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$[1, -1]$
$\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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