#30 Measure Your Calibre

You dont even need to calculate the value of A, since for every solution 'x' , there is a solution '-x' too since the equation is even in x or symmetric about x=0 , hence the sum of solutions would be zero independent of the value of A

We need to find the area satisfying both the curves

Now see $\sin(\log_2|z|)>0$ means $\log_2|z|$ lies in $(0, \pi)$ .

$\implies$ $0 < \log_2|z| < \pi$

$\implies$ $1 < |z| < 2^{\pi}$

Now see that another equation is $|z|<1$ , but from the above curve we have got $|z|>1$ . So , the area cannot be founded. It is simply $0$ .

See these curves in desmos

So $A=0$

$\implies$ $|x^2-4|+|x^2-1|=\lfloor 0 + 3 \rfloor$

$\implies$ . $|x^2-4|+|x^2-1|= 3$

Now lets try to divide them in intervals and solve these.

Case 1 When $x \geq 2$
$\implies$ $x^2-4+x^2-1 = 3 \space or \space x=2$

Case 2 In the region $2> x \geq 1$ . $\implies$ $-x^2+4+x^2-1=3$ $\implies$ $x=1$ as here for any $x$ , we can satisfy, But we need the integral one,

Case 3 In the region $1 > x \geq -1$ .

$\implies$ $-x^2+4-x^2+1=3$ $\implies$ $x=-1$ after solving

Case 4 Now again in the interval $-1 > x \geq -2$
$\implies$ $-x^2+4+x^2-1=3$ $\implies$ There is no solutions

Case 5

When $x \leq -2$

$x^2-4+x^2-1=5$

$\implies$ $x=-2$

So Adding all we get $Sum=0$