JEE Mains · Maths · STD 12 - 8. Application and integration
Let the area enclosed by the lines \(x + y =2, y =0\), \(x=0\) and the curve \(f(x)=\min \left\{x^2+\frac{3}{4}, 1+[x]\right\}\) where \([ x ]\) denotes the greatest integer \(\leq x\), be \(A\). Then the value of \(12\,A\) is \(............\).
- A \(17\)
- B \(16\)
- C \(15\)
- D \(14\)
Answer & Solution
Correct Answer
(A) \(17\)
Step-by-step Solution
Detailed explanation
\(\int \limits_0^{\frac{1}{2}}\left( x ^2+\frac{3}{4}\right) dx +\frac{1}{2} \times\left(\frac{3}{2}+\frac{1}{2}\right) \times 1\) \(=\left[\frac{ x ^3}{3}+\frac{3 x }{4}\right]_0^{\frac{1}{2}}+1\) \(A =\frac{1}{24}+\frac{3}{8}+1\) \(12 A =\frac{1}{2}+\frac{36}{8}+12\)…
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