Question

If $(\sqrt{2}+1{)}^6+(\sqrt{2}-1{)}^6=198$, then the integral part of $(\sqrt{2}+1{)}^6$ is
(Note : the integral part here refers to the greatest integer less than or equal to the number)

• A. $98$
• B. $99$
• C. $197$
• D. $196$

Answer 1

The correct option is C $197$

We know if $0<a<1$
Then $a^n<1$, where $n$ is a positive integer.
Thus for $(\sqrt{2}-1{)}^1$ which is less than $1,$
$(\sqrt{2}-1{)}^6<1$ ($\sqrt{2}-1=0.41$)
Now it is given that
$(\sqrt{2}+1{)}^6+(\sqrt{2}-1{)}^6=198$
$(\sqrt{2}+1{)}^6=198-(\sqrt{2}-1{)}^6<198$
Also, as shown above $(\sqrt{2}-1{)}^6<1$
Thus, $(\sqrt{2}+1{)}^6=198-(\sqrt{2}-1{)}^6>197$
Therefore, the greatest integer for $(\sqrt{2}+1{)}^6$ is $197.$