Differential Calculus-1
Differential Calculus-1
Allen Star Batch
Grade 12
Question:
A function is defined as $f(x) = [\tan x] + \sqrt{\tan x - [\tan x]}$ $0 \leq x 2 \\ 5x - 7 & \text{if } x \leq 2 \end{cases}$ then:
$f$ is not continuous at $x = 2$
$f$ is continuous but not differentiable at $x = 2$
$f$ is differentiable every where
$\lim_{x \to 2^+} f'(x) = 2$
Step-by-Step Solution
Key Concept: Check continuity at x=2 by verifying lim(x→2⁻)f(x) = lim(x→2⁺)f(x) = f(2), then analyze differentiability by comparing left derivative f'(2⁻) = 5 with right derivative f'(2⁺) = 2x|ₓ₌₂ = 4.
The function $f(x) = [\tan x] + \sqrt{\tan x - [\tan x]} = [t] + \sqrt{t - [t]}$ where $t = \tan x$ is defined for $0 \le t 2$, integrating: $\int_0^{1} |1+t-t| dt - \int_0^{x} (2-t) dt + \int_1^x dt = 1 + 1 - \frac{x^2}{2}$ gives $f(x) = 1 + \frac{x^2}{2}$ for $x > 2$ and $f(x) = 5x - 7$ for $x \le 2$. Checking continuity at $x = 2$: $\lim_{x \to 2^+} f(x) = 1 + 2 = 3 = f(2)$ and $f'(2^+) = 2$ while $f'(2^-) = 5$, so the function is continuous but not differentiable at $x = 2$.
Correct Answer: 1,3