Question

Let f(x) = x3+3x2+6x+2010 and g(x)=1/(x-f(1)) + 2/(x-f(2)) + 3/(x-f(3)). Then number of real solutions of g(x) = 0 is
(a) 1 (b) 2 (c) 0 (d) infinite

sry its 3x2

10 Answers

1
skygirl ·

question correct ??

u cud have written 4x3 rather than wat u wrote...

1
skygirl ·

i think square hoga ?

1
skygirl ·

kk..

11
Mani Pal Singh ·

iska ans is (A) try to find it
think the application of derivative way

106
Asish Mahapatra ·

im in XI try in sum XI methodd pls

1
skygirl ·

how can ther be one real soln ??

either two or zero //

62
Lokesh Verma ·

there is only one thng here.

f'(x) > 0 for all x..

so f(x) is an increasing function

now when x is just less than f(1) then the given function tends to -infinity

when x just tends to greater than f(1) the function tends to + infinty
when x is just less than f(2) then the given function tends to -infinity
Thus, there is a root between f(1) and f(2)
same logic will give a root between f(2) and f(3)

so the answer will be 2

I have not explained why there is no root that is nto there between - infinty and f(1) and also between f(3) and infinity.

I hope you can figure that part out :)

1
skygirl ·

but y u did all these things with f ??

i mean the question asks abt g(x)...

62
Lokesh Verma ·

this is all abut g.. look carefully

all this after the 2nd line has been done on g!

without actually saying that..

Tx for mentioning that :)

1
skygirl ·

opsie!

sry :(

time to change spects :P

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