1.

`f(x)=x|log_ex|,x lt 0` is monotonically decreasig inA. `(e,oo)`B. `(0,1//e)`C. `(-oo,-1)cup [1,oo)`D. (1,e)

Answer» Correct Answer - C
We have
`f(x)=|x log_e x|={{:(-x log_ex"," 0 lt x lt 1),(x log_ex","x ge1):}`
`f(x)={{:(-(log_ex+1)"," 0 lt x lt 1 ),((log_ex+1)","x ge1):}`
For f(x) to be decreaing ,we must have `f(x) lt 0 `
`rArr {:{(-(log_ex+1)lt0 " for " 0 lt x lt 1 ),(" "(log_e x+1)lt 0 " for " x gt 1 ):}`
`rArr {:{(log_ex+1gt0 " for " 0 lt x lt 1 ),(log_e x+1lt 0 " for " x gt 1 ):}`
`rArr {:{(x gt1//e " for "0 lt x lt 1 ),(x lt 1//e" for "x gt1):}`
`rArr x gt 1/e "for " 0 lt x lt 1 `
`rArr x in (1//e,1)`
Hence f(x) is decreasing in [1/e,1]


Discussion

No Comment Found

Related InterviewSolutions