Question

If the function $\text{F}\left( x \right)=2{{x}^{3}}-9a{{x}^{2}}+12{{a}^{2}}x+1$ has a local maximum at $x={{x}_{1}}$ and a local maximum at $x={{x}_{2}}$ such that ${{x}_{2}}=x_{1}^{2}$ then the value of ‘a’ equals.
A). 0
B). ${}^{1}/{}_{2}$.
C). 2
D). Either (a) or (c)

Answer 1

Hint: Find the maximum and minima at the given function by differentiating and double differentiating it. Then put the values in the given relation ${{x}_{2}}=x_{1}^{2}$ to find the value at a.

Complete step by step solution: Given, $\text{f}\left( x \right)=2{{x}^{3}}-9a{{x}^{2}}+12{{a}^{2}}x+1$
Differentiating with respect to $x.$
$\text{f }\!\!'\!\!\text{ }\left( x \right)=2.\dfrac{d}{dx}\left( {{x}^{3}} \right)-\dfrac{d}{dx}\left( 9a{{x}^{2}} \right)+\dfrac{d}{dx}\left( 12{{a}^{2}}x \right)+\dfrac{d}{dx}\left( 1 \right)$
Using the formula $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{(n-1)}},\text{we}\ \text{get}$
$\Rightarrow \ \text{f }\!\!'\!\!\text{ }\left( x \right)=2\left( 3{{x}^{2}} \right)-9a\left( 2x \right)+12{{a}^{2}}\left( 1 \right)+0$
$\Rightarrow \ \text{f }\!\!'\!\!\text{ }\left( x \right)=6{{x}^{2}}-18ax+12{{a}^{2}}$
Now again differentiating with respect to $x$.
$\Rightarrow \ \text{f}''\left( x \right)=\dfrac{d}{dx}\left( 6{{x}^{2}} \right)-\dfrac{d}{dx}\left( 18ax \right)+\dfrac{d}{dx}\left( 12{{a}^{2}} \right)$
$\Rightarrow \text{f}''\ \left( x \right)=12x-18a$
$\therefore \ \text{f}''\ \left( x \right)=12x-18a$
For finding maxima/minima condition is
\[\text{f }\!\!'\!\!\text{ }\left( x \right)=0\]
$\Rightarrow \ 6{{x}^{2}}-18ax+12{{a}^{2}}=0$
Now splitting middle $+am$
$\Rightarrow \ 6\left( {{x}^{2}}-3ax+2{{a}^{2}} \right)=0$
$\Rightarrow \ {{x}^{2}}-3ax+2{{a}^{2}}=0$
$\Rightarrow \ {{x}^{2}}-ax-2ax+2{{a}^{2}}=0$
$\Rightarrow \ x\left( x-a \right)-2a\left( x-a \right)=0$
Taking $\left( x-a \right)$ common
$\Rightarrow \ \left( x-a \right)\ \left( x-2a \right)=0$
$\Rightarrow \ x=a\ or\ x=2a$
For finding which one is maxima and which one is minima,
$\text{f}''\left( a \right)=12\left( a \right)-180=6a<0$
Therefore $\text{f}\left( x \right)$ is maxima at $x=a$
$\text{f}''\left( 2a \right)=12\left( 2a \right)-180$
     $24a-18a=6a>0$
Therefore $\text{f}\left( x \right)$ is minimum at $x=2a$
Now from question we know that at $x={{x}_{1}}\ ,\text{f}\left( x \right)$ has 0 local maximum
So, ${{x}_{1}}=a$
And at $x={{x}_{2}}\ \text{f}\left( x \right)$ has a local minima
So, ${{x}_{2}}=2a$
It is given,
${{x}_{2}}=x_{1}^{2}$
$\Rightarrow \ 2a={{a}^{2}}$
$\Rightarrow \ {{a}^{2}}-2a=0$
$\Rightarrow \ a\left( a-2 \right)=0$
$\Rightarrow \ a=0\ or\ a=2$
$a\ne{0},$ since in the question it is said that $a>0$

Therefore $a=2$ answer.

Note: A local maximum point an a function ‘s a point $\left( x,y \right)$ on the graph of the function whose y coordinate is large than all then co-coordinate on the graph at points “close to” $\left( x,y \right)$.
Similarly$\left( x,y \right)$s is a local minimum point if. It has locally the smallest y coordinate.