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JEE Mains · Maths · STD 12 - 8. Application and integration

ધારોકે રેખાઓ \(x+y=2, y =0, x=0\) અને વક્ર \(f(x)=\min \left\{x^2+\frac{3}{4}, 1+[x]\right\}\) વડે ધેરાયેલ ક્ષેત્રફળ \(A\) છે,જ્યાં \([x]\) એ મહત્તમ પૂર્ણાક \(\leq x\) દર્શાવે છે. તો \(12\,A\) નું મૂલ્ય \(=.......\)

  1. A \(17\)
  2. B \(16\)
  3. C \(15\)
  4. D \(14\)
Verified Solution

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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