
Which of the following represents the elements of domain of the function \[f\left( x \right) = \dfrac{1}{{\sqrt {{}^{10}{C_{x - 1}} - 3 \times {}^{10}{C_x}} }}\]?
A. 9, 10, 11
B. 9, 10, 12
C. All natural numbers
D. None of these
Answer
162.6k+ views
Hint: We know the domain of a function is always a real number. The square root of a positive number is a real number. The term which is under the square root of the given function must be a positive number. Using this condition we will find the possible values of x. The possible values of x are the domain.
Formula Used: Combination formula:
\[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]
Complete step by step solution: Given function is \[f\left( x \right) = \dfrac{1}{{\sqrt {{}^{10}{C_{x - 1}} - 3 \times {}^{10}{C_x}} }}\].
We know that the term under square root must be positive to get the real value of x. The value of \[{}^{10}{C_{x - 1}} - 3 \times {}^{10}{C_x}\] must be greater than 0.
Therefore,
\[{}^{10}{C_{x - 1}} - 3 \times {}^{10}{C_x} > 0\]
Add both sides by \[3 \times {}^{10}{C_x}\]:
\[ \Rightarrow {}^{10}{C_{x - 1}} > 3 \times {}^{10}{C_x}\]
Applying combination formula:
\[ \Rightarrow \dfrac{{10!}}{{\left( {x - 1} \right)!\left( {10 - x + 1} \right)!}} > 3 \times \dfrac{{10!}}{{x!\left( {10 - x} \right)!}}\]
\[ \Rightarrow \dfrac{{10!}}{{\left( {x - 1} \right)!\left( {11 - x} \right)!}} > 3 \times \dfrac{{10!}}{{x \times \left( {x - 1} \right)!\left( {10 - x} \right)!}}\]
Cancel out \[10!\] from both sides:
\[ \Rightarrow \dfrac{1}{{\left( {x - 1} \right)!\left( {11 - x} \right)!}} > 3 \times \dfrac{1}{{x \times \left( {x - 1} \right)!\left( {10 - x} \right)!}}\]
\[ \Rightarrow x \times \left( {x - 1} \right)!\left( {10 - x} \right)! > 3 \times \left( {x - 1} \right)!\left( {11 - x} \right)!\]
Cancel out \[\left( {x - 1} \right)!\] from both sides
\[ \Rightarrow x\left( {10 - x} \right)! > 3 \times \left( {11 - x} \right)!\]
\[ \Rightarrow x\left( {10 - x} \right)! > 3 \times \left( {11 - x} \right) \times \left( {11 - x - 1} \right)!\]
\[ \Rightarrow x\left( {10 - x} \right)! > 3 \times \left( {11 - x} \right) \times \left( {10 - x} \right)!\]
Cancel out \[\left( {10 - x} \right)!\] from both sides:
\[ \Rightarrow x > 3\left( {11 - x} \right)\]
\[ \Rightarrow x > 33 - 3x\]
\[ \Rightarrow 4x > 33\]
Divide both sides by 4
\[ \Rightarrow x > 8.25\]
The term \[{}^{10}{C_x}\] exists if the value of \[x \le 10\] and x must be an integers.
The solution of the inequalities \[x > 8.25\] and \[x \le 10\] is 9 and 10, as x is an integer.
The domain of the function is {9,10}
Option ‘D’ is correct
Note: Students often do not consider the second inequality that is \[x \le 10\]. But it is necessary to consider the second inequality and x is an integer. Otherwise, students are unable to get the correct answer.
Formula Used: Combination formula:
\[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]
Complete step by step solution: Given function is \[f\left( x \right) = \dfrac{1}{{\sqrt {{}^{10}{C_{x - 1}} - 3 \times {}^{10}{C_x}} }}\].
We know that the term under square root must be positive to get the real value of x. The value of \[{}^{10}{C_{x - 1}} - 3 \times {}^{10}{C_x}\] must be greater than 0.
Therefore,
\[{}^{10}{C_{x - 1}} - 3 \times {}^{10}{C_x} > 0\]
Add both sides by \[3 \times {}^{10}{C_x}\]:
\[ \Rightarrow {}^{10}{C_{x - 1}} > 3 \times {}^{10}{C_x}\]
Applying combination formula:
\[ \Rightarrow \dfrac{{10!}}{{\left( {x - 1} \right)!\left( {10 - x + 1} \right)!}} > 3 \times \dfrac{{10!}}{{x!\left( {10 - x} \right)!}}\]
\[ \Rightarrow \dfrac{{10!}}{{\left( {x - 1} \right)!\left( {11 - x} \right)!}} > 3 \times \dfrac{{10!}}{{x \times \left( {x - 1} \right)!\left( {10 - x} \right)!}}\]
Cancel out \[10!\] from both sides:
\[ \Rightarrow \dfrac{1}{{\left( {x - 1} \right)!\left( {11 - x} \right)!}} > 3 \times \dfrac{1}{{x \times \left( {x - 1} \right)!\left( {10 - x} \right)!}}\]
\[ \Rightarrow x \times \left( {x - 1} \right)!\left( {10 - x} \right)! > 3 \times \left( {x - 1} \right)!\left( {11 - x} \right)!\]
Cancel out \[\left( {x - 1} \right)!\] from both sides
\[ \Rightarrow x\left( {10 - x} \right)! > 3 \times \left( {11 - x} \right)!\]
\[ \Rightarrow x\left( {10 - x} \right)! > 3 \times \left( {11 - x} \right) \times \left( {11 - x - 1} \right)!\]
\[ \Rightarrow x\left( {10 - x} \right)! > 3 \times \left( {11 - x} \right) \times \left( {10 - x} \right)!\]
Cancel out \[\left( {10 - x} \right)!\] from both sides:
\[ \Rightarrow x > 3\left( {11 - x} \right)\]
\[ \Rightarrow x > 33 - 3x\]
\[ \Rightarrow 4x > 33\]
Divide both sides by 4
\[ \Rightarrow x > 8.25\]
The term \[{}^{10}{C_x}\] exists if the value of \[x \le 10\] and x must be an integers.
The solution of the inequalities \[x > 8.25\] and \[x \le 10\] is 9 and 10, as x is an integer.
The domain of the function is {9,10}
Option ‘D’ is correct
Note: Students often do not consider the second inequality that is \[x \le 10\]. But it is necessary to consider the second inequality and x is an integer. Otherwise, students are unable to get the correct answer.
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