Concept Booster -2

This one's more easy. Again take $5^{x}=y$ to get our tranformed as :

How many integral values of $a$ lie in the interval $[-1000,1000]$ such that :

$y^{2}+(a+2)y-(a+3) < 0$ is satisfied by at least one $y>0$

Now this is factorisable, this quadratic gets factorised into :

$(y-1)(y+a+3) < 0$

Now it's roots are $1,-(a+3)$

At first sight it may seem that whatever $a$ may be there exits atleast one $y>0$ .

But for $a=-4$ the roots are $1,1$ making the quadratic always positive.

Hence number of integral values are $2000$

Let $5 ^{x} = t$

One could see that can accept value >0 So the two values of t should be non-negative . So we have the quadratic inequality --

$t^{2} + (a+2)t - (a+3) <0$

So the two negative values of 't' satisfying the inequality is 0 .

(because f(t) > 0 which implies a<-3

and -b/2a <0 which implies a>-2

and D>0 (always positive here) )

The union of all cases is null set .

Hence all values of a satisfies the condition t>0 , which means atleast one real x )

Hence total cases = $\boxed {2000}$

A SIMPLE LOGICAL question! Since it is specifically asked how many values of 'a' lies in the interval [-1000,1000]. The logic behind this is just numbers on the number line, i.e (-1000)- 0 which 1000 and from 0 to 1000 which is again 1000. So on adding the two values we get 2000 that is the correct answer. No higher math required for such a simple problem