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NDA 1 2025 Maths, question 100

Topic: Integral calculus

Passage · Q99 to Q100

Let f(x)=[x2]f(x) = \left\lbrack x^{2} \right\rbrack, where [⋅]\lbrack \cdot \rbrack is the greatest integer function.

Q100

What is ∫22f(x) dx\int_{\sqrt{2}}^{2}f(x)\, dx equal to?

  1. (a) 6−3−226 - \sqrt{3} - 2\sqrt{2}Correct answer
  2. (b) 6−3−26 - \sqrt{3} - \sqrt{2}
  3. (c) 6−3+226 - \sqrt{3} + 2\sqrt{2}
  4. (d) 6+3−226 + \sqrt{3} - 2\sqrt{2}

Solution.

Answer: (a) 6−3−226 - \sqrt{3} - 2\sqrt{2}

  1. Split at x=3x = \sqrt{3}.
  2. The floor value is 2 on [2,3)\lbrack\sqrt{2},\sqrt{3}) and 3 on [3,2)\lbrack\sqrt{3},2).
  3. The integral is 2(3−2)+3(2−3)=6−3−222\left( \sqrt{3} - \sqrt{2} \right) + 3\left( 2 - \sqrt{3} \right) = 6 - \sqrt{3} - 2\sqrt{2}.