Differential Calculus-1
Differential Calculus-1
Allen Star Batch
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 $x \in \mathbb{R}$
$f(x)$ is continuous for all $x \in \mathbb{R} - \{2\}$
$f(x)$ is differentiable for all $x \in \mathbb{R}$
$f(x)$ is differentiable for all $x \in \mathbb{R} - \{2\}$
Step-by-Step Solution
Key Concept: Analyze continuity and differentiability by eliminating the parameter t and examining the piecewise function. The critical point occurs where the absolute value expressions change sign (at t=0 and t=1), which corresponds to x=-1 and x=2. Verify continuity by checking left and right limits of y(x) and differentiability by examining left and right derivatives.
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