If f(x)=alog⁡|x|+bx2+x has its extremum values at x=-1 and x=2, then

A
$a = 2, b = - 1$
B
$a = 2, b = - \frac{1}{2}$
C
$a = - 2, b = \frac{1}{2}$
D
none of these

Solution:

We have,

$f (x) = a \log | x | + b x^{2} + x \Rightarrow f^{'} (x) = \frac{a}{x} + 2 b x + 1$

Since $f (x)$ attains its extremum values at $x = - 1, 2$

$\begin{array}{l} ∴ f^{'} (- 1) = 0 and f^{'} (2) = 0 \\ \Rightarrow - a - 2 b + 1 = 0 and \frac{a}{2} + 4 b + 1 = 0 \\ \Rightarrow a = 2 and b = - 1 / 2 . \end{array}$

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