Let $f$ be a polynomial of degree 4 having real coefficients satisfying $f(0) = f'(1) = f'(-1) = 0$ and $f(0) = 4, f''\left(\frac{1}{2}\right) = -1$
Match the following:
Column 1:
(A) $f(x) = 0$ has
(B) $4 - f(x) = 0$ has
(C) $f'(x) + x - 1 = 0$ has
(D) $xf'(x) - 4f(x) = 0$ has
Column 2:
(p) root at $x = 2$
(q) root at $x = 1$
(r) 2 equal real roots
(s) no real roots
Step-by-Step Solution
Key Concept: Use derivatives and initial conditions to determine the constant $k$, then integrate to find the function.
Given $F'(x)=k(x^2-1)$ and $F''(x)=k(3x^2-1)$. From $F''\left(\frac{1}{2}\right)=k\left(\frac{3}{4}-1\right)=k=4$, we have $k=4$. Thus $F'(x)=4(x^2-1)$, integrating gives $F(x)=\frac{4x^3}{3}-4x+\mu$. Using $f(0)=4$ implies $\mu=4$, so $F(x)=x^4-2x^2+4=(x^2-1)^2+3$, which has no real roots.
Correct Answer: [A-s] [B-r] [C-q, r] [D-