Question

The range of $f\left(x\right){=}^{16-x}{C}_{2x-1}{+}^{20-3x}{C}_{4x-5}$ is

• A. $[728,1474]$
• B. $\{0,728\}$
• C. $\{728,1617\}$
• D. None of these

Answer 1

Answer: (C)

$\{728,1617\}$

For given function to be defined,

$16-x>0\Rightarrow x<\mathrm{16....}\left(i\right)$

$2x-1\ge 0\Rightarrow x\ge \frac{1}{2}...\left(ii\right)$

$20-3x>0\Rightarrow x<\frac{20}{3}...\left(iii\right)$

$4x-5\ge 0\Rightarrow x\ge \frac{5}{4}...\left(iv\right)$

$(16-x)-2x+1\ge 0\therefore x\le \frac{17}{3}...\left(v\right)$

$20-3x-4x+5\ge 0\therefore x\le \frac{25}{7}...\left(vi\right)$

Considering all the above cases,

$1+\frac{1}{4}\le x<3+\frac{4}{7}$

$\therefore x=2,3$

when $x=2,y{=}^{14}{C}_3{+}^{14}{C}_3=728$

and when $x=3,y{=}^{13}{C}_5{+}^{11}{C}_4=1287+330=1617$