Questions
The least value of the function
on the interval is
detailed solution
Correct option is D
We have F'(x)=3 sin x+4 cos x. Since sin x and cos x assume negative values in the third quadrant, we have F′(x)<0 for all x∈(5π/4, 4π/3) so F(x) assumes the least value at the point x = 4π/3.Thus the least value is F(4π/3)=∫04π/3 (3sin u+4cos u)du=(−3cos u+4sin u)04π/3=−3cos 4π3+4sin 4π3−(−3)=92−432=9−432.Similar Questions
The value of is
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