Question

Domain of the function $f(x) = \dfrac{1}{{\sqrt {4x - \left| {x_{}^2 - 10x + 9} \right|} }}$
a) $(7 - \sqrt {40.7} + \sqrt {40} )$
b) $(0.7 + \sqrt {40)} $
c) $(7 - \sqrt {40,} \infty )$
d) None of these

Answer 1

Hint: In order to find out the domain of the given function we have to assume that $4x - \left| {x_{}^2 - 10x + 9} \right| > 0$, since the mod value must be positive and the denominator value should not be zero.

Complete step-by-step answer:
Let the given function be $4x - \left| {x_{}^2 - 10x + 9} \right| > 0$
 We have to take $4x$on the right side, after that the inequality sign changes so we get
$\left| {x_{}^2 - 10x + 9} \right| < 4x$
Now we have to solve the equation two times
Firstly, $x_{}^2 - 10x + 9 < 4x$
Moving $4x$ towards left side we get-
$x_{}^2 - 10x - 4x + 9 < 0$
So, $x_{}^2 - 14x + 9 < 0$
Now, by applying shridhar Acharya formula that is $\dfrac{{ - b \pm \sqrt {b_{}^2 - 4ac} }}{2a}$ we get
Here we have a quadratic equation \[a{x^2} + bx + c{\text{ }} = 0\]
So we can write \[a = 1,{\text{ }}b = 14\] and \[c = 9\]
Substitute the above value in the formula we get,
$x = \dfrac{{14 \pm \sqrt {(14)_{}^2 - 4.1.9} }}{2}$
On squaring and\[14\] multiplying the terms we get,
\[x = \dfrac{{14 \pm \sqrt {196 - 36} }}{2}\]
On subtracting the terms we get,
$x = \dfrac{{14 \pm \sqrt {160} }}{2}$
We split the square root terms,
$x = \dfrac{{14 \pm \sqrt {40 \times 4} }}{2}$
Let us take $\sqrt 4 = 2$ and also we split $14 = 2 \times 7$ we get,
$x = \dfrac{{2 \times 7 \pm 2\sqrt {40} }}{2}$
Taking $2$ as common we get-
$x = \dfrac{{2(7 \pm \sqrt {40)} }}{2}$
Let us divided we get,
$x = (7 \pm \sqrt {40)} $
Therefore we can say that ${\text{[x - (7 - }}\sqrt {{\text{40)]}}} {\text{ [x - (7 + }}\sqrt {{\text{40}}} {\text{)]}}$$ < 0$
In this case the value of $x$ will range from [$(7 - \sqrt {40} ),(7 + \sqrt {40} )$]
In the next equation, $x_{}^2 - 10x + 9 > - 4x$
$x_{}^2 - 6x + 3_{}^2 > 0$
Applying the formula of $(a - b)_{}^2 = a_{}^2 + 2ab + b_{}^2$ we get-
$(x - 3)_{}^2 > 0$
Therefore value of $x \in R - 3$
So domain of the function is $x \in (7 - \sqrt {40} ,7 + \sqrt {40} ) - \{ 3\} $
Since the value of $x$ becomes $0$ when it reaches $3$

So, the correct answer is “Option d”.

Note: It is to be kept in mind while doing the question that the inequality sign changes when one term is moved from left side to right side of the equation.
Moreover, while determining the domain of the given function you need to find the maximum and minimum range so you have to solve the equation two times simultaneously and after that you have to take the mid-range of maximum and minimum value.