Question

If $\underset{0}{\overset{\sqrt[3]{33}}{\int }}\left[x^3\right]dx=a\cdot 3^{1/3}+b\cdot 2^{1/3}+c$, then the value of $(a-b-c)$ is
(where $[\cdot ]$ denotes the greatest integer function)

• A. $3$
• B. $4$
• C. $0$
• D. $6$

Answer 1

The correct option is B $4$

$\underset{0}{\overset{\sqrt[3]{33}}{\int }}\left[x^3\right]dx=\underset{0}{\overset{1}{\int }}\left[x^3\right]dx+\underset{1}{\overset{\sqrt[3]{32}}{\int }}\left[x^3\right]dx+\underset{\sqrt[3]{32}}{\overset{\sqrt[3]{33}}{\int }}\left[x^3\right]dx$
$=\underset{0}{\overset{1}{\int }}0dx+\underset{1}{\overset{\sqrt[3]{32}}{\int }}1dx+\underset{\sqrt[3]{32}}{\overset{\sqrt[3]{33}}{\int }}2dx$

$=0+(\sqrt[3]{32}-1)1+(\sqrt[3]{33}-\sqrt[3]{32})\times 2$

$=2\cdot 3^{1/3}-2^{1/3}-1$
$\therefore a=2,b=-1,c=-1$

$\Rightarrow (a-b-c)=4$