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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Basics of Mathematics

JEE Main · Mathematics · Basics of Mathematics

12- 68+48 2 is equal to :

12−68+482\sqrt{12-\sqrt{68+48 \sqrt{2}}} is equal to :

Options

  1. A2−3\sqrt{2}-3
  2. B2+22+\sqrt{2}
  3. C2−22-\sqrt{2}
  4. D6−286-2 \sqrt{8}

Correct answer

C. 2−22-\sqrt{2}

Step-by-step solution

12−68+482=12−(6)2+(42)2+2×6×42=12−(6+42)2=12−6−42=6−42=(2)2+(2)2−2⋅2⋅2=(2−2)2=2−2\sqrt{12-\sqrt{68+48 \sqrt{2}}}=\sqrt{12-\sqrt{(6)^2+(4 \sqrt{2})^2+2 \times 6 \times 4 \sqrt{2}}}=\sqrt{12-\sqrt{(6+4 \sqrt{2})^2}}=\sqrt{12-6-4 \sqrt{2}}=\sqrt{6-4 \sqrt{2}}=\sqrt{(2)^2+(\sqrt{2})^2-2 \cdot \sqrt{2} \cdot 2}=\sqrt{(2-\sqrt{2})^2}=2-\sqrt{2}
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