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Question: Answered & Verified by Expert
If a,bR satisfy the equation a2+4b2-4=0, then the minimum value of 2a+3b will be
MathematicsEllipseJEE Main
Options:
  • A -4
  • B -5
  • C -6
  • D -10
Solution:
2859 Upvotes Verified Answer
The correct answer is: -5

Given equation isa24+b2=1

Putting a=2cosθ, b=sinθ, we get,

2a+3b=22cosθ+3sinθ=4cosθ+3sinθ
Minimum value =-42+32=-5

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