maths msqe

[priyanka p]

 question 1  let f:[-1,1]__R be twice differentiable function at x=0, f(0)=f'(0)=0 andf''(0)=4 then the value of limit x tends to 0  {2f(x)-3f(2x)+f(4x)}/x^2  is       (a) 11   (b) 2   (c) 12    (d) none of the above.    my answer to this is option (c)i.e.  12                                                                                                                                                                                                                                                                                                                                                                                       question 2    limit n tends  to infinity {1-x^(-2n) /1+x^(-2n)},  x>0 equals  (a) 1   (b)  -1  (c) 0  (d) limit doesnot exist                  answer should be option (b) i.e -1      sir pls correct them

[duck]

hi..

q1>> its 0/0 form , so we by applying L hopital we get>>

2f'(x) - 6f'(2x) + 4f'(4x) / 2x

now again its 0/0 form , so we will apply L hopital again , we get>>

2f"(x) - 12f"(2x) + 16f"(4x) /2

by subsituting values , we get>> 24/2 = 12

[duck]

q2>>

for 0 < x < 1 , limit = -1
for x=1 , limit = 0
for x>1 , limit =1

so, limit depends upon the value of x

[duck]

but in ques , we r just given that x>0

so its none of the option..