11Is the ans (d) ?/????
Let f(x) = (a(1-sinx) + bcosx + 5)/x2 if x<0
3 if x=0
(1+(cx+dx3)/x2 )1/x if x>0
If f is continuous at x=0 , then the values of a+b+c is ?
A) 2 (B) 0 (C) -4 (D) -5
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8 Answers
62you have to find limit x tending to zero..
in the first one, it will be found by LH Rule...
Let f(x) = (a(1-sinx) + bcosx + 5)/x2 if x<0
f(x) = (-acos x - b sin x)/2x
f(x) = (asin x - b cos x)/2
so limit x tending to 0 from the -ve side will give -b/2 =3
so b=-6
not the other part... can you find the RHL using limits?
4Nishant Sir Can u help me out with (1+(cx+dx3)/x2 )1/x if x>0
part plzz
11L..HL = lim x → 0 - f(x) = lim h→ 0 f (0 -h )
= lim h → 0 a (1 - h sin h) + b cos h + 5h 2
Since f(x) is continous as h→ 0 , Numerator must be =0
therefore, a+b+5 =0 ...........................(i)
L.H.L = lim h→0 a (1 -h sin h) - (a+5) cos h +5h 2
= lim h→0 [(a+5) (1 - cos h)h 2 - a h sin hh 2 ]
= (a +5) (1/2) - a
= (5 -a) /2 = Value of the function = f(0) = 3 [bcoz f(x) is continous]
Thus, 5-a = 6 that implies a= -1
From eqn (1), -1 + b +5 =0
so b =-4
Also, R.H.L = lim x→0 + f(x) = lim h→0 f(0 +h)
= lim h→0 { 1 + (c h +d h 2) /h } 1/n
= lim h→0 { 1 + (s + d h 2 ) / h } 1/n
R.H.L is finite, so c = 0
So, therefore, a+b+c = -5