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Question

The condition for which the quadratic equation (p2−3p+2)x2+(p2−4)x+(p−1)=0 will have a graph which is a downward opening parabola is

A
pR
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B
p(1,2)
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C
p(,1) (2,)
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D
p(,1) (2,)
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Solution

The correct option is B p(1,2)
For f(x)=ax2+bx+c, the graph is a downward opening parabola when a<0.

Here we have f(x)=(p23p+2)x2+(p24)x+(p1)
So for a downward opening parabolic graph,
p23p+2<0
(p1)(p2)<0
Using wavy curve method, we get


p(1,2)

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