1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# The condition for which the quadratic equation (p2âˆ’3p+2)x2+(p2âˆ’4)x+(pâˆ’1)=0 will have a graph which is an upward opening parabola is

A
p(1,2)
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
B
pR
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
C
p(,1) (2,)
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
D
p(,1) (2,)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

## The correct option is D p∈(−∞,1) ∪ (2,∞)For f(x)=ax2+bx+c, the graph is an upward opening parabola when a>0. Here we have f(x)=(p2−3p+2)x2+(p2−4)x+(p−1) So for an upward opening parabolic graph, p2−3p+2>0 ⇒(p−1)(p−2)>0 So using wavy curve method, we get p∈(−∞,1) ∪ (2,∞)

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Graphs of Quadratic Equation for different values of D when a<0
MATHEMATICS
Watch in App
Join BYJU'S Learning Program