Surjective Function

Algebra Level 4

Let f : R → [ 1 , ∞ ) f : R\rightarrow [1,\infty) be defined by f ( x ) = x 2 − 6 a x + 3 − 2 a + 9 a 2 f(x) =x^2-6 ax+3-2 a+9 a^2 .The integral value of a a for which f ( x ) f(x) is surjective, is equal to

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The answer is 1.

Akshay Sharma by Akshay Sharma , 21, India

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2 solutions

Otto Bretscher
Mar 17, 2016

f ( x ) = ( x − 3 a ) 2 + 3 − 2 a f(x)=(x-3a)^2+3-2a , with minimal value 3 − 2 a 3-2a . We want the minimal value to be 1, so 3 − 2 a = 1 3 - 2a = 1 which gives a = 1 a=\boxed{1} .

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Prakhar Bindal
Mar 30, 2016

For a surjective functions its codomain must be equal to its range. differentiate the expression find the value of x , substitute in original and equate to 1 we get a = 1. But you should check that the discrimant of quadratic is less than zero so that its always positive for all x in its domain!

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