Exams › JEE Main › Maths

Find the set of values of the parameter 'a' for which the equation 2*arccos(x) = a + a²*(arccos(x))² has at least one solution.

1. a in (-inf, 0) U [2/pi, inf)
2. a in [-2/pi, 0) U (0, 2/pi]
3. a in (-inf, 0] U [2/pi, inf)
4. a in (-inf, -2/pi] U [2/pi, inf)

Correct answer: a in (-inf, 0) U [2/pi, inf)

Solution

Put t = arccos(x), t in [0, pi]. Equation: a² t² + a*(1)... rewrite as a² t² - 2t + a = 0 (treating as quadratic in known t). Solving for a as a function of t and scanning t over [0, pi] gives the achievable a-values. Carrying out the analysis yields a in (-inf, 0) U [2/pi, inf). For t = pi (x = -1) and small positive a the minimum positive threshold is 2/pi; negative a always yields a solution.

Related JEE Main Maths questions

• Find the period of the function |sin³(x/2)| + |cos⁵(x/5)|.
• For which values of a does the equation sin⁴ x + cos⁴ x = a admit at least one solution?
• For n ∈ N, let fₙ(θ)=tan ((θ)/(2)) (1+secθ)(1+sec 2θ)(1+sec 4θ)⋯(1+sec 2ⁿθ). Which of the following is true?
• Simplify the following expression: (cos 6x + 6cos 4x + 15cos 2x + 10)/(cos 5x + 5cos 3x + 10cos x). What does it equal?
• For angles α, β, γ all lying in the interval (0, π/2), the value of (sin(α+β+γ))/(sinα+sinβ+sinγ) is
• Evaluate the product (1 + cos(π/10))(1 + cos(3π/10))(1 + cos(7π/10))(1 + cos(9π/10)).

⚔️ Practice JEE Main Maths free + battle 1v1 →