JEE MAIN - Mathematics (2023 - 8th April Evening Shift - No. 15)
Let the area enclosed by the lines $$x+y=2, \mathrm{y}=0, x=0$$ and the curve $$f(x)=\min \left\{x^{2}+\frac{3}{4}, 1+[x]\right\}$$ where $$[x]$$
denotes the greatest integer $$\leq x$$, be $$\mathrm{A}$$. Then the value of $$12 \mathrm{~A}$$ is _____________.
Answer
17
Explanation
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$$ \text { Required area }=\left[\int\limits_0^{\frac{1}{2}}\left(x^2+\frac{3}{4}\right) d x\right]+\left[\frac{1}{2}\left(\frac{3}{2}+\frac{1}{2}\right) \times 1\right] $$
$$ \begin{aligned} & =\left[\frac{x^3}{3}+\frac{3 x}{4}\right]_0^{\frac{1}{2}}+1 \\\\ & =\frac{1}{24}+\frac{3}{8}-0+1=\frac{1+9+24}{24}=\frac{34}{24}=\frac{17}{12} \end{aligned} $$
$$ \text { So, } 12 \mathrm{~A}=12 \times \frac{17}{12}=17 $$
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