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Theory of Equations: Let Assumes Both Positive Negative Values Range

JEE Maths question with a full step-by-step solution.

Question
Let P(x)=x2−(2−p)x+p−2P(x) = x^2 - (2-p)x + p - 2. If P(x)P(x) assumes both positive and negative values ∀x∈R\forall x \in \mathbb{R}, then the range of pp is
A(−∞,2)∪(6,∞)(-\infty, 2) \cup (6, \infty)correct
B(2,6)(2, 6)
C(−∞,2)(-\infty, 2)
D(6,∞)(6, \infty)
Solution
Step 1: For P(x)P(x) (with leading coefficient 1>01 > 0) to take both positive and negative values, the discriminant must be strictly positive:
D=(2−p)2−4(p−2)>0D = (2-p)^2 - 4(p-2) > 0
p2−4p+4−4p+8>0  ⟹  p2−8p+12>0  ⟹  (p−2)(p−6)>0p^2 - 4p + 4 - 4p + 8 > 0 \implies p^2 - 8p + 12 > 0 \implies (p-2)(p-6) > 0
 Step 2: (p−2)(p−6)>0  ⟹  p<2(p-2)(p-6) > 0 \implies p < 2 or p>6p > 6. Correct answer: (1)
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