MCQ
Function $f(x) = {{4{x^2} + 1} \over x}$ is decreasing for interval
  • A
    $\left( {{{ - 1} \over 2},\,{1 \over 2}} \right)$
  • $\left[ {{1 \over 2},\, - {1 \over 2}} \right]$
  • C
    $(-1, 1)$
  • D
    $[1, -1]$

Answer

Correct option: B.
$\left[ {{1 \over 2},\, - {1 \over 2}} \right]$
b
(b) $f(x) = 4x + \frac{1}{x}$

$\frac{d}{{dx}}f(x) = \frac{d}{{dx}}\left[ {4x + \frac{1}{x}} \right] $

$= 4 - \frac{1}{{{x^2}}}$

For extremum, $\frac{{dy}}{{dx}} = 0$

==> $4 - \frac{1}{{{x^2}}} = 0$ 

==> $x = \frac{1}{2},\, - \frac{1}{2}$

$f\;\left( {\frac{1}{2}} \right) = 4.\frac{1}{2} + \frac{1}{{1/2}}$ = $2 + 2 = 4$

$f\;\left( { - \frac{1}{2}} \right) = 4.\left( { - \frac{1}{2}} \right) + \frac{1}{{ - 1/2}} = - 2 - 2 = - 4$

Hence $f(x)$ is decreasing in interval $[1/2,\, - 1/2]$.

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