| Assertion (A): If one zero of the polynomial \[p(x)=({{k}^{2}}+9){{x}^{2}}+9x+6k\] is the reciprocal of the other zero, then \[\text{k}=\text{3}\]. |
| Reason (R): If \[(x-\alpha )\] is a factor of the polynomial \[p(x),\]then a is a zero of \[p(x)\]. |
A) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
B) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
C) Assertion (A) is true but reason (R) is false.
D) Assertion (A) is false but reason (R) is true.
Correct Answer: B
Solution :
| [b] Reason is clearly true by factor theorem. |
| Let \[\alpha \] and \[\frac{1}{\alpha }\] be the zeroes of polynomial |
| \[p(x)=({{k}^{2}}+9){{x}^{2}}+9x+6k\] |
| Then, product of zeroes \[=\alpha +\frac{1}{\alpha }=\frac{6k}{{{k}^{2}}+9}\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{6k}{{{k}^{2}}+9}=1\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{k}^{2}}+9=6k\Rightarrow {{k}^{2}}-6k+9=0\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{(k-3)}^{2}}=0\Rightarrow k-3=0\Rightarrow k=3\] |
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