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JEE MAIN Mathematics QUESTION #968
Question 1
The area of the region enclosed by the curves \(y = x^2 - 4x + 4\) and \(y^2 = 16 - 8x\) is:
  • \(\dfrac{8}{3}\)✔️
  • \(\dfrac{4}{3}\)
  • 8
  • 5
Correct Answer Explanation
Rewrite: \(y=(x-2)^2\) (parabola) and \(y^2=8(2-x)\) (leftward parabola). Finding intersections and integrating between them gives area \(=\dfrac{8}{3}\).