A) \[(2,-4,\text{ }2)\]
B) (2, 4, 2)
C) \[(2,4,-2)\]
D) \[(2,-4,-2)\]
E) \[(-2,4,2)\]
Correct Answer: A
Solution :
Given that,\[\underset{x\to 1}{\mathop{\lim }}\,\,\frac{ax\text{ }+bx+c}{{{(x-1)}^{2}}}=2\] This limit will exist, if \[a{{x}^{2}}+bx+c=2{{(x-1)}^{2}}\] \[\Rightarrow \] \[a{{x}^{2}}+bx+c=2{{x}^{2}}-4x+2\] \[\Rightarrow \] \[a=2,b=-4,c=2\]You need to login to perform this action.
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