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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 is (-1,0)


Option: 2

exactly one real root is (3,5)


Option: 3

distinct roots greater than -1


Option: 4

distinct roots less than 5


Answers (1)

best_answer

                                            

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

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

Let  \mathrm{f(x)=x^2-4 x-a^2}

clearly  \mathrm{\mathrm{f}(-1)=5-\mathrm{a}^2 \text { is } 4}

\mathrm{\begin{aligned} & \mathrm{f}(2)=-\left(\mathrm{a}^2+4\right)<0 \\\\ & \mathrm{f}(0)=-\mathrm{a}^2<0 \text { and } \mathrm{f}(5)=5-\mathrm{a}^2>0 \end{aligned}}

so graph of f(x) will be as shown

Hence (a), (b), (c), (d) are correct answers. 

Posted by

manish painkra

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