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If f(x) is continuous on [0,2] , differentiable in (0,2) ,f(0)=2, f(2)=8 and f.(x) le3 for all x in (0,2), then find the value of f(1).

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Solution :Applying LMVT to f in [0,1] and again in [1,2] there exists `C_1 in (0,1)` such that
`(f(1)-f(0))/(1-0) = f.C_1 1 le 3 RARR f(1) le 5` …(i)
There exists `C_2 in (1,2)`, such that
`(f(2)-f(1))/(2-1) = f.(C_2) le 3 rArr f(1) GE 5`
Hence, (1) and (2) imply that f(1)=5


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