Let f:[a,b]→R be differentiable on [a,b] and k∈R. Let f(a)=0=f(b). Also J(x)=f′(x)+kf(x). Then
A
J(x)>0 for all x∈[a,b]
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B
J(x)<0 for all x∈[a,b]
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C
J(x)=0 has atleast one root in (a,b)
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D
J(x)=0 through (a,b)
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Solution
The correct option is DJ(x)=0 has atleast one root in (a,b) Let g(x)=ekxf(x) f(a)=0=f(b) by rolle's theorem g′(c)=0,c∈(a,b) g′(x)=ekxf′(x)+kekxf(x) g′(c)=0 ekc(f′(c)+kf(c))=0 ⇒f′(c)+kf(c)=0 for atleast one c in (a,b)