Quantitative Aptitude Ques 66

Question: If $ (m+1)=\sqrt{n}+3, $ then find out the value of $ \frac{1}{2}( \frac{m^{3}-6m^{2}+12m-8}{\sqrt{n}}-n ). $

Options:

A) 0

B) 1

C) 2

D) 3

Show Answer

Answer:

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}-8-6m^{2}+12,m $ Now, $ \frac{1}{2}( \frac{m^{3}-6m^{2}+12m-8}{\sqrt{n}}-n )=\frac{1}{2}( \frac{{n^{3/2}}}{\sqrt{n}}-n ) $ $ =\frac{1}{2}[{n^{3/2-1/2}}-n]=\frac{1}{2}\times 0=0 $

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