InterviewSolution
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InterviewSolution
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If `L=lim_(xto0)(sinx+ae^(x)+be^(-x)+clog_(e)(1+x))/(x^(3))` exists finitely, then Equation `ax^(2)+bx+c=0` has |
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Answer» Correct Answer - D `L=underset(xto0)lim(sinx+ae^(x)+be^(-x)+clog_(e)(1+x))/(x^(3))` `=underset(xto0)lim((x-(x^(3))/(3!))+a(1+(x)/(1!)+(x^(2))/(2!)+(x^(3))/(3!))+b(1-(x)/(1!)+(x^(2))/(2!)-(x^(3))/(3!))+c(x-(x^(2))/(2)+(x^(3))/(3)))/(x^(3))` `=underset(xto0)lim((a+b)+(1+a-b+c)x+((a)/(2)+(b)/(2)-(c)/(2))x^(2)+(-1(1)/(31)+(a)/(3!)-(b)/(3!)+(c)/(3))x^(3))/(x^(3))` `implies a+b=0,1+a-b+c=0,(a)/(2)+(b)/(2)-(c)/(2)=0` and `L=-(1)/(3!)+(a)/(3!)-(b)/(3!)+(c)/(3)` Solving first three equations we get `c=0, a=-1//2,b=1//2.` `:." "L=-1//3` Equation `ax^(2)+bx+c=0` reduces to `x^(2)-x=0impliesx=0,1` `||x+c|-2a|lt4b` reduces to `||x|+1|lt2` `implies-2lt|x|+1lt2` `implies0le|x|lt1` `impliesx in[-1,1]` |
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