JEE Main 2024 — 8 April, Shift 2 — Question 24

$$|x+1||x+3|-4|x+2|+5=0$$

Note that

$$|x+1||x+3|=\big|(x+1)(x+3)\big| =\big|(x+2)^2-1\big|.$$

Let

$$t=|x+2| \quad (t\ge0).$$

Then the equation becomes

$$|t^2-1|-4t+5=0.$$

Case 1:$t^2-1\ge0 \Rightarrow t\ge1$

$$t^2-1-4t+5=0$$ $$t^2-4t+4=0 \Rightarrow (t-2)^2=0$$ $$t=2$$ $$|x+2|=2 \Rightarrow x=0,-4$$

Case 2: $t^2-1<0 \Rightarrow 0\le t<1$

$$-(t^2-1)-4t+5=0$$ $$-t^2-4t+6=0 \Rightarrow t^2+4t-6=0$$ $$t=-2+\sqrt{10}\approx1.16 \notin [0,1)$$

Hence no solution from this case.

Number of distinct real roots $=2$