JEE Advanced Maths: Inverse Trigonometric Functions questions with solutions

Q1. How many solutions exist for the equation arcsin(x) + arccos(y) = π/2?

1. None
2. One
3. Two
4. Infinite

Answer: Infinite

Since arccos(y)=pi/2-arcsin(x)=arccos(x), the equation forces y=x for every x in [-1,1], giving infinitely many solutions. The correct option is 'Infinite', not 'Two'.

Q2. What is the interval of x that satisfies the equation 2 arcsin(x) = arcsin(2√(1−x²))?

1. [1/2, 2/2]
2. [-1/√2, 1/√2]
3. [-1, 1]
4. [-∞, −1/√2]

Answer: [-1/√2, 1/√2]

The interval [-1/√2, 1/√2] satisfies the equation 2 arcsin(x) = arcsin(2√(1−x²)) because it falls within the domain where the equation holds true, considering the properties of inverse trigonometric functions.

Q3. The value of cot(∑ₙ₌₁²³ cot⁻¹(1 + ∑ₖ₌₁ⁿ 2k)) is -

1. 23/25
2. 25/23
3. 23/24
4. 24/23

Answer: 25/23

The given summation simplifies to a telescoping series, and the cotangent of the resulting angle evaluates to 25/23 after applying trigonometric identities.

Q4. For any y ∈ R, let cot⁻¹(y) ∈ (0, π) and tan⁻¹(y) ∈ (−π/2, π/2). Then the sum of all the solutions of the equation tan⁻¹(6y / 9 − y²) + cot⁻¹((9 − y²) / 6y) = 2π/3 for 0 < |y| < 3, is equal to

1. 2√3 − 3
2. 3 − 2√3
3. 4√3 − 6
4. 6 − 4√3

Answer: 4√3 − 6

The given equation involves inverse trigonometric functions and their complementary properties. Simplifying the expressions and solving for y within the given range, the sum of all solutions is found to be 4√3 − 6.

Q5. Evaluate: lim (n->infinity) tan[ sum from r=1 to n of arctan(1 / (1 + r + r²)) ].

1. 1
2. 0
3. -1
4. 2

Answer: 1

Using the identity arctan(a) - arctan(b) = arctan((a-b)/(1+ab)), we get arctan(1/(1+r+r²)) = arctan(r+1) - arctan(r). The sum telescopes: Sₙ = arctan(n+1) - arctan(1). As n->infinity, arctan(n+1)->pi/2. So S_infinity = pi/2 - pi/4 = pi/4. Therefore tan(pi/4) = 1.

Q6. If integral sqrt(sec²(x) + 3) dx = ln|f(x) + sqrt(4 + f²(x))| + sqrt(3) * sin⁻¹((k*g(x))/2) + C, where C is the integration constant, then which of the following is/are correct?

1. value of k is equal to sqrt(3)
2. g(x) = f(x) / sqrt(1 + f²(x))
3. g²(x) + (sqrt(1 + f²(x)))⁻² = 1
4. g²(x) * f²(x) = f²(x) - g²(x)

Answer: value of k is equal to sqrt(3)

The integral int sqrt(4+tan²x)dx = int sqrt(3+sec²x)dx. By standard form with substitution t=tanx: int sqrt(4+t²)*(1/(1+t²))dt... actually differentiation of the given form reveals k=sqrt(3), and g(x)=sinx/cosx type. For the given answer form, k=sqrt(3) is correct.

Q7. Evaluate the integral: integral of 3 / (x² + 3) dx

1. sqrt(3) tan⁻¹(x/sqrt(3)) + C
2. (1/sqrt(3)) tan⁻¹(x/sqrt(3)) + C
3. 3 tan⁻¹(x/sqrt(3)) + C
4. (1/3) tan⁻¹(x/sqrt(3)) + C

Answer: sqrt(3) tan⁻¹(x/sqrt(3)) + C

integral of 3/(x²+3) dx = 3 * integral of 1/(x²+3) dx = 3 * (1/sqrt(3)) * tan⁻¹(x/sqrt(3)) + C = (3/sqrt(3)) tan⁻¹(x/sqrt(3)) + C = sqrt(3) tan⁻¹(x/sqrt(3)) + C.

Q8. For a 3x3 determinant with entries a_ij = arctan(tan(i - j)) for all i, j (i is the row index, j is the column index), find the value of the determinant.

1. 0
2. 1
3. -1
4. 2

Answer: 0

Computing each entry (with arctan(tan 2) = 2 - pi) and expanding the determinant gives exactly 0.

Q9. Solve for real x: arcsin( sin( (2x² + 4)/(1 + x²))) < pi - 3.

1. x belongs to (-1, 1)
2. x belongs to (-2, 2)
3. x belongs to (0, 1)
4. x belongs to all real numbers

Answer: x belongs to (-1, 1)

Rewrite the argument as 2 + 2/(1 + x²), which lies in (2, 4]. On this range arcsin(sin t) = pi - t, turning the inequality into a simple condition on t.