Exams › JEE Advanced › Maths

Find the maximum and minimum values of the expression cos²(theta) - 6 sin(theta) cos(theta) + 3 sin²(theta) + 2.

1. max = 4 + sqrt(10), min = 4 - sqrt(10)
2. max = 6 + sqrt(10), min = 6 - sqrt(10)
3. max = 5 + 3, min = 5 - 3
4. max = 4 + sqrt(13), min = 4 - sqrt(13)

Correct answer: max = 4 + sqrt(10), min = 4 - sqrt(10)

Solution

Rewriting gives 4 - cos2t - 3 sin2t, whose amplitude is sqrt(1 + 9) = sqrt(10), so the values range over 4 - sqrt(10) to 4 + sqrt(10).

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 →