Limits, Continuity & Differentiability
Continuity and Differentiability
nta_pyq_2025_jan
Grade 12
Question:
Let the function f (x) = (x + 1) ∣∣x - ax + 2∣∣ + cos |x| be not differentiable at the two points x = \alpha = 2 and 2 2 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: Apply the core result for continuity and differentiability at a point and simplify using the given constraints.
f (x) = (x + 1) ∣∣x - ax + 2∣∣ + cos |x| 2 2 (3) Notice that cos(-x) = cos x = cos |x| which means cos |x| is differentiable everywhere in x \in R 2 \Rightarrow f (x) can be non differentiable where ∣ ∣x - ax + 2∣ ∣ = 0 2 \Rightarrow x - ax + 2 = 0 \Rightarrow 4 - 2a + 2 = 0 \Rightarrow a = 3 2 \Rightarrow (x - 3x + 2) = 0 \Rightarrow x = 1, 2 \beta = 1 distance of (\alpha, \beta) from line 12x + 5y + 10 = 0 |2(12) + 5(1) + 10| 39 \Rightarrow = = 3 13 13
Correct Answer: 3