Exams › JEE Advanced › Maths

Find the minimum value of cos(theta) + cos(2*theta) over all real theta.

1. -9/8
2. -2
3. -1
4. -3/2

Correct answer: -9/8

Solution

Substituting x = cos theta gives a quadratic 2x² + x - 1 whose minimum on [-1,1] is at x = -1/4, value -9/8.

Related JEE Advanced Maths questions

• Identify the periodic function among the following:
• Given that α, β, γ, and θ are the smallest positive angles in increasing order for which their sine values equal a positive constant k, what is the result of 4 sin(α/2) + 3 sin(β/2) + 2 sin(γ/2) + sin(θ/2)?
• Consider the function fn(θ) = tan(θ/2)(1 + sec²θ)(1 + sec⁴θ)... (1 + sec²ⁿθ). Which of the following is true?
• Given that (a−b)sin(θ+ϕ) = (a+b)sin(θ−ϕ) and a tan(θ/2) − b tan(ϕ/2) = c, which of the following holds true?
• In a right-angled triangle ABC where ∠B = 90° and the sum of the two legs a and b equals 4, the triangle's area is maximized when ∠C is:
• If the altitudes p₁, p₂, and p₃ of a triangle enclose a circle with a diameter of 4/3 units, what is the minimum possible value of p₁ + p₂ + p₃?

⚔️ Practice JEE Advanced Maths free + battle 1v1 →