If the function \[f\left( x \right) = ax{e^{b{x^2}}}\] has the maximum value \[f\left( 2 \right) = 1\], then
(a) \[a = \dfrac{{\sqrt e }}{2}\] and \[b = - \dfrac{1}{8}\]
(b) \[a = - \dfrac{{\sqrt e }}{2}\] and \[b = \dfrac{1}{8}\]
(c) \[a = \dfrac{1}{8}\] and \[b = \dfrac{{\sqrt e }}{2}\]
(d) \[a = - \dfrac{{\sqrt e }}{2}\] and \[b = - \dfrac{1}{8}\]
Answer
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Hint:
Here, we need to find the value of \[a\] and \[b\]. First, we will use the given function and maximum value to form an equation. Then, we will find the first derivative of the function. We will equate this equation to 0. and simplify to get one of the required values. Using that value in the first equation and simplifying, we can find the other required value.
Formula Used: We will use the formula of :
1) The product rule of differentiation states that the derivative of the product of two functions is given as \[\dfrac{{d\left( {uv} \right)}}{{dx}} = u\dfrac{{d\left( v \right)}}{{dx}} + \dfrac{{d\left( u \right)}}{{dx}}v\].
2) The derivative of a function of the form \[{e^{f\left( x \right)}}\] is given by \[{e^{f\left( x \right)}}\dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}}\].
3) The derivative of a function of the form \[{x^n}\] is given by \[n{x^{n - 1}}\].
Complete step by step solution:
First, we will find the value of \[f\left( 2 \right)\].
Substituting \[x = 2\] in the function, we get
\[ \Rightarrow f\left( 2 \right) = a\left( 2 \right){e^{b{{\left( 2 \right)}^2}}}\]
Simplifying the expression, we get
\[ \Rightarrow f\left( 2 \right) = 2a{e^{4b}}\]
It is given that \[f\left( 2 \right) = 1\].
Therefore, we get
\[ \Rightarrow 2a{e^{4b}} = 1 \ldots \ldots \ldots \left( 1 \right)\]
To find the value of \[x\] at which a function is maximum or minimum, we equate the first derivative of the function to 0.
Differentiating both sides of the equation \[f\left( x \right) = ax{e^{b{x^2}}}\] with respect to \[x\], we get
\[\dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}} = \dfrac{{d\left( {ax{e^{b{x^2}}}} \right)}}{{dx}}\]
The product rule of differentiation states that the derivative of the product of two functions is given as \[\dfrac{{d\left( {uv} \right)}}{{dx}} = u\dfrac{{d\left( v \right)}}{{dx}} + \dfrac{{d\left( u \right)}}{{dx}}v\].
Simplifying the equation using the product rule of differentiation, we get
\[ \Rightarrow \dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}} = ax\dfrac{{d\left( {{e^{b{x^2}}}} \right)}}{{dx}} + \dfrac{{d\left( {ax} \right)}}{{dx}}{e^{b{x^2}}}\]
The derivative of a function of the form \[{e^{f\left( x \right)}}\] is given by \[{e^{f\left( x \right)}}\dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}}\].
Therefore, the equation becomes
\[ \Rightarrow \dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}} = ax \times {e^{b{x^2}}} \times \dfrac{{d\left( {b{x^2}} \right)}}{{dx}} + \dfrac{{d\left( {a{x^1}} \right)}}{{dx}}{e^{b{x^2}}}\]
The derivative of a function of the form \[f\left( {ax} \right)\] is given by \[af'\left( x \right)\].
\[ \Rightarrow \dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}} = ax \times {e^{b{x^2}}} \times b\dfrac{{d\left( {{x^2}} \right)}}{{dx}} + a\dfrac{{d\left( {{x^1}} \right)}}{{dx}}{e^{b{x^2}}}\]
The derivative of a function of the form \[{x^n}\] is given by \[n{x^{n - 1}}\].
Therefore, we get
\[\begin{array}{l} \Rightarrow \dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}} = ax \times {e^{b{x^2}}} \times b\left( {2{x^1}} \right) + a\left( {1{x^0}} \right){e^{b{x^2}}}\\ \Rightarrow \dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}} = ax \times {e^{b{x^2}}} \times b\left( {2x} \right) + a\left( 1 \right){e^{b{x^2}}}\end{array}\]
Simplifying the equation by multiplying the terms, we get
\[ \Rightarrow \dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}} = 2ab{x^2}{e^{b{x^2}}} + a{e^{b{x^2}}}\]
It is given that the maximum value of the function is at \[x = 2\].
Therefore, the first derivative is equal to 0 at \[x = 2\].
Thus, equating the first derivative to 0, we get
\[ \Rightarrow 2ab{x^2}{e^{b{x^2}}} + a{e^{b{x^2}}} = 0\]
Factoring out the term \[a{e^{b{x^2}}}\], we get
\[ \Rightarrow a{e^{b{x^2}}}\left( {2b{x^2} + 1} \right) = 0\]
Dividing both sides of the equation by \[a{e^{b{x^2}}}\], we get
\[\begin{array}{l} \Rightarrow \dfrac{{a{e^{b{x^2}}}\left( {2b{x^2} + 1} \right)}}{{a{e^{b{x^2}}}}} = \dfrac{0}{{a{e^{b{x^2}}}}}\\ \Rightarrow 2b{x^2} + 1 = 0\end{array}\]
Substituting \[x = 2\] in the equation, we get
\[ \Rightarrow 2b{\left( 2 \right)^2} + 1 = 0\]
Simplifying the expression, we get
\[\begin{array}{l} \Rightarrow 2b\left( 4 \right) + 1 = 0\\ \Rightarrow 8b + 1 = 0\end{array}\]
Subtracting 1 from both sides, we get
\[\begin{array}{l} \Rightarrow 8b + 1 - 1 = 0 - 1\\ \Rightarrow 8b = - 1\end{array}\]
Dividing both sides of the equation by 8, we get
\[ \Rightarrow \dfrac{{8b}}{8} = \dfrac{{ - 1}}{8}\]
Thus, we get
\[ \Rightarrow b = - \dfrac{1}{8}\]
Substituting \[b = - \dfrac{1}{8}\] in equation \[\left( 1 \right)\], we get
\[ \Rightarrow 2a{e^{4\left( { - \dfrac{1}{8}} \right)}} = 1\]
Multiplying the terms in the exponent, we get
\[ \Rightarrow 2a{e^{ - \dfrac{1}{2}}} = 1\]
Dividing both sides of the equation by \[2{e^{ - \dfrac{1}{2}}}\], we get
\[\begin{array}{l} \Rightarrow \dfrac{{2a{e^{ - \dfrac{1}{2}}}}}{{2{e^{ - \dfrac{1}{2}}}}} = \dfrac{1}{{2{e^{ - \dfrac{1}{2}}}}}\\ \Rightarrow a = \dfrac{1}{{2{e^{ - \dfrac{1}{2}}}}}\end{array}\]
Therefore, we get
\[\begin{array}{l} \Rightarrow a = \dfrac{1}{{2\left( {\dfrac{1}{{{e^{\dfrac{1}{2}}}}}} \right)}}\\ \Rightarrow a = \dfrac{{{e^{\dfrac{1}{2}}}}}{2}\end{array}\]
Rewriting the expression \[{e^{\dfrac{1}{2}}}\] as \[\sqrt e \], we get
\[ \Rightarrow a = \dfrac{{\sqrt e }}{2}\]
Therefore, we get the value of \[a\] and \[b\] as \[\dfrac{{\sqrt e }}{2}\] and \[ - \dfrac{1}{8}\] respectively.
Thus, the correct option is option (a).
Note:
We can rewrite \[{e^{ - \dfrac{1}{2}}}\] as \[\dfrac{1}{{{e^{\dfrac{1}{2}}}}}\] because \[{e^{ - \dfrac{1}{2}}}\] has a negative exponent, that is \[ - \dfrac{1}{2}\]. If a number \[a\] is raised to the negative power \[ - b\], then, \[{a^{ - b}}\] is equal to the reciprocal of the number \[a\] raised to the positive power \[b\], that is \[{a^{ - b}} = {\left( {\dfrac{1}{a}} \right)^b}\].
Here, we need to find the value of \[a\] and \[b\]. First, we will use the given function and maximum value to form an equation. Then, we will find the first derivative of the function. We will equate this equation to 0. and simplify to get one of the required values. Using that value in the first equation and simplifying, we can find the other required value.
Formula Used: We will use the formula of :
1) The product rule of differentiation states that the derivative of the product of two functions is given as \[\dfrac{{d\left( {uv} \right)}}{{dx}} = u\dfrac{{d\left( v \right)}}{{dx}} + \dfrac{{d\left( u \right)}}{{dx}}v\].
2) The derivative of a function of the form \[{e^{f\left( x \right)}}\] is given by \[{e^{f\left( x \right)}}\dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}}\].
3) The derivative of a function of the form \[{x^n}\] is given by \[n{x^{n - 1}}\].
Complete step by step solution:
First, we will find the value of \[f\left( 2 \right)\].
Substituting \[x = 2\] in the function, we get
\[ \Rightarrow f\left( 2 \right) = a\left( 2 \right){e^{b{{\left( 2 \right)}^2}}}\]
Simplifying the expression, we get
\[ \Rightarrow f\left( 2 \right) = 2a{e^{4b}}\]
It is given that \[f\left( 2 \right) = 1\].
Therefore, we get
\[ \Rightarrow 2a{e^{4b}} = 1 \ldots \ldots \ldots \left( 1 \right)\]
To find the value of \[x\] at which a function is maximum or minimum, we equate the first derivative of the function to 0.
Differentiating both sides of the equation \[f\left( x \right) = ax{e^{b{x^2}}}\] with respect to \[x\], we get
\[\dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}} = \dfrac{{d\left( {ax{e^{b{x^2}}}} \right)}}{{dx}}\]
The product rule of differentiation states that the derivative of the product of two functions is given as \[\dfrac{{d\left( {uv} \right)}}{{dx}} = u\dfrac{{d\left( v \right)}}{{dx}} + \dfrac{{d\left( u \right)}}{{dx}}v\].
Simplifying the equation using the product rule of differentiation, we get
\[ \Rightarrow \dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}} = ax\dfrac{{d\left( {{e^{b{x^2}}}} \right)}}{{dx}} + \dfrac{{d\left( {ax} \right)}}{{dx}}{e^{b{x^2}}}\]
The derivative of a function of the form \[{e^{f\left( x \right)}}\] is given by \[{e^{f\left( x \right)}}\dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}}\].
Therefore, the equation becomes
\[ \Rightarrow \dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}} = ax \times {e^{b{x^2}}} \times \dfrac{{d\left( {b{x^2}} \right)}}{{dx}} + \dfrac{{d\left( {a{x^1}} \right)}}{{dx}}{e^{b{x^2}}}\]
The derivative of a function of the form \[f\left( {ax} \right)\] is given by \[af'\left( x \right)\].
\[ \Rightarrow \dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}} = ax \times {e^{b{x^2}}} \times b\dfrac{{d\left( {{x^2}} \right)}}{{dx}} + a\dfrac{{d\left( {{x^1}} \right)}}{{dx}}{e^{b{x^2}}}\]
The derivative of a function of the form \[{x^n}\] is given by \[n{x^{n - 1}}\].
Therefore, we get
\[\begin{array}{l} \Rightarrow \dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}} = ax \times {e^{b{x^2}}} \times b\left( {2{x^1}} \right) + a\left( {1{x^0}} \right){e^{b{x^2}}}\\ \Rightarrow \dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}} = ax \times {e^{b{x^2}}} \times b\left( {2x} \right) + a\left( 1 \right){e^{b{x^2}}}\end{array}\]
Simplifying the equation by multiplying the terms, we get
\[ \Rightarrow \dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}} = 2ab{x^2}{e^{b{x^2}}} + a{e^{b{x^2}}}\]
It is given that the maximum value of the function is at \[x = 2\].
Therefore, the first derivative is equal to 0 at \[x = 2\].
Thus, equating the first derivative to 0, we get
\[ \Rightarrow 2ab{x^2}{e^{b{x^2}}} + a{e^{b{x^2}}} = 0\]
Factoring out the term \[a{e^{b{x^2}}}\], we get
\[ \Rightarrow a{e^{b{x^2}}}\left( {2b{x^2} + 1} \right) = 0\]
Dividing both sides of the equation by \[a{e^{b{x^2}}}\], we get
\[\begin{array}{l} \Rightarrow \dfrac{{a{e^{b{x^2}}}\left( {2b{x^2} + 1} \right)}}{{a{e^{b{x^2}}}}} = \dfrac{0}{{a{e^{b{x^2}}}}}\\ \Rightarrow 2b{x^2} + 1 = 0\end{array}\]
Substituting \[x = 2\] in the equation, we get
\[ \Rightarrow 2b{\left( 2 \right)^2} + 1 = 0\]
Simplifying the expression, we get
\[\begin{array}{l} \Rightarrow 2b\left( 4 \right) + 1 = 0\\ \Rightarrow 8b + 1 = 0\end{array}\]
Subtracting 1 from both sides, we get
\[\begin{array}{l} \Rightarrow 8b + 1 - 1 = 0 - 1\\ \Rightarrow 8b = - 1\end{array}\]
Dividing both sides of the equation by 8, we get
\[ \Rightarrow \dfrac{{8b}}{8} = \dfrac{{ - 1}}{8}\]
Thus, we get
\[ \Rightarrow b = - \dfrac{1}{8}\]
Substituting \[b = - \dfrac{1}{8}\] in equation \[\left( 1 \right)\], we get
\[ \Rightarrow 2a{e^{4\left( { - \dfrac{1}{8}} \right)}} = 1\]
Multiplying the terms in the exponent, we get
\[ \Rightarrow 2a{e^{ - \dfrac{1}{2}}} = 1\]
Dividing both sides of the equation by \[2{e^{ - \dfrac{1}{2}}}\], we get
\[\begin{array}{l} \Rightarrow \dfrac{{2a{e^{ - \dfrac{1}{2}}}}}{{2{e^{ - \dfrac{1}{2}}}}} = \dfrac{1}{{2{e^{ - \dfrac{1}{2}}}}}\\ \Rightarrow a = \dfrac{1}{{2{e^{ - \dfrac{1}{2}}}}}\end{array}\]
Therefore, we get
\[\begin{array}{l} \Rightarrow a = \dfrac{1}{{2\left( {\dfrac{1}{{{e^{\dfrac{1}{2}}}}}} \right)}}\\ \Rightarrow a = \dfrac{{{e^{\dfrac{1}{2}}}}}{2}\end{array}\]
Rewriting the expression \[{e^{\dfrac{1}{2}}}\] as \[\sqrt e \], we get
\[ \Rightarrow a = \dfrac{{\sqrt e }}{2}\]
Therefore, we get the value of \[a\] and \[b\] as \[\dfrac{{\sqrt e }}{2}\] and \[ - \dfrac{1}{8}\] respectively.
Thus, the correct option is option (a).
Note:
We can rewrite \[{e^{ - \dfrac{1}{2}}}\] as \[\dfrac{1}{{{e^{\dfrac{1}{2}}}}}\] because \[{e^{ - \dfrac{1}{2}}}\] has a negative exponent, that is \[ - \dfrac{1}{2}\]. If a number \[a\] is raised to the negative power \[ - b\], then, \[{a^{ - b}}\] is equal to the reciprocal of the number \[a\] raised to the positive power \[b\], that is \[{a^{ - b}} = {\left( {\dfrac{1}{a}} \right)^b}\].
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