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Let `f:[-pi/3,(2pi)/3]vec[0,4]`be a function defined as `f(x)=sqrt(3)sinx-cosx+2.`Then `f^(-1)(x)`is given by`sin^(-1)((x-2)/2)-pi/6``sin^(-1)((x-2)/2)+pi/6``(2pi)/3+cos^(-1)((x-2)/2)`(d) none of theseA. `sin^(-1)((x-2)/(2))-(pi)/(6)`B. `sin^(-1)((x-2)/(2))+(pi)/(6)`C. `(2pi)/(3)-cos^(-1)((x-2)/(2))`D. none of these |
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Answer» Correct Answer - B::C `f(x)=sqrt(3)sinx-cosx+2` `=2(((sqrt(3))/(2)sinx-((1)/(2))cosx)+2` `=2(cos((pi)/(6))sinx-sin((pi)/(6))cosx)+2` `2sin(x-(pi)/(6))+2` `x in[-(pi)/(3),(2pi)/(3)]impliesx-(pi)/(6)in[-(pi)/(3)-(pi)/(6),(2pi)/(3)-(pi)/(6)]` i.e., `x-(pi)/(6)in[-(pi)/(2),(pi)/(2)]` if `x-(pi)/(6)in[-(pi)/(2),(pi)/(2)]` `impliessin(x-(pi)/(6))in[-1,1]` `implies2sin(x-(pi)/(6))+2in[0.4]` `impliesf(x)` is one-one & onto `becausef^(-1)(x)` exists `y=2sin(x-(pi)/(6))+2` `(y-2)/(2)=sin(-(pi)/(6))` `x=sin^(-1)((y-2)/(2))+(pi)/(6)` i.e., `f^(-1)(x)=sin^(-1)((x-2)/(2))+(pi)/(6)` |
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