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The area enclosed by the curves y=sinx+c...

The area enclosed by the curves `y=sinx+cosx` and `y=|cosx-sinx|` over the interval `[0,pi//2]` is

A

`4(sqrt(2)-1)`

B

`2sqrt(2)(sqrt(2)-1)`

C

`2(sqrt(2)+1)`

D

`2sqrt(2)(sqrt(2)+1)`

Text Solution

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The correct Answer is:
B
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Knowledge Check

  • The minimum value of the function f(x)=(1)/(sinx+cosx) in the interval [0,(pi)/(2)] is

    A
    `(sqrt(2))/(2)`
    B
    `-(sqrt(2))/(2)`
    C
    `(2)/(sqrt(3)+1)`
    D
    `-(2)/(sqrt(3)+1)`

  • The area enclosed by y=x^(2)+cosx and its normat at x=(pi)/(2) in the first quadrant is

    A
    `(pi^(5))/(32)-(pi^(4))/(64)+(pi^(3))/(32)+1`
    B
    `(pi^(5))/(16)-(pi^(4))/(32)+(pi^(3))/(24)-1`
    C
    `(pi^(5))/(32)-(pi^(4))/(32)+(pi^(3))/(16)`
    D
    `(pi^(5))/(32)-(pi^(4))/(32)+(pi^(3))/(16)+1`

  • The area of the region whose boundaries are defined by the curves y=2cosx,y=3tanx , and the y-axis is

    A
    `1+3ln((2)/(sqrt(3)))` sq. units
    B
    `1+(3)/(2)ln3-3 ln2` sq. units
    C
    `1+(3)/(2)ln3-ln2` sq. units
    D
    `ln30ln2` sq. units

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