Find a quadratic polynomial whose zeroes are reciprocals of the zeroes of the polynomial $f(x) = ax^2 + bx + c$, where $a
eq 0, c
eq 0$.
Step-by-Step Solution
Key Concept: If $\alpha, \beta$ are zeroes of $f(x)$, the new zeroes are $1/\alpha, 1/\beta$. Find new sum $S'$ and new product $P'$.
For $f(x) = ax^2 + bx + c$, $\alpha + \beta = -\dfrac{b}{a}$ and $\alpha \beta = \dfrac{c}{a}$. [1.0 Mark]
New sum $S' = \dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{\alpha + \beta}{\alpha \beta} = \dfrac{-b/a}{c/a} = -\dfrac{b}{c}$.
New product $P' = \left(\dfrac{1}{\alpha}\right)\left(\dfrac{1}{\beta}\right) = \dfrac{1}{\alpha \beta} = \dfrac{a}{c}$. [1.0 Mark]
Required polynomial $g(x) = k[x^2 - S'x + P'] = k\left[x^2 - \left(-\dfrac{b}{c}\right)x + \dfrac{a}{c}\right] = k\left[x^2 + \dfrac{b}{c}x + \dfrac{a}{c}\right]$.
Taking $k = c$, we get $g(x) = cx^2 + bx + a$. [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Finding original sum and product: 1.0 Mark
Calculating new sum $S' = -b/c$ and new product $P' = a/c$: 1.0 Mark
Forming and simplifying required polynomial $cx^2 + bx + a$: 1.0 Mark
Correct Answer: