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The minimum value of 2sin2θ + 3cos2θ is ?
Answer & Solution
Correct Answer:
Option
C
Let x = 2sin2θ + 3cos2θ
⇒ x = 2sin2θ + 2cos2θ + cos2θ
⇒ x = 2(sin2θ + cos2θ) + cos2θ
⇒ x = 2 + cos2θ [since sin2θ + cos2θ = 1]
Therefore x will be the minimum when cosθ = 0. i.e. minimum value of x will 2
Alternative Solution:
2sin2θ + 3cos2θ
Minimum value is 2,
[If x sin2θ + y cos2θ, If x > y, then x will be always maximum value and y is minimum if y > x, vice versa will happen]
⇒ x = 2sin2θ + 2cos2θ + cos2θ
⇒ x = 2(sin2θ + cos2θ) + cos2θ
⇒ x = 2 + cos2θ [since sin2θ + cos2θ = 1]
Therefore x will be the minimum when cosθ = 0. i.e. minimum value of x will 2
Alternative Solution:
2sin2θ + 3cos2θ
Minimum value is 2,
[If x sin2θ + y cos2θ, If x > y, then x will be always maximum value and y is minimum if y > x, vice versa will happen]
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LoginThen 2+ cos2 = 2+(1)2= 2+1=3
And in alternative solution
When y>x then y is minimum
And here x=2 and y=3
Hence y>x
Then minimumvalue is y=3