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Find the maximum/minimum value of y in the functions given below.

y=(sin2x-x), where -π2xπ2

a
ymin=-34-π6 at x=-π/6 and ymax=52-π6 at x=π/6
b
ymin=-32-π4 at x=-π/4 and ymax=32-π2 at x=π/2
c
ymin=-32-π6 at x=-π/6 and ymax=32-π6 at x=π/6
d
None of the above

detailed solution

Correct option is C

y=(sin2x-x)dydx=2cos2x-1andd2ydx2=-4sin2xPutting dydx=0, we get 2cos2x-1=0cos2x=12or 2x=±60° At 2x=+60°,d2ydx2 is -ve, so value of y is  maximum. At 2x=-60°,d2ydx2 is positive. So value  of y is minimum. ymax=sin+60°-π6=32-π6 at 2x=π3or x=π/6and ymin=sin-60°+π6at x=-π6=π6-32

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