Question

The domain of the real-valued function $f\left( x \right) = \sqrt {5 - 4x - {x^2}} + {x^2}\log \left( {x + 4} \right)$ is
A. $ - 5 \le x \le 1$
B. $ - 5 \le x$ and $x \ge 1$
C. $ - 4 < x \le 1$
D. $\emptyset $

Answer 1

Hint: In this question a function is given and its domain is asked. The function is the sum of two types of function. One is square root function and the other is a product of an algebraic function and logarithmic function. Square root of an expression is valid if and only if the expression is zero or positive and the logarithm of an expression is valid if and only if the expression is greater than zero. Find the values of $x$ which satisfy these conditions and take the common values of $x$ as the required answer.

Formula Used:
Product of two expressions will be positive if and only if each of the expressions is of the same sign i.e. either both of the expressions is positive or both of them is negative.
Each square root function is defined in the domain $\left[ {0,\infty } \right)$.
The function ${x^2}$ is defined for all real values of $x$ and each logarithmic function is defined in the interval $\left( {0,\infty } \right)$.

Complete step by step solution:
The given function is $f\left( x \right) = \sqrt {5 - 4x - {x^2}} + {x^2}\log \left( {x + 4} \right)$
Clearly, $f\left( x \right)$ is a sum of two functions $\sqrt {5 - 4x - {x^2}} $ and ${x^2}\log \left( {x + 4} \right)$
Let $g\left( x \right) = \sqrt {5 - 4x - {x^2}} $, which is a square root function
And $h\left( x \right) = {x^2}\log \left( {x + 4} \right)$, which is again a product of an algebraic function ${x^2}$ and a logarithmic function $\log \left( {x + 4} \right)$
Each square root function is defined in the domain $\left[ {0,\infty } \right)$
So, $g\left( x \right)$ is defined in the domain for which $5 - 4x - {x^2} \ge 0$
Factorize the expression $5 - 4x - {x^2}$
$5 - 4x - {x^2}$
$ = 5 - 5x + x - {x^2}$
$ = 5\left( {1 - x} \right) + x\left( {1 - x} \right)$
$ = \left( {1 - x} \right)\left( {5 + x} \right)$
Now, $5 - 4x - {x^2} \ge 0$
$ \Rightarrow \left( {1 - x} \right)\left( {5 + x} \right) \ge 0$
Product of two expressions will be positive if and only if each of the expressions is of same sign i.e. either both of the expressions is positive or both of them is negative.
So, one possibility is $\left( {1 - x} \right) \ge 0$ along with $\left( {5 + x} \right) \ge 0$
And the other possibility is $\left( {1 - x} \right) \le 0$ along with $\left( {5 + x} \right) \le 0$
$\left( {1 - x} \right) \ge 0 \Rightarrow x \le 1$ and $\left( {5 + x} \right) \ge 0 \Rightarrow x \ge - 5$
Combining, we get $ - 5 \le x \le 1$
$\left( {1 - x} \right) \le 0 \Rightarrow x \ge 1$ and $\left( {5 + x} \right) \le 0 \Rightarrow x \le - 5$
Combining, we get $x \le - 5$ or $x \ge 1$
But it is not possible.
So, the one and only one solution is $ - 5 \le x \le 1$.
Hence the domain of the function $g\left( x \right)$ is $\left[ { - 5,1} \right]$.
Now, $h\left( x \right)$ is a product of an algebraic function ${x^2}$ and a logarithmic function $\log \left( {x + 4} \right)$
The function ${x^2}$ is defined for all real values of $x$ and each logarithmic function is defined in the interval $\left( {0,\infty } \right)$.
So, $h\left( x \right)$ is defined in the domain for which $x + 4 > 0$
$x + 4 > 0 \Rightarrow x > - 4$
Hence the domain of the function $h\left( x \right)$ is $\left( { - 4,\infty } \right)$.
The domain of the given function $f\left( x \right)$ is the intersection i.e. the common region of the solution regions of the functions $g\left( x \right)$ and $h\left( x \right)$.
The common domain for the function $g\left( x \right)$ and $h\left( x \right)$ is $\left( { - 4,1} \right]$ i.e. $ - 4 < x \le 1$
So, the domain of the given function is $ - 4 < x \le 1$.

Option ‘C’ is correct

Note: Domain of a function $f\left( x \right)$ is a set of values of the independent variable $x$ for which the function is defined. Domain of a function, which is given as a sum of two or more than two functions, is the common domain of the functions. Product of two expressions will be positive if and only if each of the expressions is of the same sign i.e. either both of the expressions is positive or both of them is negative. Each square root function is defined in the domain $\left[ {0,\infty } \right)$. The function ${x^2}$ is defined for all real values of $x$ and each logarithmic function is defined in the interval $\left( {0,\infty } \right)$.