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# The condition that $f\left( x \right) = a{x^3} + b{x^2} + cx + d$ has no extreme value isA) ${b^2} > 3ac$B) ${b^2} = 4ac$C) ${b^2} = 3ac$D) ${b^2} < 3ac$

Last updated date: 19th Jun 2024
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Hint: Here we are given a function of $x$ so we will first differentiate it and then put the discriminant of the equation less than zero to get the desired answer.
For a standard quadratic equation $a{x^2} + bx + c = 0$
The discriminant is given by:-
$D = {b^2} - 4ac$
If a function has no extrema, it implies that it has no maximum or minimum values.

Complete step by step solution:
We are given that:
$f\left( x \right) = a{x^3} + b{x^2} + cx + d$
Differentiating the above function with respect to $x$ we get:-
$\Rightarrow \dfrac{d}{{dx}}\left[ {f\left( x \right)} \right] = \dfrac{d}{{dx}}\left[ {a{x^3} + b{x^2} + cx + d} \right]$
$\Rightarrow \dfrac{d}{{dx}}\left[ {f\left( x \right)} \right] = \dfrac{d}{{dx}}\left( {a{x^3}} \right) + \dfrac{d}{{dx}}\left( {b{x^2}} \right) + \dfrac{d}{{dx}}\left( {cx} \right) + \dfrac{d}{{dx}}\left( d \right)$
$\Rightarrow \dfrac{d}{{dx}}\left[ {f\left( x \right)} \right] = a\dfrac{d}{{dx}}\left( {{x^3}} \right) + b\dfrac{d}{{dx}}\left( {{x^2}} \right) + c\dfrac{d}{{dx}}\left( x \right) + \dfrac{d}{{dx}}\left( d \right)$
Now we know that the derivative of a constant is zero.
Also applying the following formula:
$\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$
We get:
$\dfrac{d}{{dx}}\left[ {f\left( x \right)} \right] = a\left( {3{x^2}} \right) + b\left( {2x} \right) + c\left( 1 \right) + 0$
$\Rightarrow \dfrac{d}{{dx}}\left[ {f\left( x \right)} \right] = 3a{x^2} + 2bx + c$
Now, in order to find the extreme values we put
$\dfrac{d}{{dx}}\left[ {f\left( x \right)} \right] = 0$
Therefore we get:-
$\Rightarrow 3a{x^2} + 2bx + c = 0$
Now, For no extrema, f should have zero turning points, i.e. above quadratic have no real roots.
Therefore, the discriminant of the above equation should be less than zero
Now since for a standard quadratic equation $a{x^2} + bx + c = 0$
The discriminant is given by:-
$D = {b^2} - 4ac$
Hence,
Calculating the discriminant for above equation we get:-
$\Rightarrow D = {\left( {2b} \right)^2} - 4\left( {3a} \right)\left( c \right) < 0$
$\Rightarrow 4{b^2} - 12ac < 0$
On simplification, we get
$\Rightarrow {b^2} - 3ac < 0$
$\Rightarrow {b^2} < 3ac$

$\therefore$ Hence, option D is the correct answer.

Note:
In order to find the extema values we need to differentiate the given function and find the values of $x$ and put that value back in the function and find its value which are the extremum values.
Also, in a quadratic equation if discriminant D is greater than zero ($D > 0$ ) then its has real and distinct roots.
If discriminant is equal to zero ($D = 0$ ) then it has equal roots
If discriminant is less than zero ($D < 0$) then it has imaginary roots.
For a function to have no extrema, the equation of its derivative should have no roots
Therefore the students should put its discriminant to be less than zero in order to get correct answer.