Question
If ∣∣∣asin2θ+bsinθ cosθ+c cos2θ−12(a+c)∣∣∣≤12k, then k2 is equal to
Answer: Option A
:
A
(a) asin2θ+bsinθcosθ+ccos2θ−12(a+c)
= 12[−acos2θ+bsin2θ+ccos2θ]
= 12[bsin2θ−(a−c)cos2θ]
∵|bsin2θ−(a−c)cos2θ|≤√b2+(a−c)2
∴∣∣∣12bsin2θ−(a−c)cos2θ∣∣∣≤12√b2+(a−c)2
∴K=√b2+(a−c)2
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:
A
(a) asin2θ+bsinθcosθ+ccos2θ−12(a+c)
= 12[−acos2θ+bsin2θ+ccos2θ]
= 12[bsin2θ−(a−c)cos2θ]
∵|bsin2θ−(a−c)cos2θ|≤√b2+(a−c)2
∴∣∣∣12bsin2θ−(a−c)cos2θ∣∣∣≤12√b2+(a−c)2
∴K=√b2+(a−c)2
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