JEE Main 2025 — 29 January, Evening Shift — Question 46

$$f(x)=(x^2-1)\,|x^2-ax+2|+\cos|x|$$

Non-differentiability arises only from absolute value terms.

1. From $|x^2-ax+2|:$

$$x^2-ax+2=0$$

Given one point of non-differentiability is $\alpha=2$,

$$4-2a+2=0 \Rightarrow a=3$$

So zeros are $x=1,2$. At $x=1$, the factor $(x^2-1)=0$, hence differentiability is restored. Thus this term causes non-differentiability only at $x=2$.

2. From $\cos|x|$: $\cos|x|$ is non-differentiable at $x=0$ and the first term is non-zero there, so non-differentiability remains.

Hence,

$$\alpha=2,\quad \beta=0$$

Distance of point $(2,0)$ from the line $12x+5y+10=0$:

$$d=\frac{|12(2)+5(0)+10|}{\sqrt{12^2+5^2}} =\frac{34}{13}\approx 2.62$$ $$\boxed{3}$$