Consider the curves C1:y=sin−1⁡|sin⁡x| and C2:y=sin−1⁡|sin⁡x|2, 0≤x≤2πthen

C1y=sin−1⁡|sin⁡x|=x, 0≤x<π2π−x,π2≤x<πx−π,π≤x<3π22π−x,3π2≤x<2πRequired area =4∫0π/2 xdx=π22y=sin−1⁡|sin⁡x|2=x2, 0≤x<π2(π−x)2,π2≤x<π(x−π)2,π≤x<3π2(2π−x)2,3π2≤x<2π. The required area A is shown as shaded region in the figure. Area enclosed in between C1 and C2 =4∫01 x−x2dx+4∫1π/2 x2−xdx=43+π2π−36 sq. units.

Consider the curves C1:y=sin−1⁡|sin⁡x| and C2:y=sin−1⁡|sin⁡x|2, 0≤x≤2πthen

C1y=sin−1⁡|sin⁡x|=x, 0≤x<π2π−x,π2≤x<πx−π,π≤x<3π22π−x,3π2≤x<2πRequired area =4∫0π/2 xdx=π22y=sin−1⁡|sin⁡x|2=x2, 0≤x<π2(π−x)2,π2≤x<π(x−π)2,π≤x<3π2(2π−x)2,3π2≤x<2π. The required area A is shown as shaded region in the figure. Area enclosed in between C1 and C2 =4∫01 x−x2dx+4∫1π/2 x2−xdx=43+π2π−36 sq. units.

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Consider the curves $C_{1} : y = \sin^{- 1} | \sin x |$ and $C_{2} : y = {(\sin^{- 1} | \sin x |)}^{2}, 0 \leq x \leq 2 π$ then

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

Area bounded by C₁ with x-axis is $\frac{π^{2}}{2} sq.unit$

Area enclosed in between C₁ and C₂ is $(\frac{π^{3}}{6} - \frac{π^{2}}{2} + \frac{4}{3}) sq.unit$

The graph of C₁ and C₂ intersect on x-axis at $\frac{π^{2}}{2}, \frac{3 π}{2}$

The graph of C₁ and C₂ meets the y-axis at only one point

answer is A, B, D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} C_{1} y = \sin^{- 1} | \sin x | = \{\begin{cases} x, 0 \leq x < \frac{π}{2} \\ π - x, \frac{π}{2} \leq x < π \\ x - π, π \leq x < \frac{3 π}{2} \\ 2 π - x, \frac{3 π}{2} \leq x < 2 π \end{cases} \end{array}$

Required area $= 4 \int_{0}^{π / 2} xdx = \frac{π^{2}}{2}$

$y = {(\sin^{- 1} | \sin x |)}^{2} = \{\begin{cases} x^{2}, 0 \leq x < \frac{π}{2} \\ (π - x)^{2}, \frac{π}{2} \leq x < π \\ (x - π)^{2}, π \leq x < \frac{3 π}{2} \\ (2 π - x)^{2}, \frac{3 π}{2} \leq x < 2 π . \end{cases}$

$The required area A is shown as shaded region in the figure.$

$A r e a e n c l o s e d i n b e t w e e n C 1 a n d C 2 = 4 \int_{0}^{1} (x - x^{2}) d x + 4 \int_{1}^{π / 2} (x^{2} - x) d x = \frac{4}{3} + π^{2} [\frac{π - 3}{6}] sq. units.$