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Question

If x be real , then xx25x+9 lies between

A
1 and 111
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B
1 and 111
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C
1 and 111
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D
None of the above.
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Solution

The correct option is B 1 and 111
Let y=xx25x+9
or yx25xy+9y=xyx2(5y+1)x+9y=0
Now, x is real, so
Δ0((5y+1))24y(9y)0
or 11y2+10y+1011y210y10(11y+1)(y1)0111y1y[111,1]


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