The minimum value of expression $4 x^{2} + 2 x + 1$ is -

\frac{1}{4}

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\frac{1}{2}

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\frac{3}{4}

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\frac{- 1}{2}

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Solution

The correct option is D $\frac{- 1}{2}$
We have $y = 4 x^{2} + 2 x + 1$
$\Rightarrow {(2 x)}^{2} + 2 \times 2 x \times 1 + 1 - 2 x$
$\Rightarrow {(2 x + 1)}^{2} - 2 x$
$(2 x + 1) \geq 0$
$2 x \geq - 1$
$x \geq \frac{- 1}{2}$
The minimum value of $x$ is $\frac{- 1}{2}$

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