A) None
B) one
C) finitely many, but greater than 1
D) infinitely many
Correct Answer: B
Solution :
Given, expression is \[{{x}^{2}}-2(4k-1)+x+15{{k}^{2}}-2k-7>0\] Its discriminant, \[D={{b}^{2}}-4ac\] \[={{\{-2(4k-1)\}}^{2}}-4\times 1\times (15{{k}^{2}}-2k-7)\] \[=4{{(4k-1)}^{2}}-4(15{{k}^{2}}-2k-7)\] \[=4[{{(4k-1)}^{2}}-(15{{k}^{2}}-2k-7)]\] \[=4[16{{k}^{2}}-8k+1-15{{k}^{2}}+2k+7]\] \[=4[{{k}^{2}}-6k+8]\] \[=4[{{k}^{2}}-4k-2k+8|=4|(k-4)(k-2)]\] Now, for real values of x, \[D<0\] \[\Rightarrow \] \[(k-4)(k-2)<0\] \[\Rightarrow \] \[k<4\] or \[k>2\] \[\therefore \] Integer value of k is 3. Hence, number of integer value of k is noe.
You need to login to perform this action.
You will be redirected in
3 sec
