Question

The area of the region bounded by the curve

y = and y = sec^–1[–sin²x] (where [.] denotes the greatest integer function) is -

• A. $\frac { 1 } { 3 }$(4 – p)^3/2
• B. 8(4 – p)^3/2
• C. $\frac { 8 } { 3 }$ (4 – p)^3/2
• D. $\frac { 8 } { 3 }$ (4 – p)^1/2

Answer 1

Answer: (C)

$\frac { 8 } { 3 }$ (4 – p)^3/2

Answer 2

0 ฃ sin²x ฃ 1

 – 1 ฃ – sin²x ฃ 0 \ [– sin²x] = 0 or – 1 but sec^–1(0) is nto defined hence y = sec^–1[– sin²x] = sec^–1(–1) = now p =  x² = 16 – 4p = 4(4 – p)  x = ฑ 2 $\sqrt { 4 - \pi }$ The required area = = $\frac { 8 } { 3 }$ (4 – p)^3/2.