1

all the values of a for which the quadratic equation expression


ax2+(a-2)x-2 is negative for exactly two integral values of x may lie in.? 
1) [1,3/2)
2) [3/2,2]
3) [1,2) 
4) [-1,2)
 


10-Oct-2015 1:24 PM
~12

Answers (1)


1

f(x) = ax+ (a-2)x -2

Now, as the graph is negative for only 2 values of x, its opening should be upwards.

=> a > 0 (1st range)

As roots should be real and distinct, D > 0:
=> (a-2)2 -4(a)(-2) > 0
=> a2+4+4a > 0
=> (a+2)2 > 0
This will be true for all values of a except a =-2
Hence, a is an element of (-infinity, infinity) - {-2}

   ( -, ) - {2}  (2nd range)

Now, f(x) is negative for only 2 integral values, and as f(x) is a continuous function, it should be be negative for 2 consecutive integers.

Now, f(0) = 0 + 0 -2 = -2 <0

As f(0) < 0, that means we have found 1 value of x (=0) for which the quadratic is negative, now f(x) must be negative for either x = 1 or x = -1 only.

Now, f(-1) = a(-1)+ (a-2)(-1) -2 = a - a + 2 - 2 = 0, which is non-negative, hence f(x) must be negative for x = 1 only.

=> f(1) < 0
=> a(1)2 + (a-2)(1) - 2 < 0
=> a + a -2 -2 < 0
=> 2a - 4 < 0
=> 2a < 4
=> a < 2  (3rd range)

Now, taking the intersection of above 3 ranges that we have got for a, the answer is:

   (0,2)



(I don't actually see this answer in the options you have provided, but this is the answer that is coming. Please re-check the options, or there might be misprint from where you had seen the options.)


17-10-2015 10:41
~48

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