A) \[k\left( -p{{x}^{2}}+\frac{x}{p}+1 \right)\]
B) \[k\left( p{{x}^{2}}-\frac{x}{p}-1 \right)\]
C) \[k\left( {{x}^{2}}+px-\frac{1}{p} \right)\]
D) \[k\left( {{x}^{2}}-px+\frac{1}{p} \right)\]
Correct Answer: C
Solution :
| Given that, sum of the roots \[=-p\] |
| and product of the roots \[=\frac{-1}{p}\] |
| \[\therefore \]Required quadratic polynomial |
| \[=k\,[{{x}^{2}}-(\text{sum of the roots})x+\text{product of the roots}]\] \[=k\left[ {{x}^{2}}-(-p)x+\left( -\frac{1}{p} \right) \right]=k\left[ {{x}^{2}}+px-\frac{1}{p} \right]\] |
| where k is an arbitrary constant. |
| So, option [c] is correct. |
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