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Find the minimum integral value of k for which the equation e^(x)=kx^(2) has exactly three real distinct solutions. |
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Answer» Solution :Given equation is `e^(x)=kx^(2)`. `RARR k=e^(x)/x^(2)` Now let `f(x)=e^(x)/x^(2)` `THEREFORE f^(')(x) = ((x-2)e^(x))/(x^(3))` `f^(')(x)=0 therefore x=2`, which is the point of MINIMA. Also `f(2) =e^(2)/4` `underset(x to infty)"lim"e^(x)/x^(2)=infty` (using L' Hospital rule twice) `underset(x to infty) e^(x)/x^(2)=0` `underset(x to 0)"lim"e^(x)/x^(2)=infty` Further `f(x) gt 0, AA x in R`. From this INFORMATION, the graph of `f(x)` is as shown in the following figure. Thus, three real distinct solutions for `k lt e^(2)/4, k in I`. So, `k_("min") =2`. |
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