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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$f(x) \rightarrow \frac{\lambda\left|x^{2}-5 x+6\right|}{\mu\left(5 x-x^{2}-6\right)}, x<2$
$\quad\quad\quad\quad e^{\frac{\tan (x-2)}{x-[x]}}, \quad x>2$
$\quad\quad\quad\quad \mu \quad\quad\quad\quad x=2$
Where $[x]$ is the greatest integer less than or equal to $x$. If $f$ is continuous at $x=2$, then $\lambda+\mu$ is equal to: