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Absolute Value Function

If |4x2-16x+1...

Question

If $| 4 x^{2} - 16 x + 16 | - 7 | x - 2 | = - 3$ , then the value(s) of $x$ is/are

A

1

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B

3

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C

\frac{11}{4}

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D

\frac{5}{4}

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Solution

The correct option is D $\frac{5}{4}$
$| 4 x^{2} - 16 x + 16 | - 7 | x - 2 | = - 3 \Rightarrow 4 | x^{2} - 4 x + 4 | - 7 | x - 2 | + 3 = 0$
$\Rightarrow 4 | (x - 2)^{2} | - 7 | x - 2 | + 3 = 0 \Rightarrow 4 | (x - 2) |^{2} - 7 | x - 2 | + 3 = 0 (∵ | x^{2} | = | x |^{2})$

Let $| x - 2 | = t$
$\Rightarrow 4 t^{2} - 7 t + 3 = 0$
$\Rightarrow 4 t^{2} - 4 t - 3 t + 3 = 0$
$\Rightarrow (t - 1) (4 t - 3) = 0$
$\Rightarrow t = 1 or t = \frac{3}{4}$
$\Rightarrow | x - 2 | = 1 or | x - 2 | = \frac{3}{4}$
$\Rightarrow x - 2 = \pm 1 or x - 2 = \pm \frac{3}{4}$
$\Rightarrow x = 1, 3 or x = \frac{5}{4}, \frac{11}{4}$
$∴ x = 1, \frac{5}{4}, \frac{11}{4}, 3$

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Absolute Value Function

Standard XII Mathematics

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