MCQ
The maximum value of $f(x) = {x \over {4 + x + {x^2}}}$ on $[ - 1,\,1]$ is
- A$ - 1/4$
- ✓$ - 1/3$
- C$1/6$
- D$1/5$
Differentiate, $f'(x) = \frac{{4 + x + {x^2} - x(1 + 2x)}}{{{{(4 + x + {x^2})}^2}}}$
For maximum $f'(x) = 0$ ==> $\frac{{4 - {x^2}}}{{{{(4 + x + {x^2})}^2}}} = 0$
==> $x = 2,\, - 2$
Both values of $x $ are out of interval
$\therefore$ $f( - 1) = \frac{{ - 1}}{{4 - 1 + 1}} = \frac{{ - 1}}{4}$,
$f(1) = \frac{1}{{4 + 1 + 1}} = \frac{1}{6}$ (maximum).
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$g(x)=\left\{\begin{array}{ccc}0 & \text { if } & x < a, \\ \int_a^x f(t) d t & \text { if } & a \leq x \leq b, \\ \int_a^b f(t) d t & \text { if } & x > b .\end{array}\right.$, Then
$(A)$ $g(x)$ is continuous but not differentiable at a
$(B)$ $g(x)$ is differentiable on $R$
$(C)$ $g(x)$ is continuous but not differentiable at $b$
$(D)$ $g(x)$ is continuous and differentiable at either a or $b$ but not both