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Question

For every real number x, let f(x)=x1!+32!x2+73!x3+154!x4+...... Then the equation f(x)=0 has

A
no real solution
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B
exactly one real solution
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C
exactly two real solutions
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D
infinite number of real solutions
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Solution

The correct option is B exactly one real solution
f(x)=x1!+32!x2+73!x3+154!x4......

We can write it as
x×(11!+32!x+73!x2+154!x3......)=0
one real root is x=0
We will get other roots from 11!+32!x+73!x2+154!....=0
But we can see that the sum of the roots =
which is not possible, so the remaining roots of the equation are imaginary
Hence, only one real root x=0 exist.

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