JEE Mathematics Inverse Trigonometric Functions MCQs Set 10

Answer: (b) 1

Explanation:
1. For \( \cos^{-1}(2-x) \) to exist, \( -1\le2-x\le1 \), i.e. \( 1\le x\le3 \).
2. \( \tan^{-1}x-\cot^{-1}x=2\tan^{-1}x-\frac\pi2 \). Define \( g(x)=2\tan^{-1}x-\frac\pi2-\cos^{-1}(2-x) \).
3. \( g(1)=0-0=0 \), so \( x=1 \) is a solution; for \( x>1 \) the term \( \cos^{-1}(2-x) \) rises faster than \( 2\tan^{-1}x-\frac\pi2 \) (e.g. \( g(3)\approx0.93-\pi<0 \)), so \( g<0 \) on \( (1,3] \).
4. Hence the only solution is \( x=1 \).

Teacher's Note:
1. First find the interval on which both sides are defined, then test its end points.
2. Near x = 1 the cos inverse term has an infinite slope, which makes it grow faster than the smooth left side.

Answer: (c) 2

Explanation:
1. \( \sqrt{1+\cos2x}=\sqrt2|\cos x| \), so the equation is \( |\cos x|=\sin^{-1}(\sin x) \).
2. The right side is non-negative only for \( 0\le x\le\pi \); it equals \( x \) on \( \left[0,\frac\pi2\right] \) and \( \pi-x \) on \( \left[\frac\pi2,\pi\right] \).
3. On \( \left[0,\frac\pi2\right] \): \( \cos x=x \) has one root (about 0.739). On \( \left[\frac\pi2,\pi\right] \): \( -\cos x=\pi-x \) goes from \( 0<\frac\pi2 \) to \( 1>0 \), so it has one root.
4. Total: 2 solutions.

Teacher's Note:
1. Sketch the graphs of |cos x| and sin inverse (sin x) on [0, pi] and count the intersections.
2. For x in [-pi, 0) the right side is negative or zero while the left side is positive, so no root there.

Answer: (b) 0

Explanation:
1. The equation holds exactly when \( x^2+x+1=\lambda x+1 \) with both sides in \( [-1,1] \).
2. \( x^2+x+1\le1\Rightarrow x\in[-1,0] \) (the value is at least \( \frac34 \) always).
3. \( x(x+1-\lambda)=0\Rightarrow x=0\ \text{or}\ x=\lambda-1 \). For two distinct solutions we need \( -1\le\lambda-1<0 \), i.e. \( 0\le\lambda<1 \).
4. The only integer in this range is \( \lambda=0 \).

Teacher's Note:
1. Use sin inverse u + cos inverse v = pi/2 if and only if u = v, and then find the domain from the left side.
2. The root x = lambda - 1 must be inside [-1, 0] and different from 0.

Answer: (b) \( \frac\pi2-\beta \)

Explanation:
1. \( S_1=\sin2\beta,\ S_2=\cos2\beta,\ S_3=\cos\beta,\ S_4=-\sin\beta \).
2. \( \tan\left(\sum\tan^{-1}x_i\right)=\frac{S_1-S_3}{1-S_2+S_4}=\frac{\sin2\beta-\cos\beta}{1-\cos2\beta-\sin\beta}=\frac{\cos\beta(2\sin\beta-1)}{\sin\beta(2\sin\beta-1)}=\cot\beta \).
3. \( \cot\beta=\tan\left(\frac\pi2-\beta\right) \), so the sum is \( \frac\pi2-\beta \) (mod \( \pi \)).

Teacher's Note:
1. Use the four-term formula tan (sum) = (S1 - S3)/(1 - S2 + S4) with the symmetric functions of the roots.
2. Factor 2 sin beta - 1 from numerator and denominator; it cancels.

Answer: (c) 5

Explanation:
1. cot inverse is decreasing, so \( \cot^{-1}\frac n\pi>\frac\pi6\Leftrightarrow\frac n\pi<\cot\frac\pi6=\sqrt3 \).
2. \( n<\pi\sqrt3\approx5.44 \), so the largest natural number is \( 5 \).

Teacher's Note:
1. Remember that cot inverse reverses inequalities because it is a decreasing function.
2. pi times root 3 is about 3.1416 x 1.732 = 5.44.

Answer: (d) \( (\cot2,\infty) \)

Explanation:
1. Put \( t=\cot^{-1}x\in(0,\pi) \): \( t^2-7t+10=(t-2)(t-5)>0 \).
2. Since \( t<\pi<5 \), \( t-5<0 \), so we need \( t-2<0 \), i.e. \( t<2 \).
3. \( \cot^{-1}x<2\Rightarrow x>\cot2 \) (cot inverse is decreasing).

Teacher's Note:
1. Factorise first and discard any factor that is always negative in the range of the inverse function.
2. Because the range of cot inverse is (0, pi), the root t = 5 is outside it.

Answer: (d) 5

Explanation:
1. \( \sin^{-1}(\sin12)=12-4\pi \) and \( \cos^{-1}(\cos12)=4\pi-12 \), so the right side is \( 0 \).
2. We need \( (k-2)x^2+8x+k+4>0 \) for all \( x \): \( k>2 \) and \( 64-4(k-2)(k+4)<0 \).
3. \( (k-2)(k+4)>16\Rightarrow k^2+2k-24>0\Rightarrow(k+6)(k-4)>0 \).
4. With \( k>2 \) this gives \( k>4 \), so the least integer is \( 5 \).

Teacher's Note:
1. The right side collapses to 0 because the two inverse values are negatives of each other.
2. A quadratic is positive for all x when its leading coefficient is positive and its discriminant is negative.

Answer: (d) a unit circle with centre at the origin

Explanation:
1. \( \cos^{-1}(2xy)=\frac\pi2-\sin^{-1}x-\cos^{-1}y=\cos^{-1}x-\cos^{-1}y \).
2. Taking cosine: \( 2xy=xy+\sqrt{1-x^2}\sqrt{1-y^2} \), i.e. \( xy=\sqrt{(1-x^2)(1-y^2)} \).
3. Squaring: \( x^2y^2=1-x^2-y^2+x^2y^2\Rightarrow x^2+y^2=1 \).

Teacher's Note:
1. Take the cosine of both sides after isolating cos inverse (2xy) so that the cos inverse difference formula can be matched.
2. The final relation x squared + y squared = 1 is the unit circle.

Answer: (c) \( (\alpha+\beta)(\alpha^2+\beta^2) \)

Explanation:
1. Let \( r=\sqrt{\alpha^2+\beta^2} \). Then \( \cos\left(\tan^{-1}\frac\alpha\beta\right)=\frac\beta r \) and \( \cos\left(\tan^{-1}\frac\beta\alpha\right)=\frac\alpha r \).
2. \( \operatorname{cosec}^2\frac\theta2=\frac{2}{1-\cos\theta},\ \sec^2\frac\phi2=\frac2{1+\cos\phi} \), so the expression is \( \frac{\alpha^3}{1-\beta/r}+\frac{\beta^3}{1+\alpha/r}=r\left[\frac{\alpha^3}{r-\beta}+\frac{\beta^3}{r+\alpha}\right] \).
3. \( \frac{\alpha^3}{r-\beta}=\frac{\alpha^3(r+\beta)}{\alpha^2}=\alpha(r+\beta),\quad\frac{\beta^3}{r+\alpha}=\beta(r-\alpha) \).
4. Sum \( =r\left[\alpha r+\alpha\beta+\beta r-\alpha\beta\right]=r^2(\alpha+\beta)=(\alpha+\beta)(\alpha^2+\beta^2) \).

Teacher's Note:
1. Use cosec squared (theta/2) = 2/(1 - cos theta) and sec squared (phi/2) = 2/(1 + cos phi).
2. Rationalise using r squared - beta squared = alpha squared and r squared - alpha squared = beta squared.

Answer: (d) n is an odd integer

Explanation:
1. For \( x=n\pi-\tan^{-1}3 \): \( \tan x=-3 \), so \( \tan2x=\frac{-6}{1-9}=\frac34 \).
2. The equation becomes \( 9+\frac{\sqrt{10}}{\cos x}+1=0\Rightarrow\cos x=-\frac1{\sqrt{10}} \) (negative).
3. With \( \tan x=-3 \) and \( \cos x<0 \), \( x \) is in the second quadrant, which occurs for \( x=(2m+1)\pi-\tan^{-1}3 \): \( n \) must be odd.

Teacher's Note:
1. tan x = -3 alone allows both the second and fourth quadrants; the sign of cos x chooses the second.
2. x = n pi - tan inverse 3 is in the second quadrant for odd n and in the fourth quadrant for even n.

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

Explanation:
1. \( \sin^{-1}(\cos x)=\frac\pi2-\cos^{-1}(\cos x)=\frac\pi2-x \) because \( x\in(0,\pi) \) (Reason R is used here).
2. \( \cos^{-1}(\sin x)=\frac\pi2-\sin^{-1}(\sin x)=\frac\pi2-x \) because \( x\in\left(0,\frac\pi2\right) \).
3. Adding gives \( \pi-2x \), so A is true and R is the reason.

Teacher's Note:
1. The relation sin inverse t + cos inverse t = pi/2 converts each term to the form pi/2 minus an angle.
2. Both inner inverse functions return x itself because x lies in the first quadrant.

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

Explanation:
1. The range of \( \cot^{-1}x \) is \( (0,\pi) \) and that of \( \cos^{-1}2x \) is \( [0,\pi] \), so the left side is greater than \( \pi \).
2. It can never be 0, so there are no roots: A is true, and it follows directly from R.

Teacher's Note:
1. Ranges alone can show that an equation has no solution without solving it.
2. The strict lower bound for cot inverse makes the left side strictly greater than pi.

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

Explanation:
1. Write the equation as \( \cos^{-1}(bxy)=\sin^{-1}y-\sin^{-1}(ax) \) and take cosine: \( bxy=\sqrt{1-y^2}\sqrt{1-a^2x^2}+axy \).
2. A: \( b=0,a=1 \): \( \sqrt{1-y^2}\sqrt{1-x^2}=-xy \); squaring gives \( x^2+y^2=1 \) (p). B: \( a=b=1 \): \( \sqrt{(1-x^2)(1-y^2)}=0 \) gives \( (x^2-1)(y^2-1)=0 \) (q).
3. C: \( a=1,b=2 \): \( xy=\sqrt{(1-x^2)(1-y^2)} \) gives \( x^2+y^2=1 \) (p). D: \( a=2,b=2 \): \( \sqrt{(1-y^2)(1-4x^2)}=0 \) gives \( (4x^2-1)(y^2-1)=0 \) (s).
4. So A-p, B-q, C-p, D-s.

Teacher's Note:
1. Isolate the single cos inverse term, take cosine, and use the product-sum formula for cos (P - Q).
2. Whenever a = b the cross term cancels completely and one square-root factor must vanish.

Answer: (a) only A, B, D

Explanation:
1. (A) \( f(x)=2^{x-[x]}=2^{\{x\}} \) has period 1: periodic.
2. (B) \( \operatorname{sgn}\{x\} \) is 0 at integers and 1 elsewhere: period 1, periodic.
3. (C) \( \cos(x^2) \) oscillates faster and faster as \( |x| \) grows, so \( h \) is not periodic.
4. (D) \( \sqrt{\sin x} \) is defined where \( \sin x\ge0 \), and the pattern repeats every \( 2\pi \): periodic. So A, B, D.

Teacher's Note:
1. A function of x squared is almost never periodic because the spacing of its zeros keeps shrinking.
2. For f(x) = 2^(x - [x]) only the fractional part of x matters, which repeats every unit.

Answer: (b) \( x\in[\sin1,1] \)

Explanation:
1. First inequality: \( (\tan^{-1}x)^2-3\tan^{-1}x+2=(t-1)(t-2)>0 \) with \( t=\tan^{-1}x<\frac\pi2<2 \), so we need \( t<1 \), i.e. \( x<\tan1\approx1.56 \), which holds for every \( x\le1 \).
2. Second inequality (domain \( [-1,1] \)): \( [\sin^{-1}x]\in\{-2,-1,0,1\} \) and \( [\cos^{-1}x]\in\{0,1,2,3\} \). The only possibility is \( [\sin^{-1}x]=1 \) and \( [\cos^{-1}x]=0 \).
3. \( [\sin^{-1}x]=1\Rightarrow\sin^{-1}x\ge1\Rightarrow x\ge\sin1 \) and \( [\cos^{-1}x]=0\Rightarrow\cos^{-1}x<1\Rightarrow x>\cos1 \).
4. Since \( \sin1>\cos1 \), the solution is \( x\in[\sin1,1] \).

Teacher's Note:
1. Count the possible integer values of each greatest-integer term from the ranges of the two inverse functions.
2. The end point sin 1 is included because the greatest integer of sin inverse (sin 1) = 1 is exactly 1.

Answer: (c) g(x), h(x), f(x), k(x)

Explanation:
1. Domain of \( f \): \( [-1,1] \) (length 2). Domain of \( g \): \( 0\le2x\le1\Rightarrow\left[0,\frac12\right] \) (length \( \frac12 \)).
2. Domain of \( h \): \( -1\le2x\le1\Rightarrow\left[-\frac12,\frac12\right] \) (length 1). Domain of \( k \): \( \mathbb{R} \) (unbounded).
3. Increasing order of size: \( g,\ h,\ f,\ k \).

Teacher's Note:
1. Write each domain as an interval before comparing.
2. tan inverse is defined everywhere, so its function always has the largest domain.

Answer: (a) \( \frac{17\pi}{42}-2 \)

Explanation:
1. \( \sin^{-1}\sin\frac{22\pi}7=\sin^{-1}\sin\left(3\pi+\frac\pi7\right)=-\frac\pi7 \) and \( \cos^{-1}\cos\frac{5\pi}{3}=\cos^{-1}\cos\left(2\pi-\frac\pi3\right)=\frac\pi3 \).
2. \( \tan^{-1}\tan\frac{5\pi}{7}=\tan^{-1}\tan\left(\frac{5\pi}7-\pi\right)=-\frac{2\pi}7 \) and \( \sin^{-1}(\cos2)=\frac\pi2-2 \) (as \( 2\in[0,\pi] \)).
3. Sum \( =\pi\left(-\frac17+\frac13-\frac27+\frac12\right)-2=\frac{17\pi}{42}-2 \).

Teacher's Note:
1. Reduce each term to its own principal range and add; keep the numerical -2 separate from the multiples of pi.
2. For sin inverse (cos 2), write cos 2 = sin (pi/2 - 2) and check that pi/2 - 2 lies in [-pi/2, pi/2].

Answer: (d) \( \frac1{\sqrt3} \)

Explanation:
1. Put \( x=\tan\theta,\ 0<\theta<\frac\pi2 \): \( \frac{1-x^2}{2x}=\cot2\theta \) and \( \frac{1-x^2}{1+x^2}=\cos2\theta \).
2. Since \( 2\theta\in(0,\pi) \), \( \cot^{-1}(\cot2\theta)=2\theta \) and \( \cos^{-1}(\cos2\theta)=2\theta \), so \( 4\theta=\frac{2\pi}3\Rightarrow\theta=\frac\pi6 \).
3. \( x=\tan\frac\pi6=\frac1{\sqrt3} \) (positive, as required).

Teacher's Note:
1. The substitution x = tan theta turns both arguments into trigonometric functions of 2 theta.
2. Remember the condition x > 0, which rules out the negative root.

Answer: (c) 3

Explanation:
1. \( \tan\left(\text{left side}\right)=\frac{\frac1{1+2x}+\frac1{4x+1}}{1-\frac1{(1+2x)(4x+1)}}=\frac{6x+2}{8x^2+6x}=\frac{3x+1}{x(4x+3)} \).
2. Equate to \( \frac2{x^2} \) (for \( x\ne0 \)): \( x(3x+1)=2(4x+3)\Rightarrow3x^2-7x-6=0\Rightarrow(3x+2)(x-3)=0 \).
3. For \( x=3 \): \( \tan^{-1}\frac17+\tan^{-1}\frac1{13}=\tan^{-1}\frac29 \) holds exactly. The root \( x=-\frac23 \) is rejected since it makes the left side differ from the right side by \( \pi \).

Teacher's Note:
1. Take the tangent of both sides, but verify each root afterwards because tan removes multiples of pi.
2. For x = 3: the sum (1/7 + 1/13)/(1 - 1/91) = 20/90 = 2/9, and the product 1/91 is below 1, so the formula applies directly.

Answer: (b) \( \frac1a \)

Explanation:
1. \( \frac{b}{1+a(a+b)}=\frac{(a+b)-a}{1+(a+b)a} \), so the second term is \( \tan^{-1}(a+b)-\tan^{-1}a \).
2. For \( a+b>0 \): \( \tan^{-1}\frac{1}{a+b}=\frac\pi2-\tan^{-1}(a+b) \).
3. The sum is \( \frac\pi2-\tan^{-1}a \), and \( \tan\left(\frac\pi2-\tan^{-1}a\right)=\cot(\tan^{-1}a)=\frac1a \).

Teacher's Note:
1. Spot that the numerator b is the difference of (a + b) and a, which suggests the tan inverse subtraction formula.
2. tan of (pi/2 - A) equals cot A, so the answer is the reciprocal of a.

Answer: (c) 2

Explanation:
1. For \( \sin^{-1}\sqrt{x^2+x+1} \) to exist we need \( x^2+x+1\le1 \), i.e. \( x(x+1)\le0 \); for \( \sqrt{x(x+1)} \) to exist we need \( x(x+1)\ge0 \).
2. Both hold only when \( x(x+1)=0 \), i.e. \( x=0 \) or \( x=-1 \).
3. For each, \( \tan^{-1}0+\sin^{-1}1=\frac\pi2 \): both satisfy the equation. So there are 2 solutions.

Teacher's Note:
1. Domain restrictions can leave only isolated points; check them before attempting algebra.
2. The two conditions on x(x + 1) point in opposite directions, forcing equality.

Answer: (a) 0

Explanation:
1. Let \( s=a+b+c \) and call the three square roots \( \alpha,\beta,\gamma \). Then \( \alpha=\frac{a\sqrt s}{\sqrt{abc}} \), so \( \alpha+\beta+\gamma=\frac{s\sqrt s}{\sqrt{abc}}=\alpha\beta\gamma \).
2. \( \tan\theta=\frac{\alpha+\beta+\gamma-\alpha\beta\gamma}{1-\alpha\beta-\beta\gamma-\gamma\alpha}=\frac{0}{1-s\left(\frac1a+\frac1b+\frac1c\right)}=0 \) (the denominator is non-zero, since \( s\left(\frac1a+\frac1b+\frac1c\right)\ge9 \)).

Teacher's Note:
1. Use tan (A + B + C) = (S1 - S3)/(1 - S2) and check whether the numerator vanishes.
2. A quick numerical test with a = b = c = 1 gives three copies of tan inverse root 3 = 60 degrees, a total of 180 degrees, whose tangent is 0.

Answer: (b) \( -\frac\pi2 \)

Explanation:
1. \( x^2-2x+2=(x-1)^2+1\ge1 \), and \( \sin^{-1} \) and \( \cos^{-1} \) need an argument at most 1, so \( x=1 \).
2. At \( x=1 \): \( \sin^{-1}1+\cos^{-1}1=\frac\pi2+0=\frac\pi2 \).
3. \( a+\frac\pi2=0\Rightarrow a=-\frac\pi2 \).

Teacher's Note:
1. Complete the square to see that the argument is at least 1, which forces x = 1.
2. After finding x = 1, the equation becomes a linear equation in a.