
If the function $f\left( x \right) = {x^3} + 3\left( {a - 7} \right){x^3} + 3\left( {{a^2} - 9} \right)x - 1\,$ has a positive point of maxima, then Find the possible value of $a$
A. $a \in \left( {3,\infty } \right) \cup \left( { - \infty , - 3} \right)$
B. $a \in \left( { - \infty , - 3} \right) \cup \left( {3,\dfrac{{29}}{7}} \right)$
C. $\left( { - \infty ,7} \right)$
D. $\left( { - \infty ,\dfrac{{29}}{7}} \right)$
Answer
588.9k+ views
Hint: The immediate rate of change of a function with respect to one of its variables is the function derivative in the above problem. Using the methods of addition, subtraction, multiplication, and differentiation by a constant number, the derivative is more common for functions constructed from the primary elementary functions.
Formula used:
The composite function rule is also known as the chain rule.
\[f\left( x \right){\text{ }}\]is represent as differentiate a function,
According to the formula of differentiation,
Differentiation with respect to $x$ is given below
$f'\left( x \right) = \dfrac{{dy}}{{dx}}$
Differentiation with respect to $y$ is given below
$f'\left( y \right) = \dfrac{{dx}}{{dy}}$
Where,
$dx,dy$ is the differentiation Variable,
Also, $f'\left( x \right),f''\left( x \right),f'''\left( x \right)$ it is a continuous differentiation,
Similarly, $f'\left( y \right),f''\left( y \right),f'''\left( y \right)$
$\left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right)$
Where,
$a,b$ is an integer
Complete step-by-step answer:
Given by,
The value of $f\left( x \right) = {x^3} + 3\left( {a - 7} \right){x^3} + 3\left( {{a^2} - 9} \right)x - 1\,$
Find the corresponding the value $a$,
Let,
$f\left( x \right) = {x^3} + 3\left( {a - 7} \right){x^3} + 3\left( {{a^2} - 9} \right)x - 1\,$
The above equation can be differentiation with respect to $x$,
We get,
Already we know the formula of differentiation
So, it applies to the given equation,
$f'\left( x \right) = 3{x^2} + 6\left( {a - 7} \right){x^2} + 3\left( {{a^2} - 9} \right)$
Let $\alpha ,\beta $ be the roots of $f'\left( x \right) = 0$
Assume $\alpha $ smaller,
On simplifying,
$D > 0 \Rightarrow 36{\left( {a - 7} \right)^2} - 36\left( {{a^2} - 9} \right) > 0$
$36 - 9 = 27$
Rearranging the above equation,
Condition required, for solving the equation
$\left( i \right)D > 0\, \Rightarrow \,a < \dfrac{{29}}{7}$
The root of the given value be found,
$\left( {ii} \right)\dfrac{{ - b}}{{2a}} > 0\, \Rightarrow a < 7$
$f'\left( 0 \right) = 3\left( {{a^2} - 9} \right) > 0$
On simplifying a given equation
We know,
$\left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right)$
On substituting the value in above equation, we get,
\[\left( {a - 3} \right)\left( {a + 3} \right)\]
The range of the term ,
\[\left( {a - 3} \right)\left( {a + 3} \right) > 0\]
Rearranging,
$a < - 3\,\,$or $a > 3$
From $\left( i \right),\left( {ii} \right)$ and $\left( {iii} \right)$
We get,
$a \in \left( { - \infty , - 3} \right) \cup \left( {3,\dfrac{{29}}{7}} \right)$
Hence, option B is the correct answer.
Note: At this level, the first derivative of a function at a point determines the variance of the tangent to the graph. If the function is the value or the difference between the two functions, then the sum. Instead of using the general method of differentiation, the derivatives of some unique functions are evaluated.
Formula used:
The composite function rule is also known as the chain rule.
\[f\left( x \right){\text{ }}\]is represent as differentiate a function,
According to the formula of differentiation,
Differentiation with respect to $x$ is given below
$f'\left( x \right) = \dfrac{{dy}}{{dx}}$
Differentiation with respect to $y$ is given below
$f'\left( y \right) = \dfrac{{dx}}{{dy}}$
Where,
$dx,dy$ is the differentiation Variable,
Also, $f'\left( x \right),f''\left( x \right),f'''\left( x \right)$ it is a continuous differentiation,
Similarly, $f'\left( y \right),f''\left( y \right),f'''\left( y \right)$
$\left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right)$
Where,
$a,b$ is an integer
Complete step-by-step answer:
Given by,
The value of $f\left( x \right) = {x^3} + 3\left( {a - 7} \right){x^3} + 3\left( {{a^2} - 9} \right)x - 1\,$
Find the corresponding the value $a$,
Let,
$f\left( x \right) = {x^3} + 3\left( {a - 7} \right){x^3} + 3\left( {{a^2} - 9} \right)x - 1\,$
The above equation can be differentiation with respect to $x$,
We get,
Already we know the formula of differentiation
So, it applies to the given equation,
$f'\left( x \right) = 3{x^2} + 6\left( {a - 7} \right){x^2} + 3\left( {{a^2} - 9} \right)$
Let $\alpha ,\beta $ be the roots of $f'\left( x \right) = 0$
Assume $\alpha $ smaller,
On simplifying,
$D > 0 \Rightarrow 36{\left( {a - 7} \right)^2} - 36\left( {{a^2} - 9} \right) > 0$
$36 - 9 = 27$
Rearranging the above equation,
Condition required, for solving the equation
$\left( i \right)D > 0\, \Rightarrow \,a < \dfrac{{29}}{7}$
The root of the given value be found,
$\left( {ii} \right)\dfrac{{ - b}}{{2a}} > 0\, \Rightarrow a < 7$
$f'\left( 0 \right) = 3\left( {{a^2} - 9} \right) > 0$
On simplifying a given equation
We know,
$\left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right)$
On substituting the value in above equation, we get,
\[\left( {a - 3} \right)\left( {a + 3} \right)\]
The range of the term ,
\[\left( {a - 3} \right)\left( {a + 3} \right) > 0\]
Rearranging,
$a < - 3\,\,$or $a > 3$
From $\left( i \right),\left( {ii} \right)$ and $\left( {iii} \right)$
We get,
$a \in \left( { - \infty , - 3} \right) \cup \left( {3,\dfrac{{29}}{7}} \right)$
Hence, option B is the correct answer.
Note: At this level, the first derivative of a function at a point determines the variance of the tangent to the graph. If the function is the value or the difference between the two functions, then the sum. Instead of using the general method of differentiation, the derivatives of some unique functions are evaluated.
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