds.

Teacher's Note:
1. Check the arguments of both radicals against [0, 1]; they sum to 1, which is what makes the identity work.
2. The condition depends only on x lying between a and b, not on which of a, b is larger.

Answer: (d) \( -\frac\pi2 \)

Explanation:
1. For \( x\in\left(\frac\pi2,\pi\right) \): \( \cos^{-1}(\cos x)=x \) and \( \sin^{-1}(\sin x)=\pi-x \).
2. The sum is \( \pi \), so \( \cos\pi=-1 \).
3. \( \sin^{-1}(-1)=-\frac\pi2 \).

Teacher's Note:
1. Handle the inner functions by quadrant: x lies in [0, pi] for cos inverse but in the second quadrant for sin inverse, so pi - x appears.
2. The x terms cancel, leaving a constant.

Answer: (b) \( x>y \)

Explanation:
1. \( x=\frac{2\cdot2}{1+4}=\frac45 \) (using \( \sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta} \)).
2. Let \( \phi=\tan^{-1}\frac43 \), so \( \cos\phi=\frac35 \) and \( y=\sin\frac\phi2=\sqrt{\frac{1-\cos\phi}{2}}=\sqrt{\frac15}=\frac1{\sqrt5} \).
3. \( x=0.8>y\approx0.447 \).

Teacher's Note:
1. Use the standard sine and half-angle sine formulas in terms of a known right triangle.
2. Comparing decimals settles the inequality.

Answer: (a) \( \pi-1 \)

Explanation:
1. \( \sin^{-1}(\sin3)=\pi-3 \) (as \( 3\in\left(\frac\pi2,\pi\right) \)).
2. \( \cos^{-1}(\cos7)=7-2\pi \) (as \( 7-2\pi\approx0.72\in[0,\pi] \)).
3. \( \tan^{-1}(\tan5)=5-2\pi \) (as \( 5-2\pi\approx-1.28\in\left(-\frac\pi2,\frac\pi2\right) \)).
4. Total: \( (\pi-3)+(7-2\pi)-(5-2\pi)=\pi-1 \).

Teacher's Note:
1. Reduce each angle separately into the principal range of its own function.
2. The 2 pi terms cancel, leaving pi - 1.

Answer: (d) \( \infty \)

Explanation:
1. \( \sin^{-1}u+\cos^{-1}v=\frac\pi2 \) (with \( u,v\in[-1,1] \)) holds exactly when \( u=v \).
2. Here \( u=\frac1{1-b} \) for \( |b|<1 \) and \( v=\frac{a}{1+a/3}=\frac{3a}{3+a} \) for \( |a|<3 \).
3. \( \frac{1}{1-b}=\frac{3a}{3+a} \) and \( u\le1 \) (so \( -1<b\le0 \)) is one equation in two unknowns, so every admissible \( b \) gives a matching \( a \): infinitely many solutions.

Teacher's Note:
1. Use sin inverse u + cos inverse v = pi/2 if and only if u = v (within the domain).
2. One equation with two unknowns that has valid ranges for both gives a continuum of solutions.

Answer: (b) \( \frac\pi2 \)

Explanation:
1. Claim: \( \sin^{-1}\frac{\sqrt n-\sqrt{n-1}}{\sqrt{n(n+1)}}=\tan^{-1}\sqrt n-\tan^{-1}\sqrt{n-1} \). Indeed, with \( A=\tan^{-1}\sqrt n,\ B=\tan^{-1}\sqrt{n-1} \): \( \sin(A-B)=\frac{\sqrt n}{\sqrt{n+1}}\cdot\frac1{\sqrt n}-\frac1{\sqrt{n+1}}\cdot\frac{\sqrt{n-1}}{\sqrt n}=\frac{\sqrt n-\sqrt{n-1}}{\sqrt{n(n+1)}} \) and \( 0\le A-B<\frac\pi2 \).
2. The series telescopes: \( S_n=\tan^{-1}\sqrt n-\tan^{-1}0 \).
3. \( S_\infty=\frac\pi2-0=\frac\pi2 \).

Teacher's Note:
1. Write the n-th term as a difference of two tan inverse values and watch the terms cancel.
2. Test the first term: sin inverse (1/root 2) = pi/4 = tan inverse 1 - tan inverse 0.

Answer: (c) \( 0\le x\le1 \)

Explanation:
1. With \( u=\frac x2 \) and \( v=x \), the left side is \( \cos^{-1}\left(uv+\sqrt{1-u^2}\sqrt{1-v^2}\right) \).
2. This equals \( \cos^{-1}u-\cos^{-1}v \) only when this difference is non-negative, i.e. \( u\le v \).
3. \( \frac x2\le x\Rightarrow x\ge0 \); with the domain \( |x|\le1 \) we get \( 0\le x\le1 \).

Teacher's Note:
1. The left side is a cos inverse and so is non-negative; the right side must therefore also be non-negative.
2. cos inverse is decreasing, so cos inverse u - cos inverse v is non-negative when u is at most v.

Answer: (b) \( \frac\pi6 \)

Explanation:
1. \( \tan A=\frac{x\sqrt3}{2k-x},\ \tan B=\frac{2x-k}{k\sqrt3} \).
2. \( \tan A-\tan B=\frac{3kx-(2x-k)(2k-x)}{k\sqrt3(2k-x)}=\frac{2(x^2-kx+k^2)}{k\sqrt3(2k-x)} \).
3. \( 1+\tan A\tan B=\frac{2(x^2-kx+k^2)}{k(2k-x)} \).
4. \( \tan(A-B)=\frac1{\sqrt3}\Rightarrow A-B=\frac\pi6 \).

Teacher's Note:
1. Compute tan (A - B) with the standard formula; the common factor x squared - kx + k squared cancels.
2. The result 1/root 3 does not depend on x or k, so the angle difference is constant.

Answer: (c) 3

Explanation:
1. \( \tan^{-1}(x-1)+\tan^{-1}(x+1)=\tan^{-1}\frac{2x}{2-x^2} \) and \( \tan^{-1}3x-\tan^{-1}x=\tan^{-1}\frac{2x}{1+3x^2} \).
2. \( \frac{2x}{2-x^2}=\frac{2x}{1+3x^2}\Rightarrow x=0\ \text{or}\ 2-x^2=1+3x^2\Rightarrow x=\pm\frac12 \).
3. All three values satisfy the original equation (checked by substitution), so there are 3 solutions.

Teacher's Note:
1. Do not divide by x: the root x = 0 would be lost.
2. Substitute x = 1/2 and x = -1/2 to confirm that the product (x - 1)(x + 1) stays below 1.

Answer: (a) \( \tan^2\frac\alpha2 \)

Explanation:
1. Let \( \theta=\tan^{-1}\sqrt{\cos\alpha} \). Then \( x=\frac\pi2-\theta-\theta=\frac\pi2-2\theta \).
2. \( \sin x=\cos2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta}=\frac{1-\cos\alpha}{1+\cos\alpha} \).
3. \( =\tan^2\frac\alpha2 \).

Teacher's Note:
1. Replace cot inverse by pi/2 minus tan inverse so that only one angle remains.
2. (1 - cos alpha)/(1 + cos alpha) is the half-angle form of tan squared (alpha/2).

Answer: (b) \( \frac{(n-1)d}{1+a_1a_n} \)

Explanation:
1. Since \( d=a_{k+1}-a_k \): \( \frac{d}{1+a_ka_{k+1}}=\tan\left(\tan^{-1}a_{k+1}-\tan^{-1}a_k\right) \).
2. The sum telescopes to \( \tan^{-1}a_n-\tan^{-1}a_1 \).
3. \( \tan\left(\tan^{-1}a_n-\tan^{-1}a_1\right)=\frac{a_n-a_1}{1+a_1a_n}=\frac{(n-1)d}{1+a_1a_n} \).

Teacher's Note:
1. Write d as a_(k+1) - a_k to recognise the tan inverse difference formula.
2. a_n - a_1 equals (n - 1) d in an A.P.

Answer: (c) \( \tan^{-1}\frac xy \)

Explanation:
1. \( \frac{c_1x-y}{c_1y+x}=\frac{\frac xy-\frac1{c_1}}{1+\frac xy\cdot\frac1{c_1}} \), so the first term is \( \tan^{-1}\frac xy-\tan^{-1}\frac1{c_1} \).
2. \( \frac{c_{k+1}-c_k}{1+c_{k+1}c_k}=\frac{\frac1{c_k}-\frac1{c_{k+1}}}{1+\frac1{c_kc_{k+1}}} \), so the term is \( \tan^{-1}\frac1{c_k}-\tan^{-1}\frac1{c_{k+1}} \).
3. Everything telescopes, leaving \( \tan^{-1}\frac xy \) (the last term cancels the \( \tan^{-1}\frac1{c_n} \) left over).

Teacher's Note:
1. Divide numerator and denominator of the first term by c1 y to expose the difference formula.
2. Each term is a difference of consecutive values of tan inverse (1/c), so the sum collapses.

Answer: (d) a constant independent of x

Explanation:
1. \( \sec^{-1}\frac1{1-2x^2}=\cos^{-1}(1-2x^2) \). Put \( x=\cos\theta,\ 0\le\theta\le\frac\pi2 \): \( 1-2x^2=-\cos2\theta \), so the first term is \( \pi-\cos^{-1}(\cos2\theta)=\pi-2\theta \) (as \( 2\theta\in[0,\pi] \)).
2. \( 4\cos^{-1}\sqrt{\frac{1+\cos\theta}{2}}=4\cos^{-1}\cos\frac\theta2=2\theta \).
3. Total: \( \pi-2\theta+2\theta=\pi \), a constant.

Teacher's Note:
1. Substituting x = cos theta turns both terms into multiples of theta that cancel.
2. The restriction x is at least 0 keeps 2 theta inside [0, pi]; for negative x the cancellation fails.