JEE Mains · Maths · STD 12 - 8. Application and integration
If the area (in \(sq. units\)) of the region \(\left\{ {\left( {x,y} \right):{y^2} \le 4x,x + y \le 1,x \ge 0,y \ge 0} \right\}\) is \(a\sqrt 2 + b\), then \(a -b\) is equal to
- A \(\frac{{10}}{3}\)
- B \(6\)
- C \(\frac{{8}}{3}\)
- D \(-\frac{{2}}{3}\)
Answer & Solution
Correct Answer
(B) \(6\)
Step-by-step Solution
Detailed explanation
\(\left\{(x, y): y^{2} \leq 4 x, x+y \leq 1, x \geq 0, y \geq 0\right\}\) \(A\int\limits_0^{3 - 2\sqrt 2 } {2\sqrt x dx + \frac{1}{2}(1 - (3 - 2\sqrt 2 ))(1 - (3 - 2\sqrt 2 ))} \)…
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