The minimum possibe value of x2-6x+103-x, for x < 3, is
Explanation:
We have x2-6x+103-x
= (x-3)2+13-x
= (x-3)23-x + 13-x
= (3 - x) + 13-x
Here, since x < 3, 3 – x > 0
Also, we know that sum of a positive number and its reciprocal is always greater than or equal to 2.
⇒ (3 - x) + 13-x ≥ 2
Hence, option (c).
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