InterviewSolution
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InterviewSolution
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Let E_(1)={x in R :x ne 1 and (x)/(x-1) gt 0} and E_(2)={x in E_(1):sin^(-1)(log_(e)((x)/(x-1))) " is real number"} (Here, the inverse trigonometric function sin^(-1)x assumes values in [-(pi)/(2),(pi)/(2)].) Let f:E_(1) to R be the function defined by f(x)=log_(e)((x)/(x-1)) and g:E_(2) to Rbe the function defined byg(x)=sin^(-1)(log_(e)((x)/(x-1))) . The correct option is |
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Answer» `a to s, b to Q, C to p, d to p` `implies x in (-OO ,0) cup (1,oo)` Also `E_(2) : -1 lt log_(e)((x)/(x-1)) le 1` `implies (1)/(e) le (x)/(x-1) le e` `implies (1)/(e) le 1 +(1)/(x-1) lee` `implies (1)/(e)-1 le (1)/(x-1) le e-1` `implies (x-1) in (-oo, (e)/(1-e)] cup [(1)/(e-1),oo)` `implies x in (-oo, (e)/(1-e)] cup [(1)/(e-1),oo)` Now, `(x)/(x-1) in (0, oo)-{1} AA x in E_(1)` `implies log_(e) ((x)/(x-1)) in (-oo,oo)-{0}` `implies sin^(-1)(log_(e)((x)/(x-1))) in [-(pi)/(2),(pi)/(2)]-{0}` `("considering "log_(e) ((x)/(x-1)) in [-1,1]-{0})` |
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