A) \[{{x}^{2}}+9\]
B) \[{{x}^{2}}-4\]
C) \[{{x}^{2}}-9\]
D) \[{{x}^{2}}+4\]
Correct Answer: C
Solution :
| Given quadratic polynomial |
| \[f\left( x \right)={{x}^{2}}+x-2\] |
| \[\alpha +\beta =-1\]and \[\alpha \,.\,\beta =-2\] |
| Sum of zeroes i.e., \[\left( 2\alpha +1,\,2\beta +1 \right)\] |
| \[=2\alpha +1+2\beta +1\] |
| \[=2\left( \alpha +\beta +1 \right)\] |
| \[=2\left( -1+1 \right)=0\] |
| Product of zeroes i.e., \[\left( 2\alpha +1,\,2\beta +1 \right)\] |
| \[=\left( 2\alpha +1 \right)\left( 2\beta +1 \right)\] |
| \[=4\alpha \beta +2\left( \alpha +\beta \right)+1\] |
| \[=4\left( -2 \right)+2\left( -1 \right)+1\] |
| \[=-8-2+1=-9\] |
| The quadratic polynomial whose zeroes are |
| \[\left( 2\alpha +1 \right),\,\left( 2\beta +1 \right)\]are |
| \[={{x}^{2}}\] - (sum of zeroes) x + product of zeroes |
| \[={{x}^{2}}-0\,\,\times \,\,x+\left( -9 \right)={{x}^{2}}-9\] |
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