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Question: If x = -1 and x = 2 are extreme points of \(\text{f(x) = }\alpha \text{ log}\left| x \right|+\beta {...

If x = -1 and x = 2 are extreme points of f(x) = α log∣x∣+βx2+x\text{f(x) = }\alpha \text{ log}\left| x \right|+\beta {{x}^{2}}+x, then
a) α=−6,β=12\alpha =-6,\beta =\dfrac{1}{2}
b) α=−6,β=−12\alpha =-6,\beta =-\dfrac{1}{2}
c) α=2,β=−12\alpha =2,\beta =-\dfrac{1}{2}
d) α=2,β=12\alpha =2,\beta =\dfrac{1}{2}

Explanation

Solution

We a given function as: f(x) = α log∣x∣+βx2+x\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)=0f'(x)=0.
To find the values of α\alpha and β\beta , firstly find f′(x)=0f'(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:
f(x) = α log∣x∣+βx2+x\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)=α(1x)+2βx+1f'(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)=0f'(x)=0 at end points.
So, we can say that α(1x)+2βx+1=0\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:
⇒−α−2β+1=0 ⇒α+2β=1......(2) \begin{aligned} & \Rightarrow -\alpha -2\beta +1=0 \\\ & \Rightarrow \alpha +2\beta =1......(2) \\\ \end{aligned}
Now, put x = 2 in equation (1), we get:
⇒α2+4β+1=0 ⇒α+8β=−2......(3) \begin{aligned} & \Rightarrow \dfrac{\alpha }{2}+4\beta +1=0 \\\ & \Rightarrow \alpha +8\beta =-2......(3) \\\ \end{aligned}

To solve both equations (2) and (3), subtract equation (2) from equation (3).
we get:
⇒6β=−3 ⇒β=−12 \begin{aligned} & \Rightarrow 6\beta =-3 \\\ & \Rightarrow \beta =-\dfrac{1}{2} \\\ \end{aligned}
Put value of β=−12\beta =-\dfrac{1}{2} in equation (2), we get:
⇒α−1=1 ⇒α=2 \begin{aligned} & \Rightarrow \alpha -1=1 \\\ & \Rightarrow \alpha =2 \\\ \end{aligned}
Hence, the required values are: α=2\alpha =2 and β=−12\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:d(βx2)dx\dfrac{d\left( \beta {{x}^{2}} \right)}{dx}. So, the answer would be 2βx2\beta x. Some might miss the co-efficient while differentiating and write it as 2x2x, 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).