MCQ
The minimum value of $2{x^2} + x - 1$ is
- A$ - {1 \over 4}$
- B${3 \over 2}$
- ✓${{ - 9} \over 8}$
- D${9 \over 4}$
==> $f'(x) = 4x + 1 \Rightarrow f'(x) = 0 \Rightarrow x = - \frac{1}{4}$
$f''\,(x) = 4 = + ve$
$\therefore {[f( - 1/4)]_{\min }} = \frac{2}{{16}} - \frac{1}{4} - 1 = \frac{{ - 9}}{8}$.
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If a curve passes through the point $(1, -2)$ and has slope of the tangent at any point $(x,y)$ on it as $\frac{{{x^2} - 2y}}{x}$ then the curve also passes through the point
$\pi$
$\frac{\pi}{2}$
$\frac{\pi}{4}$
$\frac{3\pi}{4}$