JEE Mathematics Inverse Trigonometric Functions MCQs Set 02

Answer: (b) \( \frac53 \)

Explanation:
1. \( \tan^{-1}5+\tan^{-1}\tfrac15=\frac{\pi}{2} \) (reciprocal positive arguments).
2. \( \frac{\pi}{2}-\tan^{-1}\tfrac34=\cot^{-1}\tfrac34=\tan^{-1}\tfrac43 \).
3. \( \sec\left(\tan^{-1}\tfrac43\right)=\sqrt{1+\tfrac{16}{9}}=\frac53 \).

Teacher's Note:
1. Pair reciprocal arguments first: the pair gives pi/2 immediately.
2. Then pi/2 minus an angle gives its complement, which converts tan inverse of 3/4 into tan inverse of 4/3.

Answer: (c) \( \frac{\pi}{4} \)

Explanation:
1. \( 2\tan^{-1}\tfrac15=\tan^{-1}\tfrac{5}{12}\Rightarrow 4\tan^{-1}\tfrac15=\tan^{-1}\frac{2\cdot\frac5{12}}{1-\frac{25}{144}}=\tan^{-1}\tfrac{120}{119} \).
2. \( \tan^{-1}\tfrac{120}{119}-\tan^{-1}\tfrac1{70}=\tan^{-1}\frac{\frac{120}{119}-\frac1{70}}{1+\frac{120}{119\cdot70}}=\tan^{-1}\tfrac{49}{50} \).
3. \( \tan^{-1}\tfrac{49}{50}+\tan^{-1}\tfrac1{99}=\tan^{-1}\frac{\frac{49}{50}+\frac1{99}}{1-\frac{49}{4950}}=\tan^{-1}1=\frac{\pi}{4} \).

Teacher's Note:
1. This is a Machin-type identity; combine the terms in the order that keeps each product of arguments below 1.
2. The final value pi/4 confirms the arithmetic, since all intermediate arguments are positive and below 1.

Answer: (c) \( -\frac{36}{325} \)

Explanation:
1. Let \( A=\tan^{-1}\tfrac5{12} \): \( \cos A=\tfrac{12}{13},\ \sin A=\tfrac5{13} \). Let \( B=\sin^{-1}\tfrac{24}{25} \): \( \cos B=\tfrac{7}{25} \).
2. \( x=\cos(A+B)=\frac{12}{13}\cdot\frac7{25}-\frac5{13}\cdot\frac{24}{25}=\frac{84-120}{325}=-\frac{36}{325} \).
3. Check: \( A+B\approx22.6^\circ+73.7^\circ=96.3^\circ \) lies in \( [0,\pi] \), so it equals \( \cos^{-1}x \).

Teacher's Note:
1. The sum exceeds 90 degrees, which is why the cosine comes out negative.
2. Always confirm that the angle A + B lies in [0, pi] before equating it to cos inverse x.

Answer: (a) 12

Explanation:
1. \( \sec^{-1}\frac{13}{x}=\cos^{-1}\frac{x}{13} \).
2. \( \sin^{-1}\tfrac{12}{13}+\cos^{-1}\tfrac{x}{13}=\tfrac\pi2\Rightarrow\frac{x}{13}=\frac{12}{13} \) using \( \sin^{-1}y+\cos^{-1}y=\tfrac{\pi}{2} \).
3. So \( x=12 \).

Teacher's Note:
1. sec inverse of t equals cos inverse of 1/t for t >= 1.
2. Matching the argument of sin inverse and cos inverse is possible only when they are equal.

Answer: (d) 5

Explanation:
1. For the sum of two sin inverse terms to be pi/2: \( \sin^{-1}\tfrac3x=\cos^{-1}\tfrac4x \).
2. That requires \( \left(\tfrac3x\right)^2+\left(\tfrac4x\right)^2=1\Rightarrow x^2=25 \).
3. Since both arguments must be positive, \( x=5 \).

Teacher's Note:
1. sin inverse a + sin inverse b = pi/2 holds exactly when a squared + b squared = 1 with a, b >= 0.
2. The numbers 3, 4, 5 form a Pythagorean triple, which hints at the answer immediately.

Answer: (b) \( \sqrt{ab} \)

Explanation:
1. \( \tan^{-1}\tfrac ax=\tfrac\pi2-\tan^{-1}\tfrac bx=\tan^{-1}\tfrac xb \) (for a, b, x > 0).
2. \( \frac ax=\frac xb\Rightarrow x^2=ab\Rightarrow x=\sqrt{ab} \).

Teacher's Note:
1. The sum pi/2 means the two arguments are reciprocals of each other.
2. Equivalently xy = 1 for tan inverse x + tan inverse y = pi/2.

Answer: (b) 3

Explanation:
1. Put \( \frac xa=t \): \( \frac{3a^2x-x^3}{a^3-3ax^2}=\frac{3t-t^3}{1-3t^2} \).
2. \( \tan3\theta=\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta} \) with \( t=\tan\theta \).
3. So the expression is \( 3\tan^{-1}t \) in the principal range, giving \( k=3 \).

Teacher's Note:
1. The triple-angle identity for tan appears whenever you see 3t - t cubed over 1 - 3t squared.
2. The equality holds only for -1/root 3 < t < 1/root 3; the question asks for the formal value of k.

Answer: (b) \( \pm\frac{1}{\sqrt2} \)

Explanation:
1. Take tan: \( \frac{\frac{x-1}{x-2}+\frac{x+1}{x+2}}{1-\frac{(x-1)(x+1)}{(x-2)(x+2)}}=1 \).
2. Numerator \( \propto (x-1)(x+2)+(x+1)(x-2)=2x^2-4 \); denominator \( \propto (x^2-4)-(x^2-1)=-3 \).
3. \( \frac{2x^2-4}{-3}=1\Rightarrow x^2=\frac12\Rightarrow x=\pm\frac{1}{\sqrt2} \).
4. Both values satisfy the original equation (each term is positive and the product of the arguments is below 1).

Teacher's Note:
1. Always verify the roots after taking tan on both sides; extra roots can appear when the sum crosses pi/2.
2. Here the two values of x swap the two terms, which is why both work.

Answer: (c) \( \frac{1}{\sqrt2} \)

Explanation:
1. \( \cot^{-1}\tfrac43=\tan^{-1}\tfrac34 \).
2. \( \tan^{-1}\tfrac34+\tan^{-1}\tfrac17=\tan^{-1}\frac{\frac34+\frac17}{1-\frac3{28}}=\tan^{-1}1=\frac\pi4 \).
3. \( \cos^{-1}x=\frac\pi4\Rightarrow x=\frac1{\sqrt2} \).

Teacher's Note:
1. Convert cot inverse of a positive number into tan inverse of its reciprocal.
2. Then add the two tan inverse terms with the standard formula.

Answer: (a) \( |a|\le\frac{1}{\sqrt2} \)

Explanation:
1. The right side lies in \( [-\pi,\pi] \) but \( \sin^{-1}x\in\left[-\frac\pi2,\frac\pi2\right] \).
2. So we need \( -\frac\pi2\le2\sin^{-1}a\le\frac\pi2\Rightarrow|\sin^{-1}a|\le\frac\pi4 \).
3. Hence \( |a|\le\sin\frac\pi4=\frac{1}{\sqrt2} \).

Teacher's Note:
1. The range of sin inverse limits the doubled angle on the right-hand side.
2. The solution is then x = sin(2 sin inverse a) = 2a root(1 - a squared).

Answer: (a) n

Explanation:
1. Each term is at most \( \frac\pi2 \).
2. The sum of n terms equals \( \frac{n\pi}{2} \) only if every term equals \( \frac\pi2 \).
3. So \( x_k=1 \) for all k and \( \sum x_k=n \).

Teacher's Note:
1. When a sum of bounded quantities reaches the maximum possible value, every term must attain its own maximum.
2. The same logic applies to sums of cos inverse terms that equal zero.

Answer: (b) \( \frac{x+z}{1-xz} \)

Explanation:
1. A.P. means \( 2\tan^{-1}y=\tan^{-1}x+\tan^{-1}z \).
2. Take tan of both sides: \( \frac{2y}{1-y^2}=\frac{x+z}{1-xz} \).

Teacher's Note:
1. For terms in A.P., twice the middle equals the sum of the extremes.
2. Then use tan 2A and tan (A + B) formulas on the two sides.

Answer: (c) \( \left[\frac\pi4,\frac{3\pi}{4}\right] \)

Explanation:
1. The expression equals \( \frac\pi2+\tan^{-1}x \) with domain \( x\in[-1,1] \).
2. \( \tan^{-1}x\in\left[-\frac\pi4,\frac\pi4\right] \) on this domain (the end values are attained).
3. Range \( =\left[\frac\pi4,\frac{3\pi}{4}\right] \).

Teacher's Note:
1. Simplify first using sin inverse x + cos inverse x = pi/2, then restrict the domain to [-1, 1].
2. The closed bracket is used because x = 1 and x = -1 are both in the domain.

Answer: (c) \( [-2,-1]\cup[1,2] \)

Explanation:
1. \( -1\le\log_2\frac{x^2}{2}\le1\Rightarrow\frac12\le\frac{x^2}{2}\le2 \).
2. \( 1\le x^2\le4\Rightarrow x\in[-2,-1]\cup[1,2] \).

Teacher's Note:
1. Take 2 to the power of each side; the base is greater than 1, so the inequality direction is unchanged.
2. Solving x squared between 1 and 4 gives two intervals, symmetrical about 0.

Answer: (d) 0.28

Explanation:
1. \( \cos2\theta=2\cos^2\theta-1=2(0.64)-1=0.28 \).

Teacher's Note:
1. Use cos 2 theta = 2 cos squared theta - 1; the inverse function itself need not be evaluated.
2. Check: cos inverse 0.8 is about 36.9 degrees, so twice that is 73.7 degrees and cos 73.7 degrees is about 0.28.

Answer: (a) x

Explanation:
1. Let \( \theta=\sin^{-1}\sqrt{\frac{1-x}{2}} \), so \( \sin^2\theta=\frac{1-x}{2} \).
2. \( \cos2\theta=1-2\sin^2\theta=1-(1-x)=x \).

Teacher's Note:
1. Using cos 2 theta = 1 - 2 sin squared theta removes the square root completely.
2. This is the half-angle form of the identity cos inverse x = 2 sin inverse root((1 - x)/2).

Answer: (a) \( 0,\ \frac12 \)

Explanation:
1. \( \sin^{-1}x-\left(\tfrac\pi2-\sin^{-1}x\right)=\sin^{-1}(x-1)\Rightarrow2\sin^{-1}x-\tfrac\pi2=\sin^{-1}(x-1) \).
2. Take sine: \( -\cos(2\sin^{-1}x)=x-1\Rightarrow-(1-2x^2)=x-1\Rightarrow2x^2-x=0 \).
3. \( x=0\text{ or }x=\tfrac12 \).
4. Check: \( x=0:\ -\tfrac\pi2=\sin^{-1}(-1) \); \( x=\tfrac12:\ \tfrac\pi6-\tfrac\pi3=-\tfrac\pi6=\sin^{-1}(-\tfrac12) \). Both work.

Teacher's Note:
1. Substitute cos inverse x = pi/2 - sin inverse x to have a single inverse function on the left.
2. Taking sine of both sides can introduce extra roots, so always test the values obtained.

Answer: (b) \( -\frac1{12} \)

Explanation:
1. Then \( \sin^{-1}(6\sqrt3x)=-\tfrac\pi2-\sin^{-1}(6x) \), so \( 6\sqrt3x=-\cos(\sin^{-1}6x)\Rightarrow108x^2=1-36x^2 \).
2. \( 144x^2=1\Rightarrow x=\pm\frac1{12} \). The sum is negative, so \( x<0 \) and \( x=-\frac1{12} \).
3. Check: \( \sin^{-1}(-\tfrac12)+\sin^{-1}(-\tfrac{\sqrt3}{2})=-\tfrac\pi6-\tfrac\pi3=-\tfrac\pi2 \).

Teacher's Note:
1. Isolate one inverse term, take sine and square; then reject the root that fails the sign condition.
2. Verifying by direct substitution is the safest final step.

Answer: (c) \( \frac16 \)

Explanation:
1. Taking tan: \( \frac{5x}{1-6x^2}=1\Rightarrow6x^2+5x-1=0 \).
2. \( (6x-1)(x+1)=0\Rightarrow x=\tfrac16\text{ or }-1 \).
3. For \( x=-1 \) both terms are negative, so the sum cannot be \( \frac\pi4 \). Hence \( x=\tfrac16 \).

Teacher's Note:
1. A positive sum requires x > 0 here, since each tan inverse has the sign of x.
2. Reject roots that give the wrong sign even when they satisfy the algebraic equation.

Answer: (b) \( \frac\pi4 \)

Explanation:
1. \( \tan(\cdot)=\frac{a(c+a)+b(b+c)}{(b+c)(c+a)-ab}=\frac{ac+a^2+b^2+bc}{bc+ab+c^2+ac-ab} \).
2. With \( a^2+b^2=c^2 \): numerator \( =c^2+ac+bc \), denominator \( =c^2+ac+bc \), ratio \( =1 \).
3. Both terms are positive and the product is below 1, so the sum is \( \frac\pi4 \).

Teacher's Note:
1. Use a squared + b squared = c squared once, in the numerator.
2. Both fractions are less than 1, so the sum lies in (0, pi/2) and equals pi/4.

Answer: (a) \( x=y=z \)

Explanation:
1. From the first A.P.: \( 2y=x+z \). From the second: \( \frac{2y}{1-y^2}=\frac{x+z}{1-xz} \).
2. Substituting \( x+z=2y \): \( \frac{2y}{1-y^2}=\frac{2y}{1-xz} \).
3. So either \( y=0 \) or \( y^2=xz \). With \( 2y=x+z \) the second gives \( (x-z)^2=0 \), i.e. \( x=y=z \); the case \( y=0 \) gives \( x=-z \) and the same sequence reduces to all zero.

Teacher's Note:
1. A sequence that is in both A.P. and G.P. is constant.
2. The case y = 0 leads to the same conclusion, because the tan inverse A.P. then forces x = z.

Answer: (b) \( [-2,1] \)

Explanation:
1. \( -1\le\frac{2x+1}{3}\le1\Rightarrow-3\le2x+1\le3\Rightarrow-2\le x\le1 \).

Teacher's Note:
1. The argument of sin inverse must lie in [-1, 1] and both end points are allowed.
2. Multiply the inequality by 3 first, then subtract 1 and divide by 2.

Answer: (c) \( \left[0,\frac12\right] \)

Explanation:
1. The square root needs \( 2x\ge0 \) and cos inverse needs \( \sqrt{2x}\le1\Rightarrow2x\le1 \).
2. Hence \( 0\le x\le\frac12 \).

Teacher's Note:
1. Combine the domain of the square root with that of cos inverse.
2. The argument of the inverse function is never negative here, so only the upper bound matters.

Answer: (b) \( 0\le x\le1 \)

Explanation:
1. Put \( x=\cos\theta,\ \theta\in[0,\pi] \): \( 2x^2-1=\cos2\theta \).
2. \( \cos^{-1}(\cos2\theta)=2\theta \) only if \( 2\theta\in[0,\pi]\Rightarrow\theta\in\left[0,\tfrac\pi2\right] \).
3. Hence \( x=\cos\theta\in[0,1] \).

Teacher's Note:
1. The doubled angle must stay within the principal range [0, pi] of cos inverse.
2. For x in [-1, 0] the identity changes to 2 pi - 2 cos inverse x.

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

Explanation:
1. Divide numerator and denominator by a and put \( \frac ba=t^2 \): \( \frac{a-b}{a+b}=\frac{1-t^2}{1+t^2} \).
2. \( \cos^{-1}\frac{1-t^2}{1+t^2}=2\tan^{-1}t \) for \( t\ge0 \).
3. So the value is \( 2\tan^{-1}\sqrt{b/a} \).

Teacher's Note:
1. The formula cos inverse((1 - t squared)/(1 + t squared)) = 2 tan inverse t is the standard one to remember.
2. Here t squared is the ratio b/a, so t = root(b/a).