Test Index

AP EAPCET 24-Aug-2021 Shift 1 Paper

Section: Mathematics

© examsnet.com

Question : 18 of 160

Marks: +1, -0

The solution set of

{}^5C_{x-1} > 2 \cdot ({}^5C_x)

is

$\{1,2,5\}$
$\{2,3,5\}$
$\{5\}$
$\{1,3,5\}$

Solution:

{}^5C_{x-1} > 2 \cdot {}^5C_x

Inequality defines, when

x-1>0 ; \therefore x-1

and

x

are positive integers

x>0

x-1<5

\Rightarrow x<5

Now,

{}^5C_{x-1} > 2 \cdot {}^5C_x

\frac{{}^5C_{x-1}}{{}^5C_x} > 2

\frac{\frac{5!}{(5-x+1)!(x-1)!}}{\frac{5!}{(5-x)!x!}} > 2 \\ {}^{n}C_{r} = \frac{n!}{(n-r)!r!}

\Rightarrow \frac{(5-x)! x (x-1)}{(5-x+1)(5-x)! (x-1)!} > 2

\Rightarrow \frac{x}{5-x+1} > 2

\Rightarrow \frac{x}{6-x} - 2 > 0

\Rightarrow \frac{x-12+2x}{6-x} > 0

\Rightarrow \frac{3x-12}{6-x} > 0

\Rightarrow \frac{3(x-4)}{x-6} < 0

Critical points are

x=4,6

∴

x \in I^{+}

\Rightarrow x=5

is only one possible value.

∴ Solution set

= \{5\}

© examsnet.com

Go to Question:

Prev Question

Next Question