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STD 12 - 6. Application of derivatives — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 6. Application of derivatives1 Mark
MCQ
If $f(x) = {1 \over {4{x^2} + 2x + 1}}$, then its maximum value is
  • ✓
    $4/3$
  • B
    $2/3$
  • C
    $1$
  • D
    $3/4$
✓

Answer

Correct option: A.
$4/3$
a
(a) $f(x) = \frac{1}{{4{x^2} + 2x + 1}}$==> $f'(x) = \frac{{ - (8x + 2)}}{{{{(4{x^2} + 2x + 1)}^2}}}$

Put $f'(x) = 0$ ==> $8x + 2 = 0$ ==> $x = - 1/4$.

$f''\,(x) = \frac{{ - [{{(4{x^2} + 2x + 1)}^2}8 - (8x + 2)\,2\,(4{x^2} + 2x + 1)(8x + 2)]}}{{{{(4{x^2} + 2x + 1)}^4}}}$

$f''( - 1/4) = - ve$ (point of maxima)

$\therefore$ $f{( - 1/4)_{{\rm{max}}{\rm{.}}}} = \frac{1}{{4 \times \frac{1}{{16}} - 2 \times \frac{1}{4} + 1}} = \frac{4}{3}$.

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