1.

Find the minimum integral value of k for which the equation e^(x)=kx^(2) has exactly three real distinct solutions.

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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