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Find a, for which \( f(x) = x^2-2ax+6\) is strictly increasing when x > 0.

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If the function is increasing or decreasing f'(x) =0 Given f(x) =x^2-2ax+6 Differentiating w.r.t x we get, 2x-2a=0 This implies x-a=0 Or x =a Now substituting for x in f(x) we get A^2-2a^2+6=0 That is -a^2-2a^2+6=0 This implies a^2=6 Or a=√6
answered Nov 30, 2013 by vijayalakshmi_ramakrishnans

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