Question

If x = -1 and x = 2 are extreme points of $\text{f(x) = }\alpha \text{ log}\left| x \right|+\beta {{x}^{2}}+x$, then
a) $\alpha =-6,\beta =\dfrac{1}{2}$
b) $\alpha =-6,\beta =-\dfrac{1}{2}$
c) $\alpha =2,\beta =-\dfrac{1}{2}$
d) $\alpha =2,\beta =\dfrac{1}{2}$

Answer 1

Hint: We a given function as: $\text{f(x) = }\alpha \text{ log}\left| x \right|+\beta {{x}^{2}}+x$. It is said that x = -1 and x = 2 are extreme points of the given function. So, at extreme points $f'(x)=0$.
To find the values of $\alpha $and $\beta $, firstly find $f'(x)=0$ for x = -1 and x = 2 and then solve both the linear equations in two variables.

Complete step by step answer:
As we have a give function:
$\text{f(x) = }\alpha \text{ log}\left| x \right|+\beta {{x}^{2}}+x$
Now, differentiate the function with respect to x, we get:
$f'(x)=\alpha \left( \dfrac{1}{x} \right)+2\beta x+1$
$\left\{ \dfrac{d\left( \log x \right)}{dx}=\dfrac{1}{x};\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}} \right\}$
Since $f'(x)=0$ at end points.
So, we can say that $\alpha \left( \dfrac{1}{x} \right)+2\beta x+1=0$ for x = -1 and x =2.

Now, put x = -1 in equation (1), we get:
$\begin{align}
  & \Rightarrow -\alpha -2\beta +1=0 \\
 & \Rightarrow \alpha +2\beta =1......(2) \\
\end{align}$
Now, put x = 2 in equation (1), we get:
$\begin{align}
  & \Rightarrow \dfrac{\alpha }{2}+4\beta +1=0 \\
 & \Rightarrow \alpha +8\beta =-2......(3) \\
\end{align}$

To solve both equations (2) and (3), subtract equation (2) from equation (3).
we get:
$\begin{align}
  & \Rightarrow 6\beta =-3 \\
 & \Rightarrow \beta =-\dfrac{1}{2} \\
\end{align}$
Put value of $\beta =-\dfrac{1}{2}$ in equation (2), we get:
$\begin{align}
  & \Rightarrow \alpha -1=1 \\
 & \Rightarrow \alpha =2 \\
\end{align}$
Hence, the required values are: $\alpha =2$ and $\beta =-\dfrac{1}{2}$

So, the correct answer is “Option C”.

Note: While differentiating the given function with respect to x, always take care to multiply the coefficient of that quantity with the variable. As in the given question, we have:$\dfrac{d\left( \beta {{x}^{2}} \right)}{dx}$. So, the answer would be $2\beta x$. Some might miss the co-efficient while differentiating and write it as $2x$, which is a wrong method according to the differentiation rule.
Also, while solving the linear equations in two variables, we can use elimination method or substitution method (whichever makes the process easier).