Question

65. Given that se s 27, find the maximum value of f(0) = 6+2 cose .............J03 WOOOOOOOO Cole 106 + 2 cost 99013

JEE/Engineering Exams

Maths

Solution

103

4.0 (1 ratings)

( f(theta)=frac{4}{6+2 cos theta} quad frac{pi}{2} leq theta leq 2 pi ) f(o) will be max, when ( 6+2000 ) will be min. because ( 6+2 cos theta ) is in ofenominator ( therefore ) ( 6+2 cos theta ) is min. when cose is min. in given domain, ( cos theta ) min. is -1 at ( theta=pi ) ( therefore quad(6+2 cos theta)_{text {runn. }}=6-2=4 ) ( left(frac{4}{6+2 cos theta}right)_{max }=1 cdot tan ^{prime} )