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

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.