Let Δ be the area of the region x , y ∈ ℝ 2 : x 2 + y 2 ≤ 21 , y 2 ≤ 4 x , x…

Let Δ be the area of the region x,y2:x2+y221,y24x,x1. Then 12Δ-21sin-127 is equal to
  1. 23-13
  2. 3-23
  3. 23-23
  4. 3-43

Solution

Plotting the diagram of x2+y221 & y24x for x1 we get,

Now from above diagram required Area Δ=2132xdx+232121-x2dx

Δ=8333-1+21sin-127-63

So, 12Δ-21sin-127

=128333-1+21sin-127-63-21sin-127

=128333-1-63

=23-832

=3-43

Asked in: JEE Main 2023 (29 Jan Shift 1)

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