23f(x)=ax2 - (3+2a)x + 6
= ax2 -2ax - 3x + 6
= ax(x-2) -3(x-2)
=(ax-3)(x-2)
f(x) >0 for 3 values of x < 0
i.e (ax-3)(x-2) > 0 ...(1)
look below ( bipin sir's post) for further part of d ans
If set of values of 'a' for which f(x) = ax2 - (3+2a)x + 6 , a≠0 is positive for exactly 3
distinct negative integral values of x is (c,d] then the value of (c2 + 16 d2)
is equal to ----------??
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9 Answers
1What i've got is that f(x) can never be less than 0.
for f(x) to be greater than 0 the given quadratic has to have a discriminant <0.
so (3+2a)2 < 4.a.6
so we get (3-2a)2 < 0.
this is impossible as the square of a number can never be -ve.
So, for the given condition to hold..... c=0 and d = 0.
Hence answer is 0.
But i'm damn sure that i've gone wrong somewhere .
But where ?
23
1(ax-3)(x-2) > 0 for exactly 3 -ve integral x
clearly a<0 otherwise there will be many -ve integral x satisfying above inequality
(x - 3a) (x - 2) < 0
x belongs to (3a,2)
since x takes exactly 3 -ve integral values (it can be x = -1, -2, -3 ) :
-4 <= 3/a < -3
a belongs to (-1, -3/4]
so, c2+16d2 = 10
23sir u killed it b4 me and within no time !!!
but sir y cant x be -4,-5,-6 ?
and y did u take the max possible -ve values i.e -1,-2,-3
1Bipin Sir , could you please explain the part " x must vary upto 2 continuously " ?
And moreover I'm facing problem with #3 line 4.
Bipin Sir...please help.
1@aveek : #3 line4 : that discriminant < 0 only when coefficient of x2 is +ve to make the quadratic +ve
suppose x belongs to (-5,2) then there are four -ve integral values i.e. x = -4, -3, -2, -1
so, (3/a , 2) ranges from (-4,2) to [-3,2) (to have exactly three of them x = -3, -2, -1)