IIT JEE 2008 P1 Q11

7 Answers

1
Aditya ·

d coz the expression is statement1 is zero.

11
virang1 Jhaveri ·

OPTION
C)

1
neil.dhruva ·

answer is B

11
rkrish ·

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

11
rkrish ·

Never forget your principles (especially 1st principles)[4][4]

11
Mani Pal Singh ·

BUT YAAR USKO KOI RELATION NAHIN HA S1 SE

TO IT HAVE TO BE B[1]

11
rkrish ·

@mani....

I have edited post #5 (see the part in [i ]....[/i ])....now do u see any relation ??

Your Answer

Close [X]