A) -4
B) 1
C) 4
D) 0
Correct Answer: C
Solution :
[c] : Let \[\alpha ,\beta \] be the roots of \[{{x}^{2}}-2\sqrt{2}kx+2{{e}^{2\log k}}-1=0\]. \[\therefore \]\[\alpha \beta =2{{e}^{2\log k}}-1\] ...(i) Product of roots \[(\alpha \beta )\] is given 31 Hence from (i), \[2{{e}^{2\log k}}-1=31\] or \[2{{e}^{\log {{k}^{2}}}}-1=31\] or \[2{{k}^{2}}=32\]or\[{{k}^{2}}=16\] or \[k=\pm 4\]. As log(-4) is not defined so k = 4.You need to login to perform this action.
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