Question

Domain of the function $f\left( x \right) = \dfrac{1}{{\sqrt {4x - \left| {{x^2} - 10x + 9} \right|} }}$ is
A) $\left( {7 - \sqrt {40} ,7 + \sqrt {40} } \right)$
B) $\left( {0,7 + \sqrt {40} } \right)$
C) $\left( {7 - \sqrt {40} ,\infty } \right)$
D) None of these

Answer 1

Hint: Substitute the given values in the formula ${\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca$ and simplify it. It will give the polynomial of degree three. Use the hit and trial method to find the factors of p. The other two factors of p will not be real. So discard them. Now find the value of $\dfrac{z}{y}$ by multiplying numerator and denominator by x. Then, substitute the values and solve it to get the result.

Complete step-by-step answer:
Since, the value in the root cannot be less than zero and the denominator cannot be zero. So,
$4x - \left| {{x^2} - 10x + 9} \right| > 0$
Move absolute function on the right side of the equation,
$4x > \left| {{x^2} - 10x + 9} \right|$
By definition of absolute value,
$ - 4x < {x^2} - 10x + 9 < 4x$
So, the two inequalities are,
${x^2} - 10x + 9 > - 4x$ ….. (1)
${x^2} - 10x + 9 < 4x$ ….. (2)
Solve equation (1),
${x^2} - 10x + 9 > - 4x$
Add $4x$ on both sides,
${x^2} - 6x + 9 > 0$
Factorize the terms,
${x^2} + 2 \times - 3 \times x + {\left( { - 3} \right)^2} > 0$
As we know that ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$. Then,
${\left( {x - 3} \right)^2} > 0$
Then,
$x \in \left( { - \infty ,3} \right) \cup \left( {3,\infty } \right)$ ….. (3)
Now, solve equation (2),
${x^2} - 10x + 9 < 4x$
Subtract $4x$ from both sides,
${x^2} - 14x + 9 < 0$
Add 40 on both sides,
${x^2} - 14x + 49 < 40$
Factorize the terms,
${x^2} + 2 \times - 7 \times x + {\left( { - 7} \right)^2} < 40$
As we know that ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$. Then,
${\left( {x - 7} \right)^2} < 40$
Take square root on both sides,
$ - \sqrt {40} < x - 7 < \sqrt {40} $
Add 7 on all sides of the equation,
$7 - \sqrt {40} < x < 7 + \sqrt {40} $
Then,
$x \in \left( {7 - \sqrt {40} ,7 + \sqrt {40} } \right)$ ….. (4)
For domain of the function $f\left( x \right)$, take intersection of equation (3) and (4),
$x \in \left( {7 - \sqrt {40} ,3} \right) \cup \left( {3,7 + \sqrt {40} } \right)$

Hence, option (d) is the correct answer.

Note: The students might make mistakes to take the intersection of the 2 equations to find the domain of the function.
A function is a relation for which each value from the set of the first components of the ordered pairs is associated with exactly one value from the set of second components of the ordered pair.
The domain of a function is the set of all possible inputs for the function.