JEE Mathematics Inverse Circular Functions MCQs Set 03

Answer: (c) both \( \cot^{-1}\alpha \) and \( \cot^{-1}\beta \) are real

Explanation:
1. \( 6x^2-11x+3=(3x-1)(2x-3) \), so the roots are \( \frac13 \) and \( \frac32 \).
2. \( \sin^{-1} \) and \( \cos^{-1} \) are not defined for \( \frac32 \), but \( \cot^{-1} \) is defined for every real number.

Teacher's Note:
1. Find the roots first.
2. cot inverse has domain R.

Answer: (c) \( x\in(-\infty,-1]\cup[1,\infty) \)

Explanation:
1. \( \tan^{-1}x \) is real for all x, but \( \sec^{-1}x \) is real only when \( |x|\ge1 \).
2. So f is real for \( x\in(-\infty,-1]\cup[1,\infty) \).

Teacher's Note:
1. Take the intersection of the two domains.
2. Only sec inverse restricts x.

Answer: (d) \( \frac1{\sqrt2} \)

Explanation:
1. \( \cos^{-1}x=\sin^{-1}x \) and \( \cos^{-1}x+\sin^{-1}x=\frac\pi2 \), so each equals \( \frac\pi4 \).
2. \( x=\sin\frac\pi4=\frac1{\sqrt2} \). The value \( -\frac1{\sqrt2} \) gives \( \cos^{-1}x=\frac{3\pi}4\ne-\frac\pi4 \), so it is rejected.

Teacher's Note:
1. Use the identity for the sum of the two inverse functions.
2. Check negative values against the principal range.

Answer: (d) x is a multiple of \( \pi \)

Explanation:
1. \( \operatorname{cosec}^{-1}t \) is real only for \( |t|\ge1 \).
2. Since \( |\cos x|\le1 \), we need \( |\cos x|=1 \), i.e. \( x=n\pi \).

Teacher's Note:
1. The modulus of cos x cannot exceed 1.
2. So the only possibility is equality.

Answer: (c) \( \frac\pi3 \)

Explanation:
1. \( \sin\frac{7\pi}6=-\frac12 \), so \( -\sin\frac{7\pi}6=\frac12 \).
2. \( \cos^{-1}\frac12=\frac\pi3 \), which lies in \( [0,\pi] \).

Teacher's Note:
1. Evaluate the sine first.
2. The principal value of cos inverse is in 0 to pi.

Answer: (a) \( \frac\pi6 \)

Explanation:
1. \( \sin^{-1}\frac{\sqrt3}2=\frac\pi3 \) and \( \cos\frac\pi3=\frac12 \).
2. \( \sin^{-1}\frac12=\frac\pi6 \).

Teacher's Note:
1. Work from the inside outwards.

Answer: (d) none of these

Explanation:
1. \( \frac1{\sqrt2}(\cos t-\sin t)=\cos\left(t+\frac\pi4\right) \) with \( t=\frac{9\pi}{10} \), giving \( \cos\frac{23\pi}{20} \).
2. \( \frac{23\pi}{20} \) is outside \( [0,\pi] \). Using \( \cos\theta=\cos(2\pi-\theta) \), the principal value is \( 2\pi-\frac{23\pi}{20}=\frac{17\pi}{20} \).
3. \( \frac{17\pi}{20} \) is not among the first three options.

Teacher's Note:
1. Write the bracket as a single cosine.
2. Reduce the angle into 0 to pi.

Answer: (a) \( \frac1{\sqrt2} \)

Explanation:
1. \( \frac{15\pi}4=4\pi-\frac\pi4 \), so \( \tan\frac{15\pi}4=-1 \).
2. \( \tan^{-1}(-1)=-\frac\pi4 \) and \( \cos\left(-\frac\pi4\right)=\frac1{\sqrt2} \).

Teacher's Note:
1. Reduce the angle by a multiple of pi.
2. Cosine is an even function.

Answer: (b) \( \sin^{-1}\left(\sin\frac{7\pi}5\right) \)

Explanation:
1. \( \sin\frac{7\pi}5=\sin\left(\pi+\frac{2\pi}5\right)=-\sin\frac{2\pi}5=\sin\left(-\frac{2\pi}5\right) \), and \( -\frac{2\pi}5 \) lies in \( \left[-\frac\pi2,\frac\pi2\right] \).
2. A principal value of cos inverse or sec inverse is never negative, so (a) and (c) are impossible.

Teacher's Note:
1. Negative principal values occur only for sin inverse and tan inverse here.

Answer: (b) \( 10 \)

Explanation:
1. Each \( \cos^{-1}x_i\ge0 \), so the sum is 0 only if every term is 0.
2. Then every \( x_i=1 \) and the sum is 10.

Teacher's Note:
1. A sum of non-negative terms is zero only if each term is zero.

Answer: (b) \( 2n \)

Explanation:
1. Each term is at most \( \frac\pi2 \), so the maximum possible sum is \( 2n\cdot\frac\pi2=n\pi \).
2. Equality forces every \( x_i=1 \), so the sum is \( 2n \).

Teacher's Note:
1. Compare with the maximum of the sum.

Answer: (d) none of these

Explanation:
1. \( \cos^{-1}\left(-\frac12\right)=\frac{2\pi}3 \) and \( \sin^{-1}\frac{\sqrt3}2=\frac\pi3 \).
2. The sum is \( \pi \), which is not among the first three options.

Teacher's Note:
1. Evaluate each term in its principal range.

Answer: (d) none of these

Explanation:
1. \( \tan\left(2\tan^{-1}\frac15\right)=\frac{2/5}{1-1/25}=\frac5{12} \).
2. \( \tan\left(\theta-\frac\pi4\right)=\frac{5/12-1}{1+5/12}=-\frac7{17} \).
3. The value is negative, so none of the given positive values is correct.

Teacher's Note:
1. Use the double angle formula for tan.
2. Then use tan(A - B).

Answer: (c) \( \pi+\tan^{-1}\frac{a+b}{1-ab} \)

Explanation:
1. The sum lies in \( \left(\frac\pi2,\pi\right) \) because both angles are positive and \( ab>1 \).
2. \( \tan^{-1}\frac{a+b}{1-ab} \) is negative (as \( 1-ab\lt 0 \)), so we add \( \pi \) to bring it into that interval.

Teacher's Note:
1. The formula changes by pi when ab is greater than 1.

Answer: (b) \( (-1,1) \)

Explanation:
1. Put \( x=\sin\theta \) with \( \theta\in\left(-\frac\pi2,\frac\pi2\right) \). Then \( \frac x{\sqrt{1-x^2}}=\tan\theta \), and \( \tan^{-1}\tan\theta=\theta \).
2. The expression needs \( 1-x^2>0 \), so the identity holds for \( -1\lt x\lt 1 \).

Teacher's Note:
1. The denominator vanishes at x = plus or minus 1.

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

Explanation:
1. For \( a=\frac x2,\ b=x \): \( \cos^{-1}a-\cos^{-1}b=\cos^{-1}\left(ab+\sqrt{1-a^2}\sqrt{1-b^2}\right) \) only when \( \cos^{-1}a-\cos^{-1}b\in[0,\pi] \), i.e. \( \cos^{-1}a\ge\cos^{-1}b \).
2. Since \( \cos^{-1} \) is decreasing, this means \( a\le b \), i.e. \( \frac x2\le x \), so \( x\ge0 \). With \( |x|\le1 \), the set is \( 0\le x\le1 \).

Teacher's Note:
1. The right side of the identity is never negative.
2. So cos inverse of a must not be smaller than cos inverse of b.

Answer: (b) \( \sin^{-1}x-\frac\pi6 \)

Explanation:
1. Let \( \sin^{-1}x=A \). The bracket is \( \sin A\cos\frac\pi6-\cos A\sin\frac\pi6=\sin\left(A-\frac\pi6\right) \).
2. For \( \frac12\le x\le1 \), \( A\in\left[\frac\pi6,\frac\pi2\right] \), so \( A-\frac\pi6\in\left[0,\frac\pi3\right] \), inside the principal range.
3. Hence \( f(x)=\sin^{-1}x-\frac\pi6 \).

Teacher's Note:
1. Recognise the sine of a difference.
2. Check that the angle stays in the principal range.

Answer: (b) \( x\in\left[-\frac1{\sqrt2},\frac1{\sqrt2}\right] \)

Explanation:
1. With \( x=\sin\theta \): the right side is \( \sin^{-1}(\sin2\theta) \), which equals \( 2\theta \) only if \( 2\theta\in\left[-\frac\pi2,\frac\pi2\right] \).
2. So \( \theta\in\left[-\frac\pi4,\frac\pi4\right] \), i.e. \( x\in\left[-\frac1{\sqrt2},\frac1{\sqrt2}\right] \).

Teacher's Note:
1. Substitute x = sin theta.
2. 2 theta must be in the principal range.

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

Explanation:
1. For \( x\ge1 \), \( \sin^{-1}\frac{2x}{1+x^2}=\pi-2\tan^{-1}x \), so the sum is the constant \( \pi \).
2. For \( |x|\le1 \), the second term is \( 2\tan^{-1}x \) and the sum \( 4\tan^{-1}x \) varies with x, so (b) and (c) are not correct.

Teacher's Note:
1. Split the domain at x = 1.
2. Use x = tan theta.

Answer: (a) \( \frac\pi4 \)

Explanation:
1. \( \tan^{-1}2+\tan^{-1}3=\pi+\tan^{-1}\frac{5}{1-6}=\pi-\frac\pi4=\frac{3\pi}4 \), since the product is greater than 1.
2. The third angle is \( \pi-\frac{3\pi}4=\frac\pi4 \).

Teacher's Note:
1. Add the two angles using the tan formula with the pi correction.
2. Angles of a triangle add to pi.

Answer: (c) \( \frac\pi2 \)

Explanation:
1. \( \tan^{-1}1=\frac\pi4 \).
2. \( \tan^{-1}\frac12+\tan^{-1}\frac13=\tan^{-1}\frac{5/6}{1-1/6}=\tan^{-1}1=\frac\pi4 \).
3. The total is \( \frac\pi2 \).

Teacher's Note:
1. Combine the last two terms first.

Answer: (b) \( \frac{17}6 \)

Explanation:
1. \( \tan\left(\cos^{-1}\frac45\right)=\frac34 \) and \( \tan\left(\sin^{-1}\frac2{\sqrt{13}}\right)=\frac23 \).
2. \( \tan(A+B)=\frac{3/4+2/3}{1-1/2}=\frac{17/12}{1/2}=\frac{17}6 \).

Teacher's Note:
1. Convert each inverse function into a tangent.

Answer: (a) \( \frac\pi4 \)

Explanation:
1. \( 2\tan^{-1}\frac13=\tan^{-1}\frac{2/3}{1-1/9}=\tan^{-1}\frac34 \).
2. \( \tan^{-1}\frac34+\tan^{-1}\frac17=\tan^{-1}\frac{25/28}{1-3/28}=\tan^{-1}1=\frac\pi4 \).

Teacher's Note:
1. Use the double angle formula first.