Definite Integration Question 412
Question: On the interval $ [ \frac{5\pi }{3},\frac{7\pi }{4} ], $ the greatest value of the function $ f(x)=\int _{5\pi /3}^{x}{(6\cos t-2\sin t)dt=} $
Options:
A) $ 3\sqrt{3}+2\sqrt{2}+1 $
B) $ 3\sqrt{3}-2\sqrt{2}-1 $
C) Does not exist
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ f’(x)=(6\cos x-2\sin x)1-0 $
$ =2[3\cos x-\sin x]>0 $ in $ [ \frac{5\pi }{3},\frac{7\pi }{4} ] $
$ f(x) $ is an increasing function, hence $ f(x) $ has greatest value at $ x=\frac{7\pi }{4} $
Greatest $ f(x)=f( \frac{7\pi }{4} )=[6\sin t+2\cos t] _{5\pi /3}^{7\pi /4} $
$ =-6\frac{1}{\sqrt{2}}+\frac{2}{\sqrt{2}}+6\frac{\sqrt{3}}{2}-1=3\sqrt{3}-2\sqrt{2}-1 $ .
BETA
