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

Hint: A combination is a mathematical technique that determines the number of possible arrangements in a collection of numbers where the order of the selection does not matter. Given by the formula \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\], where n is the number of items and r is the number of items to choose at a time.
In the given question, find the ranges of x from its lower limit to its higher limit by following the above-discussed properties of the combination.

Complete step by step answer:
 Given function is \[f\left( x \right) = {}^{16 - x}{C_{2x - 1}} + {}^{20 - 3x}{C_{4x - 5}}\]
The formula for combination is given as\[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\], where
$\Rightarrow$ \[n > 0\]and \[n \in N\]
$\Rightarrow$ \[r \geqslant 0\]and \[r \in N\]
Also \[n \geqslant r\]
Hence we can write in the first term of function \[{}^{16 - x}{C_{2x - 1}}\]
\[
\Rightarrow 16 - x > 0 \\
\Rightarrow x < 16 - - - - (i) \\
 \]
Also
\[
\Rightarrow 2x - 1 \geqslant 0 \\
\Rightarrow x \geqslant \dfrac{1}{2} - - - - (ii) \\
 \]
And
\[
\Rightarrow 16 - x \geqslant 2x - 1 \\
\Rightarrow 16 + 1 \geqslant 2x + x \\
\Rightarrow 3x \leqslant 17 \\
\Rightarrow x \leqslant \dfrac{{17}}{3} - - - - (iii) \\
 \]
So from equation (i), (ii) and (iii), the point of intersection for the range of x is given as:
\Rightarrow \[\dfrac{1}{2} \leqslant x \leqslant \dfrac{{17}}{3} - - - - (iv)\]
Hence we get the range of the first term.
Now check for the first term of the function\[{}^{20 - 3x}{C_{4x - 5}}\],
\[
\Rightarrow 20 - 3x > 0 \\
\Rightarrow 3x < 20 \\
\Rightarrow x < \dfrac{{20}}{3} - - - - (v) \\
 \]
Also
\[
\Rightarrow 4x - 5 \geqslant 0 \\
\Rightarrow 4x \geqslant 5 \\
\Rightarrow x \geqslant \dfrac{5}{4} - - - - (vi) \\
 \]
And
\[
\Rightarrow 20 - 3x \geqslant 4x - 5 \\
\Rightarrow 20 + 5 \geqslant 4x + 3x \\
\Rightarrow 25 \geqslant 7x \\
\Rightarrow x \leqslant \dfrac{{25}}{7} - - - - (vii) \\
 \]
So from equation (v), (vi) and (vii), the point of intersection for the range of x is given as:
$\Rightarrow$ \[\dfrac{5}{4} \leqslant x \leqslant \dfrac{{25}}{7} - - - - (viii)\]
Now form equation (iv) and (viii) for the intersection of x for the function comparing the range of both terms, we get
$\Rightarrow$ \[\dfrac{5}{4} \leqslant x \leqslant \dfrac{{25}}{7} - - - - (ix)\]
So we get the domain range of the function.
As we know, the terms of the functions should be Natural number; hence we can write
$\Rightarrow$ \[16 - x \in N\]
And this is possible only if x is an integer \[x \in I\], and if x is an integer then all the numbers will be a natural number,
Now consider for the domain range of x from the equation (ix), it gives a natural number if it is an integer
$\Rightarrow$ \[\dfrac{5}{4} \leqslant x \leqslant \dfrac{{25}}{7}\]
So the integer number which lies in this range is {2, 3}
So if\[x = 2\]the range of the function
\[
\Rightarrow f\left( x \right) = {}^{16 - x}{C_{2x - 1}} + {}^{20 - 3x}{C_{4x - 5}} \\
\Rightarrow f\left( 2 \right) = {}^{16 - 2}{C_{2 \times 2 - 1}} + {}^{20 - 3 \times 2}{C_{4 \times 2 - 5}} \\
   = {}^{14}{C_3} + {}^{14}{C_3} \\
 \]
Where \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]
Hence we can write
\[{}^{14}{C_3} = \dfrac{{14!}}{{3!\left( {11} \right)!}} = \dfrac{{14 \times 13 \times 12}}{{3 \times 2}} = 364\]
Therefore
\[
  f\left( 2 \right) = {}^{14}{C_3} + {}^{14}{C_3} \\
   = 364 + 364 \\
   = 728 \\
 \]
Now when\[x = 3\]the range of the function
\[
\Rightarrow f\left( x \right) = {}^{16 - x}{C_{2x - 1}} + {}^{20 - 3x}{C_{4x - 5}} \\
\Rightarrow f\left( 2 \right) = {}^{16 - 3}{C_{2 \times 3 - 1}} + {}^{20 - 3 \times 3}{C_{4 \times 3 - 5}} \\
   = {}^{13}{C_5} + {}^{11}{C_7} \\
 \]
Where \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]
Hence we can write
\[{}^{13}{C_5} = \dfrac{{13!}}{{5!\left( 8 \right)!}} = \dfrac{{13 \times 12 \times 11 \times 10 \times 9}}{{5 \times 4 \times 3 \times 2}} = 1287\]
\[{}^{11}{C_7} = \dfrac{{11!}}{{4!\left( 7 \right)!}} = \dfrac{{11 \times 10 \times 9 \times 8}}{{4 \times 3 \times 2}} = 330\]
Therefore
\[
  f\left( 2 \right) = {}^{13}{C_5} + {}^{11}{C_7} \\
   = 1287 + 330 \\
   = 1617 \\
 \]

Hence the range of \[f\left( x \right) = \left\{ {728,1617} \right\}\], so Option C is correct.

Note: The range of a function is a set of outputs achieved when it is applied to its whole set of outputs. The range of a set of data is the difference between the highest and the lowest values in the set. As the solution is a bit complex and involves many calculations, it is advised to the students to be careful while applying the inequalities. Even a change in the single sign in equality can lead to wrong answers.