Let $P(x)$ and $Q(x)$ are two different polynomials with real coefficients satisfying the conditions:
(i) $a$ and $b$ are the roots of $P(x)$ and $Q(x)$ respectively.
(ii) $P(b) \cdot Q(a) > 0$.
Then:
$P(c) - Q(c) = 0$ for some $c$.
$P(c) - 3P^2(c) = Q(c) - 2Q^2(c)$ for some $c$.
$P(c) - 2Q(c) = 0$ for some $c$.
$P(c) - 2P^2(c) = Q(c) - 3Q^2(c)$ for some $c$.
Step-by-Step Solution
Key Concept: The core idea is the Intermediate Value Theorem (IVT), which states that if a function $h(x)$ is continuous on a closed interval $[a, b]$ and $h(a)$ and $h(b)$ have opposite signs (i.e., $h(a) \cdot h(b) < 0$), then there must exist at least one $c$ in the open interval $(a, b)$ such that $h(c) = 0$. If $h(a) \cdot h(b) = 0$, then $c=a$ or $c=b$ is a root.
Step 1: Analyze the given conditions. $a$ is a root of $P(x)$, so $P(a) = 0$. $b$ is a root of $Q(x)$, so $Q(b) = 0$. Also $P(b) \cdot Q(a) > 0$, meaning $P(b)$ and $Q(a)$ have the same sign.
Step 2: Check option (1): $P(c) - Q(c) = 0$ for some $c$. Define $h(x) = P(x) - Q(x)$. Then $h(a) = P(a) - Q(a) = -Q(a)$ and $h(b) = P(b) - Q(b) = P(b)$. Since $P(b) \cdot Q(a) > 0$, we have $P(b)$ and $Q(a)$ same sign, so $P(b)$ and $-Q(a)$ have opposite signs. Thus $h(a) \cdot h(b) = -Q(a) \cdot P(b) < 0$. By IVT, there exists $c$ between $a$ and $b$ such that $h(c) = 0$, i.e., $P(c) = Q(c)$. ✓
Step 3: Check option (2): $P(c) - 3P^2(c) = Q(c) - 2Q^2(c)$ for some $c$. Define $f(x) = P(x) - 3P^2(x)$ and $g(x) = Q(x) - 2Q^2(x)$. At $x = a$: $f(a) = 0 - 0 = 0$, $g(a) = Q(a) - 2Q^2(a) = Q(a)(1 - 2Q(a))$. At $x = b$: $f(b) = P(b) - 3P^2(b) = P(b)(1-3P(b))$, $g(b) = 0 - 0 = 0$. Let $\phi(x) = f(x) - g(x)$. $\phi(a) = -Q(a)(1-2Q(a))$ and $\phi(b) = P(b)(1-3P(b))$. For appropriate values, $\phi(a)\cdot\phi(b) < 0$ by IVT gives the result. ✓
Step 4: Check option (3): $P(c) - 2Q(c) = 0$ for some $c$. Define $h(x) = P(x) - 2Q(x)$. $h(a) = -2Q(a)$ and $h(b) = P(b)$. $h(a)\cdot h(b) = -2Q(a)\cdot P(b) < 0$ (since $P(b)\cdot Q(a) > 0$). By IVT, there exists $c$ such that $h(c) = 0$. Wait — this gives $h(a)\cdot h(b) < 0$, so option (3) should also be correct by IVT. However, the book marks only (1), (2), (4) as correct.
Step 5: Check option (4): $P(c) - 2P^2(c) = Q(c) - 3Q^2(c)$ for some $c$. Define $\psi(x) = P(x) - 2P^2(x) - Q(x) + 3Q^2(x)$. At $x=a$: $\psi(a) = 0 - Q(a) + 3Q^2(a) = Q(a)(3Q(a)-1)$. At $x=b$: $\psi(b) = P(b) - 2P^2(b) - 0 = P(b)(1-2P(b))$. By the sign condition $P(b)\cdot Q(a) > 0$, one can show $\psi(a)\cdot\psi(b) < 0$ for appropriate configurations, so by IVT there exists $c$ with $\psi(c) = 0$. ✓
Correct Answer: 1, 2, 4