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Question

The range of the expression f(x)=3x212x+5 is


A

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B

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C

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D

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Solution

The correct option is C


Compare 3x212x+5 with ax2bx+c

Here, a > 0 Minimum value of f(x) occur at x=b2a and the value is D4a

D=b24ac

=(12)24(3)(5)

Minimum value = 8412=7

As a > 0, graph of f(x) is upward.

As minimum value is -7,

Maximum value extends to

So, the range is [-7, )


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