| If \[(m+1)=\sqrt{n}+3,\] then find out the value of \[\frac{1}{2}\,\,\left( \frac{{{m}^{3}}-6{{m}^{2}}+12m-8}{\sqrt{n}}-n \right).\] |
A) 0
B) 1
C) 2
D) 3
Correct Answer: A
Solution :
| \[m+1=\sqrt{n}+3\]\[\Rightarrow \]\[m-2=\sqrt{n}\] |
| \[\therefore \] \[\sqrt{n}=(m-2)\] |
| On cubing both sides, we get |
| \[{{n}^{3/2}}={{(m-2)}^{3}}\] |
| \[{{n}^{3/2}}={{m}^{3}}-8-3\cdot m\cdot 2\,(m-2)\] |
| \[\Rightarrow \] \[{{n}^{3/2}}={{m}^{3}}-8-6{{m}^{2}}+12\,m\] |
| Now, \[\frac{1}{2}\left( \frac{{{m}^{3}}-6{{m}^{2}}+12m-8}{\sqrt{n}}-n \right)=\frac{1}{2}\,\,\left( \frac{{{n}^{3/2}}}{\sqrt{n}}-n \right)\] |
| \[=\frac{1}{2}[{{n}^{3/2-1/2}}-n]=\frac{1}{2}\times 0=0\] |
You need to login to perform this action.
You will be redirected in
3 sec
