A) 2
B) \[-2\]
C) 1
D) \[-1\]
Correct Answer: A
Solution :
\[f(x)=({{k}^{2}}+4){{x}^{2}}+13x+4k\] Now, let \[\alpha \] and \[\beta \] be the roots, then according to the question, \[\alpha =\frac{1}{\beta }\Rightarrow \alpha \beta =1\] Now, we know that \[\alpha \beta =\frac{4k}{{{k}^{2}}+4}\] \[\Rightarrow \] \[1=\frac{4k}{{{k}^{2}}+4}\Rightarrow {{k}^{2}}-4k+4=0\] \[\Rightarrow \] \[{{(k-2)}^{2}}=0\,\,\,\Rightarrow k-2=0\Rightarrow k=2\]
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