JEE Mathematics Inverse Trigonometric Functions MCQs Set 03

Answer: (d) \( \frac{3}{29} \)

Explanation:
1. Let \( A=\cos^{-1}\tfrac1{5\sqrt2} \): \( \cos A=\tfrac1{5\sqrt2},\ \sin A=\sqrt{1-\tfrac1{50}}=\tfrac7{5\sqrt2} \), so \( \tan A=7 \).
2. Let \( B=\sin^{-1}\tfrac4{\sqrt{17}} \): \( \cos B=\tfrac1{\sqrt{17}} \), so \( \tan B=4 \).
3. \( \tan(A-B)=\frac{7-4}{1+28}=\frac3{29} \).

Teacher's Note:
1. Convert each inverse function into a tangent first, then apply the tan(A - B) formula.
2. Both angles are acute, so all square roots take the positive sign.

Answer: (b) 1

Explanation:
1. Put \( x=\sin\theta,\ 0\le\theta\le\tfrac\pi6 \). The braces become \( \frac{\sin\theta+\cos\theta}{\sqrt2}=\sin\left(\theta+\tfrac\pi4\right) \).
2. \( \theta+\tfrac\pi4\in\left[\tfrac\pi4,\tfrac{5\pi}{12}\right]\subset\left[-\tfrac\pi2,\tfrac\pi2\right] \), so \( \sin^{-1}(\cdot)=\theta+\tfrac\pi4 \).
3. The bracket equals \( \theta+\tfrac\pi4-\theta=\tfrac\pi4 \) and \( \tan\tfrac\pi4=1 \).

Teacher's Note:
1. The restriction on x is exactly what keeps theta + pi/4 inside the principal range of sin inverse.
2. Without the restriction the answer could change.

Answer: (c) 0

Explanation:
1. Maximum of each square: \( (\sin^{-1}a)^2\le\tfrac{\pi^2}4,\ (\cos^{-1}b)^2\le\pi^2,\ (\sec^{-1}c)^2\le\pi^2,\ (\operatorname{cosec}^{-1}d)^2\le\tfrac{\pi^2}4 \).
2. The maxima add to \( \tfrac{\pi^2}{4}+\pi^2+\pi^2+\tfrac{\pi^2}{4}=\tfrac{5\pi^2}{2} \), so every term is at its maximum.
3. Required value \( =\tfrac{\pi^2}4-\pi^2+\pi^2-\tfrac{\pi^2}4=0 \).

Teacher's Note:
1. A sum that equals the sum of the individual maxima forces each term to be at its maximum.
2. This works because each squared term is bounded using the principal range of its inverse function.

Answer: (d) \( \frac{\pi ab+c(a-b)}{a+b} \)

Explanation:
1. Let \( s=\sin^{-1}x \), so \( \cos^{-1}x=\tfrac\pi2-s \). Then \( as-b\left(\tfrac\pi2-s\right)=c\Rightarrow s=\frac{c+b\pi/2}{a+b} \).
2. \( as+b\cos^{-1}x=(a-b)s+\tfrac{b\pi}2 \).
3. \( =\frac{(a-b)(c+\frac{b\pi}2)+\frac{b\pi}2(a+b)}{a+b}=\frac{\pi ab+c(a-b)}{a+b} \).

Teacher's Note:
1. Express everything in terms of a single unknown s = sin inverse x using the complementary identity.
2. Check with a = b: the given relation fixes s = pi/4 + c/(2a) and the answer reduces to pi a/2, as it should.

Answer: (b) \( \tan^{-1}2 \)

Explanation:
1. General term \( \cot^{-1}\left(n^2+\tfrac34\right)=\tan^{-1}\frac{1}{1+\left(n^2-\tfrac14\right)} \).
2. \( =\tan^{-1}\frac{\left(n+\tfrac12\right)-\left(n-\tfrac12\right)}{1+\left(n+\tfrac12\right)\left(n-\tfrac12\right)}=\tan^{-1}\left(n+\tfrac12\right)-\tan^{-1}\left(n-\tfrac12\right) \).
3. The series telescopes: \( S=\lim_{n\to\infty}\tan^{-1}\left(n+\tfrac12\right)-\tan^{-1}\tfrac12=\tfrac\pi2-\tan^{-1}\tfrac12=\tan^{-1}2 \).

Teacher's Note:
1. Write n squared + 3/4 as 1 + (n + 1/2)(n - 1/2) to see the telescoping difference.
2. The last step uses tan inverse 2 + tan inverse (1/2) = pi/2.

Answer: (c) 9

Explanation:
1. Let \( r=\sin(\cos^{-1}x) \). The G.P. sum is \( \frac1{1-r}=2\Rightarrow r=\tfrac12 \).
2. \( \cos^{-1}x=\tfrac\pi6\approx0.52 \) (which lies in \( (0,1) \)), so \( x=\tfrac{\sqrt3}2 \).
3. \( 12x^2=12\cdot\tfrac34=9 \).

Teacher's Note:
1. Check the stated condition 0 < cos inverse x < 1: pi/6 is about 0.52, so the solution is valid (the other choice 5 pi/6 is not).
2. The series converges only if |r| < 1, and r = 1/2 satisfies this.

Answer: (a) 1

Explanation:
1. \( \log_2x\ge0\Rightarrow x\ge1 \). For \( x\ge1 \): \( \sin^{-1}\frac{2x}{1+x^2}=\pi-2\tan^{-1}x \).
2. The bracket becomes \( \pi-2\tan^{-1}x+2\tan^{-1}x=\pi \).
3. \( \log_{1/\pi}\pi=-1 \), so the modulus is \( 1 \).

Teacher's Note:
1. For x greater than 1 the identity 2 tan inverse x = sin inverse(2x/(1 + x squared)) is replaced by pi minus it.
2. The base 1/pi is less than 1, so the logarithm of pi is negative; the modulus makes it 1.

Answer: (c) 1

Explanation:
1. Domain: \( -1\le1-x\le1 \) and \( -1\le x\le1 \), so \( 0\le x\le1 \), hence \( \cos^{-1}x\in\left[0,\tfrac\pi2\right] \).
2. Let \( \theta=\cos^{-1}x \): \( 1-x=\cos\left(\tfrac\pi2+2\theta\right)=-\sin2\theta\le0 \) but \( 1-x\ge0 \).
3. So both sides must be zero: \( x=1 \) (then \( \theta=0 \)). Check: \( \cos^{-1}0-0=\tfrac\pi2 \). Exactly one solution.

Teacher's Note:
1. After restricting x to [0, 1], the sign of both sides forces them to be zero.
2. Always begin with the domain; it removes most of the work in such equations.

Answer: (a) \( x^2=\frac{\sqrt5-1}{2} \)

Explanation:
1. Let \( \theta=\cos^{-1}x=\tan^{-1}x \), so \( \cos\theta=\tan\theta=x \) and \( x>0 \).
2. \( \cos\theta=\frac{\sin\theta}{\cos\theta}\Rightarrow\sin\theta=\cos^2\theta=1-\sin^2\theta \), so \( \sin^2\theta+\sin\theta-1=0\Rightarrow\sin\theta=\frac{\sqrt5-1}{2} \).
3. Then \( x^2=\cos^2\theta=\sin\theta=\frac{\sqrt5-1}{2} \). Also \( \sin(\cos^{-1}x)=\frac{\sqrt5-1}{2} \) and \( \tan(\cos^{-1}x)=x\approx0.786\ne\frac{\sqrt5-1}{2} \), so only (a) is correct.

Teacher's Note:
1. Use sin theta = cos squared theta, which comes from setting cos theta equal to tan theta.
2. The golden-ratio number (root 5 - 1)/2 equals sin theta, not tan theta.

Answer: (b) -1

Explanation:
1. Domains require \( x^2+x+1\le1 \) and \( x^2+x\ge0 \), so \( x^2+x=0 \).
2. \( x=0 \) or \( x=-1 \). Check: \( 2\sin^{-1}1+\cos^{-1}0=\pi+\tfrac\pi2=\tfrac{3\pi}2 \) for both.
3. Sum of solutions \( =-1 \).

Teacher's Note:
1. Two domain conditions that point in opposite directions leave a single value of x squared + x.
2. Both roots give the same inverse-function values, so both are valid.

Answer: (b) 2

Explanation:
1. For \( x\in[-1,1] \) the left side is increasing from \( -\tfrac{3\pi}4\approx-2.36 \) to \( \tfrac{3\pi}4\approx2.36 \).
2. So \( -2.36\le2k+1\le2.36 \), i.e. the odd integers \( -1 \) and \( 1 \) qualify.
3. Thus \( k=-1 \) or \( k=0 \): 2 values.

Teacher's Note:
1. The sum of two increasing functions is increasing, so its range is the interval between its end values.
2. 3 pi/4 is about 2.36, which is less than 3, so the odd integers 3 and -3 are excluded.

Answer: (d) \( -\frac12 \)

Explanation:
1. \( \sin(\cot^{-1}t)=\frac1{\sqrt{1+t^2}} \) and \( \cos(\tan^{-1}x)=\frac1{\sqrt{1+x^2}} \).
2. \( 1+(1+x)^2=1+x^2\Rightarrow(1+x)^2=x^2\Rightarrow2x+1=0 \), so \( x=-\tfrac12 \).

Teacher's Note:
1. Both sides are positive for every real argument, so squaring is safe.
2. The quadratic terms cancel, leaving a linear equation.

Answer: (c) 1

Explanation:
1. Let \( A=\sin^2\theta-2\sin\theta+3=(\sin\theta-1)^2+2 \), so \( 2\le A\le6 \).
2. \( \tan^{-1}A+\cot^{-1}B=\tfrac\pi2\Rightarrow\cot^{-1}B=\cot^{-1}A \) gives \( B=A \), i.e. \( 5^{\sec^2\beta}+1=A \).
3. But \( 5^{\sec^2\beta}+1\ge6 \) and \( A\le6 \), so both equal 6: \( \sin\theta=-1 \).
4. \( \cos^2\theta-\sin\theta=0+1=1 \).

Teacher's Note:
1. Compare the minimum of one side with the maximum of the other; equality at a single point fixes the variables.
2. sec squared beta is at least 1, so 5 to that power is at least 5.

Answer: (a) \( \phi \) (empty set)

Explanation:
1. \( \sin^{-1}(\sin4)=\sin^{-1}\sin(\pi-4)=\pi-4\approx-0.86 \) (as \( \pi-4\in\left[-\tfrac\pi2,\tfrac\pi2\right] \)).
2. The quadratic is positive for all x only if \( \lambda^2-4(\pi-4)<0\Rightarrow\lambda^2<4(\pi-4)<0 \).
3. No real \( \lambda \) satisfies this (the quadratic always has real roots because its constant term is negative). The set is empty.

Teacher's Note:
1. A quadratic with a negative constant term is negative at x = 0, so it cannot be positive for all x.
2. The key step is evaluating sin inverse(sin 4) correctly as pi - 4, which is negative.

Answer: (d) \( 5\pi-16 \)

Explanation:
1. \( \sin^{-1}(\cos2)=\tfrac\pi2-2 \) and \( \cos^{-1}(\sin2)=2-\tfrac\pi2 \) (both angles are in the correct principal ranges).
2. \( \tan^{-1}(\cot4)=\tfrac{3\pi}2-4 \) and \( \cot^{-1}(\tan4)=\tfrac{3\pi}2-4 \); these two cancel.
3. \( \sec^{-1}(\operatorname{cosec}6)=\tfrac{5\pi}2-6 \) and \( \operatorname{cosec}^{-1}(\sec6)=6-\tfrac{3\pi}2 \).
4. Total: \( \left(\tfrac\pi2-2\right)-\left(2-\tfrac\pi2\right)+0+\left(\tfrac{5\pi}2-6\right)-\left(6-\tfrac{3\pi}2\right)=5\pi-16 \).

Teacher's Note:
1. Shift each angle by a multiple of 2 pi or reflect it until it lies in the principal range of the outer inverse function.
2. Numerically the total is about -0.292, matching 5 pi - 16.

Answer: (c) 3

Explanation:
1. \( y=\cos^{-1}(\cos x)\ge0 \), so the first equation gives \( y=\sin x \) with \( \sin x\ge0 \).
2. \( \cos^{-1}(\cos x)=|x| \) on \( [-\pi,\pi] \) and \( =2\pi-|x| \) for \( \pi\le|x|\le2\pi \).
3. On \( [0,\pi] \): \( x=\sin x \) only at \( x=0 \); on \( [-\pi,0] \) the sine is negative; on \( [\pi,2\pi] \) the sine is not positive, so only \( x=\pm2\pi \) (where both are 0) remain.
4. Solutions: \( x=-2\pi,\ 0,\ 2\pi \), three in all.

Teacher's Note:
1. Sketch y = cos inverse(cos x), the zig-zag wave between 0 and pi, against the positive arches of sin x.
2. Both curves are zero at x = 0 and at the end points, which accounts for all three common points.

Answer: (d) \( \tan\left(\tfrac12\cos^{-1}\tfrac{\sqrt5}3\right) \)

Explanation:
1. \( \tan^{-1}3+\tan^{-1}\tfrac13=\tfrac\pi2 \), so the first number is \( 1 \) (rational).
2. \( \cos\left(\tfrac\pi2-\sin^{-1}\tfrac34\right)=\sin\left(\sin^{-1}\tfrac34\right)=\tfrac34 \) (rational).
3. With \( \theta=\sin^{-1}\tfrac{\sqrt{63}}8 \), \( \cos\theta=\tfrac18 \). Then \( \sin\tfrac\theta4 \) works out to \( \tfrac1{2\sqrt2} \), so the logarithm is \( -\tfrac32 \) (rational).
4. \( \tan\tfrac\varphi2=\frac{\sin\varphi}{1+\cos\varphi}=\frac{2/3}{1+\sqrt5/3}=\frac{2}{3+\sqrt5}=\frac{3-\sqrt5}{2} \), which is irrational.

Teacher's Note:
1. For part (c) use the half-angle formula twice: cos theta = 1/8 gives cos(theta/2) = 3/4, and then sin squared(theta/4) = (1 - 3/4)/2 = 1/8.
2. Only the last expression involves root 5 in its final value, so it is the irrational one.

Answer: (b) 0 and \( \frac12 \)

Explanation:
1. Domain: \( -1\le x\le1 \) and \( -1\le1-x\le1 \), so \( 0\le x\le1 \).
2. Take sine of \( \sin^{-1}x+\sin^{-1}(1-x)=\cos^{-1}x \): \( x\sqrt{1-(1-x)^2}+(1-x)\sqrt{1-x^2}=\sqrt{1-x^2} \), i.e. \( x\sqrt{2x-x^2}=x\sqrt{1-x^2} \).
3. So \( x=0 \) or \( 2x-x^2=1-x^2\Rightarrow x=\tfrac12 \).
4. Check: \( x=0:\ 0+\tfrac\pi2=\tfrac\pi2 \); \( x=\tfrac12:\ \tfrac\pi6+\tfrac\pi6=\tfrac\pi3=\cos^{-1}\tfrac12 \). Both valid; \( x=1 \) fails (\( \tfrac\pi2\ne0 \)).

Teacher's Note:
1. Take cosine rather than sine if that gives cleaner algebra, but always verify afterwards.
2. The domain restriction 0 <= x <= 1 rules out x = 2.

Answer: (b) \( \frac12 \)

Explanation:
1. Take tan of both sides: \( \frac{(x-1)+x+(x+1)-(x-1)x(x+1)}{1-\left[(x-1)x+x(x+1)+(x-1)(x+1)\right]}=3x \).
2. \( \frac{4x-x^3}{2-3x^2}=3x\Rightarrow4x-x^3=6x-9x^3\Rightarrow8x^3-2x=0 \), so \( x=0,\ \pm\tfrac12 \).
3. Check: \( x=\tfrac12:\ \tan^{-1}(-\tfrac12)+\tan^{-1}\tfrac12+\tan^{-1}\tfrac32=\tan^{-1}\tfrac32 \) (true); \( x=0 \) gives \( -\tfrac\pi4+0+\tfrac\pi4=0 \).
4. Sum of squares \( =0+\tfrac14+\tfrac14=\tfrac12 \).

Teacher's Note:
1. The three-term tan formula is (S1 - S3)/(1 - S2) in terms of the sums of the arguments, their pair products and their product.
2. Verify each root, because the tan formula is valid only when no term makes the denominator zero or changes the branch.

Answer: (b) \( \pi \)

Explanation:
1. \( 3\tan^{-1}\tfrac12=\tan^{-1}\frac{\frac32-\frac18}{1-\frac34}=\tan^{-1}\tfrac{11}{2} \) and \( 2\tan^{-1}\tfrac15=\tan^{-1}\tfrac5{12} \).
2. Since \( \tfrac{11}2\cdot\tfrac5{12}>1 \): \( \tan^{-1}\tfrac{11}2+\tan^{-1}\tfrac5{12}=\pi-\tan^{-1}\tfrac{142}{31} \).
3. \( \sin^{-1}\frac{142}{65\sqrt5}=\tan^{-1}\tfrac{142}{31} \) because \( \sqrt{(65\sqrt5)^2-142^2}=\sqrt{21125-20164}=31 \).
4. Total \( \pi \).

Teacher's Note:
1. Because the product of the two arguments exceeds 1, the sum is pi minus a tan inverse, not a plain tan inverse.
2. Check numerically: 3(0.4636) + 2(0.1974) + 0.9990... = 3.1416.

Answer: (d) 53

Explanation:
1. \( \sin^{-1}(\sin8)=3\pi-8,\ \tan^{-1}(\tan10)=10-3\pi,\ \cos^{-1}(\cos12)=4\pi-12 \).
2. \( \sec^{-1}(\sec9)=9-2\pi,\ \cot^{-1}(\cot6)=6-\pi,\ \operatorname{cosec}^{-1}(\operatorname{cosec}7)=7-2\pi \).
3. \( y=(3\pi-8)-(10-3\pi)+(4\pi-12)-(9-2\pi)+(6-\pi)-(7-2\pi)=13\pi-40 \).
4. \( a=13,\ b=-40\Rightarrow a-b=53 \).

Teacher's Note:
1. For each term, shift the argument by a multiple of the period (2 pi or pi) until it falls in the principal range.
2. The sum of the pi coefficients is 3 + 3 + 4 + 2 - 1 + 2 = 13 and the constants add to -40.

Answer: (a) \( \left\{\frac{1\pm\sqrt{17}}{2}\right\} \)

Explanation:
1. \( \cot^{-1}t\in\left(0,\tfrac\pi2\right] \) requires \( t\ge0 \) for every x, so the quadratic must satisfy \( 16-4(\alpha^2-\alpha)\le0 \), i.e. \( \alpha^2-\alpha-4\ge0 \).
2. For onto we need the quadratic to take every value in \( [0,\infty) \), so its minimum must be exactly 0: \( \alpha^2-\alpha-4=0 \).
3. \( \alpha=\frac{1\pm\sqrt{17}}2 \).

Teacher's Note:
1. Range inside the codomain gives an inequality, and onto-ness forces equality at the minimum.
2. The minimum of x squared + 4x + c is c - 4, which gives the condition alpha squared - alpha - 4 = 0.

Answer: (a) \( 2-\sqrt3,\ \sqrt3 \)

Explanation:
1. Let \( \theta=\tan^{-1}x \) and treat the ranges of x separately.
2. Case \( 03. Case \( x>1 \): both terms equal \( \pi-2\theta \), so \( 2\pi-4\theta=\tfrac{2\pi}3\Rightarrow\theta=\tfrac\pi3 \) and \( x=\sqrt3 \).
4. Case \( -15. Solutions: \( x=2-\sqrt3,\ \sqrt3 \).

Teacher's Note:
1. Splitting into the ranges |x| < 1 and |x| > 1 is needed because the tan inverse of 2x/(x squared - 1) changes form there.
2. Numerically, for x = root 3 the two terms are pi/3 and pi/3, and for x = 2 - root 3 they are 5 pi/6 and -pi/6; both add to 2 pi/3.

Answer: (a) \( \frac32 \)

Explanation:
1. Put \( x=\cos\theta,\ \tfrac{2\pi}3\le\theta\le\pi \). Then \( 3x-4x^3=-\cos3\theta \) and \( 4x^3-3x=\cos3\theta \).
2. \( 2\pi\le3\theta\le3\pi\Rightarrow0\le3\theta-2\pi\le\pi \), so \( \cos^{-1}(\cos3\theta)=3\theta-2\pi \).
3. \( \sin^{-1}(-\cos3\theta)=-\sin^{-1}\sin\left(\tfrac{5\pi}2-3\theta\right)=-\left(\tfrac{5\pi}2-3\theta\right) \) since \( \tfrac{5\pi}{2}-3\theta\in\left[-\tfrac\pi2,\tfrac\pi2\right] \).
4. \( f=3\theta-\tfrac{5\pi}2+3\theta-2\pi=6\theta-\tfrac{9\pi}2=6\cos^{-1}x-\tfrac{9\pi}2 \), so \( a=6,\ b=-\tfrac92,\ a+b=\tfrac32 \).

Teacher's Note:
1. Substituting x = cos theta turns both terms into functions of 3 theta; the range of 3 theta decides the branch.
2. Check with x = -1: f = sin inverse 1 + cos inverse(-1) = pi/2 + pi = 3 pi/2, and 6 pi - 9 pi/2 = 3 pi/2.