TargetIIT

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

9.

Assuming u meant [im]http://codecogs.izyba.com/gif.latex?lim_%7Bx%5Crightarrow%202%7Df%28x%29%3D....[/im]

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

3) log [ss]1[b].[/b]5[/ss][cot[p]-1[/p]x - sgn(e[p]x[/p]) ] = 2

Does it have real solutions???

(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]

(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

@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?

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?

Ok fine I agree with ur ans

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)