Multiple choice

I. lf ${f}({x})$ be a quadratic expression which is positive for all real ${x}$ and ${g}({x})={f}({x})+{f'}({x})+{f''}({x})$ for any real ${x}$, then ${g}({x})>0$. II. lf the equations ${p}{x}^{2}-7{x}+3{p}=0$ and $2{x}^{2}+{q}{x}+6=0$ have the same roots, then ${pq}=-14$.Which of the above statement(s) is/are true?

  1. Only I

  2. Only II

  3. Both I and II

  4. Neither I nor II
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

I: If f(x) is a quadratic with positive values, its discriminant is negative. g(x) is also a quadratic with the same leading coefficient, and its discriminant can be shown to be negative, so g(x) > 0. II: If roots are the same, coefficients are proportional. 2/p = q/-7 = 6/3p. Solving gives p=1, q=-14, so pq=-14.

AI explanation

Statement I is true because if f(x) is a positive quadratic, its leading coefficient is positive and its discriminant is negative. When you add f(x), f'(x), and f''(x) to form g(x), the resulting quadratic also has a positive leading coefficient and a discriminant that remains negative, ensuring g(x) is positive for all real x. Statement II is true because if two quadratic equations have the exact same roots, their coefficients must be proportional. Setting the ratios of the coefficients of px^2 - 7x + 3p = 0 and 2x^2 + qx + 6 = 0 equal gives p/2 = -7/q = 3p/6, which simplifies directly to pq = -14. Both statements are correct.

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