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Let S be the set of positive integral values of a for which ax2+2(a+1)x+9a+4x2-8x+32<0, x R. Then, the number of elements in S is :
(1) 0
x2-8x+32=(x-4)2+16>0 for every x
⇒ need ax2+2(a+1)x+9a+4<0 for all x R
∵ a quadratic is negative for all x only if its leading coefficient is negative, we need a<0
But a must be a positive integer ⇒ no such a
n(S)=0
Answer: (1)
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