OPTION
C)
7 Answers
rkrish
·2009-04-05 01:28:05
Abhi only A is left !!![4]
But actualy the ans. is A
f'(x) = g'(x)sin x + g(x)cos x
f'(0) = g(0) ..................St 2 is true
f"(0) = Lt h-->0 [ { f'(0+h) - f '(0) }/h ]
= Lt h-->0 [ { g'(h)sin (h) + g(h)cos (h) - g(0) }/h ]
=Lt h-->0 [ { g'(h) + g(h)cot (h) - g(0)sin (h) } / {h/sin (h)} ]
=Lt h-->0 [ { g'(h) + g(h)cot (h) - g(0)cosec (h) } ]
=Lt h-->0 [g'(h)] + Lt h-->0 [ g(h)cot (h) - g(0)cosec (h) } ]
= Lt x-->0 [ g(x)cot (x) - g(0)cosec (x) } ] ..........St 1 is also true and St 2 is correct reason for it.
Hence ans is A
rkrish
·2009-04-05 03:41:43
@mani....
I have edited post #5 (see the part in [i ]....[/i ])....now do u see any relation ??