JNU ECOM 2011 QUES 57-66

[Arushi :))]

QUES 57
If f(x) and g(x) are differentiable functions such that f'(x)=3x and g'(x)=2*x^2, then the limit
[(f(x)+g(x))-(f(1)+g(1))]/(x-1) as x approaches 1 is equal to
a) 5
b) 0
c) 20
d) None of the above
Answer-a
QUES 58
Functions g and h are given by; g(x)= (x-1)^1/2 and h(x)= (x^2)+1. The composite function (goh)(x) is given by:
a) x
b) |x|
c) (x)^1/2
d) None of the above
Answer-b
QUES 59

Answer-c
QUES 60

Answer-
QUES 61

Answer-d
QUES 62
If x and y are any real numbers such that 0<x<2<y, which of the following is necessarily true?
a) x<xy/2<y
b) 0<xy<2x
c) x<xy<2
d) xy<y
<b>answer-a
QUES 63
A number of the form 213xy, where x and y are digits, has a remainder less than 10 when divided by 100. The sum of all the digits in the above number is equal to 13. The digit y is:
a)5
b)7
c)6
d)8
Answer-b
QUES 64

Answer-c
QUES 65

Answer-
QUES 66
If (1.56)^x=2, then x=
a) log1.56/log2
b)log2/log1.56
c)2/log1.56
d)log2/1.56
Answer-b

[Arushi :))]

How to do questions like 65?
And any idea about 60?

[kangkan]

Hi arushi.. f(x-k) will shift the curve k units to the right..hence f(x-2) will be 2 units to the right..so the new roots will be 0,2,5...i can paste the graph if required

and the income distribution is log normal

[Arushi :))]

haan please paste . i would be grateful to you :)

[kangkan]

Here you go..forgive the crappy drawing skills :)

[Arushi :))]

hehe.... doesn't matter yar :) Thanks :)

[The Villain]

Arushi could u pls explaib quest 57

[Arushi :))]

@ Ron
f'(x) & g'(x) are given . Integrate them to find out f(x) and g(x) . Find f(1), g(1)
and find the limit as x tends to 1 :)

[jack]

f'(x)=3x, then f(x)=3x^2/2
g'(x)=2x^2. then g(x)=2x^3/3
put the values in the given function
(3x^2/2+2x^3/3-3/2-2/3)/(x-1)
use l' hopital rule
then we get,
(3x+2X^2)/-1
put x approaches 1
3+2
5
so option A will be correct





















































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[The Villain]

Thanxx

[suvidya patel]

plz help in qno 64

[The Villain]

Q64...since both the discreminants are negative so these roots are imaginary..and hence cannot intersect...ans c

[riyaf]

can someone explain Q63 ?

[Noel]

@riyaf..sum of digits is 13.hence 6+x+y=13
so x+y=7
now on dividing by 100 the remainder is less than 10.so the possible values for the number range from 21301 to 21309..but x+y=7.so y=7

[riyaf]

thanks :)