JEE Main 2014 (Offline)Question 1
If $x=-1$ and $x=2$ are extreme points of $f(x)=\alpha\log|x|+\beta x^{2}+x$ then
Differentiation and its Application
Solution
Concept. Extreme points of a differentiable function are zeros of $f'$. Formulas used. (1) $\dfrac{d}{dx}\log|x|=\dfrac1x$Step 1 - Derivative. By (1), $f'(x)=\frac{\alpha}{x}+2\beta x+1$. Step 2 - Equations. $$f'(-1)=-\alpha-2\beta+1=0,\qquad f'(2)=\frac\alpha2+4\beta+1=0.$$ Step 3 - Solve. From the first, $\alpha=1-2\beta$. Then $\frac12-\beta+4\beta+1=0$, so $\beta=-\frac12$ and $\alpha=2$. Answer: (A)