NIMCET 2011 — Mathematics PYQ
NIMCET | Mathematics | 2011∫01/2x−x2dx will be equal to:
Choose the correct answer:
- A.
π/2
(Correct Answer) - B.
π
- C.
π/3
- D.
π/4
π/2
Explanation
1. Rearrange the expression inside the square root:
x−x2=−(x2−x)
=−[x2−x+(21)2−(21)2]
=−[(x−21)2−(21)2]
=(21)2−(x−21)2
2. Substitute back into the integral:
∫01/2(21)2−(x−21)2dx
3. Use the standard integral formula ∫a2−x2dx=sin−1(ax):
[sin−1(1/2x−1/2)]01/2
[sin−1(2x−1)]01/2
4. Apply the limits:
=sin−1(2(1/2)−1)−sin−1(2(0)−1)
=sin−1(0)−sin−1(−1)
=0−(−π/2)
=π/2
Correct Option: (a) π/2
Explanation
1. Rearrange the expression inside the square root:
x−x2=−(x2−x)
=−[x2−x+(21)2−(21)2]
=−[(x−21)2−(21)2]
=(21)2−(x−21)2
2. Substitute back into the integral:
∫01/2(21)2−(x−21)2dx
3. Use the standard integral formula ∫a2−x2dx=sin−1(ax):
[sin−1(1/2x−1/2)]01/2
[sin−1(2x−1)]01/2
4. Apply the limits:
=sin−1(2(1/2)−1)−sin−1(2(0)−1)
=sin−1(0)−sin−1(−1)
=0−(−π/2)
=π/2
Correct Option: (a) π/2