Question
$2{x^2} + x - 1$ का न्यूनतम मान है
==>$f'(x) = 4x + 1 \Rightarrow f'(x) = 0 \Rightarrow x = - \frac{1}{4}$
$f''\,(x) = 4 = $ धनात्मक,
$\therefore {[f( - 1/4)]_{\min }} = \frac{2}{{16}} - \frac{1}{4} - 1 = \frac{{ - 9}}{8}$.
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$ A = \{ (x,\,y):y = \frac{1}{x},\,0 \ne x \in R\} $
$B = \{ (x,\,y):y = - x,\,\,x \in R\} $, तब