Q1. [1]
If $\alpha$, $\beta$ are the zeroes of the polynomial $p(x) = 4x^2 - 3x - 7$, then $\left(\dfrac{1}{\alpha} + \dfrac{1}{\beta}\right)$ is equal to :
- (a) $\dfrac{7}{3}$
- (b) $\dfrac{-7}{3}$
- (c) $\dfrac{3}{7}$
- (d) $\dfrac{-3}{7}$
Previously asked in CBSE board exam
2023 30/4/1 Q17
Generated by claude-sonnet-4-6 · 2026-06-15 10:35 · grounding rag
Model Answer
(d) $\dfrac{-3}{7}$
For $p(x) = 4x^2 - 3x - 7$: $\alpha + \beta = \dfrac{3}{4}$, $\alpha\beta = \dfrac{-7}{4}$.
$$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{3/4}{-7/4} = \frac{-3}{7}$$
Source: Chapter 2, Section 2.3
---
Explanation
- Use the standard formulae: sum of zeroes $= -b/a$ and product of zeroes $= c/a$.
- Rewrite $\frac{1}{\alpha}+\frac{1}{\beta}$ as $\frac{\alpha+\beta}{\alpha\beta}$ — this is the key algebraic step examiners look for.
- Here $a=4,\ b=-3,\ c=-7$, giving the answer $-3/7$. Don't forget the negative sign.
If a question refers to an image, map, graph or diagram that is not shown here, open the Study Guide single page app, go to
Library and find the actual CBSE question paper. The original papers are also available on the CBSE website:
cbse.gov.in.