InterviewSolution
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InterviewSolution
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`f(x)=sin[x]+[s in x],0A. `{{:(0",",0ltxlt1),(1+sin1",",1lexlt(pi)/(4)):}`B. `{{:((1)/(sqrt(2))",",0ltxlt(pi)/(4)),(1+(1)/(2)+(1)/(sqrt2)+(sqrt3)/(2)",",(pi)/(4)lexlt(pi)/(4)):}`C. `{{:(0",",0ltxlt1),(sin1",",1lexlt(pi)/(2)):}`D. `{{:(0",",0ltxlt(pi)/(2)),(1",",(pi)/(4)ltxlt1),(sin1",",1lexle(pi)/(4)):}` |
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Answer» Correct Answer - C `0ltxlt(pi)/(2)` `therefore" "[x]={{:(0,"if,0ltxlt1),(1,if,1lexlt(pi)/(2)):}` `rArr" "sin[x]={{:(sin0=0,if,0ltxlt1),(sin1,if,1lexlt(pi)/(2)):}` we have `0sinxlt1` when `0ltxlt(pi)/(2).` `therefore" "[sinx]=0" for "0ltxlt(pi)/(2)` `therefore" "sin[x]+[sinx]=({:(0,if,0ltxlt1),(sin1,if,1lexlt(pi)/(2)):}` |
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