1question correct ??
u cud have written 4x3 rather than wat u wrote...
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
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10 Answers
11iska ans is (A) try to find it
think the application of derivative way
62there 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 :)
1but y u did all these things with f ??
i mean the question asks abt g(x)...
62this 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 :)