JEE Mathematics Inverse Trigonometric Functions MCQs Set 07

Answer: (b) \( \left\{\frac\pi4,\frac{3\pi}4\right\} \)

Explanation:
1. sin⁻¹ x needs |x| ≤ 1 and sec⁻¹ x needs |x| ≥ 1, so the domain is {-1, 1}.
2. \( f(1)=\frac\pi2+\frac\pi4+0=\frac{3\pi}4 \).
3. \( f(-1)=-\frac\pi2-\frac\pi4+\pi=\frac\pi4 \).
4. The range is the set \( \{\frac\pi4,\frac{3\pi}4\} \).

Teacher's Note:
1. The domain here has only two points, so the range has two values, not an interval.

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

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

Teacher's Note:
1. Exponentiate with base 2 (an increasing function) to remove the logarithm. The square gives two intervals.

Answer: (b) g, h, f, k

Explanation:
1. Domains: f is [-1, 1]; g is [0, 1/2]; h is [-1/2, 1/2]; k is all real numbers.
2. [0, 1/2] is contained in [-1/2, 1/2], which is contained in [-1, 1], which is contained in R.
3. So the order is g, h, f, k.

Teacher's Note:
1. Find each domain from the condition that the argument lies in [-1, 1]. Then compare by inclusion.

Answer: (c) 0.28

Explanation:
1. \( \cos2\theta=2\cos^2\theta-1 \) with \( \cos\theta=0.8 \).
2. \( =2(0.64)-1=0.28 \).

Teacher's Note:
1. The double-angle formula in terms of cosine alone needs no other trigonometric value.

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

Explanation:
1. \( \sin\frac{22\pi}7=\sin\left(3\pi+\frac\pi7\right)=-\sin\frac\pi7 \), so the first term is \( -\frac\pi7 \).
2. \( \cos^{-1}\left(\cos\frac{5\pi}3\right)=\cos^{-1}\left(\cos\frac\pi3\right)=\frac\pi3 \).
3. \( \tan\frac{5\pi}7=\tan\left(-\frac{2\pi}7\right) \), so the third term is \( -\frac{2\pi}7 \).
4. \( \cos2=\sin\left(\frac\pi2-2\right) \) with \( \frac\pi2-2\approx-0.43 \) in range, giving \( \frac\pi2-2 \).
5. Sum = \( \pi\left(-\frac17+\frac13-\frac27+\frac12\right)-2=\frac{17\pi}{42}-2 \).

Teacher's Note:
1. Reduce each angle to the principal range of its inverse function separately. Be careful with the sign of each reduced angle.

Answer: (c) \( x \)

Explanation:
1. Let \( \sin t=\sqrt{\frac{1-x}2} \).
2. \( \cos2t=1-2\sin^2t=1-(1-x)=x \).

Teacher's Note:
1. Use cos 2t = 1 - 2 sin² t, which removes the square root.

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

Explanation:
1. The left side is \( 2\sin^{-1}x-\frac\pi2 \).
2. Taking sine: \( \sin(2\sin^{-1}x-\frac\pi2)=-\cos(2\sin^{-1}x)=-(1-2x^2)=2x^2-1 \), which must equal x - 1.
3. \( 2x^2-x=0 \), so x = 0 or x = 1/2.
4. Check x = 1: left side is π/2 but right side is 0, so reject. x = 0 and x = 1/2 both satisfy.

Teacher's Note:
1. Check every root in the original equation because taking sines can create extra roots. Both x = 0 and x = 1/2 pass.

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

Explanation:
1. Both terms must be negative (a positive term would make the sum exceed -π/2), and for two negative sines with sum -π/2 we need \( a^2+b^2=1 \) where \( a=6x,\ b=6\sqrt3x \).
2. \( 36x^2+108x^2=1 \), so \( x^2=\frac1{144} \).
3. Since x is negative, \( x=-\frac1{12} \). Then 6x = -1/2 and 6 √3 x = -√3/2, giving -π/6 - π/3 = -π/2.

Teacher's Note:
1. Both terms must be negative, so x is negative. Verify by substituting the value: -π/6 + (-π/3) = -π/2.

Answer: (b) \( \frac16 \)

Explanation:
1. Taking tangents: \( \frac{5x}{1-6x^2}=1 \), so \( 6x^2+5x-1=0 \).
2. The roots are \( x=\frac16 \) and \( x=-1 \).
3. For x = -1 both terms are negative and the sum is -3pi/4, so reject it.
4. x = 1/6.

Teacher's Note:
1. Always check the roots because the tangent addition formula can hide a π shift. Here the negative root gives the wrong angle.

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

Explanation:
1. Put \( x=\tan\theta \) with \( \theta\in(0,\frac\pi2) \). Then \( \frac{1-x^2}{2x}=\cot2\theta \) and \( \frac{1-x^2}{1+x^2}=\cos2\theta \).
2. Since \( 2\theta\in(0,\pi) \), both inverse functions equal \( 2\theta \).
3. \( 4\theta=\frac{2\pi}3 \), so \( \theta=\frac\pi6 \) and \( x=\frac1{\sqrt3} \).

Teacher's Note:
1. The substitution x = tan θ turns both fractions into double-angle forms. Check that 2 θ stays in (0, π).

Answer: (c) 3

Explanation:
1. Add the left side: \( \frac{\frac1{1+2x}+\frac1{4x+1}}{1-\frac1{(1+2x)(4x+1)}}=\frac{6x+2}{8x^2+6x} \).
2. Equate to \( \frac2{x^2} \): \( x^2(6x+2)=2(8x^2+6x) \), so \( 3x^2-7x-6=0 \) (x not zero).
3. x = 3 or x = -2/3.
4. x = -2/3 makes both left terms negative while the right side is positive, so it fails.
5. Check x = 3: tan⁻¹(1/7) + tan⁻¹(1/13) = tan⁻¹(2/9). Hence x = 3.

Teacher's Note:
1. Quadratic roots must be tested in the original equation because of the principal value branch.

Answer: (d) \( \frac1a \)

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. The sum is \( \tan^{-1}\frac1{a+b}+\tan^{-1}(a+b)-\tan^{-1}a=\frac\pi2-\tan^{-1}a \).
3. \( \tan\left(\frac\pi2-\tan^{-1}a\right)=\frac1a \).

Teacher's Note:
1. Write the numerator b as (a + b) - a to recognise a difference of inverse tangents. Reciprocal terms add to π/2.

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

Explanation:
1. The product of the arguments is \( \frac{ab}{(b+c)(c+a)} \) which is less than 1, so the sum is \( \tan^{-1}\frac{a(c+a)+b(b+c)}{(b+c)(c+a)-ab} \).
2. With \( c^2=a^2+b^2 \) the numerator is \( a^2+b^2+c(a+b)=c^2+c(a+b)=c(a+b+c) \) and the denominator is \( c^2+c(a+b)=c(a+b+c) \).
3. The ratio is 1, so the sum is \( \frac\pi4 \).

Teacher's Note:
1. Substitute the Pythagoras relation to make numerator and denominator identical.

Answer: (a) 2

Explanation:
1. Both terms are positive for x > 1 and the product of arguments \( \frac{(x+1)(x-1)}{(x-1)x}=\frac{x+1}x>1 \) so the sum is \( \pi+\tan^{-1}\frac{\frac{x+1}{x-1}+\frac{x-1}x}{1-\frac{x+1}x} \).
2. The fraction simplifies to \( -\frac{2x^2-x+1}{x-1} \); setting it equal to -7 gives \( 2x^2-x+1=7(x-1) \), i.e. \( (x-2)^2=0 \).
3. So x = 2. Check: tan⁻¹ 3 + tan⁻¹(1/2) = 1.7127 = π - tan⁻¹ 7.

Teacher's Note:
1. The π term signals that the product of the arguments exceeds 1. The resulting quadratic has a double root.

Answer: (d) 2

Explanation:
1. sin⁻¹ needs \( x^2+x+1\le1 \), i.e. \( x(x+1)\le0 \).
2. The square root needs \( x(x+1)\ge0 \).
3. So \( x(x+1)=0 \), giving x = 0 or x = -1.
4. Both satisfy: tan⁻¹ 0 + sin⁻¹ 1 = π/2. So there are 2 solutions.

Teacher's Note:
1. Domain restrictions alone pin down the solutions. Always test them in the equation.

Answer: (b) \( x=\pi y \)

Explanation:
1. \( x=\frac\pi4+\frac{2\pi}3-\frac\pi6=\frac{3\pi}4 \).
2. \( y^2=\frac{1+\cos(\cos^{-1}\frac18)}2=\frac{1+\frac18}2=\frac9{16} \), so \( y=\frac34 \) (positive since the half angle is acute).
3. So \( x=\pi y \).

Teacher's Note:
1. Use the half-angle formula for y and take the positive root. Compare x and y by their ratio.

Answer: (c) 0

Explanation:
1. Let s = a + b + c and x, y, z be the three arguments. Then x = a √(s)/√(abc), and similarly for y, z.
2. \( x+y+z=\frac{s\sqrt s}{\sqrt{abc}} \) and \( xyz=\sqrt{\frac{s^3}{abc}} \), so \( x+y+z-xyz=0 \).
3. Also \( xy+yz+zx=s\left(\frac1a+\frac1b+\frac1c\right)\ne1 \).
4. \( \tan\theta=\frac{x+y+z-xyz}{1-(xy+yz+zx)}=0 \).

Teacher's Note:
1. Use the tangent of a sum of three angles. The numerator vanishes, while the denominator is not zero.

Answer: (c) x = y = z

Explanation:
1. From the second A.P.: \( 2\tan^{-1}y=\tan^{-1}x+\tan^{-1}z \), so \( \frac{2y}{1-y^2}=\frac{x+z}{1-xz} \).
2. From the first A.P.: \( x+z=2y \).
3. So \( \frac{2y}{1-y^2}=\frac{2y}{1-xz} \), giving \( y^2=xz \) (y positive).
4. With \( y=\frac{x+z}2 \): \( (x+z)^2=4xz \), so \( (x-z)^2=0 \) and x = y = z.

Teacher's Note:
1. Positivity avoids the trivial case y = 0. The two conditions force x = z.

Answer: (d) \( [\tan\sin\cos1,\ \tan\sin\cos\sin1] \)

Explanation:
1. The outer value lies in \( [-\frac\pi2,\frac\pi2] \), so [.] = 1 means it lies in \( [1,\frac\pi2] \).
2. \( \sin^{-1}u\ge1 \) gives \( u\ge\sin1 \) with \( u\le1 \).
3. \( u=\cos^{-1}v\ge\sin1 \) gives \( v\le\cos\sin1 \), and \( u\le1 \) gives \( v\ge\cos1 \).
4. \( v=\sin^{-1}w \), so \( \sin\cos1\le w\le\sin\cos\sin1 \).
5. \( w=\tan^{-1}x \), so \( x\in[\tan\sin\cos1,\tan\sin\cos\sin1] \).

Teacher's Note:
1. Work from the outside in, reversing each function. The end points are included because the greatest integer function is 1 at equality.

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

Explanation:
1. \( x(3-x)\ge2 \) gives \( x^2-3x+2\le0 \), so \( 1\le x\le2 \).
2. sin⁻¹ needs \( |x|\le1 \), so x = 1.
3. Each of the 10 terms is \( \sin^{-1}1=\frac\pi2 \), so the sum is \( 5\pi \).

Teacher's Note:
1. The two conditions together force a single value. Then add 10 equal terms.

Answer: (a) \( \frac32 \)

Explanation:
1. \( \tan^{-1}x=\tan^{-1}y \) gives x = y.
2. Then \( 2\cos^{-1}x=\frac\pi2 \), so \( x=\cos\frac\pi4=\frac1{\sqrt2} \).
3. \( x^2+xy+y^2=3x^2=\frac32 \).

Teacher's Note:
1. Use the second condition first to reduce to one unknown.

Answer: (d) \( x\sqrt{1+x^2} \)

Explanation:
1. \( \cos(\cot^{-1}x)=\frac x{\sqrt{1+x^2}} \) and \( \sin(\cot^{-1}x)=\frac1{\sqrt{1+x^2}} \).
2. The bracket inside the braces is \( \frac{x^2+1}{\sqrt{1+x^2}}=\sqrt{1+x^2} \).
3. Squared minus 1: \( 1+x^2-1=x^2 \), whose root is x.
4. Result: \( x\sqrt{1+x^2} \).

Teacher's Note:
1. Use a right triangle with legs x and 1. The expression simplifies in two clean steps.

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

Explanation:
1. \( \sin^{-1}u+\cos^{-1}u=\frac\pi2 \) is defined only for \( |u|\le1 \).
2. \( u=(x-1)^2+1\ge1 \), so u = 1 and x = 1.
3. The equation becomes \( a+\frac\pi2=0 \), so \( a=-\frac\pi2 \).

Teacher's Note:
1. The domain restriction gives the only possible x. Then solve for a.

Answer: (c) 8

Explanation:
1. Let \( s=\sin^{-1}x \). The expression is \( s^2+(\frac\pi2-s)^2=2s^2-\pi s+\frac{\pi^2}4 \).
2. Its minimum is at \( s=\frac\pi4 \) (which is in range), with value \( \frac{\pi^2}8 \).
3. So k = 8.

Teacher's Note:
1. Convert to a quadratic in sin⁻¹ x and complete the square. The vertex lies inside the allowed range.

Answer: (c) -1

Explanation:
1. Let \( t=\tan^{-1}x \). Then \( t^2+(\frac\pi2-t)^2=\frac{5\pi^2}8 \).
2. \( 16t^2-8\pi t-3\pi^2=0 \), so \( t=\frac{3\pi}4 \) or \( t=-\frac\pi4 \).
3. \( \frac{3\pi}4 \) is outside \( (-\frac\pi2,\frac\pi2) \), so \( \tan^{-1}x=-\frac\pi4 \) and x = -1.

Teacher's Note:
1. Use cot⁻¹ x = π/2 - tan⁻¹ x. Reject the root outside the principal range.

Answer: (b) 0 (the equation has no real solution)

Explanation:
1. The equation requires \( \sin^{-1}(x^2-6x+12)=\frac\pi3 \), so \( x^2-6x+12=\frac{\sqrt3}2 \).
2. But \( x^2-6x+12=(x-3)^2+3\ge3 \), which is outside the domain [-1, 1] of sin⁻¹.
3. So there is no real solution.

Teacher's Note:
1. Check the argument against the domain before solving. Complete the square to see its minimum.

Answer: (c) \left[-1,\frac1{\√2}\right)

Explanation:
1. \( \cos^{-1}x=\frac\pi2-\sin^{-1}x \), so the inequality becomes \( \sin^{-1}x<\frac\pi4 \).
2. Since sin⁻¹ is increasing, \( x<\frac1{\sqrt2} \).
3. With the domain \( x\ge-1 \), \( x\in[-1,\frac1{\sqrt2}) \).

Teacher's Note:
1. Use the complementary identity to get a single inequality. Remember the lower end of the domain.