A) 1
B) 2
C) 3
D) more than 3
Correct Answer: D
Solution :
| [d] Let \[p(x)=a{{x}^{2}}+bx+c\] be the required polynomial whose zeroes are \[-2\]and 5. |
| \[\therefore \] Sum of zeroes \[=-\frac{b}{a}=-2+5\] |
| \[=\frac{3}{1}=-\frac{(-3)}{1}\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\frac{b}{a}=-\frac{(-3)}{1}\] .(1) |
| and product of zeroes \[=\frac{c}{a}=-2\times 5=\frac{-10}{1}\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{c}{a}=\frac{-10}{1}\] .(2) |
| From eqs. (1) and (2), \[a=1,\,b=-3\]and \[c=-10\] |
| \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,p(x)={{x}^{2}}-3x-10\] |
| But we know that, if we multiply/divide any polynomial by any arbitrary constant, then the zeroes of polynomial never change. |
| \[\therefore \,\,\,\,\,\,p(x)=k{{x}^{2}}-3kx-10k,\]where k is a real number and \[p(x)=\frac{{{x}^{2}}}{k}-\frac{3}{k}x-\frac{10}{k},\]is a non-zero real number. |
| Thus, the required number of polynomial is finite i.e., more than 3. Hence, out of given options, option [d] is correct. |
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