The smallest integral value of p for which the inequality (p−3)x2−2px+3(p−2)>0 is satisfied for all real values of x is
A
8
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B
7
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C
6
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D
5
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Solution
The correct option is D7 f(x)>0 when D<0 and p>3 ⇒4p2−12(p−3)(p−2)<0 ⇒2p2−15p+18>0 ⇒(p−6)(2p−3)>0 ⇒p<32 or p>6 Hence, smallest integral value of p is 7.