If the zeroes of the polynomial $f(x) = x^3 - 3px^2 + qx - r$ are in Arithmetic Progression, prove that $2p^3 - pq + r = 0$.
(Note: This classic NCERT Exemplar HOTS problem uses the zero relationship property $\alpha + \beta + \gamma = -b/a$.)
Step-by-Step Solution
Key Concept: Let zeroes be $a - d, a, a + d$. Sum of zeroes $= 3a = -(-3p) = 3p \Rightarrow a = p$. Since $a=p$ is a zero, $f(p) = 0$.
Stepwise Solution:
Let the zeroes of $f(x)$ be in AP: $\alpha = a - d, \beta = a, \gamma = a + d$. [1.0 Mark]
Sum of zeroes: $\alpha + \beta + \gamma = (a - d) + a + (a + d) = 3a$. [1.0 Mark]
From the polynomial $f(x) = x^3 - 3px^2 + qx - r$, sum of zeroes $= -\dfrac{\text{coeff of } x^2}{\text{coeff of } x^3} = -\dfrac{-3p}{1} = 3p$. [1.0 Mark]
Equating: $3a = 3p \Rightarrow a = p$. [0.5 Mark]
Since $a = p$ is one of the zeroes of $f(x)$, $f(p) = 0$.
Substituting $x = p$ into $f(x)$:
$p^3 - 3p(p^2) + q(p) - r = 0 \Rightarrow p^3 - 3p^3 + pq - r = 0 \Rightarrow -2p^3 + pq - r = 0 \Rightarrow 2p^3 - pq + r = 0$. Proved! [1.5 Marks]
Marking Scheme:
• Assuming zeroes in AP as $(a-d, a, a+d)$: 1.0 Mark
• Sum of zeroes $= 3a$: 1.0 Mark
• Relating sum of zeroes to coefficient $3p$ and deducing $a = p$: 1.5 Marks
• Substituting $x = p$ into $f(x)=0$ and proving $2p^3 - pq + r = 0$: 1.5 Marks
Correct Answer: