Theory of EquationsmediumFree

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)
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Theory of Equations · easy
The values of α\alpha and β\beta for which the quadratic equation x2+2x+2+e2α−cos⁡β=0x^2 + 2x + 2 + e^{2\alpha} - \cos\beta = 0 has a real solution is
Theory of Equations · easy
If the vertex of the quadratic expression y=2x2−4x+6y = 2x^2 - 4x + 6 is (m,n)(m, n), then the quadratic equation whose vertex is (2m,n2)\left(2m, \dfrac{n}{2}\right) is
Theory of Equations · easy
If α,β\alpha, \beta are the roots of x2−p(x+1)−c=0x^2 - p(x+1) - c = 0, c≠1c \neq 1, then the value of (α+1)(β+1)(\alpha+1)(\beta+1) is equal to
Theory of Equations · easy
Let α\alpha and β\beta be the roots of the equation x2−(p−2)x−(p−1)=0x^2 - (p-2)x - (p-1) = 0, p∈Rp \in \mathbb{R}. If α2+β2\alpha^2 + \beta^2 is least, then pp equals

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath