Multiple choice

If $\alpha, \beta$ are the roots of the quadratic equation $(p^2+p+1)x^2+(p-1)x+p^2=0$ such that unity lies between the roots then the set of values of $p$ is:

  1. $\phi$
  2. $p\:\:\epsilon\:\:(-\infty,-1)U(0,\infty)$
  3. $p\:\:\epsilon\:\: (-1,0)$
  4. $(-1,1)$
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C Correct answer
Explanation

For unity to lie between the roots of f(x) = ax^2 + bx + c, a * f(1) < 0. f(1) = (p^2+p+1) + (p-1) + p^2 = 2p^2 + 2p. So (p^2+p+1)(2p^2+2p) < 0. Since p^2+p+1 is always positive, 2p(p+1) < 0, which holds for p in (-1, 0).

AI explanation

For unity to lie strictly between the roots of the quadratic equation, the function evaluated at x = 1 must have the opposite sign of the leading coefficient. Here, the leading coefficient is p^2 + p + 1, which is always positive for all real numbers since its discriminant is negative. Substituting x = 1 into the expression yields 1^2(p^2 + p + 1) + (p - 1) + p^2 < 0, which simplifies to 2p^2 + 2p < 0. Factoring out 2p gives 2p(p + 1) < 0, meaning p must lie in the interval (-1, 0).

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