1

^{2}+(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)

0

f(x) = ax^{2 }+ (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

=> a^{2}+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)^{2 }+ (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.)