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Let. f: R ---R be given by f(x) = x^2 + 3. Find (i) {x:f(x)=28}
(ii) The pre- images of 39 and 2 under 'f'.



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Answer 1

As f: R → R, the range and domain are real numbers.

(i) f(x) = 28 = x² + 3

x² = 25

x = +5 or -5

Therefore for range 28, the pre-image is 5 and -5

(ii) f(x) = 39

x² + 3 = 39

x² = 36

x = +6 or -6

Therefore, pre-image of 39 is either +6 or -6 under f

For pre-image of 2,

f(x) = x² + 3

2 = x² + 3

-1 = x²

x = i

But since ' i ' is a complex number, Therefore f(x) is not defined for it.

Hence, 2 is not an image under f. Therefore, the function will not have a pre-image of 2 under f.

Answer 2

Step-by-step explanation:

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