InterviewSolution
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InterviewSolution
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क्या `f(x)={{:(x",","यदि ",x le1),("5,","यदि ",xgt1):}` द्वारा परिभाषित फलन `f, x=0,x=1` तथा x = 2 पर सतत है? |
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Answer» `f(x)={{:(x",","यदि ",x le1),("5,","यदि ",xgt1):}` x = 0 पर `L.H.L.=underset(xrarr0^(-))(lim)f(x)" माना "x=0-h` `=underset(hrarr0)(lim)f(0-h)" "rArr 0-h rarr0` `=underset(hrarr0)(lim)(0-h)=0" "rArr" "hrarr0` f(0)=0 `R.H.L.=underset(xrarr0^(+))(lim)f(x)" माना "x=0+h` `=underset(hrarr0)(lim)f(0+h)" "rArr 0+h rarr0` `=underset(hrarr0)(lim)(0+h)=0" "rArr" "hrarr0` `because" L.H.L. "=f(0)=" R.H.L."` `therefore" "f(x),x=0` पर सतत है | `f(1)=1` `L.H.L.=underset(xrarr1^(-))(lim)f(x)" माना "x=1-h` `=underset(hrarr0)(lim)f(1-h)" "rArr1-h rarr1` `=underset(hrarr0)(lim)(1-h)=1-0 =1" "rArr" "hrarr0` `R.H.L.=underset(xrarr1^(+))(lim)f(x)" माना "x=1+h` `=underset(hrarr0)(lim)f(1+h)" "rArr1+h rarr1` `=underset(hrarr0)(lim)5=5" "rArr" "hrarr0` `because" "L.H.L. ne R.H.L.` `therefore" "f(x),x = 1 ` पर सतत नहीं है। x = 2 पर f(2) = 5 `L.H.L. = underset(xrarr2^(-))(lim)f(x)" माना "x=2-h` `=underset(hrarr0)(lim)f(2-h)" "rArr 2-h rarr2` `=underset(hrarr0)(lim)5=5" "rArr" "hrarr0` `R.H.L. = underset(xrarr2^(+))(lim)f(x)" माना "x=2+h` `=underset(hrarr0)(lim)f(2+h)" "rArr 2+h rarr2` `=underset(hrarr0)(lim)5=5" "rArr" "hrarr0` `because R.H.L. = f(2) = R.H.L.` `therefore f(x), x=2 `पर सतत है। |
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