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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Let $P$ and $Q$ be the points of intersection of the curve $C$ and the $y$-axis. If normals at $P$ and $Q$ on the curve $C$ intersect $x$-axis at points $R$ and $S$ respectively, then the length of the line segment $RS$ is
| Class | $ 0-4$ | $ 4-8$ | $ 8-12$ | $ 12-16$ | $ 16-20$ |
| Frequency | $ 3$ | $ 9$ | $ 10$ | $ 8$ | $ 6$ |