Exams › JEE Main › Maths › Inverse Trigonometric Functions
100 questions with worked solutions.
Q1. For the function f(x) = log(x + √(x² + 1)), which of the following describes it correctly?
Answer: It is an odd function
f(-x) = log(-x + sqrt(x^2+1)) = log(1/(x + sqrt(x^2+1))) = -log(x + sqrt(x^2+1)) = -f(x). Since f(-x) = -f(x), the function is odd.
Q2. For the function f(x) = log((1 + x)/(1 − x)), which of the following relations is satisfied?
Answer: f(x1) + f(x2) = f((x1 + x2)/(1 + x1x2))
Since f(x) = log((1+x)/(1-x)) = 2 artanh(x), the addition formula gives f(x1) + f(x2) = f((x1+x2)/(1+x1*x2)). Direct substitution confirms this identity.
Q3. Find the domain of the function f(x)=sin⁻¹{log2(1/2 x²)}.
Answer: (-2,-1]∪[1,2]
For arcsin to be defined: -1 <= log2(x^2/2) <= 1, i.e. 1/2 <= x^2/2 <= 2, giving 1 <= x^2 <= 4, so 1 <= |x| <= 2. Including the endpoints, the domain is [-2,-1] U [1,2].
Q4. For which values of x does the inequality sin⁻¹(sin 5) > x² − 4x hold?
Answer: x ∈ (2 − √(9 − 2π), 2 + √(9 − 2π))
Since 5 lies in (3π/2, 2π), sin^-1(sin5)=5-2π. The inequality x^2-4x < 5-2π becomes x^2-4x-(5-2π)<0, i.e. x^2-4x-(9-2π)+4<0, whose roots are 2 ± sqrt(9-2π). So x ∈ (2 - sqrt(9-2π), 2 + sqrt(9-2π)).
Q5. Determine the range of the function f(x) = sin⁻¹x + tan⁻¹x + sec⁻¹x.
Answer: {π/4, 3π/4}
sin^-1 needs |x|<=1 and sec^-1 needs |x|>=1, so the domain is just x=+-1. At x=1: pi/2+pi/4+0 = 3pi/4. At x=-1: -pi/2-pi/4+pi = pi/4. Range = {pi/4, 3pi/4}.
Q6. If A = tan⁻¹ ((x√(3))/(2K-x)) and B = tan⁻¹ ((2x-K)/(K√(3))), then what is the value of A - B?
Answer: 30°
Writing tan A = x*sqrt3/(2K-x) and tan B = (2x-K)/(K*sqrt3), then tan(A-B) = (tanA - tanB)/(1+tanA tanB) reduces to 1/sqrt3, so A - B = 30 degrees.
Q7. If the inequality |cos⁻¹ ((1-x²)/(1+x²)) |<(π)/(3) holds, then which of the following is true?
Answer: x ∈ (-(1)/(√(3)), (1)/(√(3)))
Using arccos((1-x^2)/(1+x^2)) = 2 arctan|x|, the condition 2 arctan|x| < pi/3 gives arctan|x| < pi/6, so |x| < tan(pi/6) = 1/sqrt(3). Since the bound is strict, x lies in the open interval (-1/sqrt(3), 1/sqrt(3)).
Q8. If cot⁻¹(n/π) < π/6, where n ∈ N, then the greatest possible value of n is:
Answer: None of these
cot^-1(n/pi) < pi/6 means n/pi > cot(pi/6) = sqrt(3), so n > pi*sqrt(3) ~ 5.44. Since n can be arbitrarily large, there is no greatest value -> None of these.
Q9. If ∑_(i=1)²ⁿ cos⁻¹(x_i)=0, then the value of ∑_(i=1)²ⁿ x_i is
Answer: 2n
Each cos^-1(x_i) lies in [0, pi] and is >= 0. A sum of 2n non-negative terms equals 0 only if every term is 0, so each x_i = 1. Therefore sum of x_i = 2n.
Q10. If cos⁻¹ ((1-x²)/(1+x²))+cos⁻¹ ((1-y²)/(1+y²))=(π)/(2), with xy<1, which relation must hold?
Answer: x + y + xy = 1
Since arccos((1-x^2)/(1+x^2))=2 arctan x, the equation becomes 2 arctan x + 2 arctan y = pi/2, so arctan x + arctan y = pi/4. Then (x+y)/(1-xy)=1, giving x+y+xy=1.
Q11. If cos⁻¹x + cos⁻¹y = (2π)/(7), then sin⁻¹x + sin⁻¹y equals
Answer: (5π)/(7)
Since sin^-1 t + cos^-1 t = pi/2 for each variable, sin^-1 x + sin^-1 y = pi - (cos^-1 x + cos^-1 y) = pi - 2pi/7 = 5pi/7.
Q12. If sin⁻¹ a + sin⁻¹ b + sin⁻¹ c = π, then what is the value of a√(1-a²) + b√(1-b²) + c√(1-c²)?
Answer: 2abc
Put A=asin a, B=asin b, C=asin c with A+B+C=pi. Each term a*sqrt(1-a^2)=sinA cosA=(1/2)sin2A. Sum=(1/2)(sin2A+sin2B+sin2C)=(1/2)(4 sinA sinB sinC)=2abc.
Q13. Let α = sin⁻¹(√3/2) + sin⁻¹(1/3) and β = cos⁻¹(√3/2) + cos⁻¹(1/3). Which of the following is true?
Answer: α < β
alpha = pi/3 + arcsin(1/3), beta = pi/6 + arccos(1/3). Using arcsin(1/3)+arccos(1/3)=pi/2, alpha-beta = 2*arcsin(1/3) - pi/3 = -0.37 < 0. So alpha < beta (and alpha+beta = pi, not 2pi).
Answer: 1
The series for the inverse sine converges to arcsin(x) and the series for the inverse cosine converges to arccos(x). Since arcsin(x) + arccos(x) equals π/2 for any x in the interval (0, 1), the correct value of x that satisfies this condition is 1.
Answer: 2
The sum of the inverse trigonometric functions simplifies to 2π, which corresponds to k=2. Each term contributes to the total angle measure, and their combined value indicates that k must equal 2.
Answer: 2π
Reading the terms as cot^-1((ab+1)/(a-b)) + cot^-1((bc+1)/(b-c)) + cot^-1((ca+1)/(c-a)) with 0<a<b<c, the first two denominators are negative. Using cot^-1((xy+1)/(y-x)) = cot^-1 x - cot^-1 y and the principal range (0,pi), the negative-argument terms each contribute an extra pi, so the telescoping sum totals 2*pi.
Answer: 2
The equation sin⁻¹x + sin⁻¹y + sin⁻¹z = π implies that x, y, and z are related in such a way that they can be expressed in terms of sine functions that sum to π. This leads to a specific relationship among their powers, resulting in the identity simplifying to a form where k must equal 2 to maintain equality.
Answer: π/2
The infinite sum converges to (C0)/(2) because each term in the series represents the inverse sine of a value that approaches the limit of 1 as n increases, leading to the cumulative angle approaching (C0)/(2).
Q19. If sin⁻¹(x-1)+cos⁻¹(x-3)+tan⁻¹ ((x)/(2-x²))=cos⁻¹k+π, then what is the value of k?
Answer: 1/√2
sin^-1(x-1) needs x in [0,2]; cos^-1(x-3) needs x in [2,4]; their intersection is x=2. At x=2: pi/2 + pi + tan^-1(2/(2-4)) = pi/2 + pi - pi/4 = 5pi/4. Setting = cos^-1(k) + pi gives cos^-1(k) = pi/4, so k = 1/sqrt(2).
Q20. Evaluate the value of tancos⁻¹(1/√(82)) - sin⁻¹(5/√(26)).
Answer: 2/23
The expression involves the tangent of the difference of two inverse trigonometric functions. By using the identities for cosine and sine, we can simplify the expression to find that the value of tan(cos⁻¹(1/sqrt{82}) - sin⁻¹(5/sqrt{26})) equals 2/23.
Q21. If (a)/(b)=tan x>1 and -(π)/(2)<x<(π)/(2), then tan⁻¹ ((acos x-bsin x)/(bcos x+asin x)) is equal to:
Answer: tan⁻¹(a/b)-x
Dividing through gives (a/b - tan x)/(1 + (a/b)tan x) = tan(tan^-1(a/b) - x), so the expression equals tan^-1(a/b) - x. (Note the correct value appears at both index 1 and index 3; stored index 0 'tan^-1(a/b - x)' is wrong.)
Q22. Evaluate the expression: tan( tan⁻¹x + tan⁻¹y + tan⁻¹z) − cot( cot⁻¹x + cot⁻¹y + cot⁻¹z).
Answer: 0
Let S = tan^-1 x + tan^-1 y + tan^-1 z. Then cot^-1 x + cot^-1 y + cot^-1 z = 3pi/2 - S, and cot(3pi/2 - S) = tan S. So the expression is tan S - tan S = 0.
Answer: Statement 1 is true, Statement 2 is true, and Statement 2 correctly explains Statement 1.
For x,y in (0,1) with x>y: asin(x)>x>y>atan(y), so Statement 2 is true. With x=1/sqrt(e) ~ 0.61 > y=1/e ~ 0.37, Statement 2 directly gives Statement 1, so it correctly explains it -> option (a).
Q24. Find the differential coefficient of tan⁻¹(2x/(1-x²)) with respect to sin⁻¹(2x/(1+x²)).
Answer: 1
tan^-1(2x/(1-x^2)) = 2 arctan x and sin^-1(2x/(1+x^2)) = 2 arctan x for |x|<1, so each differentiates to 2/(1+x^2). The required derivative is their ratio, which is 1.
Q25. If f(x)=cot⁻¹ ((x^x-x^(-x))/(2)), then the value of f'(1) is
Answer: -1
Let u(x)=(x^x - x^-x)/2, with u(1)=0 and u'(1)=( (x^x(1+ln x)) + (x^-x(1+ln x)) )/2 at x=1 = (1+1)/2 = 1. Then f'(x) = -u'/(1+u^2), so f'(1) = -1/(1+0) = -1.
Q26. For which values of a does the equation sin⁻¹x = 2 sin⁻¹a admit a solution?
Answer: |a| ≤ 1/√2
The equation sin⁻¹x = 2 sin⁻¹a requires that the values of sin⁻¹a must be within the range of the arcsine function, which is defined for inputs between -1 and 1. The maximum value of 2 sin⁻¹a occurs when |a| is at most 1/√2, ensuring that the output remains within the valid range of the arcsine function.
Q27. If cos⁻¹x - cos⁻¹(y/2) = α, then the value of 4x² - 4xycosα + y² is
Answer: 4sin²α
The expression can be derived using the cosine of the angle difference identity and the properties of the cosine function. By substituting the values and simplifying, it can be shown that the result equals 4 times the square of the sine of the angle alpha, confirming option C as correct.
Answer: (−π/2, π/2)
The function f(x) = tan⁻¹((−1 + 2x)/(1 − x²)) is defined for x in the interval (−1, 1) and maps to the interval (−π/2, π/2) because the arctangent function outputs values in this range. For f to be both injective (one-to-one) and surjective (onto), B must cover the entire range of the function, which is (−π/2, π/2).
Q29. If sin⁻¹(x/5) + cosec⁻¹(5/4) = π/2, then what is the value of x?
Answer: 3
The equation states that the sum of the inverse sine of x/5 and the inverse cosecant of 5/4 equals π/2. This implies that sin⁻¹(x/5) is the complementary angle to cosec⁻¹(5/4), which means sin⁻¹(x/5) = π/2 - cosec⁻¹(5/4). Since cosec⁻¹(5/4) corresponds to sin(θ) = 4/5, we find that x/5 must equal 3/5, leading to x = 3.
Answer: [0, π/2)
The function f(x) is defined where each component is valid: 4^(-x²) is defined for all x, cos⁻¹((x/2) - 1) requires (x/2) - 1 to be in [0, 1], which restricts x to [0, 2], and log(cos x) is defined where cos x > 0, limiting x to (−π/2, π/2). The intersection of these conditions gives the interval [0, π/2).
Q31. Evaluate the expression cot⁻¹(cosec⁻¹(5/3) + tan⁻¹(2/3)).
Answer: 6/17
cosec^-1(5/3) is an angle A with sin A=3/5, so tan A=3/4; tan B=2/3. tan(A+B)=(3/4+2/3)/(1-(3/4)(2/3))=(17/12)/(1/2)=17/6. Hence cot(A+B)=6/17.
Q32. If tan⁻¹ y = tan⁻¹ x + tan⁻¹ ((2x)/(1-x²)), with |x| < (1)/(√(3)), then one possible value of y is:
Answer: (3x-x³)/(1-3x²)
For |x|<1, tan-1(2x/(1-x^2)) = 2 tan-1 x, so the RHS = 3 tan-1 x = tan-1((3x-x^3)/(1-3x^2)). Hence y = (3x-x^3)/(1-3x^2).
Answer: 9/(1+9x³)
The derivative of the function involves applying the chain rule and the derivative of the arctangent function, leading to a simplification that reveals the factor of sqrt{x} and the function g(x) as 9/(1+9x³), which matches the required form.
Q34. If cos⁻¹ ((2)/(3x))+cos⁻¹ ((3)/(4x))=(π)/(2), where x>(3)/(4), then the value of x is:
Answer: (√(145))/(12)
From cos^-1(2/3x)+cos^-1(3/4x)=pi/2, we get cos^-1(2/3x)=sin^-1(3/4x), so 2/(3x)=sqrt(1-9/(16x^2)). Squaring: 4/(9x^2)+9/(16x^2)=1 -> 145/(144x^2)=1 -> x=sqrt(145)/12.
Q35. Let y=sec(tan⁻¹x). The value of (dy)/(dx) when x=1 is:
Answer: (1)/(√(2))
To find ( rac{dy}{dx}) for ( y = ext{sec}( an⁻¹x)), we first express ( y) in terms of ( x) using trigonometric identities. By differentiating and evaluating at ( x = 1, we find that ( rac{dy}{dx} = rac{1}{ ext{sec}( an⁻¹(1)) imes ext{tan}( an⁻¹(1))} = rac{1}{ rac{ ext{sqrt}(2)}{2} imes 1} = rac{1}{ rac{ ext{sqrt}(2)}{2}} = rac{1}{ ext{sqrt}(2)}), confirming option A as correct.
Q36. The function f(x) = tan⁻¹(sin x + cos x) is an increasing function in
Answer: (-π/2, π/4)
The function f(x) = tan⁻¹(sin x + cos x) is increasing in the interval (-π/2, π/4) because the derivative of the function is positive in this range, indicating that the function's output increases as x increases.
Answer: (0, 2π/3)
For x in (0,pi/2), f(x)=arctan(tan(pi/4 + x/2))=pi/4 + x/2, so f'(x)=1/2. At x=pi/6, f=pi/3 and the normal slope is -2: y-pi/3=-2(x-pi/6). Setting x=0 gives y=pi/3+pi/3=2pi/3, so it passes through (0, 2pi/3).
Q38. If y = sec(tan⁻¹ x), then dy/dx at x = 1 is equal to
Answer: 1/sqrt(2)
To find dy/dx, we first differentiate y = sec(tan⁻¹ x) using the chain rule and implicit differentiation. At x = 1, the calculations yield dy/dx = 1/sqrt(2), confirming that this is the correct answer.
Q39. If for x ∈ (0, 1/4), the derivative of tan⁻¹((6x√x)/(1 − 9x³)) is √x · g(x), then g(x) equals:
Answer: 9/(1 + 9x³)
The correct option is derived from applying the chain rule and quotient rule to differentiate the given function, leading to the simplification that results in g(x) being expressed as 9/(1 + 9x³), which matches the form required by the problem.
Q40. If x = √(2^(cosec⁻¹ t)) and y = √(2^(sec⁻¹ t)) (|t| ≥ 1), then dy/dx is equal to -
Answer: −y/x
x^2 = 2^(cosec^-1 t) and y^2 = 2^(sec^-1 t), and cosec^-1 t + sec^-1 t = pi/2, so x^2 y^2 = 2^(pi/2) = constant. Differentiating ln(x^2) + ln(y^2) = const gives 2/x + (2/y)(dy/dx) = 0, so dy/dx = -y/x.
Q41. If f(x) = sin⁻¹((2x 3ˣ)/(1 + 9ˣ)), then f′(-1/2) equals -
Answer: √3 loge √3
With t=3^x, f(x)=sin^-1(2*3^x/(1+9^x))=2 arctan(3^x), so f'(x)=2*3^x ln3/(1+9^x). At x=-1/2: 3^x=1/sqrt(3), 9^x=1/3, giving f'=(sqrt(3)/2)ln3 = sqrt(3) loge(sqrt(3)).
Q42. All x satisfying the inequality (cot⁻¹ x)² − 7(cot⁻¹ x) + 10 > 0, lie in the interval:
Answer: (cot 2, ∞)
The inequality represents a quadratic function in terms of cot⁻¹ x, and solving it reveals that the roots are at cot 2 and cot 5. The quadratic opens upwards, indicating that the values satisfying the inequality are those greater than the larger root, which is cot 2.
Answer: 4 sin² α
The expression 4x² - 4xy cos α + y² can be rewritten using the identity for the cosine of the angle difference, leading to the conclusion that it simplifies to 4 sin² α, which is consistent with the relationship between the angles and their sine values.
Q44. The value of sin⁻¹(12/13) − sin⁻¹(3/5) is equal to:
Answer: π/2 − sin⁻¹(56/65)
The expression sin⁻¹(12/13) − sin⁻¹(3/5) can be simplified using the identity sin⁻¹(a) - sin⁻¹(b) = sin⁻¹(a√(1-b²) - b√(1-a²)), which leads to the result π/2 − sin⁻¹(56/65), confirming that this option correctly represents the difference of the two inverse sine values.
Answer: (4) is a singleton
The equation tan⁻¹(2x) + tan⁻¹(3x) = π/4 has a unique solution for x ≥ 0 because the sum of two increasing functions can only equal a constant value at one point in this domain.
Q46. If f(x) = logₑ((1 - x)/(1 + x)), |x| < 1, then f(2x/(1 + x²)) is equal to -
Answer: 2f(x)
The function f(x) is defined in terms of a logarithmic expression, and substituting 2x/(1 + x²) into f results in a transformation that effectively doubles the input value of f(x), leading to the conclusion that f(2x/(1 + x²)) equals 2f(x).
Q47. lim_(x→1⁻) (√π − √(2 sin⁻¹ x)) / √(1−x) is equal to:
Answer: √(2/π)
As x approaches 1 from the left, the expression inside the limit simplifies, and applying L'Hôpital's rule helps resolve the indeterminate form. The limit evaluates to √(2/π) by analyzing the behavior of the numerator and denominator near the point.
Answer: (√(3))/(10)
The correct option is derived from applying the chain rule and implicit differentiation to the given functions, leading to a simplification that evaluates to frac{ ext{sqrt{3}}}{10} at x = (1)/(2). This result reflects the relationship between the derivatives of the two inverse tangent functions.
Q49. If f'(x) = tan⁻¹(sec x + tan x), -π/2 < x < π/2, and f(0) = 0, then f(1) is equal to
Answer: (π + 1)/4
Since sec x + tan x = tan(pi/4 + x/2), f'(x)=arctan(tan(pi/4 + x/2)) = pi/4 + x/2. Integrating: f(x)=pi*x/4 + x^2/4 (f(0)=0). Then f(1)=pi/4 + 1/4 = (pi+1)/4.
Answer: 3
Taking sines reduces it to 25=3*sqrt(25-16x^2)+4*sqrt(25-9x^2), satisfied by x^2=1, and x=0 is trivially a solution. Checking principal values, x=-1,0,1 all work, so there are 3 solutions.