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The range of $y=2x^{2}-4x+9$ when $0\leq x\leq 3$ is
Option: 1
$\left [ 7,15 \right ]\\$

Option: 2
$\left [ 9,15 \right ]\\$

Option: 3
$\left [ 7,9 \right ]$

Option: 4
None of these

Answers (1)

$a=2\Rightarrow a> 0$

Now only some values of $x$ are given, so range is not $\left [ \frac{-D}{4a},\infty \right )$

For such questions, we use graphs

Vertex: $\left ( \frac{-b}{2a},\frac{-D}{4a} \right )$

$=\left ( \frac{4}{4},\frac{-\left ( 16-4\cdot 2\cdot 9 \right )}{4\cdot 2} \right )$

$=\left ( 1,7 \right )$

$0\leq x\leq 3$

Clearly from graph, minimum value $=7$

Maximum value = higher of $f\left ( 0 \right )\: or\: f\left ( 3 \right )$

$f\left ( 0 \right )=9\: and\: f\left ( 3 \right )=2\cdot 3^{2}-4\cdot 3+9=15$

Range : $\left [ 7,15 \right ]$

Posted by

Ramraj Saini

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