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`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) |
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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] |
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