JEE Mathematics Inverse Trigonometric Functions MCQs Set 13

Answer: (b) \( [-2,0]\cup[2,4] \)

Explanation:
1. We need ||x - 1| - 2| <= 1, so 1 <= |x - 1| <= 3.
2. |x - 1| >= 1 gives x <= 0 or x >= 2; |x - 1| <= 3 gives -2 <= x <= 4.
3. Together: x in [-2, 0] U [2, 4].

Teacher's Note:
1. Remove the outer modulus first: -1 <= |x - 1| - 2 <= 1.
2. Then split the absolute value.

Answer: (b) {0, -3}

Explanation:
1. The first term needs x(x + 3) >= 0.
2. The second needs 0 <= x^2 + 3x + 1 <= 1, so x^2 + 3x <= 0.
3. Both hold only when x^2 + 3x = 0, so x = 0 or x = -3.

Teacher's Note:
1. Write x^2 + 3x = x(x + 3) in both conditions.
2. Both inequalities together force equality.

Answer: (b) {-1, 1}

Explanation:
1. cos^-1 x needs |x| <= 1.
2. cosec^-1 x needs |x| >= 1.
3. Both together give |x| = 1, so the domain is {-1, 1}.

Teacher's Note:
1. Intersect the domains of the three functions.
2. cot^-1 is defined for all x.

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

Explanation:
1. sin^-1 needs |x| <= 1 and sec^-1 needs |x| >= 1, so x = 1 or x = -1.
2. f(1) = pi/2 + pi/4 + 0 = 3pi/4.
3. f(-1) = -pi/2 - pi/4 + pi = pi/4. So the range is the two-element set {pi/4, 3pi/4}.

Teacher's Note:
1. Find the domain first; it has only two points.
2. sec^-1(-1) = pi.

Answer: (b) x is a multiple of pi

Explanation:
1. cosec^-1(t) is real only for |t| >= 1.
2. |cos x| >= 1 means cos x = +1 or -1.
3. This holds exactly when x is a multiple of pi.

Teacher's Note:
1. Recall the domain of cosec^-1.
2. |cos x| <= 1 always.

Answer: (b) \( \left[-\frac14,\frac12\right] \)

Explanation:
1. We need -1 <= 2x <= 1 for sin^-1(2x) to exist.
2. We also need sin^-1(2x) >= -pi/6, i.e. 2x >= -1/2, so x >= -1/4.
3. Combining: -1/4 <= x <= 1/2.

Teacher's Note:
1. sin^-1 is increasing, so the inequality keeps its direction.
2. sin(-pi/6) = -1/2.

Answer: (b) x = 2

Explanation:
1. tan(pi/4) = 1 and sin^-1(1) = pi/2.
2. So sin^-1(sqrt(3)/x) = pi/2 - pi/6 = pi/3.
3. Hence sqrt(3)/x = sin(pi/3) = sqrt(3)/2, giving x = 2.

Teacher's Note:
1. Simplify the first term.
2. Take the sine of both sides.

Answer: (b) 2

Explanation:
1. The domain needs 4 - x^2 >= 0 and -1 <= x^2 - 5 <= 1.
2. Here cot^-1 of a non-negative number is at most pi/2 and cos^-1 is at most pi, so the sum is at most 3pi/2, with equality only when both are maximal.
3. This requires sqrt(4 - x^2) = 0 and x^2 - 5 = -1, i.e. x^2 = 4. So x = 2 or -2: two solutions.

Teacher's Note:
1. Find the largest possible value of each term.
2. Equality in a sum of maxima needs both maxima to occur.

Answer: (b) \( 2\pi-x \)

Explanation:
1. cos^-1(cos x) = x only for x in [0, pi].
2. For x in [pi, 2pi], 2pi - x lies in [0, pi] and cos(2pi - x) = cos x.
3. So cos^-1(cos x) = 2pi - x.

Teacher's Note:
1. Bring the angle into [0, pi] keeping the cosine.
2. cos(2pi - x) = cos x.

Answer: (b) \( \frac\pi3 \)

Explanation:
1. cos^-1 x = pi/2 - sin^-1 x and cos^-1 y = pi/2 - sin^-1 y.
2. Adding: cos^-1 x + cos^-1 y = pi - (sin^-1 x + sin^-1 y).
3. This is pi - 2pi/3 = pi/3.

Teacher's Note:
1. Use sin^-1 t + cos^-1 t = pi/2 for both x and y.
2. Add the two identities.

Answer: (b) \( \frac\pi4\le\theta\le\frac\pi2 \)

Explanation:
1. sin^-1 x + cos^-1 x = pi/2, so theta = pi/2 - tan^-1 x.
2. The domain of the inverse sine and cosine restricts x to [0, 1].
3. For x in [0, 1], tan^-1 x lies in [0, pi/4], so theta lies in [pi/4, pi/2].

Teacher's Note:
1. Remember the restriction x <= 1.
2. tan^-1 is increasing.

Answer: (b) 2

Explanation:
1. For |ln x| >= 0, cot^-1(|ln x|) = pi/2 - tan^-1(|ln x|), so the equation becomes tan^-1(e^-x) = tan^-1(|ln x|).
2. Hence e^-x = |ln x|, with x > 0.
3. On (0, 1) the left side is positive and finite while -ln x goes from infinity down to 0: exactly one crossing. On (1, infinity) ln x rises from 0 while e^-x falls from 1/e: exactly one crossing. Total 2 solutions.

Teacher's Note:
1. Convert cot^-1 of a non-negative number into tan^-1.
2. Sketch e^-x and |ln x|.

Answer: (b) \( \frac{-3}{4} \)

Explanation:
1. Let beta = sin^-1(3/5), so sin beta = 3/5, cos beta = 4/5, tan beta = 3/4.
2. cos^-1(3/5) = pi/2 - sin^-1(3/5) = pi/2 - beta.
3. The angle is 2 beta + pi/2 - beta = beta + pi/2, and cot(beta + pi/2) = -tan beta = -3/4.

Teacher's Note:
1. Note that 3/5 and 4/5 are the sides of the 3-4-5 triangle.
2. The two angles are complementary.

Answer: (b) Statement 1 is true and statement 2 is false

Explanation:
1. tan^2(sec^-1 2) = sec^2 - 1 = 4 - 1 = 3; cot^2(cosec^-1 3) = cosec^2 - 1 = 9 - 1 = 8; the sum is 11, so statement 1 is true.
2. The correct identities are sec^2 - tan^2 = 1 and cosec^2 - cot^2 = 1, not the sums, so statement 2 is false.

Teacher's Note:
1. Use tan^2 = sec^2 - 1 and cot^2 = cosec^2 - 1.
2. Compare statement 2 with the true identities.

Answer: (b) pi

Explanation:
1. x^3 + 3x is strictly increasing, so there is exactly one real root; since tan 2 < 0 (2 radians lies in the second quadrant), alpha < 0.
2. With cot^-1 t = pi/2 - tan^-1 t, the sum cot^-1 alpha + cot^-1(1/alpha) = pi - (tan^-1 alpha + tan^-1(1/alpha)).
3. For alpha < 0, tan^-1 alpha + tan^-1(1/alpha) = -pi/2, so the sum is 3pi/2 and the expression equals 3pi/2 - pi/2 = pi.

Teacher's Note:
1. Find the sign of the root first.
2. cot^-1 takes values in (0, pi), and cot^-1 t = pi/2 - tan^-1 t.

Answer: (b) \( \text{maximum value of }x^2+y^2\text{ is }\frac{49}3 \)

Explanation:
1. Let a = sqrt(x)/2 in [0, 1]. Then sqrt(1 - x/4) = sqrt(1 - a^2), so sin^-1 a + sin^-1 sqrt(1 - a^2) = pi/2.
2. Hence tan^-1 y = 2pi/3 - pi/2 = pi/6 and y = 1/sqrt(3), so y^2 = 1/3.
3. With 0 <= x <= 4: x^2 + y^2 ranges from 1/3 (at x = 0) to 16 + 1/3 = 49/3 (at x = 4).

Teacher's Note:
1. The two inverse sines are complementary when a is non-negative.
2. Find the allowed values of x from the square roots.

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

Explanation:
1. For x > 0, tan^-1 x + tan^-1(1/x) = pi/2.
2. Both terms are odd functions of x, so for x < 0 the sum changes sign.
3. Hence the value is -pi/2.

Teacher's Note:
1. Check with x = -1: the sum is -pi/4 - pi/4.
2. tan^-1 is an odd function.

Answer: (b) \( \frac1{\sqrt5} \)

Explanation:
1. sin^-1 x = pi/2 - cot^-1(1/2) = tan^-1(1/2).
2. So sin(sin^-1 x) = sin(tan^-1(1/2)): in the right triangle with legs 1 and 2 the hypotenuse is sqrt(5).
3. Therefore x = 1/sqrt(5).

Teacher's Note:
1. Use pi/2 - cot^-1 t = tan^-1 t.
2. Draw the triangle with opposite 1 and adjacent 2.

Answer: (b) \( \frac{-7}{17} \)

Explanation:
1. tan(2 tan^-1(1/5)) = (2/5)/(1 - 1/25) = 5/12.
2. tan(A - pi/4) = (5/12 - 1)/(1 + 5/12) = (-7/12)/(17/12).
3. So the value is -7/17.

Teacher's Note:
1. Use the double angle formula for tangent.
2. Then the subtraction formula with tan(pi/4) = 1.

Answer: (b) \( \left[0,\frac\pi4\right] \)

Explanation:
1. Put u = x/sqrt(3). The fraction (sqrt(3)x - 3x)/(3 sqrt(3) + x^2) simplifies to u(1 - sqrt(3))/(sqrt(3) + u^2), which is the tangent of tan^-1(u/sqrt(3)) - tan^-1(u).
2. So the first term is tan^-1(u/sqrt(3)) - tan^-1 u, and adding tan^-1 u gives f(x) = tan^-1(u/sqrt(3)) = tan^-1(x/3).
3. For 0 <= x <= 3, x/3 lies in [0, 1], so f lies in [0, pi/4].

Teacher's Note:
1. Write the fraction as a difference of two inverse tangents.
2. Substitute u = x/sqrt(3).

Answer: (b) Both are true and 2 does not explain 1

Explanation:
1. Statement 1: the sum is pi/2 only when (a/x)(b/x) = 1 with x > 0, giving x = sqrt(ab): true.
2. Statement 2: tan of the sum is (m(n + m) + n(n - m)) / (n(n + m) - m(n - m)) = (m^2 + n^2)/(m^2 + n^2) = 1, so the sum is pi/4: true.
3. The two statements are about different identities, so statement 2 does not explain statement 1.

Teacher's Note:
1. Product of the tangents must be 1 for a sum of pi/2.
2. Use the addition formula for tangents in statement 2.

Answer: (b) \( 4\sin^2\alpha \)

Explanation:
1. Let A = cos^-1 x and B = cos^-1(y/2), so cos A = x, cos B = y/2 and A - B = alpha.
2. cos(A - B) = cos A cos B + sin A sin B gives cos alpha = x(y/2) + sqrt(1 - x^2) sqrt(1 - y^2/4).
3. Rearranging and squaring: (cos alpha - xy/2)^2 = (1 - x^2)(1 - y^2/4), which simplifies to x^2 + y^2/4 - xy cos alpha = sin^2 alpha. Multiplying by 4 gives the result 4 sin^2 alpha.

Teacher's Note:
1. Isolate the square-root product before squaring.
2. The cross terms in x^2 y^2 cancel.

Answer: (b) \( \frac2x \)

Explanation:
1. Let t = (1/2) cos^-1 x. tan(pi/4 + t) + tan(pi/4 - t) = (1 + tan t)/(1 - tan t) + (1 - tan t)/(1 + tan t).
2. This equals 2(1 + tan^2 t)/(1 - tan^2 t) = 2/cos(2t).
3. cos(2t) = cos(cos^-1 x) = x, so the value is 2/x.

Teacher's Note:
1. Combine the two fractions.
2. (1 + tan^2 t)/(1 - tan^2 t) = 1/cos 2t.

Answer: (b) \( \frac\pi2+\frac x2 \)

Explanation:
1. For pi/2 < x < pi: sqrt(1 + sin x) = sin(x/2) + cos(x/2) and sqrt(1 - sin x) = sin(x/2) - cos(x/2), because x/2 lies in (pi/4, pi/2).
2. The fraction becomes 2 sin(x/2) / (-2 cos(x/2)) = -tan(x/2).
3. cot^-1(-tan(x/2)) = pi - cot^-1(tan(x/2)) = pi - (pi/2 - x/2) = pi/2 + x/2.

Teacher's Note:
1. Write 1 +- sin x as (sin(x/2) +- cos(x/2))^2.
2. Mind the sign of sin(x/2) - cos(x/2) for the given range.

Answer: (b) \( \left(\frac1{\sqrt2},1\right] \)

Explanation:
1. With range [pi/2, 3pi/2], sin^-1 x = y means sin y = x with y in [pi/2, 3pi/2], where sine decreases from 1 to -1.
2. The condition y < 3pi/4 means y in [pi/2, 3pi/4).
3. Then x = sin y lies in (sin(3pi/4), sin(pi/2)] = (1/sqrt(2), 1].

Teacher's Note:
1. Sine is decreasing on [pi/2, 3pi/2].
2. sin(3pi/4) = 1/sqrt(2).

Answer: (b) \( \pi-x \)

Explanation:
1. With range [pi/2, 3pi/2] - {pi}, we need an angle in that range with the same cosecant as x.
2. For x in (0, pi/2), the angle pi - x lies in (pi/2, pi) and has the same sine. For x in (-pi/2, 0), the angle pi - x lies in (pi, 3pi/2) and again has sine equal to sin x.
3. So cosec^-1(cosec x) = pi - x throughout.

Teacher's Note:
1. Use sin(pi - x) = sin x.
2. Check that pi - x lies in the new range for both signs of x.

Answer: (b) \( \left[\frac{3\pi}4,\frac{5\pi}4\right] \)

Explanation:
1. With range (pi/2, 3pi/2), tan^-1 t = pi + arctan t, where arctan is the usual principal value.
2. For t = -x in [-1, 1], arctan t lies in [-pi/4, pi/4].
3. So tan^-1(-x) lies in [pi - pi/4, pi + pi/4] = [3pi/4, 5pi/4].

Teacher's Note:
1. Relate the new tan^-1 to the usual one by adding pi.
2. The usual arctan of [-1, 1] is [-pi/4, pi/4].