JEE Mathematics Inverse Trigonometric Functions MCQs Set 11

Answer: (a) Only A and B

Explanation:
1. With \( x=\tan\theta \), \( \frac{2x}{1+x^2}=\sin2\theta \), and \( \sin^{-1}(\sin2\theta)=2\theta \) only when \( |2\theta|\le\frac\pi2 \), i.e. \( |x|\le1 \). So A is true.
2. With \( x=\sin\theta \), \( \sin^{-1}(\sin2\theta)=2\theta \) needs \( |\theta|\le\frac\pi4 \), i.e. \( |x|\le\frac1{\sqrt2} \). So B is true.
3. For \( x>1 \) the right side equals \( \pi-2\tan^{-1}x \) and for \( x<-1 \) it equals \( -\pi-2\tan^{-1}x \), so C and D are false.

Teacher's Note:
1. The double-angle identities hold only while the doubled angle stays inside the principal range of the outer function.
2. Outside that range add or subtract pi, which is why the sign of the restriction matters.

Answer: (b) Only A, B and C

Explanation:
1. \( \sin\left(2\tan^{-1}\tfrac13\right)=\frac{2/3}{1+1/9}=\frac35 \) and the angle is acute, so A is true.
2. \( \cot^{-1}2=\tan^{-1}\tfrac12 \) and \( \frac\pi4-\tan^{-1}\frac12=\tan^{-1}\frac{1-\frac12}{1+\frac12}=\tan^{-1}\frac13 \), so B is true.
3. \( \frac12\sin^{-1}\frac45=\tan^{-1}\frac12 \) (because \( \sin\left(2\tan^{-1}\tfrac12\right)=\frac45 \)), so C reduces to B and is true. D gives \( \frac\pi2-\cot^{-1}3=\tan^{-1}3\neq\tan^{-1}\tfrac13 \), false.

Teacher's Note:
1. Convert every term to an inverse tangent and compare the tangents of both sides.
2. Option D equals tan inverse 3, not tan inverse of one-third.

Answer: (c) Only A, B and C

Explanation:
1. Let \( t=\tan\theta \). Taking tangent of both sides: \( t=\frac{2t^2-\frac t3}{1+\frac{2t^3}{3}} \).
2. Clearing fractions: \( 2t^4-6t^2+4t=0\Rightarrow t(t-1)^2(t+2)=0 \), so \( t=0,1,-2 \).
3. Each is consistent with the original equation (for example \( t=-2 \) gives \( \theta=\tan^{-1}8+\tan^{-1}\frac23\approx2.034 \) with \( \tan\theta=-2 \)), while \( t=2 \) is not a root.

Teacher's Note:
1. Take the tangent of both sides after rewriting the equation as an identity in a single variable t.
2. Always substitute the roots back, because squaring-free tangent equations can still pick up extraneous angles.

Answer: (b) Only B and C

Explanation:
1. At \( x=\frac{8\pi}9 \): \( \sin\frac{11\pi}9=-\sin\frac{2\pi}9 \), so \( \cos^{-1}\left(\sin\left(-\frac{2\pi}9\right)\right)=\frac\pi2+\frac{2\pi}9=\frac{13\pi}{18} \) (B true, A false).
2. At \( x=-\frac{7\pi}4 \): \( \sin\frac{7\pi}{12}=\cos\frac\pi{12} \), so the exponent is \( \frac\pi{12} \) (C true).
3. At \( x=\frac{7\pi}4 \): \( \sin\frac{25\pi}{12}=\sin\frac\pi{12} \), exponent \( \frac\pi2-\frac\pi{12}=\frac{5\pi}{12} \), so D is false.

Teacher's Note:
1. Reduce the sine argument to a standard angle, then use cos inverse of sin t = pi/2 - (principal value of sin inverse of sin t).
2. Check each option separately; the exponents differ only in the final fraction.

Answer: (b) Only A and C

Explanation:
1. \( f(x)=\frac\pi2 \) only for \( x\in[-1,1] \); elsewhere \( f \) is undefined.
2. \( -\frac12 \) and \( \frac1{1+k^2}\in(0,1] \) lie in the domain, so A and C hold.
3. \( k^2-2k+3=(k-1)^2+2\ge2 \) and \( -2 \) are outside the domain, so B and D are meaningless.

Teacher's Note:
1. The identity sin inverse + cos inverse = pi/2 carries the restriction that x lies in [-1,1].
2. Complete the square to see that the argument in B never enters the domain.

Answer: (a) Only A and B

Explanation:
1. The domain requires \( 0\le x\le1 \). Write \( \sin^{-1}(1-x)=\frac\pi2-2\sin^{-1}x \).
2. Take sine: \( 1-x=\cos(2\sin^{-1}x)=1-2x^2\Rightarrow x(2x-1)=0 \), so \( x=0 \) or \( x=\frac12 \).
3. Check: \( x=0 \) gives \( 0+\frac\pi2=\frac\pi2 \); \( x=\frac12 \) gives \( \frac\pi6+\frac\pi6=\frac\pi3=\cos^{-1}\frac12 \). For \( x=1 \): \( \frac\pi2+0\ne0 \).

Teacher's Note:
1. Isolate one inverse term, take sine, and always verify the roots in the original equation.
2. Domain restrictions remove x = 2 immediately.

Answer: (d) 5

Explanation:
1. On one period the two sides agree exactly when \( x\in\left[0,\frac\pi2\right] \) (both equal \( x \)); on \( \left(\frac\pi2,2\pi\right) \) they differ except at \( x=2\pi \) which starts the next block.
2. So the solution set is \( \bigcup_k\left[2k\pi,2k\pi+\frac\pi2\right] \).
3. In \( [0,5\pi] \) the integers are \( 0,1 \) (from \( [0,1.57] \)), \( 7 \) (from \( [6.28,7.85] \)), \( 13,14 \) (from \( [12.57,14.14] \)): five in all.

Teacher's Note:
1. Draw the graphs of sin inverse of sin x and cos inverse of cos x; they coincide only on the first quarter of each 2 pi block.
2. Count integers inside each interval carefully, including 0.

Answer: (a) Only A and C

Explanation:
1. Taking tangent: \( \frac{\frac{x-1}{x-2}+\frac{x+1}{x+2}}{1-\frac{(x-1)(x+1)}{(x-2)(x+2)}}=1 \).
2. The numerator is \( \frac{2x^2-4}{x^2-4} \) and the denominator is \( \frac{-3}{x^2-4} \); so \( 2x^2-4=-3\Rightarrow x^2=\frac12 \).
3. Both \( x=\pm\frac1{\sqrt2} \) satisfy the original equation (the sum of the two inverse tangents is \( \frac\pi4 \) with no extra \( \pi \)).

Teacher's Note:
1. After taking the tangent the denominators cancel, leaving a simple equation in x squared.
2. Verify that the sum of the two angles lies in the principal range so that no multiple of pi is hidden.

Answer: (d) All of A, B, C and D

Explanation:
1. \( x^2+2x+2=(x+1)^2+1\ge1 \), and \( \sin^{-1} \) requires an argument at most 1, so \( x=-1 \) and \( \sin^{-1}1=\frac\pi2 \).
2. The inequality becomes \( \tan^{-1}\left(1+3-k^2\right)>0\Rightarrow4-k^2>0\Rightarrow k\in(-2,2) \).
3. Each of the intervals \( (-1,0),(0,1),(1,2),(0,2) \) lies inside \( (-2,2) \), so all four statements hold.

Teacher's Note:
1. The domain of sin inverse collapses the quadratic argument to a single point.
2. After x = -1, the remaining condition is a simple inequality in k.

Answer: (b) Only A, B and C

Explanation:
1. \( \tan^{-1}(\tan4)=4-\pi\approx0.86 \) and \( \cot^{-1}(\cot4)=4-\pi \), so (A) \( \cos0.86>0 \) and (B) \( \sin0.86>0 \).
2. \( \cos^{-1}(\cos5)=2\pi-5\approx1.28<\frac\pi2 \), so (C) the tangent is positive.
3. \( \sin^{-1}(\sin4)=\pi-4\approx-0.86 \) and the cotangent of a negative acute angle is negative, so D is not positive.

Teacher's Note:
1. Reduce each inverse-of-trig value to the principal range first, then read the sign of the outer function.
2. An angle in the fourth quadrant gives a negative cotangent.

Answer: (b) Only A, B and D

Explanation:
1. \( \tan\left|\tan^{-1}x\right|=\left|\tan\left(\tan^{-1}x\right)\right|=|x| \) (A) and similarly \( \sin|\sin^{-1}x|=|x| \) (D).
2. \( \cot^{-1}x\in(0,\pi) \) is already non-negative, so \( \cot\left|\cot^{-1}x\right|=x \) (B).
3. (C) holds only for \( |x|<\frac\pi2 \); for example \( x=2 \) gives \( \tan^{-1}(|\tan2|)\approx1.14\ne2 \). So C is false.

Teacher's Note:
1. The absolute value passes through tan and sin because they are odd functions, but cot inverse is never negative.
2. A relation with tan inverse of tan x is valid only on the principal interval.

Answer: (b) Only A, B and D

Explanation:
1. (A) The infimum of \( e^x \) is \( 0 \) and \( [0]=0=\left[e^{-\pi/2}\right]=[0.21] \), so the statement of inequality is wrong.
2. (B) For \( x\in(0,\frac\pi2) \), \( \sin(\sin x)\in(0,1) \) and \( \cos(\cos x)\in(0.54,1) \), so both floors are 0 and \( 0>0 \) is wrong.
3. (C) \( \sin1\approx0.841>\cos0.841\approx0.666 \) is correct. (D) \( \cos^{-1}\frac14\approx1.32>\sin^{-1}\frac14\approx0.25 \), so D is incorrect.

Teacher's Note:
1. Evaluate each side numerically to a couple of decimals; the greatest integer function hides small differences.
2. Note that cos inverse of a small positive number is large, while sin inverse of it is small.

Answer: (a) \( -\frac\pi3 \)

Explanation:
1. With \( \theta=\cos^{-1}x\in\left[\frac\pi3,\frac{2\pi}3\right] \): \( 3\theta\in[\pi,2\pi] \) and \( \cos^{-1}(\cos3\theta)=2\pi-3\theta \).
2. Hence \( a=2\pi,\ b=-3 \) and \( \frac ab=-\frac{2\pi}3 \).
3. \( \sin\left(-\frac{2\pi}3\right)=-\frac{\sqrt3}2\Rightarrow\sin^{-1}\left(-\frac{\sqrt3}2\right)=-\frac\pi3 \).

Teacher's Note:
1. On [pi, 2 pi] the function cos inverse of cos phi equals 2 pi minus phi.
2. Always finish by returning to the principal range of sin inverse.

Answer: (d) 3

Explanation:
1. Now \( \theta\in\left[0,\frac\pi3\right) \) so \( 3\theta\in[0,\pi) \) and \( \cos^{-1}(\cos3\theta)=3\theta \), giving \( a=0,\ b=3 \).
2. \( \lim_{y\to0}3\cos y=3 \).

Teacher's Note:
1. Here 3 theta stays inside [0, pi), so cos inverse of cos 3 theta is simply 3 theta.
2. Compare this with the two other intervals: a and b change with the position of x.

Answer: (d) \( \cos\frac{\pi^2}4 \)

Explanation:
1. As in the previous step, \( p=2 \) gives \( u=\cos^{-1}x=\frac{\pi^2}4 \) (this is less than \( \pi \), so it is admissible).
2. Therefore \( x=\cos\frac{\pi^2}4 \).

Teacher's Note:
1. After solving for u, take the cosine of both sides to recover x.
2. The value pi squared over 4 is about 2.47, which lies in [0, pi].

Answer: (c) \( \frac12 \)

Explanation:
1. With \( p=2 \), \( v=\left(\sin^{-1}y\right)^2=\frac{\pi^2}4 \) so \( \sin^{-1}y=\pm\frac\pi2 \).
2. Hence \( y=\pm1 \), and \( y=\frac12 \) gives \( \left(\frac\pi6\right)^2 \), which is not \( \frac{\pi^2}4 \).

Teacher's Note:
1. Because v equals pi squared over 4, sin inverse y must equal plus or minus pi over 2.
2. The value one half is not one of the two allowed values of y.

Answer: (a) A-S, B-R, C-P, D-Q

Explanation:
1. (A) Each \( \cos^{-1}\le\pi \), so all equal \( \pi \): \( \lambda=\mu=\gamma=-1 \) and \( \lambda\mu+\mu\gamma+\gamma\lambda=3 \) (S). (B) every \( \cos^{-1}x_i=0 \), so \( x_i=1 \) and the sum is 10 (R).
2. (C) The \( 2n \) angles each at most \( \frac\pi2 \) add to \( n\pi \) only if each is \( \frac\pi2 \): every \( x_i=1 \), sum \( 2n \) (P).
3. (D) \( \sin\left(\sin^{-1}x-\frac\pi6\right)=\frac{\sqrt3}2x-\frac12\sqrt{1-x^2} \), and for \( |x|\le\frac12 \) the angle \( \sin^{-1}x-\frac\pi6\in\left[-\frac\pi3,0\right] \) is principal, so the value is \( \sin^{-1}x-\frac\pi6 \) (Q).

Teacher's Note:
1. Extreme-value arguments: sums of bounded inverse functions pin each term to its maximum or minimum.
2. Use the sine subtraction formula backwards for part D and then check the principal range.

Answer: (a) A-R, B-S, C-P, D-Q

Explanation:
1. (A) \( \sin\left(\frac\pi3+\frac\pi6\right)=\sin\frac\pi2=1 \) (R). (B) \( \cos\left(\frac{5\pi}6+\frac\pi6\right)=\cos\pi=-1 \) (S).
2. (C) \( \tan\frac\varphi2=\sqrt{\frac{1-\cos\varphi}{1+\cos\varphi}} \) with \( \cos\varphi=\frac{\sqrt5}3 \) gives \( \frac{3-\sqrt5}{2} \) (P).
3. (D) With \( \tan\theta=\frac43 \): \( \tan2\theta=\frac{8/3}{1-16/9}=-\frac{24}7 \) (Q).

Teacher's Note:
1. Evaluate each row independently and then match values.
2. The half-angle tangent formula avoids finding the angle itself.

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

Explanation:
1. R is the standard result for positive \( x,y \) with \( xy>1 \) (the fraction is negative so a \( \pi \) must be added).
2. With \( x=2,y=3 \): \( \pi+\tan^{-1}\frac{5}{-5}=\pi-\frac\pi4=\frac{3\pi}4 \), so A is true and R explains it.

Teacher's Note:
1. The addition formula for inverse tangent changes by pi when the product of the two numbers exceeds 1.
2. Both angles are acute but their sum exceeds pi/2, so the sum lies in the second quadrant.

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

Explanation:
1. For \( t\in(0,1) \): \( \tan^{-1}t<t<\sin^{-1}t \). So if \( x>y \): \( \sin^{-1}x>\sin^{-1}y>\tan^{-1}y \), and R is true.
2. Here \( \frac1{\sqrt e}\approx0.607>\frac1{\sqrt\pi}\approx0.564 \), so A follows directly from R (numerically \( 0.652>0.514 \)).

Teacher's Note:
1. Compare through the intermediate function: sin inverse is increasing and sin inverse y exceeds tan inverse y on (0,1).
2. Check the numbers: 1/root e is larger than 1/root pi.

Answer: (c) 5

Explanation:
1. Let \( p=\cot^{-1}x \), \( \tan^{-1}x=\frac\pi2-p \). The left side factors as \( (p-3)\left(\tan^{-1}x+2-\frac\pi2\right)=(p-3)(2-p) \).
2. \( (p-3)(2-p)>0\Rightarrow2<p<3 \), so \( 2<\cot^{-1}x<3\Rightarrow\cot3<x<\cot2 \) (cot inverse is decreasing).
3. Thus \( a=\cot3,\ b=\cot2 \) and \( \cot^{-1}a+\cot^{-1}b=3+2=5 \).

Teacher's Note:
1. Substitute p for cot inverse x and use tan inverse x = pi/2 - p to factor the expression.
2. The inequality direction flips when passing from the angle back to x because cot inverse is decreasing.

Answer: (b) 2

Explanation:
1. Domains force \( a^6+1\le1 \) and \( a^4+1\le1 \), so \( a=0 \).
2. \( x=\sin^{-1}1+\cos^{-1}1-\tan^{-1}1=\frac\pi2+0-\frac\pi4=\frac\pi4 \) and \( \sec^2\frac\pi4=2 \).

Teacher's Note:
1. Even powers plus 1 are at least 1, so the arguments of sin inverse and cos inverse can only equal 1 when a = 0.
2. Evaluate each inverse function at the single admissible value.

Answer: (c) 9

Explanation:
1. Let \( s=\sin(\cos^{-1}x) \); as \( 0<\cos^{-1}x<1<\frac\pi2 \), \( 0<s<1 \). The geometric series gives \( \frac1{1-s}=2\Rightarrow s=\frac12 \).
2. \( \cos^{-1}x=\frac\pi6\approx0.52 \) (inside \( (0,1) \)), so \( x=\frac{\sqrt3}2 \) and \( 12x^2=9 \).

Teacher's Note:
1. Sum the infinite geometric series with common ratio sin(cos inverse x).
2. Choose the angle pi/6 (not 5 pi/6) because cos inverse x was given to be less than 1.

Answer: (c) 4

Explanation:
1. The domain is \( [-1,1] \). Each term is increasing there, because \( \frac d{dx}\left(x^2+4x+1\right)=2x+4>0 \).
2. \( p=f(-1)=-\frac\pi2-\frac\pi2+1-4+1=-\pi-2 \) and \( q=f(1)=\frac\pi2+\frac\pi2+1+4+1=\pi+6 \).
3. \( p+q=4 \).

Teacher's Note:
1. First find the domain, then check monotonicity so the range is given by the end values.
2. The pi terms cancel in p + q.

Answer: (b) 3

Explanation:
1. Domain conditions: \( x(x+3)\ge0 \) and \( 0\le x^2+3x+1\le1 \), the latter meaning \( x^2+3x\le0 \).
2. Together \( x^2+3x=0 \), so \( x=0 \) or \( x=-3 \) (both give \( \cos^{-1}1 \) which is defined).
3. Sum \( 0+(-3)=-3 \), absolute value \( 3 \).

Teacher's Note:
1. Combine the two domain inequalities; they squeeze x squared + 3x to exactly zero.
2. The question asks for the absolute value of the sum.

Answer: (b) \( [0,1) \)

Explanation:
1. With \( \theta=\tan^{-1}x\in\left(-\frac\pi2,\frac\pi2\right) \): \( \cos^{-1}(\cos\theta)=|\theta|\in\left[0,\frac\pi2\right) \).
2. \( p=\sin|\theta|\in[0,1) \).

Teacher's Note:
1. Cos inverse of cos theta returns the absolute value of theta for theta in (-pi, pi).
2. The upper end 1 is never reached because |theta| stays strictly below pi/2.

Answer: (a) \( (0,1) \)

Explanation:
1. Put \( x=\cos\theta \): \( \frac{2x^2-1}{2x\sqrt{1-x^2}}=\frac{\cos2\theta}{\sin2\theta}=\cot2\theta \) (as \( \sin\theta\ge0 \)).
2. \( \cot^{-1}(\cot2\theta)=2\theta \) only if \( 2\theta\in(0,\pi) \), i.e. \( \theta\in\left(0,\frac\pi2\right) \), which means \( x\in(0,1) \).

Teacher's Note:
1. After the substitution the fraction becomes cot 2 theta, so the question is when cot inverse of cot returns its argument.
2. For theta beyond pi/2 the value is 2 theta minus pi, which does not match.

Answer: (b) \( \left[-\frac\pi4,\frac\pi4\right] \)

Explanation:
1. Let \( g(x)=\sin^{-1}x-\tan^{-1}x \) on \( [-1,1] \). It is odd, and \( g'(x)=\frac1{\sqrt{1-x^2}}-\frac1{1+x^2}\ge0 \) because \( \sqrt{1-x^2}\le1\le1+x^2 \).
2. So \( g \) increases from \( g(-1)=-\frac\pi4 \) to \( g(1)=\frac\pi2-\frac\pi4=\frac\pi4 \).

Teacher's Note:
1. An odd, increasing function on [-1,1] takes all values between its end values.
2. Compute the end values at x = 1 and x = -1.

Answer: (a) \( 2\pi+3 \)

Explanation:
1. The left side lies in \( [0,\pi] \), so \( [x]\in\{0,1,2,3\} \).
2. \( [x]=0,1,2 \) give \( x=0,1,2 \) (there \( \cos^{-1}\cos x=x \)). For \( [x]=3 \): \( x=3 \) and also \( x=2\pi-3\approx3.28\in[3,4) \) (where the left side is \( 2\pi-x \)).
3. Sum \( 0+1+2+3+(2\pi-3)=2\pi+3 \).

Teacher's Note:
1. Because cos inverse of cos x is between 0 and pi, only four integer values of [x] are possible.
2. In [3, 4) the graph of cos inverse of cos x changes direction at pi, giving a second root.