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The minimum value of (2sin2θ+3cos2θ)(2 \sin^2\theta + 3 \cos^2\theta) is

Solution

Correct Option: 3
  1. Given expression: 2sin2θ+3cos2θ2\sin^2 \theta + 3\cos^2 \theta
  1. Using identity sin2θ+cos2θ=1\sin^2 \theta + \cos^2 \theta = 1:

Therefore, sin2θ=1cos2θ\sin^2 \theta = 1 - \cos^2 \theta

  1. Substituting:

2(1cos2θ)+3cos2θ2(1 - \cos^2 \theta) + 3\cos^2 \theta

=22cos2θ+3cos2θ= 2 - 2\cos^2 \theta + 3\cos^2 \theta

=2+cos2θ= 2 + \cos^2 \theta

  1. Since cos2θ\cos^2 \theta varies between 0 and 1:
  • When cos2θ=0\cos^2 \theta = 0, expression = 2
    • When cos2θ=1\cos^2 \theta = 1, expression = 3
  1. Therefore:

Minimum value = 2 (when cos2θ=0\cos^2 \theta = 0)

Maximum value = 3 (when cos2θ=1\cos^2 \theta = 1)

The minimum value is 2.

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