JEE Mathematics Inverse Trigonometric Functions MCQs Set 18

Answer: (c) \( \left[\frac12,1\right] \)

Explanation:
1. Let \( t=\cos^{-1}x\in[0,\pi] \), so \( x=\cos t,\ \sqrt{1-x^2}=\sin t \). Then \( \frac x2+\frac12\sqrt{3-3x^2}=\cos t\cos\frac\pi3+\sin t\sin\frac\pi3=\cos\left(t-\frac\pi3\right) \), and \( \cos^{-1}\left[\cos\left(t-\frac\pi3\right)\right]=\left|t-\frac\pi3\right| \).
2. If \( 0\le x\le\frac12 \), then \( \cos^{-1}x\ge\frac\pi3 \) and the second term is \( \cos^{-1}x-\frac\pi3 \); the equation becomes \( 2\cos^{-1}x-\frac\pi3=\frac\pi3 \), which holds only for \( x=\frac12 \).
3. If \( \frac12\le x\le1 \), then \( \cos^{-1}x\le\frac\pi3 \) and the second term is \( \frac\pi3-\cos^{-1}x \); the sum is \( \frac\pi3 \) for every such \( x \), so the identity holds on \( \left[\frac12,1\right] \).

Teacher's Note:
1. The formula \( \cos^{-1}x\pm\cos^{-1}y \) changes with the position of \( x \) relative to the given angle, so split into cases.
2. Test a value: \( x=0.8 \) gives \( 0.6435+0.4036=1.0471=\frac\pi3 \).

Answer: (b) \( a=1 \) and \( b=-\frac13 \)

Explanation:
1. \( \sin^{-1}u+\cos^{-1}v=\frac\pi2 \) holds when \( u=v \), both in \( [-1,1] \).
2. The first series is a G.P. with ratio \( -\frac a3 \) (needs \( |a|<3 \)): sum \( \frac{a}{1+\frac a3}=\frac{3a}{a+3} \). The second has sum \( \frac1{1-b} \) for \( |b|<1 \).
3. Test the options: \( a=-3 \) and \( b=1 \) give divergent series; \( a=1,\ b=-\frac13 \) give \( \frac34=\frac1{4/3}=\frac34 \), which is true; \( a=\frac16,\ b=\frac12 \) give \( \frac3{19}\ne2 \).

Teacher's Note:
1. Sum the infinite G.P.s only when the common ratio has modulus less than 1.
2. \( \sin^{-1}u+\cos^{-1}u=\frac\pi2 \) is the key identity.

Answer: (b) 1

Explanation:
1. Put \( x=0 \) and \( x=\pi \): \( a_1+a_3+a_5=0 \) and \( a_1-a_3+a_5=0 \), so \( a_3=0 \) and \( a_1+a_5=0 \).
2. Put \( x=\frac\pi2 \) and \( \frac{3\pi}2 \): \( a_1+a_2-a_5=0 \) and \( a_1-a_2-a_5=0 \), so \( a_2=0 \) and \( a_1=a_5 \).
3. With \( a_1+a_5=0 \) and \( a_1=a_5 \), \( a_1=a_5=0 \). The equation reduces to \( a_4\sin2x=0 \), true for all \( x \) only if \( a_4=0 \).
4. So only the zero tuple works: the number of 5-tuples is 1.

Teacher's Note:
1. The functions \( 1,\sin x,\cos x,\sin2x,\cos2x \) are linearly independent, so every coefficient must vanish.
2. Choose values of \( x \) that make several terms zero or equal.

Answer: (d) odd values of \( n \)

Explanation:
1. From \( x=n\pi-\tan^{-1}3 \): \( \tan x=-3 \), so \( \tan2x=\frac{2(-3)}{1-9}=\frac34 \) and \( 12\tan2x=9 \).
2. The equation becomes \( 9+\frac{\sqrt{10}}{\cos x}+1=0 \), so \( \cos x=-\frac{\sqrt{10}}{10}=-\frac1{\sqrt{10}} \).
3. \( \tan x=-3 \) and \( \cos x<0 \) place \( x \) in the second quadrant. \( x=n\pi-\tan^{-1}3 \) is in the second quadrant (angle \( \pi-\tan^{-1}3 \), plus multiples of \( 2\pi \)) only for odd \( n \).

Teacher's Note:
1. Find \( \tan x,\ \tan2x \) first, then determine the sign of \( \cos x \) from the quadrant.
2. Odd \( n \) gives \( \cos x=-\frac1{\sqrt{10}} \); even \( n \) gives \( +\frac1{\sqrt{10}} \).

Answer: (d) \( S_\infty=\cot^{-1}\frac12 \)

Explanation:
1. The \( r \)th term is \( \cot^{-1}\frac{4r^2+3}4=\tan^{-1}\frac1{1+r^2-\frac14} \) (check: \( r=1 \) gives \( \frac74 \)).
2. Since \( 1+\left(r+\frac12\right)\left(r-\frac12\right)=r^2+\frac34 \), the term equals \( \tan^{-1}\left(r+\frac12\right)-\tan^{-1}\left(r-\frac12\right) \).
3. The sum telescopes: \( S_n=\tan^{-1}\left(n+\frac12\right)-\tan^{-1}\frac12=\tan^{-1}\frac{4n}{2n+5} \). So (a), (b), (c) are incorrect.
4. \( S_\infty=\tan^{-1}\infty-\tan^{-1}\frac12=\frac\pi2-\tan^{-1}\frac12=\tan^{-1}2=\cot^{-1}\frac12 \), so (d) is correct.

Teacher's Note:
1. Write each term as a difference of two inverse tangents so the sum telescopes.
2. Check each option separately; \( S_n=\cot^{-1}\frac{2n+5}{4n} \) is the correct finite sum.

Answer: (a) (i), (ii) and (iii) only

Explanation:
1. Put \( x=\tan\theta \). Then \( y=\tan4\theta=\frac{2\tan2\theta}{1-\tan^22\theta} \), and \( \tan2\theta=\frac{2x}{1-x^2} \).
2. \( y=\frac{4x(1-x^2)}{(1-x^2)^2-4x^2}=\frac{4x(1-x^2)}{1-6x^2+x^4} \).
3. \( y \) is not finite when \( x^4-6x^2+1=0 \), i.e. \( x^4=6x^2-1 \) and \( x^2=3\pm2\sqrt2 \). So (i), (ii), (iii) are correct and (iv) is not.

Teacher's Note:
1. The denominator \( 1-6x^2+x^4 \) vanishing is the condition for \( y \) to be infinite.
2. \( x^2=3\pm2\sqrt2 \) are the roots of \( t^2-6t+1=0 \).

Answer: (b) \( -1,\ 1 \) and \( 2 \)

Explanation:
1. The equation holds when \( x^2+x+1=ax+1 \) with \( -1\le x^2+x+1\le1 \) (domain of \( \sin^{-1} \)).
2. \( x^2+x+1\le1\Rightarrow x(x+1)\le0\Rightarrow-1\le x\le0 \). Also \( x(x+1-a)=0 \), so \( x=0 \) or \( x=a-1 \).
3. For a second, different solution: \( -1\le a-1<0\Rightarrow0\le a<1 \). The only integer in this range is 0.
4. Hence \( a \) cannot be \( -1,\ 1 \) or \( 2 \).

Teacher's Note:
1. Reduce to the equality of arguments, then impose the domain restrictions.
2. Exactly two solutions means \( x=0 \) and a distinct root \( x=a-1 \) inside the domain.

Answer: (b) (ii), (iii) and (iv) only

Explanation:
1. \( \tan^{-1}A+\cot^{-1}B=\frac\pi2 \) requires \( A=B \): \( \sin^2\theta+2\sin\theta+2=4^{\sec^2\phi}+1 \).
2. LHS \( =(\sin\theta+1)^2+1\le5 \), while RHS \( \ge4+1=5 \) because \( \sec^2\phi\ge1 \).
3. So both sides equal 5: \( (\sin\theta+1)^2=4\Rightarrow\sin\theta=1 \) and \( \sec^2\phi=1\Rightarrow\cos\phi=\pm1 \).

Teacher's Note:
1. Compare the maximum of one side with the minimum of the other.
2. Note \( \sin\theta=-1 \) gives LHS \( =1 \), not 5.

Answer: (a) (i), (ii) and (iv) only

Explanation:
1. \( \tan^{-1}(x-1)+\tan^{-1}x=\tan^{-1}3x-\tan^{-1}(x+1) \), so \( \frac{2x-1}{1-x^2+x}=\frac{2x-1}{1+3x^2+3x} \).
2. Hence \( (2x-1)\left[(1+3x^2+3x)-(1-x^2+x)\right]=0 \), i.e. \( (2x-1)(4x^2+2x)=0 \), so \( x=0,\ \pm\frac12 \) (roots \( \frac12,-\frac12,0 \) all verified by substitution).
3. Sum \( =0 \) (true); \( \alpha\beta+\beta\gamma+\gamma\alpha=-\frac14 \) (true); product \( =0\ne1 \) (false); the largest difference of two roots is \( \frac12-\left(-\frac12\right)=1 \) (true).

Teacher's Note:
1. Cross-multiplying leads to a cubic that factorises with \( (2x-1) \) common.
2. Check the roots in the original equation, since inverse tangent identities can add extraneous roots.

Answer: (a) 1

Explanation:
1. Domains: \( 0\le x^2+x+1\le1 \) and \( 0\le x^2+x\le1 \). Together \( x^2+x+1\le1 \) and \( x^2+x\ge0 \) give \( x^2+x=0 \), so \( x=0 \) or \( x=-1 \).
2. For \( x=0 \) or \( x=-1 \): LHS \( =2\sin^{-1}1+\cos^{-1}0=\pi+\frac\pi2=\frac{3\pi}2 \). Both are solutions.
3. The sum of solutions is \( -1=-p \), so \( p=1 \).

Teacher's Note:
1. Domain restrictions can pin the variable down completely.
2. Verify each candidate in the original equation.

Answer: (a) 1

Explanation:
1. \( \cos^{-1}\left(-\frac{\sqrt3}2\right)=\pi-\cos^{-1}\frac{\sqrt3}2=\pi-\frac\pi6=\frac{5\pi}6 \).
2. The bracket becomes \( \cos\left(\frac{5\pi}6+\frac\pi6\right)=\cos\pi=-1 \).
3. So the value is \( -(-1)=1 \).

Teacher's Note:
1. Use \( \cos^{-1}(-x)=\pi-\cos^{-1}x \).
2. Mind the negative sign outside the cosine.

Answer: (a) 1

Explanation:
1. Both series are infinite G.P.s with ratios \( -\frac x2 \) and \( -\frac{x^2}2 \) (both of modulus below 1 because \( 0<|x|<\sqrt2 \)).
2. Sums: \( \frac x{1+\frac x2} \) and \( \frac{x^2}{1+\frac{x^2}2} \). Since \( \sin^{-1}u+\cos^{-1}v=\frac\pi2 \) needs \( u=v \): \( \frac x{1+\frac x2}=\frac{x^2}{1+\frac{x^2}2} \).
3. Cross-multiplying: \( x+\frac{x^3}2=x^2+\frac{x^3}2\Rightarrow x=x^2 \). As \( x\ne0 \), \( x=1 \).

Teacher's Note:
1. Identify the G.P. ratio before summing to infinity.
2. Reject \( x=0 \) using the given condition \( 0<|x| \).

Answer: (a) 3

Explanation:
1. \( \operatorname{cosec}^{-1}\frac54=\sin^{-1}\frac45 \).
2. So \( \sin^{-1}\frac x5=\frac\pi2-\sin^{-1}\frac45=\cos^{-1}\frac45 \).
3. \( \cos^{-1}\frac45=\sin^{-1}\frac35 \) (3-4-5 triangle), hence \( \frac x5=\frac35 \) and \( x=3 \).

Teacher's Note:
1. Convert \( \operatorname{cosec}^{-1} \) to \( \sin^{-1} \) by taking the reciprocal.
2. \( \sin^{-1}u+\cos^{-1}u=\frac\pi2 \).

Answer: (a) 4

Explanation:
1. \( x=\cos\left(\alpha+\cos^{-1}\frac y2\right)=\frac y2\cos\alpha-\sin\alpha\sqrt{1-\frac{y^2}4} \).
2. So \( 2x-y\cos\alpha=-\sin\alpha\sqrt{4-y^2} \).
3. Squaring: \( 4x^2-4xy\cos\alpha+y^2\cos^2\alpha=\sin^2\alpha(4-y^2) \), so \( 4x^2-4xy\cos\alpha+y^2=4\sin^2\alpha \) (using \( \cos^2\alpha+\sin^2\alpha=1 \)). Hence \( k=4 \).

Teacher's Note:
1. Isolate the square root before squaring.
2. Use \( \cos(A+B) \) with \( \cos\left(\cos^{-1}\frac y2\right)=\frac y2 \).

Answer: (b) 1

Explanation:
1. The largest value of \( \sin^{-1}x \) is \( \frac{5\pi}2 \) (at \( x=1 \)) and the largest value of \( \cos^{-1}y \) is \( 3\pi \) (at \( y=-1 \)).
2. Their greatest possible sum is \( \frac{5\pi}2+3\pi=\frac{11\pi}2 \), which equals the required value.
3. So equality needs both maxima simultaneously: \( x=1,\ y=-1 \). There is exactly one solution.

Teacher's Note:
1. When the required value equals the maximum possible sum, each term must attain its own maximum.
2. Always use the stated range of the functions, not the usual one.

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

Explanation:
1. Let \( \theta=\sin^{-1}x\in\left[-\frac\pi2,\frac\pi2\right] \). Then \( 2x\sqrt{1-x^2}=\sin2\theta \), and \( \sin^{-1}(\sin2\theta)=2\theta \) only if \( 2\theta\in\left[-\frac\pi2,\frac\pi2\right] \).
2. That gives \( -\frac\pi4\le\theta\le\frac\pi4 \), so \( x=\sin\theta\in\left[-\frac1{\sqrt2},\frac1{\sqrt2}\right] \).

Teacher's Note:
1. \( \sin^{-1}(\sin t)=t \) only when \( t\in\left[-\frac\pi2,\frac\pi2\right] \).
2. Check: for \( x=0.8 \) the two sides differ (1.854 against 1.287).

Answer: (a) \( -1 \)

Explanation:
1. \( \sin^{-1}\sqrt{x^2+x+1}=\frac\pi2-\tan^{-1}\sqrt{x^2+x}=\cot^{-1}\sqrt{x^2+x}=\sin^{-1}\frac1{\sqrt{x^2+x+1}} \).
2. Hence \( \sqrt{x^2+x+1}=\frac1{\sqrt{x^2+x+1}} \), so \( x^2+x+1=1 \) and \( x(x+1)=0 \), giving \( x=0 \) or \( x=-1 \) (both satisfy \( x^2+x\ge0 \)).
3. The negative solution is \( -1 \).

Teacher's Note:
1. Convert \( \cot^{-1}t \) into \( \sin^{-1}\frac1{\sqrt{1+t^2}} \).
2. Check the domain \( x(x+1)\ge0 \).

Answer: (a) \( -\frac12 \)

Explanation:
1. \( \sin[\cot^{-1}(1+x)]=\frac1{\sqrt{1+(1+x)^2}} \) and \( \cos(\tan^{-1}x)=\frac1{\sqrt{1+x^2}} \) (for \( 1+x>0 \)).
2. Equating: \( 1+(1+x)^2=1+x^2\Rightarrow2x+1=0 \).
3. So \( x=-\frac12 \). Check: \( \sin[\cot^{-1}\frac12]=\frac2{\sqrt5}=\cos(\tan^{-1}(-\frac12)) \).

Teacher's Note:
1. Express both sides as algebraic functions using a right triangle.
2. Check that the principal-value assumptions hold for the solution found.

Answer: (a) \( \frac{25}{23} \)

Explanation:
1. \( 1+\sum_{k=1}^n2k=1+n(n+1) \), so each term is \( \cot^{-1}[1+n(n+1)]=\tan^{-1}\frac1{1+n(n+1)}=\tan^{-1}(n+1)-\tan^{-1}n \).
2. The sum telescopes: \( \sum_{n=1}^{23}=\tan^{-1}24-\tan^{-1}1=\tan^{-1}\frac{24-1}{1+24}=\tan^{-1}\frac{23}{25} \).
3. \( \cot\left(\tan^{-1}\frac{23}{25}\right)=\frac{25}{23} \).

Teacher's Note:
1. Use \( \tan^{-1}a-\tan^{-1}b=\tan^{-1}\frac{a-b}{1+ab} \) to telescope.
2. \( 1+n(n+1) \) is exactly the denominator \( 1+(n)(n+1) \) in that formula.

Answer: (d) 1

Explanation:
1. Each of \( (\sin^{-1}x)^2 \) and \( (\sin^{-1}y)^2 \) is at most \( \frac{\pi^2}4 \). Their sum equals \( \frac{\pi^2}2 \) only if each equals \( \frac{\pi^2}4 \).
2. So \( \sin^{-1}x=\pm\frac\pi2 \) and \( \sin^{-1}y=\pm\frac\pi2 \), i.e. \( x=\pm1,\ y=\pm1 \).
3. Then \( x^3+y^3\in\{-2,0,2\} \), so it can never be 1.

Teacher's Note:
1. When a sum of two bounded quantities equals the sum of their maxima, each must be at its maximum.
2. List all sign combinations of \( x,y \).

Answer: (a) \( -2 \)

Explanation:
1. Each of \( (\cos^{-1}x)^2 \) and \( (\cos^{-1}y)^2 \) is at most \( \pi^2 \), so both must equal \( \pi^2 \).
2. Then \( \cos^{-1}x=\cos^{-1}y=\pi \), i.e. \( x=y=-1 \).
3. \( x^5+y^5=-1-1=-2 \).

Teacher's Note:
1. The range of \( \cos^{-1} \) is \( [0,\pi] \), so the maximum of each square is \( \pi^2 \).
2. Equality with the sum of maxima forces both extremes.

Answer: (a) 0

Explanation:
1. \( (\sin^{-1}x)^2\le\frac{\pi^2}4 \) and \( (\cos^{-1}y)^2\le\pi^2 \), so the product is at most \( \frac{\pi^4}4 \), with equality only if both are at their maxima.
2. So \( \sin^{-1}x=\pm\frac\pi2 \) and \( \cos^{-1}y=\pi \), giving \( x=\pm1 \) and \( y=-1 \).
3. \( |x-y|=|1-(-1)|=2 \) or \( |-1-(-1)|=0 \). Among the options, 0 is possible.

Teacher's Note:
1. The product reaches its maximum only when both factors do.
2. Compute \( |x-y| \) for both sign choices of \( x \).

Answer: (a) 1

Explanation:
1. The difference of two values in \( \left[-\frac\pi2,\frac\pi2\right] \) has modulus \( \pi \) only when one is \( \frac\pi2 \) and the other \( -\frac\pi2 \).
2. So \( (x,y)=(1,-1) \) or \( (-1,1) \).
3. \( x^y=1^{-1}=1 \) or \( (-1)^1=-1 \). Among the options, only 1 is possible.

Teacher's Note:
1. The maximum possible difference of two principal values of \( \sin^{-1} \) is \( \pi \).
2. Evaluate \( x^y \) for both pairs.