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  • If is a point on a diameter of the circle <img alt="\mathrm{x^{2}+y^{2}=4}" src="https://entrancecorner.onc

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 two  real  root in ( -2, -1]


Option: 2

exactly  two real root in [2, 5]


Option: 3

distinct  roots greater than 1


Option: 4

equal  roots less than -1


Answers (1)

best_answer

Since \left ( a,0 \right ) 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(5)}

\mathrm{f(0)=-a^{2}>0\: and \: f(5)=5-a^{2}>0}

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


Hence (C) is the correct answer.

Posted by

shivangi.bhatnagar

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