Question

If the function $f\left( x \right)=2{{x}^{3}}-9a{{x}^{2}}+12{{a}^{2}}x+1$ has a local maximum at $x={{x}_{1}}$ and a local minimum at $x={{x}_{2}}$ such that ${{x}_{2}}={{x}_{1}}^{2}$ then the value of a is equal to
$\begin{align}
  & a)0 \\
 & b)\dfrac{1}{2} \\
 & c)2 \\
 & d)\text{ either (a) or (c)} \\
\end{align}$

Answer 1

Hint: Now we are given with $f\left( x \right)=2{{x}^{3}}-9a{{x}^{2}}+12{{a}^{2}}x+1$ . Now we know that the conditions for extrema is $f'\left( x \right)=0$ hence using this we find the points of extrema
Now for an extremum point we have $f''\left( x \right)<0\Rightarrow $function has a maxima at that point
And similarly for $f''\left( x \right)>0\Rightarrow $function has a minimum at that point. Hence using this condition we can determine which point is maximum and which point is minimum. And hence find a with the help of condition ${{x}_{2}}={{x}_{1}}^{2}$ where, a local maximum is at $x={{x}_{1}}$ and a local minimum is at $x={{x}_{2}}$

Complete step-by-step answer:
Now consider the given function $f\left( x \right)=2{{x}^{3}}-9a{{x}^{2}}+12{{a}^{2}}x+1$
Now we know that the condition for extrema of $f\left( x \right)$ is $f'\left( x \right)=0$ .
Now this extrema is either a minimum or a maximum.
Hence first let us find the extrema of the function $f\left( x \right)=2{{x}^{3}}-9a{{x}^{2}}+12{{a}^{2}}x+1$
Differentiating the function and using the formula $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$ we get
$f'\left( x \right)=2\left( 3{{x}^{2}} \right)+9a\left( 2x \right)+12{{a}^{2}}$
$f'\left( x \right)=6{{x}^{2}}-18ax+12{{a}^{2}}$
Now let us write – 18ax as – 18ax = – 6ax – 12ax.
Hence we get
$f'\left( x \right)=6{{x}^{2}}-6ax-12ax+12{{a}^{2}}$
Now taking 6 and 12 common we get
$\begin{align}
  & f'\left( x \right)=6x\left( x-a \right)-12a\left( x-a \right) \\
 & f'\left( x \right)=\left( 6x-12a \right)\left( x-a \right) \\
\end{align}$
Now equating $f'\left( x \right)=0$ we will get the conditions for extrema.
$f'\left( x \right)=6\left( x-2a \right)\left( x-a \right)=0$
Now we know that if $a.b=0$ then $a=0$ or $b=0$
Hence $f'\left( x \right)=6\left( x-2a \right)\left( x-a \right)=0$ means
$\left( x-2a \right)=0$ or $\left( x-a \right)=0$
These are our conditions for extrema.
Hence we get extrema at x = 2a and x = a. ……………………. (1)
Now again consider the equation $f'\left( x \right)=6{{x}^{2}}-18ax+12{{a}^{2}}$
Let us again differentiate the equation and using the formula $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$ we get $f''\left( x \right)=6\left( 2x \right)-18a\left( 1 \right)$
$f''\left( x \right)=12x-18a$
Now at x = a we have $f''\left( a \right)=12a-18a=-6a$
And if x = 2a we have $f''\left( 2a \right)=24a-18a=6a$
Hence at x = a we have $f''\left( x \right)< 0$ and at x = 2a we have $f''\left( x \right)> 0$
Now for an extremum point we have $f''\left( x \right)< 0\Rightarrow $function has a maxima at that point
And similarly for $f''\left( x \right)> 0\Rightarrow $function has minima at that point
Hence we get x = 2a is the local minimum and x = a is the local maximum.
Now hence we were given that $f\left( x \right)$ has local maximum at $x={{x}_{1}}$ and a local minimum at $x={{x}_{2}}$ such that ${{x}_{2}}={{x}_{1}}^{2}$
Hence we have $2a={{a}^{2}}$
Now dividing the equation by a we get
$2=a$
Hence the value of a is 2.

So, the correct answer is “Option c”.

Note: Now note that we can also check if the point of extrema is minimum or maximum by substituting the point of extrema in $f\left( x \right)$ and checking which point has greater value and which has lesser value. If it has greater value then it is a maximum. If it has lower value then it is a minimum