Question

If $(6+\sqrt{35}{)}^n=I+F$ when $I$ is odd and $0<F<1$, then $(I+F)(I-F)=$

• A. $1$
• B. $\frac{1}{2}$
• C. $2$
• D. $4$

Answer 1

Answer: (A)

$1$

$I+F=(6+\sqrt{35}{)}^n$ ...(i)
Therefore
$I-F=(6-\sqrt{35}{)}^n$ ...(ii)
Multiplying i and ii we get
$(I+F)(I-F)$
$=(6+\sqrt{35}{)}^n(6-\sqrt{35}{)}^n$
$=(36-35{)}^n$
$=1$