Question

The area of the region bounded by the curve y = $\frac { 16 - x ^ { 2 } } { 4 }$ and y = sec^–1 [– sin² x] (where [.] denotes the greatest integer function) -

• 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 £ sin² x £ 1 Ž – 1 £ – sin² x £ 0 \ [– sin² x] = 0 or – 1

let sec^–1 is not defined hence y = sec^–1 [– sin² x] = sec^–1 (–1) =

p Now p = Ž x² = 16 – 4p = 4(4 – p) Ž x = ± 2$\sqrt { 4 - \pi }$The required area =

$= \frac { 8 } { 3 }$ (4 – p)^3/2