JEE Mathematics Inverse Trigonometric Functions MCQs Set 12

Answer: (a) 9

Explanation:
1. Let \( t=|x| \). The argument \( A=2-\frac2{t^2+5t+3} \) lies in \( \left[\frac43,2\right)\subset[0,\pi] \), so \( \sin^{-1}\cos A=\frac\pi2-A=\frac\pi2-2+\frac2{t^2+5t+3} \).
2. The right side is \( \frac{2-18t}{9t}+\frac\pi2=\frac2{9t}-2+\frac\pi2 \).
3. \( \frac2{t^2+5t+3}=\frac2{9t}\Rightarrow t^2-4t+3=0\Rightarrow t=1,3 \), so \( x=\pm1,\pm3 \) and the product is \( (1)(-1)(3)(-3)=9 \).

Teacher's Note:
1. Reduce both sides to rational functions of |x| by using the principal values of the inverse functions.
2. Each positive value of |x| gives two values of x, one of each sign.

Answer: (b) \( |x|<1 \)

Explanation:
1. Let \( A=\frac{2x^2+4}{1+x^2}=2+\frac2{1+x^2}\in(2,4] \). For \( A\in\left[\frac\pi2,\frac{3\pi}2\right] \) we have \( \sin^{-1}(\sin A)=\pi-A \).
2. \( \pi-A<\pi-3\Rightarrow A>3\Rightarrow\frac2{1+x^2}>1\Rightarrow x^2<1 \).

Teacher's Note:
1. Because A lies between 2 and 4, it sits in the second and third quadrant range where sin inverse of sin A equals pi minus A.
2. Then the inequality reduces to a simple condition on x squared.

Answer: (d) \( a\in(-\infty,\frac18)\cup(2,\infty) \)

Explanation:
1. Let \( t=2^{\pi/\sin^{-1}x} \). Since \( \sin^{-1}x\in\left[-\frac\pi2,\frac\pi2\right]\setminus\{0\} \), \( \frac\pi{\sin^{-1}x}\in(-\infty,-2]\cup[2,\infty) \), so \( t\in\left(0,\frac14\right]\cup[4,\infty) \).
2. The inequality is \( (t-4)(t-2a)<0 \). If \( a>2 \), pick \( t\in(4,2a) \). If \( a\le2 \), we need \( t\in(2a,4) \) with \( t\le\frac14 \): possible iff \( 2a<\frac14 \).
3. So \( a<\frac18 \) or \( a>2 \).

Teacher's Note:
1. Substitute t for the exponential and factor the quadratic as (t - 4)(t - 2a).
2. The attainable values of t are two separate pieces, so test which piece can lie between the two roots.

Answer: (a) 3

Explanation:
1. \( \cot^{-1} \) is decreasing, so the condition is \( \frac{n^2-10n+26}{2\sqrt3}<\cot\frac\pi6=\sqrt3 \).
2. \( n^2-10n+26<6\Rightarrow n^2-10n+20<0\Rightarrow5-\sqrt5<n<5+\sqrt5 \), i.e. \( 2.76<n<7.24 \).
3. The least natural number is \( 3 \).

Teacher's Note:
1. cot inverse reverses inequalities: a larger angle means a smaller argument.
2. Solve the quadratic inequality and take the first natural number above 2.76.

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

Explanation:
1. (A) With \( x=\cos\theta,\ \theta\in[0,\pi]\setminus\{\frac\pi2\} \): \( \tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{\sqrt{1-x^2}}x \) on \( [-1,1]\setminus\{0\} \).
2. (B) \( \cot^{-1}x=\theta\Rightarrow\tan\theta=\frac1x \), same domain \( x\ne0 \). (C) \( \sin\left(\tan^{-1}x\right)=\frac x{\sqrt{1+x^2}} \) for all real \( x \).
3. (D) Both equal \( \frac1{\sqrt{1+x^2}} \) for all real \( x \). All four pairs coincide on a common domain.

Teacher's Note:
1. Use a right triangle (or the sign of the angle's range) to convert each composition to an algebraic expression.
2. Check that the domains of the two functions in each pair are the same.

Answer: (b) Only A, B and C

Explanation:
1. For infinite \( y \) the angle \( 4\theta=\pm\frac\pi2,\pm\frac{3\pi}2 \) where \( \theta=\tan^{-1}x\in\left(-\frac\pi2,\frac\pi2\right) \), so \( \theta=\pm\frac\pi8,\pm\frac{3\pi}8 \).
2. \( \tan\frac\pi8=\sqrt2-1\Rightarrow x^2=3-2\sqrt2 \) and \( \tan\frac{3\pi}8=\sqrt2+1\Rightarrow x^2=3+2\sqrt2 \) (A and B).
3. These are the roots of \( u^2-6u+1=0 \), i.e. \( x^4=6x^2-1 \) (C), not D.

Teacher's Note:
1. Infinity corresponds to tan inverse y being an odd multiple of pi/2, so 4 theta must equal such a multiple.
2. The two values of x squared are the roots of one quadratic in x squared; read off its coefficients.

Answer: (c) Only A, C and D

Explanation:
1. The domain needs \( x^2+x+1\le1 \), i.e. \( -1\le x\le0 \). Then \( \cos^{-1}(\lambda x+1)=\frac\pi2-\sin^{-1}(x^2+x+1)=\cos^{-1}(x^2+x+1) \), so \( \lambda x+1=x^2+x+1 \).
2. That gives \( x=0 \) or \( x=\lambda-1 \), and \( x=\lambda-1\in[-1,0) \) requires \( 0\le\lambda<1 \) (at \( \lambda=1 \) both roots coincide).
3. Exactly two solutions occur for \( \lambda\in[0,1) \), so the integer \( \lambda=0 \) is possible while \( -1,1,2 \) are not.

Teacher's Note:
1. Reduce to an equation without inverse functions but keep the domain restriction x in [-1, 0].
2. Check the endpoints lambda = 0 and lambda = 1 separately.

Answer: (b) Only A and D

Explanation:
1. \( \sin^{-1}x+\sin^{-1}y=\frac\pi2 \) gives \( x^2+y^2=1 \) with \( x,y\ge0 \).
2. \( \sin2x=\cos2y=\sin\left(\frac\pi2-2y\right) \) and \( x,y\in[0,1] \) force \( 2x=\pi-\left(\frac\pi2-2y\right) \), i.e. \( x-y=\frac\pi4 \) (the case \( x+y=\frac\pi4 \) contradicts \( x^2+y^2=1 \)).
3. \( (x+y)^2=2-\frac{\pi^2}{16} \) so \( x+y=\sqrt{2-\frac{\pi^2}{16}} \); then \( x=\frac\pi8+\sqrt{\frac12-\frac{\pi^2}{64}} \) and \( y=\sqrt{\frac12-\frac{\pi^2}{64}}-\frac\pi8 \) (numerically \( x\approx0.981,\ y\approx0.195 \)).

Teacher's Note:
1. Both unknowns are numbers in radians, so equate the angles of the sines carefully using the allowed ranges.
2. Use the sum and difference of x and y to solve the pair of equations.

Answer: (b) 4

Explanation:
1. With \( u=kx \), the function \( g(u)=\sin^{-1}|\sin u|+\cos^{-1}(\cos u) \) equals \( 2u \) on \( \left[0,\frac\pi2\right] \), \( \pi \) on \( \left[\frac\pi2,\frac{3\pi}2\right] \) and \( 4\pi-2u \) on \( \left[\frac{3\pi}2,2\pi\right] \) (period \( 2\pi \)).
2. The longest constant stretch has length \( \pi \) in \( u \), i.e. \( \frac\pi k \) in \( x \).
3. \( \frac\pi k=\frac\pi4\Rightarrow k=4 \).

Teacher's Note:
1. Draw one period of g(u): it rises, stays flat at pi for a stretch of length pi, then falls.
2. The flat piece has length pi in u, so divide by k to get its length in x.

Answer: (b) an odd function

Explanation:
1. In the new system, with \( S(x) \) and \( C(x) \) denoting the usual principal values, \( \sin^{-1}x=\pi-S(x) \) (the angle in \( \left[\frac\pi2,\frac{3\pi}2\right] \) with the same sine) and \( \cos^{-1}x=2\pi-C(x) \) (the angle in \( [\pi,2\pi] \) with the same cosine).
2. With \( \cos^{-1}x=2\pi-C(x)=2\pi-\left(\frac\pi2-S(x)\right)=\frac{3\pi}2+S(x) \).
3. \( f=3(\pi-S)-2\left(\frac{3\pi}2+S\right)=-5S(x) \), and \( S \) is odd, so \( f \) is odd.

Teacher's Note:
1. Express everything through the usual sin inverse, which is an odd function.
2. The constants 3 pi cancel exactly.

Answer: (a) \( -\frac{63}8\pi^3 \)

Explanation:
1. In the new system, with \( S(x) \) and \( C(x) \) denoting the usual principal values, \( \sin^{-1}x=\pi-S(x) \) (the angle in \( \left[\frac\pi2,\frac{3\pi}2\right] \) with the same sine) and \( \cos^{-1}x=2\pi-C(x) \) (the angle in \( [\pi,2\pi] \) with the same cosine).
2. Let \( s=S(x)\in\left[-\frac\pi2,\frac\pi2\right] \). The expression is \( h(s)=(\pi-s)^3-\left(\frac{3\pi}2+s\right)^3 \), which is decreasing in \( s \).
3. Minimum at \( s=\frac\pi2 \): \( \left(\frac\pi2\right)^3-(2\pi)^3=\frac{\pi^3}8-8\pi^3=-\frac{63}8\pi^3 \).

Teacher's Note:
1. Both cubes are monotonic in s, with opposite signs, so the difference is monotonic and the minimum is at an endpoint.
2. Evaluate at s = pi/2, which is x = 1.

Answer: (a) \( [4\pi,5\pi] \)

Explanation:
1. \( \cos^{-1}x=\frac{5\pi}2-\sin^{-1}x \).
2. With \( \sin^{-1}x\in\left[-\frac{5\pi}2,-\frac{3\pi}2\right] \) the range becomes \( \left[\frac{5\pi}2+\frac{3\pi}2,\ \frac{5\pi}2+\frac{5\pi}2\right]=[4\pi,5\pi] \).

Teacher's Note:
1. Subtract the interval from the constant 5 pi/2 and remember that the end points swap.
2. The length of the interval stays pi.

Answer: (d) -1

Explanation:
1. The left side is at most \( \frac\pi2+\pi+\pi=\frac{5\pi}2 \), while \( t^2-\sqrt{2\pi}t+3\pi=\left(t-\sqrt{\frac\pi2}\right)^2+\frac{5\pi}2\ge\frac{5\pi}2 \).
2. So equality must hold everywhere: \( x=1,\ y=-1,\ z=-1,\ t=\sqrt{\pi/2} \).
3. Therefore \( x+y+z=1-1-1=-1 \).

Teacher's Note:
1. Compare the largest possible value of the left side with the smallest possible value of the right side.
2. Equality forces every variable to its extreme.

Answer: (c) \( \pi \)

Explanation:
1. The left side is at most \( \frac\pi2+\pi+\pi=\frac{5\pi}2 \), while \( t^2-\sqrt{2\pi}t+3\pi=\left(t-\sqrt{\frac\pi2}\right)^2+\frac{5\pi}2\ge\frac{5\pi}2 \).
2. So equality must hold everywhere: \( x=1,\ y=-1,\ z=-1,\ t=\sqrt{\pi/2} \).
3. The minimum of \( 1,-1,-1 \) is \( -1 \) and \( \cos^{-1}(-1)=\pi \).

Teacher's Note:
1. After finding x, y, z take the smallest of the three numbers.
2. Cos inverse of -1 is pi.

Answer: (a) A-q, B-s, C-r, D-p

Explanation:
1. Let \( s=\sin^{-1}x \) and \( c=\cos^{-1}x=\frac\pi2-s \). The breakpoints are \( s=\frac\pi2-1\approx0.57 \) (at \( x=\cos1 \)), \( s=\frac\pi4 \) (at \( x=\frac1{\sqrt2} \)) and \( s=1 \) (at \( x=\sin1 \)).
2. Evaluating the four numbers in each interval (for example \( x=0.3,0.6,0.8,0.95 \)) gives the orderings \( t_4>t_3>t_1>t_2 \), \( t_3>t_4>t_1>t_2 \), \( t_2>t_1>t_4>t_3 \), \( t_1>t_2>t_4>t_3 \) respectively.
3. Hence A-q, B-s, C-r, D-p.

Teacher's Note:
1. Both bases and exponents are between 0 and pi/2 and the sign of s - c, and whether each base exceeds or is below 1, decides the order.
2. Test one convenient x inside each interval; the order does not change within an interval.

Answer: (c) -2

Explanation:
1. Each square is at most \( \pi^2 \), so \( \cos^{-1}x=\cos^{-1}y=\pi \) and \( x=y=-1 \).
2. \( x^5+y^5=-1-1=-2 \).

Teacher's Note:
1. The maximum value of cos inverse is pi, and two squares of pi add to 2 pi squared.
2. Odd powers keep the sign of -1.

Answer: (c) 0 or 2

Explanation:
1. \( |\sin^{-1}x|\le\frac\pi2 \) and \( \cos^{-1}y\le\pi \), so the product of the two squares is at most \( \frac{\pi^2}4\cdot\pi^2=\frac{\pi^4}4 \): equality needs \( x=\pm1 \) and \( y=-1 \).
2. Then \( |x-y|=|1+1|=2 \) (for \( x=1 \)) or \( |-1+1|=0 \) (for \( x=-1 \)).

Teacher's Note:
1. Find the largest possible value of each factor; their product reaches the stated value only at the extremes.
2. Two sign choices for x lead to two values of |x - y|.

Answer: (a) \( \{1,-1\} \)

Explanation:
1. The difference of two numbers in \( \left[-\frac\pi2,\frac\pi2\right] \) has modulus \( \pi \) only if they are the end points: \( (x,y)=(1,-1) \) or \( (-1,1) \).
2. \( x^y=1^{-1}=1 \) or \( (-1)^1=-1 \).

Teacher's Note:
1. The only way to get a difference of pi is to take the two extreme values of sin inverse.
2. Substitute each ordered pair.

Answer: (a) A-p, B-q, C-p, D-s

Explanation:
1. Rewrite as \( \sin^{-1}(ax)+\cos^{-1}y=\sin^{-1}(bxy) \) and take sine: \( axy+\sqrt{1-a^2x^2}\sqrt{1-y^2}=bxy \), so \( \sqrt{1-a^2x^2}\sqrt{1-y^2}=(b-a)xy \).
2. (A) \( a=1,b=0 \): \( (1-x^2)(1-y^2)=x^2y^2\Rightarrow x^2+y^2=1 \) (p). (C) \( a=1,b=2 \): the same equation \( x^2+y^2=1 \) (p).
3. (B) \( a=b=1 \): \( \sqrt{1-x^2}\sqrt{1-y^2}=0 \), so \( (x^2-1)(y^2-1)=0 \) (q). (D) \( a=b=2 \): \( \sqrt{1-4x^2}\sqrt{1-y^2}=0 \), so \( (4x^2-1)(y^2-1)=0 \) (s).

Teacher's Note:
1. Isolate the product of square roots after taking the sine of both sides; when a equals b it must vanish.
2. For b - a = 0 each factor under the root gives a pair of lines.

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

Explanation:
1. As \( 5\in\left[\frac{3\pi}2,\frac{5\pi}2\right] \), \( \sin^{-1}(\sin5)=5-2\pi \): R is true.
2. \( 5-2\pi>x^2-4x\Rightarrow(x-2)^2<9-2\pi \), which holds on the whole interval \( \left(2-\sqrt{9-2\pi},2+\sqrt{9-2\pi}\right) \): A is true, and uses R.

Teacher's Note:
1. Reduce sin inverse of sin 5 by subtracting 2 pi, since 5 lies in [3 pi/2, 5 pi/2].
2. Complete the square on the quadratic side.

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

Explanation:
1. Both \( \cos(\sin^{-1}x) \) and \( \sin(\cos^{-1}x) \) equal \( \sqrt{1-x^2} \), so R is true.
2. With \( u=\sqrt{1-x^2}\in[0,1] \) the sum is \( \sin^{-1}u+\cos^{-1}u=\frac\pi2 \): A is true, and R is what makes both arguments the same.

Teacher's Note:
1. Both compositions equal the square root of 1 minus x squared.
2. Once the two arguments coincide, the identity sin inverse + cos inverse = pi/2 applies.

Answer: (d) A is false, but R is true.

Explanation:
1. With \( b=-a \) and \( t=ax \): \( \cos^{-1}(-t)=\pi-\cos^{-1}t \), so the left side is \( \sin^{-1}t+\pi-\cos^{-1}t=2\sin^{-1}t+\frac\pi2 \).
2. Setting it equal to \( \frac{3\pi}2 \) gives \( \sin^{-1}t=\frac\pi2 \), i.e. \( t=1 \) or \( x=\frac1a \): a solution exists, so A is false. R is the standard identity, true.

Teacher's Note:
1. Replace b by minus a and use cos inverse of a negative number = pi minus cos inverse of the positive one.
2. The extreme value t = 1 makes the equation solvable.

Answer: (b) 2

Explanation:
1. Put \( x=\sin A,\ y=\sin B,\ z=\sin C \) with \( A+B+C=\pi \) (each angle in \( \left[0,\frac\pi2\right] \)).
2. For angles summing to \( \pi \) one has the identity \( \sum\sin^4A+4\prod\sin^2A=2\sum\sin^2A\sin^2B \) (verified by direct substitution of the relation \( \sin C=\sin(A+B) \)).
3. Hence the ratio is \( 2 \) (also confirmed numerically for several angle triples).

Teacher's Note:
1. Treat the three numbers as sines of angles of a triangle.
2. The identity can be tested with a simple triple such as angles 90, 60 and 30 degrees.

Answer: (c) 8

Explanation:
1. \( \tan^{-1}a-\tan^{-1}b=\tan^{-1}\frac18\Rightarrow\frac{a-b}{1+ab}=\frac18 \) (with \( a>b \)), so \( ab-8a+8b+1=0 \), i.e. \( (a+8)(8-b)=65 \).
2. With \( a+8\ge9 \): \( a+8=13,\ 8-b=5 \) gives \( (5,3) \) and \( a+8=65,\ 8-b=1 \) gives \( (57,7) \).
3. \( \frac{5+3+57+7}9=\frac{72}9=8 \).

Teacher's Note:
1. Convert the equation to (a + 8)(8 - b) = 65 and list factor pairs of 65.
2. Natural numbers force the larger factor to be at least 9.

Answer: (a) 1

Explanation:
1. \( \cot^{-1}2n^2=\tan^{-1}\frac2{4n^2}=\tan^{-1}(2n+1)-\tan^{-1}(2n-1) \) because \( 1+(2n+1)(2n-1)=4n^2 \).
2. The series telescopes to \( \lim_{n\to\infty}\tan^{-1}(2n+1)-\tan^{-1}1=\frac\pi2-\frac\pi4=\frac\pi4 \), so \( k=1 \).

Teacher's Note:
1. Write 1 over 2 n squared as 2 over (4 n squared) and use the tan inverse difference formula.
2. Telescoping sums leave only the first and last terms.

Answer: (c) 4

Explanation:
1. Let \( t=\tan^{-1}x\in\left(-\frac\pi2,\frac\pi2\right) \) and \( c=\frac\pi2 \). Then \( f=t^3+(c-t)^3=3c\left(t-\frac c2\right)^2+\frac{c^3}4 \).
2. Minimum at \( t=\frac\pi4 \): \( a=\frac{\pi^3}{32} \). As \( t\to-\frac\pi2 \): \( f\to3\cdot\frac\pi2\cdot\frac{9\pi^2}{16}+\frac{\pi^3}{32}=\frac{7\pi^3}8 \) (not attained), so \( b=\frac{7\pi^3}8 \).
3. \( \frac b{7a}=\frac{7\pi^3/8}{7\pi^3/32}=4 \).

Teacher's Note:
1. Write the sum of two cubes with a fixed sum c as a quadratic in t.
2. The open end of the range comes from the excluded value t = -pi/2.

Answer: (b) 1

Explanation:
1. Shift by \( 2\pi \): for \( x'\in\left[\frac\pi4,\frac{7\pi}4\right] \), \( f=\cos^{-1}(\cos x') \) is the triangular wave and \( g=\frac\pi2-f \), so \( f-g=2f-\frac\pi2\ge0 \) (since \( f\ge\frac\pi4 \) there).
2. \( \int f\,dx=2\cdot\frac12\left(\pi^2-\frac{\pi^2}{16}\right)=\frac{15\pi^2}{16} \) and \( \int\frac\pi2dx=\frac\pi2\cdot\frac{3\pi}2=\frac{3\pi^2}4 \).
3. Area \( =\frac{15\pi^2}8-\frac{3\pi^2}4=\frac{9\pi^2}8 \), so \( a=9,\ b=8 \) and \( a-b=1 \).

Teacher's Note:
1. Shift the interval by 2 pi to use the principal intervals, and note that g = pi/2 - f.
2. The curves meet only at the end points of the given interval.

Answer: (c) 4

Explanation:
1. \( \sqrt{(x^2+1)^2-4x^2}=\sqrt{(x^2-1)^2}=|x^2-1| \). With \( t=|x| \) the equation is \( \tan^{-1}t=|t^2-1| \).
2. On \( [0,1] \): \( 1-t^2-\tan^{-1}t \) decreases from \( 1 \) to \( -\frac\pi4 \): one root. On \( (1,\infty) \): \( t^2-1-\tan^{-1}t \) increases from \( -\frac\pi4 \) to \( \infty \): one root.
3. So two positive values of \( t \), each giving \( x=\pm t \): four solutions.

Teacher's Note:
1. Simplify the radical to |x squared - 1| and work with t = |x|.
2. Check the sign change on [0, 1] and on (1, infinity) separately, using monotonicity.