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#1 Posted 10:44pm 27-12-09 SIMPLE ONE THOUGH....MULTIPLE ANS QS 1) Lt [ss]x--->2[/ss] [(f(x) -9) / (x-2)] = 3 Then Lt[ss]x-->2[/ss]f(x) is ? (A) 2 (b)5 (C) 9 (D)12
But there is suffering in life, and there are defeats. No one can avoid them. But it's better to lose some of the battles in the struggles for your dreams than to be defeated without ever knowing what you're fighting for.
Edited on 10:51pm 27-12-09 |
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#2 Posted 10:45pm 27-12-09 Re: SIMPLE ONE THOUGH....9. |
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#3 Posted 10:46pm 27-12-09 Re: SIMPLE ONE THOUGH....Assuming u meant [im]http://codecogs.izyba.com/gif.latex?lim_%7Bx%5Crightarrow%202%7Df%28x%29%3D....[/im] |
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#4 Posted 10:47pm 27-12-09 Re: SIMPLE ONE THOUGH....yes f(x)→9 then only we get 0/0 form |
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#5 Posted 10:50pm 27-12-09 Re: SIMPLE ONE THOUGH....2) f : R---->R is defined as f(x) = { x^2 +kx+3 for x>=0 { 2kx+3 for x<0 If f(x) is injective then 'k' = (A) 2 (B)5 (C) 9 (d)12
But there is suffering in life, and there are defeats. No one can avoid them. But it's better to lose some of the battles in the struggles for your dreams than to be defeated without ever knowing what you're fighting for.
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#6 Posted 10:55pm 27-12-09 Re: SIMPLE ONE THOUGH....3) log [ss]1[b].[/b]5[/ss][cot[p]-1[/p]x - sgn(e[p]x[/p]) ] = 2 Does it have real solutions???
But there is suffering in life, and there are defeats. No one can avoid them. But it's better to lose some of the battles in the struggles for your dreams than to be defeated without ever knowing what you're fighting for.
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#7 Posted 0:30pm 28-12-09 Re: SIMPLE ONE THOUGH....(3) yes log will be defined when cot[p]-1[/p]x > 1 (=sgne^x) for which there are real solutions as range of cot[p]-1[/p]x is [0,3.14]
sadly life is not masti in IITK with MTH 101 making first yr hell fr all freshers (no exception)... even got seniors repeating the course |
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#8 Posted 0:42pm 28-12-09 Re: SIMPLE ONE THOUGH....(2) as the parabola is concave upward, and if k<0 then the st line is downward and the parabola is upward so it wont be one-one sory my mistake. now after x=0 f'(x) > 0 for the parabola => 2x+k>0 for all x>0 => k>0 so a,b,c,d
sadly life is not masti in IITK with MTH 101 making first yr hell fr all freshers (no exception)... even got seniors repeating the course Edited on 0:51pm 28-12-09 |
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#9 Posted 9:12pm 28-12-09 Re: SIMPLE ONE THOUGH....@Ashish : This is my explanation for 2nd qs which's contradicting 3 of u're options f(x) is injective so the quad eqn shud have no distinct roots So D≤0 k[p]2[/p] ≤ 24 So ans shud b only A na?
But there is suffering in life, and there are defeats. No one can avoid them. But it's better to lose some of the battles in the struggles for your dreams than to be defeated without ever knowing what you're fighting for.
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#10 Posted 6:54pm 29-12-09 Re: SIMPLE ONE THOUGH....no D need not be less than zero, what if both the roots are less than zero?? and the value of f(0) = 3 then the function will be injective wouldnt it?
sadly life is not masti in IITK with MTH 101 making first yr hell fr all freshers (no exception)... even got seniors repeating the course |
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#11 Posted 1:21pm 30-12-09 Re: SIMPLE ONE THOUGH....Ok fine I agree with ur ans
But there is suffering in life, and there are defeats. No one can avoid them. But it's better to lose some of the battles in the struggles for your dreams than to be defeated without ever knowing what you're fighting for.
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#12 Posted 9:08pm 30-12-09 Re: SIMPLE ONE THOUGH....for the first one, when you say it should be 0/0 form, you are assuming 1. continuity at x=2 2. differentiability at x =2 3. and Lt[ss]x→2[/ss] f'(x) |
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