Question

Read the statements given below:
I. If α, β are the zeros of the polynomial x² − p(x + 1) −c, then (a + 1)(β + 1) = 1 − c.

II. If α, β are the zeros of the polynomial x² + px + q, then the polynomial having $\frac{1}{\mathrm{\alpha }},\frac{1}{\mathrm{\beta }}$ as zeros is qx² + px + 1.
III. When x³ + 3x² − 5x + 4 is divided by (x + 1), then the remainder is 9.

Which of the above statements is false?

(a) I only
(b) II only
(c) III only
(d) I and III both

Answer 1

$$\begin{gathered} \left(\mathrm{c}\right)\mathrm{III}\mathrm{only} \\ \mathrm{The}\mathrm{g}\text{iven polynomial is}{x}^2-px-(p-c). \\ \therefore \mathrm{\alpha }+\mathrm{\beta }=p\text{and}\mathrm{\alpha \beta }\text{=}-(p-c) \\ =>(\mathrm{\alpha }+1)(\mathrm{\beta }+1)=(\mathrm{\alpha }+\mathrm{\beta })+\mathrm{\alpha \beta }+1=p-(p-c)+1=1+c \\ \therefore \text{I is true}\text{.} \\ \begin{array}{l}\left(\text{II}\right)\mathrm{\alpha }\mathrm{and}\mathrm{\beta }\text{are the zeroes of}{x}^2+px+q.\\ \end{array} \\ \mathrm{\alpha }+\mathrm{\beta }=-p\text{and}\mathrm{\alpha \beta }\text{=}q \\ \frac{1}{\mathrm{\alpha }}+\frac{1}{\mathrm{\beta }}=\frac{\mathrm{\alpha }+\mathrm{\beta }}{\mathrm{\alpha \beta }}=\frac{-p}{q}\text{and}\frac{1}{\mathrm{\alpha }}\times \frac{1}{\mathrm{\beta }}=\frac{1}{q} \\ \therefore \mathrm{R}\text{equired polynomial=}{x}^2-\frac{p}{q}x+\frac{1}{q},\text{i}\text{.e}\text{.,}q{x}^2-px+1 \\ \therefore \text{II is true}\text{.} \\ \left(\text{III}\right)\text{Let}p\left(x\right)=x^3+3x^2-5x+4\text{be divided by}(x+1)\text{.} \\ \mathrm{Then}\text{remainder=}p(-1)={(-1)}^3+3\times {(-1)}^2-5\times (-1)+4 \\ =-1+3+5+4=11 \\ \therefore \text{III is false}\text{.} \end{gathered}$$