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Question

The minimum value of √(3sinx−4cosx−10)(3sinx+4cosx−10) is

A
9
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B
7
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C
5
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D
6
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Solution

The correct option is C 7Given,√(3sinx−4cosx−10)(3sinx+4cosx−10)Simplifying, we get5√(35sinx−45cosx−2)(35sinx+45cosx−2)Put sinθ=45⟹cosθ=35=5√(sinxcosθ−sinθcosx−2)(sinxcosθ+sinθcosx−2)=5√(sin(x−θ)−2)(sin(θ+x)−2) ..................... sin(A−B)=sinAcosB−sinBcosA and sin(A+B)=sinAcosB+sinBcosA=5√(2−sin(x−θ))(2−sin(θ+x))=5√4−2[sin(x−θ)+sin(x+θ)]+sin(x−θ).sin(x+θ)=5√4−4sin(x).cos(θ)+cos2θ−cos2x2Hence the minimum value is achieved whenx=π2.Thus f(π2)=√(3sin(π2)−4cos(π2)−10)(3sin(π2)+4cos(π2)−10)=√(3−10)×(3−10)=√−7×−7=√49=7.

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