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B
1+In x<x<1+x In x
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C
x <1 + x In x<1+In x <x
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D
1+In x <1+x In x<x
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Solution
The correct option is B
1+In x<x<1+x In x
Let f(x) =1+x In x in (1, x) ∴f′(x)=(1+Inx)>0(∵x>1)
By LMVT, for 1 < c < x f′(c)=f(x)−f(1)x−1=(1+xInx−1)x−1=xInxx−1>0
or x In x > x -1 or x In x+1 > x
Also, similarly by LMVT
We get, 1 + In x < x
Hence, 1+In x < x < 1 + x In x.