If \[f(x) = \ln ({x^2} + 7\left| x \right| + 10)\] is a single valued real function then the range of\[f(x)\] in its natural domain will be
A) \[[0, + \infty ]\]
B) \[[\ln 10, + \infty ]\]
C) \[[0,10]\]
D) \[R\]
Answer
642.6k+ views
Hint:
Here we will form a condition as the log function is present in the given equation. Then we will analyze and solve the equation. Simplifying the equation further we will get the natural domain of the given function\[f(x)\].
Complete step by step solution:
The given function is\[f(x) = \ln ({x^2} + 7\left| x \right| + 10)\]
Firstly we will form a condition as the log function is present. So, from the given expression we can say that the log term is positive as log can only take the positive value. Therefore, we get
\[{x^2} + 7\left| x \right| + 10 > 0\]………………. \[\left( 1 \right)\]
This condition is valid for both the positive and negative value of the variable \[x\].
Now from the equation \[\left( 1 \right)\] we can see that it have a term of
\[{x^2}\]. So, whatever the value of\[x\] is i.e. positive or negative. The value of this term i.e. \[{x^2}\] is always positive.
Similarly from the equation \[\left( 1 \right)\] we can see that it have a term of\[7\left| x \right|\]. So, whatever the value of \[x\] is i.e. positive or negative. The value of this term i.e. \[7\left| x \right|\] is always positive because the variable \[x\] is in the mod function.
So by this we came to know that the value of the expression \[{x^2} + 7\left| x \right| + 10\] is always positive.
So the upper limit of the function is positive infinity i.e. \[ + \infty \].
Now we have to find out the lower limit of the function. So, by putting the value of\[x = 0\] in the equation \[\left( 1 \right)\] we will get the lower limit of the function. Therefore, we get
\[ \Rightarrow {x^2} + 7\left| x \right| + 10 = {0^2} + 7 \times \left| 0 \right| + 10 = 10\]
So, \[ \Rightarrow {x^2} + 7\left| x \right| + 10 \ge 10\]
Applying natural log to both sides, we get
\[ \Rightarrow \ln \left( {{x^2} + 7\left| x \right| + 10} \right) \ge \ln \left( {10} \right)\]
\[ \Rightarrow f(x) \ge \ln \left( {10} \right)\]
So,\[\ln \left( {10} \right)\] is the lower limit of the function and \[ + \infty \] is the upper limit of the function.
Hence, the natural domain of the function\[f(x)\] is \[[\ln 10, + \infty ]\]
So, option B is the correct option.
Note:
We should know that the natural domain of a function is the range of the function where its value can lie. Also we should know that the value of any term inside the log function is always positive; it can never be negative. Also we have to keep in mind the type of bracket which should be used to show the natural domain of the function. Either the brackets may be open bracket or closed bracket.
Open bracket is used to show the natural domain of the function when the end points are not included in it. For example: \[(2,15)\] which means \[2 < x < 15\].
Closed bracket is used to show the natural domain of the function when the end points are included in it. For example: \[[2,15]\] which means \[2 \le x \le 15\].
Here we will form a condition as the log function is present in the given equation. Then we will analyze and solve the equation. Simplifying the equation further we will get the natural domain of the given function\[f(x)\].
Complete step by step solution:
The given function is\[f(x) = \ln ({x^2} + 7\left| x \right| + 10)\]
Firstly we will form a condition as the log function is present. So, from the given expression we can say that the log term is positive as log can only take the positive value. Therefore, we get
\[{x^2} + 7\left| x \right| + 10 > 0\]………………. \[\left( 1 \right)\]
This condition is valid for both the positive and negative value of the variable \[x\].
Now from the equation \[\left( 1 \right)\] we can see that it have a term of
\[{x^2}\]. So, whatever the value of\[x\] is i.e. positive or negative. The value of this term i.e. \[{x^2}\] is always positive.
Similarly from the equation \[\left( 1 \right)\] we can see that it have a term of\[7\left| x \right|\]. So, whatever the value of \[x\] is i.e. positive or negative. The value of this term i.e. \[7\left| x \right|\] is always positive because the variable \[x\] is in the mod function.
So by this we came to know that the value of the expression \[{x^2} + 7\left| x \right| + 10\] is always positive.
So the upper limit of the function is positive infinity i.e. \[ + \infty \].
Now we have to find out the lower limit of the function. So, by putting the value of\[x = 0\] in the equation \[\left( 1 \right)\] we will get the lower limit of the function. Therefore, we get
\[ \Rightarrow {x^2} + 7\left| x \right| + 10 = {0^2} + 7 \times \left| 0 \right| + 10 = 10\]
So, \[ \Rightarrow {x^2} + 7\left| x \right| + 10 \ge 10\]
Applying natural log to both sides, we get
\[ \Rightarrow \ln \left( {{x^2} + 7\left| x \right| + 10} \right) \ge \ln \left( {10} \right)\]
\[ \Rightarrow f(x) \ge \ln \left( {10} \right)\]
So,\[\ln \left( {10} \right)\] is the lower limit of the function and \[ + \infty \] is the upper limit of the function.
Hence, the natural domain of the function\[f(x)\] is \[[\ln 10, + \infty ]\]
So, option B is the correct option.
Note:
We should know that the natural domain of a function is the range of the function where its value can lie. Also we should know that the value of any term inside the log function is always positive; it can never be negative. Also we have to keep in mind the type of bracket which should be used to show the natural domain of the function. Either the brackets may be open bracket or closed bracket.
Open bracket is used to show the natural domain of the function when the end points are not included in it. For example: \[(2,15)\] which means \[2 < x < 15\].
Closed bracket is used to show the natural domain of the function when the end points are included in it. For example: \[[2,15]\] which means \[2 \le x \le 15\].
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

What do you mean by retardation What is its SI uni class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Difference between physical and chemical change class 11 chemistry CBSE

