Let y=f(x)andy=g(x) be two differentiable function in [0,2] such that f(0)=3,f(2)=5,g(0)=1 and g(2)=2. If there exists atleast one cϵ(0,2) such that f′(c)=kg′(c), then k must be
A
2
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B
3
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C
12
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D
1
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Solution
The correct option is A 2 Leth(x)=f(x)−kg(x) and h′(c)=0
So Rolle's theorem is applicable. ⇒h(0)=h(2) ⇒3−k=5−2k⇒k=2