Answer
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Hint: Here we will use the property of functions \[f\left( x \right)\] and \[g\left( x \right)\] which states that we can do the composition between these two functions, which means that we can plug \[g\left( x \right)\] into \[f\left( x \right)\]. This is written as \[\left( {fog} \right)\left( x \right)\] , pronounced as
\[f\] compose \[g\] of \[x\].
\[\left( {fog} \right)\left( x \right) = f\left( {g\left( x \right)} \right)\].
Complete step-by-step answer:
Step (1): For statement (A):
It is given that \[f\left( x \right) = \log x\] and \[g\left( x \right) = {x^3}\]. Now, we need to check if \[f\left[ {g\left( a \right)} \right] + f\left[ {g\left( b \right)} \right] = f\left[ {g\left( {ab} \right)} \right]\].
For calculating the LHS side of the equation \[f\left[ {g\left( a \right)} \right] + f\left[ {g\left( b \right)} \right] = f\left[ {g\left( {ab} \right)} \right]\], we will substitute the values of functions \[f\left( x \right) = \log x\] and \[g\left( x \right) = {x^3}\] in it.
\[ \Rightarrow f\left[ {g\left( a \right)} \right] = \log \left( {{a^3}} \right)\] (\[\because \]\[g\left( x \right) = {x^3}\] ) …….. (1)
\[ \Rightarrow f\left[ {g\left( b \right)} \right] = \log \left( {{b^3}} \right)\] (\[\because \]\[g\left( x \right) = {x^3}\] ) …………. (2)
Now, by substituting the values from (1) and (2) in the LHS side of the equation \[f\left[ {g\left( a \right)} \right] + f\left[ {g\left( b \right)} \right] = f\left[ {g\left( {ab} \right)} \right]\], we get:
\[ \Rightarrow f\left[ {g\left( a \right)} \right] + f\left[ {g\left( b \right)} \right] = \log {a^3} + \log {b^3}\]
But we know that \[\log {a^3} + \log {b^3} = \log {a^3}{b^3} = \log {\left( {ab} \right)^3}\] , by putting this value in the expression \[f\left[ {g\left( a \right)} \right] + f\left[ {g\left( b \right)} \right]\], we get:
\[ \Rightarrow f\left[ {g\left( a \right)} \right] + f\left[ {g\left( b \right)} \right] = \log {\left( {ab} \right)^3}\]
Now for calculating the RHS side of the equation \[f\left[ {g\left( a \right)} \right] + f\left[ {g\left( b \right)} \right] = f\left[ {g\left( {ab} \right)} \right]\], we will substitute the values of \[f\left( x \right) = \log x\] and \[g\left( x \right) = {x^3}\] in it.
\[ \Rightarrow f\left[ {g\left( {ab} \right)} \right] = \log {\left( {ab} \right)^3}\] (\[\because \]\[f\left( x \right) = \log x\] and \[g\left( x \right) = {x^3}\])
So, for the expression \[f\left[ {g\left( a \right)} \right] + f\left[ {g\left( b \right)} \right] = f\left[ {g\left( {ab} \right)} \right]\], LHS= RHS, hence the statement is true.
Step 2: Statement (2):
We need to check if every trigonometric function is an even function.
Now we know that a function is said to be even if \[f\left( { - x} \right) = f\left( x \right)\]. For checking this we will take an example as shown below:
For checking if \[\sin \left( x \right)\]is an even function or not, we need to prove that \[\sin \left( { - x} \right) = \sin \left( x \right)\] but that is not true. Because \[\sin \left( { - x} \right) = - \sin \left( x \right)\]. Hence the function is not even.
It is proved that all trigonometric functions are not even. So, statement (B) is false.
Option C which states that statement A is correct and statement B is false is correct.
Note: In these types of questions, students’ needs to remember that for two different functions \[f\left( x \right)\] and \[g\left( x \right)\]:
\[\left( {fog} \right)\left( x \right) = f\left( {g\left( x \right)} \right)\]
Also, you should remember that all trigonometric functions are not even. A function is said to be an odd function if for any number \[x\] , \[f\left( { - x} \right) = - f\left( x \right)\]. And a function is said to be even for any number \[x\] , \[f\left( { - x} \right) = f\left( x \right)\].
Sine and tangent are odd functions. But cosine is an even function.
\[f\] compose \[g\] of \[x\].
\[\left( {fog} \right)\left( x \right) = f\left( {g\left( x \right)} \right)\].
Complete step-by-step answer:
Step (1): For statement (A):
It is given that \[f\left( x \right) = \log x\] and \[g\left( x \right) = {x^3}\]. Now, we need to check if \[f\left[ {g\left( a \right)} \right] + f\left[ {g\left( b \right)} \right] = f\left[ {g\left( {ab} \right)} \right]\].
For calculating the LHS side of the equation \[f\left[ {g\left( a \right)} \right] + f\left[ {g\left( b \right)} \right] = f\left[ {g\left( {ab} \right)} \right]\], we will substitute the values of functions \[f\left( x \right) = \log x\] and \[g\left( x \right) = {x^3}\] in it.
\[ \Rightarrow f\left[ {g\left( a \right)} \right] = \log \left( {{a^3}} \right)\] (\[\because \]\[g\left( x \right) = {x^3}\] ) …….. (1)
\[ \Rightarrow f\left[ {g\left( b \right)} \right] = \log \left( {{b^3}} \right)\] (\[\because \]\[g\left( x \right) = {x^3}\] ) …………. (2)
Now, by substituting the values from (1) and (2) in the LHS side of the equation \[f\left[ {g\left( a \right)} \right] + f\left[ {g\left( b \right)} \right] = f\left[ {g\left( {ab} \right)} \right]\], we get:
\[ \Rightarrow f\left[ {g\left( a \right)} \right] + f\left[ {g\left( b \right)} \right] = \log {a^3} + \log {b^3}\]
But we know that \[\log {a^3} + \log {b^3} = \log {a^3}{b^3} = \log {\left( {ab} \right)^3}\] , by putting this value in the expression \[f\left[ {g\left( a \right)} \right] + f\left[ {g\left( b \right)} \right]\], we get:
\[ \Rightarrow f\left[ {g\left( a \right)} \right] + f\left[ {g\left( b \right)} \right] = \log {\left( {ab} \right)^3}\]
Now for calculating the RHS side of the equation \[f\left[ {g\left( a \right)} \right] + f\left[ {g\left( b \right)} \right] = f\left[ {g\left( {ab} \right)} \right]\], we will substitute the values of \[f\left( x \right) = \log x\] and \[g\left( x \right) = {x^3}\] in it.
\[ \Rightarrow f\left[ {g\left( {ab} \right)} \right] = \log {\left( {ab} \right)^3}\] (\[\because \]\[f\left( x \right) = \log x\] and \[g\left( x \right) = {x^3}\])
So, for the expression \[f\left[ {g\left( a \right)} \right] + f\left[ {g\left( b \right)} \right] = f\left[ {g\left( {ab} \right)} \right]\], LHS= RHS, hence the statement is true.
Step 2: Statement (2):
We need to check if every trigonometric function is an even function.
Now we know that a function is said to be even if \[f\left( { - x} \right) = f\left( x \right)\]. For checking this we will take an example as shown below:
For checking if \[\sin \left( x \right)\]is an even function or not, we need to prove that \[\sin \left( { - x} \right) = \sin \left( x \right)\] but that is not true. Because \[\sin \left( { - x} \right) = - \sin \left( x \right)\]. Hence the function is not even.
It is proved that all trigonometric functions are not even. So, statement (B) is false.
Option C which states that statement A is correct and statement B is false is correct.
Note: In these types of questions, students’ needs to remember that for two different functions \[f\left( x \right)\] and \[g\left( x \right)\]:
\[\left( {fog} \right)\left( x \right) = f\left( {g\left( x \right)} \right)\]
Also, you should remember that all trigonometric functions are not even. A function is said to be an odd function if for any number \[x\] , \[f\left( { - x} \right) = - f\left( x \right)\]. And a function is said to be even for any number \[x\] , \[f\left( { - x} \right) = f\left( x \right)\].
Sine and tangent are odd functions. But cosine is an even function.
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