JEE Mathematics Inverse Circular Functions MCQs Set 04

Answer: (c) \( \frac\pi4 \)

Explanation:
1. \( \cot^{-1}3=\tan^{-1}\frac13 \). If \( \operatorname{cosec}\theta=\sqrt5 \), then \( \sin\theta=\frac1{\sqrt5} \) and \( \tan\theta=\frac12 \).
2. \( \tan^{-1}\frac13+\tan^{-1}\frac12=\tan^{-1}\frac{5/6}{5/6}=\frac\pi4 \).

Teacher's Note:
1. Express both terms as tan inverse.

Answer: (a) \( x \)

Explanation:
1. Let \( \theta=\cos^{-1}x \), so \( \tan\theta=\frac{\sqrt{1-x^2}}x \).
2. \( \sin(\cot^{-1}t)=\frac1{\sqrt{1+t^2}} \). With \( t=\frac{\sqrt{1-x^2}}x \): \( 1+t^2=\frac1{x^2} \), so the value is \( x \).

Teacher's Note:
1. Write the tangent in terms of x.
2. Use sin(cot inverse t) = 1 over root of (1 + t squared).

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

Explanation:
1. \( \tan^{-1}\frac13+\tan^{-1}\frac15=\tan^{-1}\frac{8/15}{14/15}=\tan^{-1}\frac47 \).
2. \( \tan^{-1}\frac17+\tan^{-1}\frac18=\tan^{-1}\frac{15/56}{55/56}=\tan^{-1}\frac3{11} \).
3. \( \tan^{-1}\frac47+\tan^{-1}\frac3{11}=\tan^{-1}\frac{(44+21)/77}{(77-12)/77}=\tan^{-1}1=\frac\pi4 \).

Teacher's Note:
1. Add the terms in pairs.

Answer: (c) \( 11 \)

Explanation:
1. \( \tan^2(\sec^{-1}2)=\sec^2(\sec^{-1}2)-1=4-1=3 \).
2. \( \cot^2(\operatorname{cosec}^{-1}3)=\operatorname{cosec}^2(\cdot)-1=9-1=8 \).
3. The total is \( 11 \).

Teacher's Note:
1. Use the identities tan squared = sec squared - 1 and cot squared = cosec squared - 1.

Answer: (c) \( 2 \)

Explanation:
1. The left side is \( \sqrt2|\cos x| \), so the equation is \( |\cos x|=\sin^{-1}(\sin x) \), which needs \( \sin^{-1}(\sin x)\ge0 \), i.e. \( 0\le x\le\pi \).
2. On \( \left[0,\frac\pi2\right] \): \( \cos x=x \) has exactly one root (about 0.739). On \( \left[\frac\pi2,\pi\right] \): \( -\cos x=\pi-x \) has exactly one root (about 2.40).
3. So there are 2 solutions.

Teacher's Note:
1. Simplify the root using 1 + cos 2x = 2 cos squared x.
2. Plot or solve on each side of pi over 2.

Answer: (c) two

Explanation:
1. The second term needs \( x^2+x+1\le1 \), i.e. \( x(x+1)\le0 \). The first term needs \( x(x+1)\ge0 \).
2. So \( x(x+1)=0 \), giving \( x=0 \) or \( x=-1 \).
3. Both satisfy the equation: \( \tan^{-1}0+\sin^{-1}1=\frac\pi2 \).

Teacher's Note:
1. Find the common domain first.

Answer: (b) two

Explanation:
1. For \( y>0 \), \( \cos^{-1}\frac y{\sqrt{1+y^2}}=\tan^{-1}\frac1y \), and \( \sin^{-1}\frac3{\sqrt{10}}=\tan^{-1}3 \).
2. \( \tan^{-1}x+\tan^{-1}\frac1y=\tan^{-1}3\Rightarrow\frac{xy+1}{y-x}=3\Rightarrow(x-3)(y+3)=-10 \).
3. With \( y+3\ge4 \) dividing 10: \( y+3=5 \) gives \( (x,y)=(1,2) \); \( y+3=10 \) gives \( (x,y)=(2,7) \). Both check.
4. So there are two solutions.

Teacher's Note:
1. Convert every term to tan inverse.
2. Factorise the resulting equation.

Answer: (c) one

Explanation:
1. \( \frac{5x}{1-6x^2}=1\Rightarrow6x^2+5x-1=0\Rightarrow x=\frac16 \) or \( x=-1 \).
2. For \( x=-1 \) both terms are negative, so the sum cannot be \( \frac\pi4 \) (and the product \( 6x^2>1 \) changes the formula).
3. Only \( x=\frac16 \) works, so there is one solution.

Teacher's Note:
1. Check each root in the original equation.

Answer: (d) \( -1\le x\lt \frac1{\sqrt2} \)

Explanation:
1. Using \( \cos^{-1}x=\frac\pi2-\sin^{-1}x \), the inequality becomes \( \sin^{-1}x\lt \frac\pi4 \).
2. Therefore \( x\lt \frac1{\sqrt2} \), together with the domain \( x\ge-1 \).

Teacher's Note:
1. Use the complementary relation.

Answer: (b) \( 5 \)

Explanation:
1. \( \cot^{-1} \) is a decreasing function, so \( \cot^{-1}t>\frac\pi6 \) means \( t\lt \cot\frac\pi6=\sqrt3 \).
2. \( \frac n\pi\lt \sqrt3\Rightarrow n\lt \sqrt3\pi\approx5.44 \), so the maximum natural n is 5.

Teacher's Note:
1. cot inverse is decreasing, which reverses the inequality.

Answer: (a) \( \phi \)

Explanation:
1. \( \sin^{-1}(\sin4)=\pi-4\lt 0 \) because \( 4\notin\left[-\frac\pi2,\frac\pi2\right] \).
2. For the quadratic to be positive for all x we need \( k^2-4(\pi-4)\lt 0 \), i.e. \( k^2\lt 4(\pi-4)\lt 0 \), which is impossible.
3. So the set is empty.

Teacher's Note:
1. Evaluate the constant term first.
2. It is negative.

Answer: (a) \( f\left(-\frac12\right) \) and \( f\left(\frac1{1+k^2}\right),\ k\in\mathbb R \), only

Explanation:
1. \( \sin^{-1}x+\cos^{-1}x=\frac\pi2 \) holds exactly for \( x\in[-1,1] \), where f is defined.
2. \( -\frac12 \) and \( \frac1{1+k^2}\in(0,1] \) are in the domain. \( k^2-2k+3=(k-1)^2+2\ge2 \) and \( -2 \) are outside it, so f is not defined there.

Teacher's Note:
1. Check whether each argument lies in minus 1 to 1.

Answer: (a) \( \tan^{-1}x \) and \( \operatorname{cosec}^{-1}x \), only

Explanation:
1. \( \tan^{-1}\frac32 \) is real.
2. \( \operatorname{cosec}^{-1}\frac32 \) is real since \( \left|\frac32\right|\ge1 \).
3. \( \cos^{-1}2x=\cos^{-1}3 \) is not real, since \( |3|>1 \).

Teacher's Note:
1. Check each domain at the given value.

Answer: (c) A, B and C

Explanation:
1. \( \tan\frac{5\pi}4=1 \), so \( \alpha=\frac\pi4 \). \( \tan\frac{2\pi}3=-\sqrt3 \), so \( \beta=\tan^{-1}\sqrt3=\frac\pi3 \).
2. (A) \( \frac\pi4\lt \frac\pi3 \) true. (B) \( 4\cdot\frac\pi4-3\cdot\frac\pi3=0 \) true. (C) \( \frac\pi4+\frac\pi3=\frac{7\pi}{12} \) true.

Teacher's Note:
1. Find the values of alpha and beta first.

Answer: (c) both A and B

Explanation:
1. \( \cos^{-1}(\sin\theta)=\frac\pi2-\sin^{-1}(\sin\theta) \).
2. For \( x=\frac{8\pi}9 \): \( \theta=\frac{11\pi}9 \), \( \sin^{-1}\sin\theta=\pi-\theta=-\frac{2\pi}9 \), giving \( \frac\pi2+\frac{2\pi}9=\frac{13\pi}{18} \). (A) is true.
3. For \( x=-\frac{7\pi}4 \): \( \theta=-\frac{17\pi}{12}\equiv\frac{7\pi}{12} \), \( \sin^{-1}\sin\theta=\frac{5\pi}{12} \), giving \( \frac\pi2-\frac{5\pi}{12}=\frac\pi{12} \). (B) is true.

Teacher's Note:
1. Reduce the angle into the range of sin inverse.

Answer: (c) both A and B

Explanation:
1. The bracket equals \( \cos\left(\cos^{-1}x-\frac\pi3\right) \), so the second term is \( \left|\cos^{-1}x-\frac\pi3\right| \).
2. For \( x=\frac23 \): \( \cos^{-1}\frac23\lt \frac\pi3 \), so the second term is \( \frac\pi3-\cos^{-1}x \) and \( f=\frac\pi3 \). (A) is true.
3. For \( x=\frac13 \): \( \cos^{-1}\frac13>\frac\pi3 \), so \( f=2\cos^{-1}\frac13-\frac\pi3 \). (B) is true.

Teacher's Note:
1. Write the bracket as a single cosine.
2. Decide the sign of the angle inside the modulus.

Answer: (a) \( \frac{2x}{1-x^2} \) if \( 0\le x\lt 1 \), and not finite if \( x>1 \)

Explanation:
1. For \( 0\le x\lt 1 \), both terms equal \( 2\tan^{-1}x \), so the expression is \( \tan(2\tan^{-1}x)=\frac{2x}{1-x^2} \).
2. For \( x>1 \), \( \sin^{-1}\frac{2x}{1+x^2}=\pi-2\tan^{-1}x \) and \( \cos^{-1}\frac{1-x^2}{1+x^2}=2\tan^{-1}x \); half the sum is \( \frac\pi2 \), where tan is not finite.

Teacher's Note:
1. Treat the cases x less than 1 and x greater than 1 separately.

Answer: (c) A, B and C

Explanation:
1. Put \( x=\cos\theta \): \( \frac{\sin\theta}{1+\cos\theta}=\tan\frac\theta2 \), so the value is \( \frac\theta2=\frac12\cos^{-1}x \). (A) true.
2. \( \cos\frac\theta2=\sqrt{\frac{1+x}2} \) and \( \sin\frac\theta2=\sqrt{\frac{1-x}2} \), and \( \frac\theta2\in\left(0,\frac\pi4\right) \). So (B) and (C) are true too.

Teacher's Note:
1. Substitute x = cos theta.

Answer: (b) \( \frac3{\sqrt5} \)

Explanation:
1. \( \sin\left(\cos^{-1}\frac1{\sqrt5}\right)=\frac2{\sqrt5} \), which is positive.
2. \( \tan(\sec^{-1}x) \) is positive only for \( x\ge1 \) and equals \( \sqrt{x^2-1} \). So \( x^2-1=\frac45\Rightarrow x=\frac3{\sqrt5} \).
3. The options with \( |x|\lt 1 \) are not in the domain, and for \( x\lt 0 \) the tangent would be negative.

Teacher's Note:
1. Find the right side first.
2. Check the sign of tan for sec inverse x.

Answer: (d) none of these

Explanation:
1. Using \( \cos^{-1}t=\frac\pi2-\sin^{-1}t \), the second equation becomes \( \sin^{-1}y-\sin^{-1}x=\frac\pi3 \).
2. Adding to the first: \( \sin^{-1}y=\frac\pi2 \), so \( y=1 \), and \( \sin^{-1}x=\frac\pi6 \), so \( x=\frac12 \).
3. There is exactly one pair \( \left(\frac12,1\right) \), so the answer is none of these.

Teacher's Note:
1. Convert cos inverse to sin inverse.

Answer: (b) \( \left\{0,\frac12\right\} \)

Explanation:
1. The domain is \( 0\le x\le1 \). With \( s=\sin^{-1}x \), the left side is \( \frac\pi2-2s \), which lies in \( \left[-\frac\pi2,\frac\pi2\right] \).
2. Taking sine: \( \cos2s=1-x\Rightarrow1-2x^2=1-x\Rightarrow x=0 \) or \( x=\frac12 \).
3. Both roots check, so the solution set is \( \left\{0,\frac12\right\} \).

Teacher's Note:
1. Take sine of both sides and check the roots.

Answer: (b) 1

Explanation:
1. \( \tan^{-1}\frac{1}{2n^2}=\tan^{-1}\frac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)}=\tan^{-1}(2n+1)-\tan^{-1}(2n-1) \).
2. The sum telescopes to \( \lim\tan^{-1}(2N+1)-\tan^{-1}1=\frac\pi2-\frac\pi4=\frac\pi4 \).
3. So \( \tan t=1 \).

Teacher's Note:
1. Write each term as a difference of two inverse tangents.
2. The series then telescopes.