Given that\[f'\](2)=6 and \[{f}'(1)=4)=\], then \[\underset{h\to 0}{\mathop{\lim }}\,\frac{f(2h+2+{{h}^{2}})-f(2)}{f(h-{{h}^{2}}+1)-f(1)}=\] [IIT Screening 2003]

Given that\[f'\](2)=6 and \[{f}'(1)=4)=\], then \[\underset{h\to 0}{\m

1.	Given that\[f'\](2)=6 and \[{f}'(1)=4)=\], then \[\underset{h\to 0}{\mathop{\lim }}\,\frac{f(2h+2+{{h}^{2}})-f(2)}{f(h-{{h}^{2}}+1)-f(1)}=\] [IIT Screening 2003]
A.	Does not exist
B.	Is equal to ? 3/2
C.	Is equal to 3/2
D.	Is equal to 3
Answer» E.

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Given that\[f'\](2)=6 and \[{f}'(1)=4)=\], then \[\underset{h\to 0}{\mathop{\lim }}\,\frac{f(2h+2+{{h}^{2}})-f(2)}{f(h-{{h}^{2}}+1)-f(1)}=\] [IIT Screening 2003]

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