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6. Application of derivatives — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths6. Application of derivatives1 Mark
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
    $1/5$
✓

Answer

Correct option: B.
$ - 1/3$
b
(c) $f(x) = \frac{x}{{4 + x + {x^2}}}$

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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