Let [t] denote the largest integer less than or equal to t. If _0… - Vidyadip

MCQ

Let $[t]$ denote the largest integer less than or equal to $t$. If

$\int_0^3\left(\left[x^2\right]+\left[\frac{x^2}{2}\right]\right) d x=a+b \sqrt{2}-\sqrt{3}-\sqrt{5}+c \sqrt{6}-\sqrt{7},$ where $a, b, c \in z$, then $a+b+c$ is equal to.........

A
$21$
B
$12$
C
$29$
✓
$23$

✓

Answer

Correct option: D.

$23$

d
$ \int_0^3\left[x^2\right] d x+\int_0^3\left[\frac{x^2}{2}\right] d x $

$ =\int_0^1 0 d x+\int_1^{12} 1 d x+\int_{\sqrt{2}}^{\sqrt{3}} 2 d x$

$ +\int_{\sqrt{3}}^2 3 \mathrm{dx}+\int_2^{\sqrt{5}} 4 \mathrm{dx}+\int_{\sqrt{5}}^{\sqrt{6}} 5 \mathrm{dx} $

$ +\int_{\sqrt{6}}^{\sqrt{7}} 6 \mathrm{dx}+\int_{\sqrt{7}}^{\sqrt{8}} 7 \mathrm{dx}+\int_{\sqrt{8}}^3 8 \mathrm{dx} $

$ +\int_0^{\sqrt{2}} 0 \mathrm{dx}+\int_{\sqrt{2}}^2 1 \mathrm{dx} $

$ +\int_2^{\sqrt{6}} 2 \mathrm{dx}+\int_{\sqrt{6}}^{\sqrt{8}} 3 \mathrm{dx}+\int_{\sqrt{8}}^3 4 \mathrm{dx}=31-6 \sqrt{2}-\sqrt{3}-\sqrt{5} $

$ -2 \sqrt{6}-\sqrt{7} $

$ \mathrm{a}=31 \quad b=-6 \quad c=-2 $

$ \mathrm{a}+\mathrm{b}+\mathrm{c}=31-6-2=23$

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