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Question

Let a function f defined from RR as f(x)={x+p2,x22px+5,x>2.
If the function is surjective, then sum of all possible integral value of p in [-100, 100] is

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Solution

We have,
f(x)={x+p2,x22px+5,x>2.
Case I: p>0
Range of f(x) will be = (,2]+p2=(,2+p2] for all x2 and 2p(2,)+5=(4p+5,) for all x>2
Since, the given function f(x) is an onto function.
2+p24p+5p24p30p2+7 or p27 (which is not possible as p > 0p2+7>4Hence,p can take values 5, 6, 7,....., 100.Required sum for p > 0=5+6+7+....+100=505010=5040
Case II: p<0
Range of f(x) will be = (,2]+p2=(,2+p2] for all x2 and 2p(2,)+5=(,4p+5) for all x>2
Function cannot take all the values in the set R and hence, will not be an onto function.
Therefore, p < 0 is not possible.
Required Sum = 5040

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