Limits, Continuity & Differentiability
Non-Differentiability via Absolute Value
nta_pyq_2025_apr
Grade 12
Question:
Let the function $f(x) = (x^2+1)|x^2 - ax + 2| + \cos|x|$ be not differentiable at the two points $x = \alpha = 2$ and $x = \beta$. Then the distance of the point $(\alpha, \beta)$ from the line $12x + 5y + 10 = 0$ is equal to:
Step-by-Step Solution
Key Concept: $\cos|x|$ is differentiable everywhere. Non-differentiability comes from $|x^2-ax+2|=0$. At $x=\alpha=2$: $4-2a+2=0\Rightarrow a=3$. Then find the other zero of $x^2-3x+2=0$.
$4-2a+2=0\Rightarrow a=3$. $x^2-3x+2=(x-1)(x-2)\Rightarrow\beta=1$. Distance of $(2,1)$ from $12x+5y+10=0$: $\frac{|24+5+10|}{13}=\frac{39}{13}=3$.
Correct Answer: 3