
Which of the following is an odd function:
A.$\sin {x^2}$
B.$\dfrac{{\left( {{a^x} + 1} \right)}}{{\left( {{a^x} - 1} \right)}}$
C.${x^2}\left| x \right|$
D.None of these
Answer
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Hint: We will firstly define what are odd functions. Then, we will check each and every option if they satisfy the given condition that if f (x) is an odd function then, f (-x) = - f(x). If they satisfy, then they will be called as an odd function and if not, then they will be called as an even function.
Complete step-by-step answer:
We are given options and we need to determine if they are odd functions or not.
Definition of an odd function: In mathematics, odd functions are those functions which satisfy some certain symmetry relation with respect to the additive inverses.
Or in other simpler words, a function having a graph symmetric with respect to origin is an odd function and a function is an odd function if and only if it satisfies the given condition:
i.e., if f (-x) = - f(x).
We will check every option if they satisfy this condition of odd functions.
Option(A): $\sin {x^2}$
Let f (x) = $\sin {x^2}$
Then, f (-x) = $\sin {\left( { - x} \right)^2}$= $\sin {x^2}$= f (x)
Hence, we can say that option(A) is an even function not an odd function.
Option(B): $\dfrac{{\left( {{a^x} + 1} \right)}}{{\left( {{a^x} - 1} \right)}}$
Let g (x) = $\dfrac{{\left( {{a^x} + 1} \right)}}{{\left( {{a^x} - 1} \right)}}$
Then, g (-x) = $\dfrac{{\left( {{a^{ - x}} + 1} \right)}}{{\left( {{a^{ - x}} - 1} \right)}}$
$ \Rightarrow $g (-x) = $\dfrac{{\left( {{a^{\dfrac{1}{x}}} + 1} \right)}}{{\left( {{a^{\dfrac{1}{x}}} - 1} \right)}} = \dfrac{{1 + {a^x}}}{{1 - {a^x}}} = - \dfrac{{{a^x} + 1}}{{{a^x} - 1}}$ = – g (x)
Hence, g (x) is an odd function. Therefore, option(B) is correct.
Option(C): ${x^2}\left| x \right|$
Let h (x) = ${x^2}\left| x \right|$
Then, h (-x) = ${\left( { - x} \right)^2}\left| { - x} \right| = {x^2}\left| x \right|$ = h (x)
Therefore, option(C) is also an even function not an odd function.
Hence, our answer will be option(B).
Note: In such questions, you should know about the concepts used. This question can also be solved using graphical methods by proving the symmetry of the graph. Also, you should check each and every option as such questions can have more than one answer.
Complete step-by-step answer:
We are given options and we need to determine if they are odd functions or not.
Definition of an odd function: In mathematics, odd functions are those functions which satisfy some certain symmetry relation with respect to the additive inverses.
Or in other simpler words, a function having a graph symmetric with respect to origin is an odd function and a function is an odd function if and only if it satisfies the given condition:
i.e., if f (-x) = - f(x).
We will check every option if they satisfy this condition of odd functions.
Option(A): $\sin {x^2}$
Let f (x) = $\sin {x^2}$
Then, f (-x) = $\sin {\left( { - x} \right)^2}$= $\sin {x^2}$= f (x)
Hence, we can say that option(A) is an even function not an odd function.
Option(B): $\dfrac{{\left( {{a^x} + 1} \right)}}{{\left( {{a^x} - 1} \right)}}$
Let g (x) = $\dfrac{{\left( {{a^x} + 1} \right)}}{{\left( {{a^x} - 1} \right)}}$
Then, g (-x) = $\dfrac{{\left( {{a^{ - x}} + 1} \right)}}{{\left( {{a^{ - x}} - 1} \right)}}$
$ \Rightarrow $g (-x) = $\dfrac{{\left( {{a^{\dfrac{1}{x}}} + 1} \right)}}{{\left( {{a^{\dfrac{1}{x}}} - 1} \right)}} = \dfrac{{1 + {a^x}}}{{1 - {a^x}}} = - \dfrac{{{a^x} + 1}}{{{a^x} - 1}}$ = – g (x)
Hence, g (x) is an odd function. Therefore, option(B) is correct.
Option(C): ${x^2}\left| x \right|$
Let h (x) = ${x^2}\left| x \right|$
Then, h (-x) = ${\left( { - x} \right)^2}\left| { - x} \right| = {x^2}\left| x \right|$ = h (x)
Therefore, option(C) is also an even function not an odd function.
Hence, our answer will be option(B).
Note: In such questions, you should know about the concepts used. This question can also be solved using graphical methods by proving the symmetry of the graph. Also, you should check each and every option as such questions can have more than one answer.
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