Q114 / 160 · AP EAMCET 2022 (05 Jul Shift 2) · Mathematics

[.] represents a greatest integer function.If $\int_{\sqrt{3}}^{\sqrt{18}}[x] d x=a+b \sqrt{2}+c \sqrt{3}$, then $\mathrm{a}+\mathrm{b}+\mathrm{c}=$

  • A
    $0$
  • B
    $1$
  • C
    $-1$
  • Correct answer
    D
    $2$

Solution

$\int_{\sqrt{3}}^{\sqrt{18}}[x] d x=$ $\int_{\sqrt{3}}^{\sqrt{4}}[x] d x+\int_2^3[x] d x$ $+\int_3^4[x] d x+\int_4^{\sqrt{18}}[x] d x$ $=\int_{\sqrt{3}}^{\sqrt{4}} 1 d x+\int_2^3 2 d x+$ $\int_3^4 3 d x+\int_4^{\sqrt{18}} 4 d x$ $=(2-\sqrt{3})+(2)+(3)+4(\sqrt{18}-4)$ $=2-\sqrt{3}+2+3+4 \sqrt{18}-16$ $=-9-\sqrt{3}+12 \sqrt{2}$ $\begin{aligned} & =12 \sqrt{2}-\sqrt{3}-9 \\ & =b \sqrt{2}+c \sqrt{3}+9\end{aligned}$ Where $a=-9, b=12, c=-1$ $a+b+c=2$

Hence, the correct answer is option: 'D'.

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