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The sum of all the integral values of p such that the equation 3 2 x+12 x-3=p, x R, has at least one solution, is:
(3) -75
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p=3 (1- 2x )+12 x-3=-3t2+12t, where t= x [-1, 1]
Let g(t)=-3t2+12t; g'(t)=-6t+12>0 on [-1, 1] (vertex at t=2)
g is increasing on [-1, 1]
Range of p =[g(-1), g(1)]=[-15, 9]
Integral p: -15, -14, , 9
Sum =(-15-14- -10)+(-9+ +9)=-75+0=-75
Answer: (3)
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