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QJEEM23Maa803: JEE Main 2023 Mathematics - Class 12 JEEM23Maa803 Solution: Let the area enclosed by the lines and the curve where denotes…

Let the area enclosed by the lines x+y=2,y=0,x=0x+y=2, \mathrm{y}=0, x=0 and the curve f(x)=min⁡{x2+34,1+[x]}f(x)=\min \left\{x^{2}+\frac{3}{4}, 1+[x]\right\} where [x][x] denotes the greatest integer ≤x\leq x , be A\mathrm{A} . Then the value of 12 A12 \mathrm{~A} is _____________.

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JEE Main 2023 (Online) 8th April Evening Shift Mathematics - Area Under The Curves Question 36 English Explanation

 Required area =[∫012(x2+34)dx]+[12(32+12)×1]\text { Required area }=\left[\int\limits_0^{\frac{1}{2}}\left(x^2+\frac{3}{4}\right) d x\right]+\left[\frac{1}{2}\left(\frac{3}{2}+\frac{1}{2}\right) \times 1\right]

amp;=[x33+3x4]012+1amp;=124+38−0+1=1+9+2424=3424=1712\begin{aligned}& =\left[\frac{x^3}{3}+\frac{3 x}{4}\right]_0^{\frac{1}{2}}+1 \\\\& =\frac{1}{24}+\frac{3}{8}-0+1=\frac{1+9+24}{24}=\frac{34}{24}=\frac{17}{12}\end{aligned}

 So, 12 A=12×1712=17\text { So, } 12 \mathrm{~A}=12 \times \frac{17}{12}=17

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