The minimum value of (3sinx−4cosx−10)(3sinx+4cosx−10) is
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a
1
b
2
c
3
d
7.00
answer is D.
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Detailed Solution
f(x)=9sin2x−16cos2x−10(3sinx−4cosx)−10(3sinx+4cosx)+100=25sin2x−60sinx+84=(5sinx−6)2+48∴ the minimum value of f(x) occurs when sinx=1∴ minimum value of f(x) is 7.