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The range of the function `f(x)=int_(-1)^(1)(sinxdt)/(1+2tcosx+t^(2))` isA. `[-(pi)/2,(pi)/2]`B. `[0,pi]`C. `{0,pi}`D. `{-(pi)/2,(pi)/2}` |
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Answer» Correct Answer - D We have `f(x)=int_(-1)^(1)(sinx dt)/(sin^(2)x+(t-cosx)^(2))` `=(sinx)/(sinx)tan^(-1)((t-cosx)/(sinx))|_(-1)^(1)` `=tan^(-1)((1-cosx)/(sinx))-tan^(-1)((-1-cosx)/(sinx))` `=tan^(-1) (tanx//2) +tan^(-1) (cotx//2)` Now we know that `tan^(-1)x+tan^(-1)1/x={((pi)/2,xgt0),(-(pi)/2,xlt0):}` or `tan^(-1)("tan"x/2)+tan^(-1)(1/("tan"x/2))={((pi)/2, "tan"(pi)/2gt0),(-(pi)/2,"tan"x/2lt0):}` Hence range of `f(x)` is `{-(pi)/2,(pi)/2}` |
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