Question

If $x=2+i$ and $\frac{1}{0!8!}+\frac{1}{2!6!}+\frac{1}{2!4!4!}=\frac{2^{a-1}}{m!}$ then value of $x^3-5x^2+9x+10$ equals

• A. $m$
• B. $a$
• C. $a-m$
• D. $a+m$

Answer 1

Answer: (D)

$a+m$

From the given we have
$2(\frac{1}{0!8!}+\frac{1}{2!6!}+\frac{1}{2!4!4!})=\frac{2^a}{m!}$
$\Rightarrow \frac{1}{8!}[\frac{2\cdot 8!}{0!8!}+\frac{2\cdot 8!}{2!6!}+\frac{8!}{2!4!4!}]=\frac{2^a}{m!}$
$\Rightarrow \frac{1}{8!}[2\cdot {{}^{8}C}_0+2\cdot {{}^{8}C}_2+{{}^{8}C}_2]=\frac{2^a}{m!}$
$\Rightarrow \frac{1}{8!}[{{}^{8}C}_0+{{}^{8}C}_2+{{}^{8}C}_4+{{}^{8}C}_6+{{}^{8}C}_8]=\frac{2^a}{m!}$
$\Rightarrow \frac{2^7}{8!}=\frac{2^a}{m!}$
$\therefore a=7,m=8$
$x=2+i\Rightarrow x^2-4x+5=0$
$x^3-5x^2+9x+10=(x-1)(x^2-4x+15)+15$
$0+a+m=a+m$
$(a=7,m=8)$