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Question

If $$b>a$$, then the equation $$(x-a)(x-b)-1=0$$, has?


A
Both roots in [a,b]
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B
Both roots in [,a]
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C
Both roots in [b,]
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D
One root in [α,a] and other in [b,α]
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Solution

The correct option is C One root in $$[-\alpha, a]$$ and other in $$[b, \alpha]$$
$$(x-a)(x-b)-1=0$$
$$\Rightarrow { x }^{ 2 }-\left( a+b \right) x-ab-1=0$$
then discriminant is $${ \left( a+b \right)  }^{ 2 }-4\left( ab-1 \right) ={ \left( a-b \right)  }^{ 2 }+4>0$$,
it has two real roots further $$f(a)=-1$$ and $$f(b)=-1$$ but $$b>a$$ i.e $$a$$ and $$b$$ are distinct as coefficient of $${ x }^{ 2 }$$ is position (it is $$1$$), minima of $$f(x)$$ is between $$a$$ and $$b$$
Hence, one root will be in the interval $$\left( -\alpha ,a \right) $$ and the other root will be in the interval $$\left( b,\alpha  \right) $$.

Mathematics

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