Exams › JEE Main › Maths

Find the maximum and minimum values of the expression 2 sin(theta + pi/6) + sqrt(3) cos(theta - pi/6).

1. max = sqrt(13), min = -sqrt(13)
2. max = 5, min = -5
3. max = sqrt(7), min = -sqrt(7)
4. max = 2 + sqrt(3), min = -(2 + sqrt(3))

Correct answer: max = sqrt(13), min = -sqrt(13)

Solution

Expanding gives (3sqrt3/2) sin(theta) + (5/2) cos(theta)... reducing to amplitude sqrt(13); hence max = sqrt(13), min = -sqrt(13).

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 →