2010_Question1_a

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We want k to be such that kx=(x-1)^2 has real solutions. This equation is equivalent to x^2-(2+k)x+1=0. The quadratic formula then tells us that the roots of this are at \frac{-b\pm\sqrt{b^2-4ac}}{2a} = \frac{2+k\pm\sqrt{(k^2+4k+4)-4}}{2} = \frac{2+k\pm\sqrt{k^2+4k}}{2}. To have two real roots it is necessary and sufficient to have the square root term in this to be real, i.e. to have k^2+4k\geqslant0, which is equivalent to asking that k be either greater than or equal to zero, or that k be less than or equal to -4. So the answer is (c).

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