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QJEEM23Mac269: JEE Main 2023 Mathematics - Class 12 JEEM23Mac269 Solution: Let denote the greatest integer function. If ∫ 0 2.4 [ x 2 ] d…

Let [t][t] denote the greatest integer function. If ∫02.4[x2]dx=α+β2+γ3+δ5, then α+β+γ+δ\alpha+\beta+\gamma+\delta is equal to __________.

Step-by-Step Verified Solution

∫02.4[x2]dxamp;=∫01[x2]dx+∫12[x2]dxamp;+∫23[x2]dx+∫32[x2]dx+∫25[x2]dx+∫52.4[x2]dx\begin{aligned}\int\limits_0^{2.4}\left[x^2\right] d x & =\int\limits_0^1\left[x^2\right] d x+\int\limits_1^{\sqrt{2}}\left[x^2\right] d x \\\\& +\int\limits_{\sqrt{2}}^{\sqrt{3}}\left[x^2\right] d x+\int\limits_{\sqrt{3}}^2\left[x^2\right] d x+\int\limits_2^{\sqrt{5}}\left[x^2\right] d x+\int\limits_{\sqrt{5}}^{2.4}\left[x^2\right] d x\end{aligned}

amp;=∫00.dx+∫121⋅dx+∫232dx+∫323dx+∫254dx+∫52.45dxamp;=0+[x]12+2[x]23+3[x]32+4[x]25+5[x]52.4amp;=2−1+23−22+6−33+45−8+12−55amp;=−2−3−5+9amp;∴α=9,β=−1,γ=−1,δ=−1amp; So, α+β+γ+δ=9−1−1−1=6\begin{aligned}& =\int_0 0 . d x+\int_1^{\sqrt{2}} 1 \cdot d x+\int_{\sqrt{2}}^{\sqrt{3}} 2 d x+\int_{\sqrt{3}}^2 3 d x+\int_2^{\sqrt{5}} 4 d x+\int_{\sqrt{5}}^{2.4} 5 d x \\\\& = 0+[x]_1^{\sqrt{2}}+2[x]_{\sqrt{2}}^{\sqrt{3}}+3[x]_{\sqrt{3}}^2+4[x]_2^{\sqrt{5}}+5[x]_{\sqrt{5}}^{2.4} \\\\& =\sqrt{2}-1+2 \sqrt{3}-2 \sqrt{2}+6-3 \sqrt{3}+4 \sqrt{5}-8+12-5 \sqrt{5} \\\\& =-\sqrt{2}-\sqrt{3}-\sqrt{5}+9 \\\\& \therefore \alpha=9, \beta=-1, \gamma=-1, \delta=-1 \\\\& \text { So, } \alpha+\beta+\gamma+\delta=9-1-1-1=6\end{aligned}

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