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Q.

If x is real, then xx2−5x+9 lies between

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a

-1 and -1/11

b

1 and -1/11

c

1 and 1/11

d

none of these

answer is B.

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Detailed Solution

Let xx2−5x+9=yor yx2−5yx+9y=xor yx2−(5y+1)x+9y=0Now, x is real, soD≥0⇒ (−(5y+1))2−4⋅y⋅(9y)≥0or −11y2+10y+1≥0or 11y2−10y−1≤0or (11y+1)(y−1)≤0or −111≤y≤1

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