Let ƒ and g be real valued functions defined on interval (–1, 1) such that g′′(x) is continuous, g(0) ≠0,
g′(0) = 0, g′′(0) ≠0, and ƒ(x) = g(x) sin x.
STATEMENT - 1
lim x→0 [g(x) cot x – g(0) cosec x] = ƒ′′(0)
and
STATEMENT - 2
ƒ′(0) = g (0).
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for
Statement-1
Please tell which option is correct with explanation !!
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4 Answers
1since interval is (-1,1)
since g(0),g''(o)!=0;
i took g(x)=cosx
1limx-->0( g(x)cosx-g(0))/sinx
itz 0/0 form ........
thus diff we get
limx-->0 [g'(x)cosx-g(x)sinx]/cosx
sincef(x)=g(x)sinx
f'(x)=g'(x)sinx+g(x)cosx
f''(x)=g''(x)sinx+2g'(x0cosx-g(x)sinx
f''(0)=0
thus that limx-->0 is =0
statmnt 1 is true
satmnt2 says............f'(x)=g'(x)sinx+g(x)cosx
f'(0)=g(0)
stsnt 2 is nt d corrct xpannaitio of stsmnt 1
thus ansewr is b)