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Question

The number of integral values of k, for which (x2x+1)(kx23kx5)<0 is

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Solution

(x2x+1)(kx23kx5)<0
x2x+1>0 as D<0
When k=0, we get
5(x2x+1)<0x2x+1>0
Which is true.

Now,
kx23kx5<0
Therefore,
k<0(1)D<09k2+20k<0k(9k+20)<0k(209,0)(2)
From (1) and (2),we get
k(209,0)

Hence, the integral values are k=2,1,0

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