A) \[2f'(c)=g'(c)\]
B) \[2f'(c)=3g'(c)\]
C) \[f'(c)=g'(c)\]
D) \[f'(c)=2g'(c)\]
Correct Answer: D
Solution :
Given, \[f(0)=2,g(1)=2,g(0)=0,\text{ }f(1)=6\] Let, \[F(x)=f(x)-2g(x)\] \[F(0)=f(0)-2g(0)\] \[F(0)=2-2\times 0\] \[F(0)=2\] ? (1) \[F(1)=F(1)-2g(1)\] \[F(1)=6-2\times 2\] \[F(1)=2\] ? (2) F(x) is continuous and differentiable in [0, 1]. \[F(0)=F(1)\] So, according to Rolle?s theorem, there is at least are root between 0 and 1. At which \[F\prime (x)=0.\] \[f\prime (x)-2g\prime (x)=0\] \[f\prime (c)-2g\prime (c)=0\] \[f\prime (c)=2g\prime (c)\]You need to login to perform this action.
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