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Q.

For all real values of x,|12x4x2+9|

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a

≤1

b

≤2

c

>1

d

>2

answer is A.

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Detailed Solution

Let,12x4x2+9=y4yx2−12x+9y=0As x is real,  D=144−4⋅4y⋅9y≥0⇒1−y2≥0 ⇒ y2≤1 ∴ |y|≤1 Hence, |12x4x2+9|≤1
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