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If (a, 0) is a point on a diameter of the circle \mathrm{x^{2}+y^{2}=4}, then \mathrm{x^{2}-4 x-a^{2}=0} has

Option: 1

exactly one  real  root in ( -2, -1]


Option: 2

exactly  one real root in [2, 5]


Option: 3

distinct  roots greater than 1


Option: 4

distinct  roots less than -1


Answers (1)

best_answer

Since \mathrm{(a, 0)} is a point on the diameter of the circle \mathrm{x^{2}+y^{2}=4},

so maximum value of \mathrm{a^{2}} is 4

Let \mathrm{f(x)=x^{2}-4 x-a^{2}}
clearly \mathrm{f(-1)=5-a^{2}>0, f(2)=-\left(a^{2}+4\right)<0}
\mathrm{f(0)=-a^{2}<0\: and \: f(5)=5-a^{2}>0}

so graph of \mathrm{f(x)} will be as shown

Hence (A) is the correct answer.

 

 

Posted by

Anam Khan

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