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  • If (a, 0) is a point on a diameter of the circle <img alt="\mathrm{x^2+y^2=4 \text { then } x^2-4 x-a^2=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5E2&plus;y%5E2%3D4

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

 

Option: 1

exactly one  real  root in ( -1, 0]

 


Option: 2

exactly  one real root in [ 2, 5]

 


Option: 3

distinct  roots greater than –1


Option: 4

distinct  roots less than 5

 


Answers (1)

best_answer

\mathrm{\text { Since }(a, 0) \text { is a point on the diameter of the }}

\mathrm{\text { circle } x^2+y^2=4}

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

\mathrm{\text { Let } f(x)=x^2-4 x-a^2}

\mathrm{\text { Clearly } f(-1)=5-a^2>0, f(2)=-\left(a^2+4\right)<0}

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

Graph  of f(x) will be as shown

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