Question

the set values of'a 'for which the $f\left(x\right)=a{x}^2+2x(1-a)-4$ is negative for exactly three integral value of $x$, is

Answer 1

Given:

$f\left(x\right)=a{x}^2+2x(1-a)-4$ is negative for exactly three integral value of $x$.

it is clear that the graph of the function open upwards.

$\Rightarrow a>0$----------(1)

As roots should be real and distinct, $D>0$

$\Rightarrow \left(2\right(1-a){)}^2-4(a\left)\right(-4)>0$

$\Rightarrow 4(1-a{)}^2+16a>0$

$\Rightarrow (1-a{)}^2+4a>0$

$\Rightarrow 1+a^2-2a+4a>0$

$\Rightarrow 1+a^2+2a>0$

$\Rightarrow (1+a{)}^2>0$

$\Rightarrow a>-1$______(2)

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

Now, $f\left(0\right)=0+0-2=-2<0$------(3)
As $f\left(0\right)<0$, that means we have found $1$ value of $x=0$ for which the quadratic is negative, now $f\left(x\right)$ must be negative for either $x=1$ or $x=-1$ only.

Now,

$f(-1)=a(-1{)}^2+2(-1\left)\right(1-a)-4<0$

$\Rightarrow f(-1)=a-2+2a-4<0$

$\Rightarrow f(-1)=3a-6<0$

$\Rightarrow a<2$------(4)

$f\left(1\right)=a(1{)}^2+2(1\left)\right(1-a)-4<0$

$\Rightarrow f\left(1\right)=a+2-2a-4<0$

$\Rightarrow f\left(1\right)=-(a+2)<0$

$\Rightarrow a>-2$ ------(5)

Now, taking the intersection of above four ranges that we have got for a, the answer is $[1,2)$.

$\therefore x\in [1,2)$.