Limits, Continuity & Differentiability
Differential Calculus-1
star_batch_jee_advanced_2025
Grade 12
Question:
If $y = f(x)$ defined parametrically by $x = 2t - |t - 1|$ and $y = 2t^2 + t|t|$, then:
f(x) is continuous for all x ∈ ℝ
f(x) is continuous for all x ∈ ℝ - {2}
f(x) is differentiable for all x ∈ ℝ
f(x) is differentiable for all x ∈ ℝ - {2}
Step-by-Step Solution
Key Concept: Parametric equations with absolute values require case-by-case analysis based on the sign of expressions inside the absolute value.
Given $x = 2t - |t - 1|$ and $y = 2t^2 + |t|$. For $t < 0$: $x = 3t - 1$ and $y = 2t^2 - t^2 = t^2$, giving $y = \frac{1}{9}(x+1)^2$. For $0 \leq t < 1$: $x = 3t - 1$ and $y = 3t^2$, giving $y = \frac{1}{3}(x+1)^2$. For $t \geq 1$: $x = t + 1$ and $y = 3(x-1)^2$. These represent three parabolic curves parameterized by $t$ in different domains.
Correct Answer: 1,4