Question

If `L=lim_(xto0)(sinx+ae^(x)+be^(-x)+clog_(e)(1+x))/(x^(3))` exists finitely, then
The solutions set of `||x+c|-2a|lt4b` is
A. `{:(a,b,c,d),(s,r,q,p):}`
B. `{:(a,b,c,d),(q,s,r,p):}`
C. `{:(a,b,c,d),(s,r,p,q):}`
D. `{:(a,b,c,d),(s,p,q,r):}`

Answer 1

Correct Answer - C
`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]`