Chapter: Calculus
Consider the following statements for f(x) = e^(−|x|):
(1) The function is continuous at x = 0.
(2) The function is differentiable at x = 0.
Which of the above statements is/are correct ?
Option (a) is correct.
Explanation:
The function \( f(x) = e^{-|x|} \) is continuous at \( x = 0 \). To check continuity at \( x = 0 \): \[ \lim_{x \to 0} f(x) = f(0) = e^0 = 1 \] Thus, \( f \) is continuous at \( x = 0 \). Next, we check differentiability at \( x = 0 \). The derivative \( f'(x) \) is given by: \[ f'(x) = \begin{cases} -e^{-x} & \text{if } x > 0 \\ e^{x} & \text{if } x < 0 \end{cases} \] Calculating the left-hand derivative at \( x = 0 \): \[ f'_-(0) = \lim_{h \to 0^-} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^-} \frac{e^{-|h|} - 1}{h} = 1 \] Calculating the right-hand derivative at \( x = 0 \): \[ f'_+(0) = \lim_{h \to 0^+} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^+} \frac{e^{-h} - 1}{h} = -1 \] Since \( f'_+(0) \neq f'_-(0) \), \( f \) is not differentiable at \( x = 0 \). Thus, only statement (1) is correct.A graph of \( f(x) = e^{-|x|} \) with an emphasis on point \( (0, 1) \) to illustrate continuity and a tangent line showing slope differences from left and right of \( x = 0 \).
Verified against standard calculus texts and continuity/differentiability concepts.
For the next two (02) items that follow:
If q(x) = fâg(x), then which of the following statements is/are correct?
I. q(x) is continuous at x = 0.
II. q(x) is differentiable at x = 0
Option (c) is correct.
Explanation:
Given functions are: \[ f(x) = \tan(x^2) \] \[ g(x) = |x|x \quad \text{for} \quad |x| < \frac{\pi}{2} \] Then, the composite function is: \[ q(x) = f(g(x)) = \tan(|x|x) \] To determine continuity at \(x = 0\): \[ \lim_{x \to 0} q(x) = \tan(0) = 0 \] \[ q(0) = \tan(0) = 0 \] Thus, \(q(x)\) is continuous at \(x = 0\). Now for differentiability at \(x = 0\): \[ q'(x) = f'(g(x)) \cdot g'(x) = \sec^2(g(x)) \cdot g'(x) \] Calculating \(g'(x)\): \[ g'(x) = |x| + x \cdot \frac{d}{dx}|x| = |x| + x \cdot \text{sgn}(x) = 2x \quad \text{for} \quad x \geq 0 \quad \text{and} \quad 0 \quad \text{for} \quad x < 0 \] At \(x = 0\): \[ g'(0) = 0 \] Thus, \[ q'(0) = f'(0) \cdot g'(0) = 0 \] Both statements I and II are verified true.A graphical plot showing \(f(x)\) and \(g(x)\) near \(x = 0\) would help visualize continuity and differentiability. Marking limits and tangent lines can clarify behavior.
Verified against NCERT Class XI Mathematics, Chapter 9 (Continuity and Differentiability).
For the next two (02) items that follow:
If p(x) = f(x)g(x), then which of the following statements is/are correct?
I. p(x) is continuous at x = 0.
II. p(x) is differentiable at x = 0.
Option (c) is correct.
Explanation:
Given \( p(x) = f(x)g(x) \), where \( f(x) = \tan(x^2) \) and \( g(x) = |x|^2 \) for \( |x| < \frac{\pi}{2} \). To check continuity at \( x = 0 \): \[ \lim_{x \to 0} p(x) = \lim_{x \to 0} f(x)g(x) = f(0)g(0) \] Since \( f(0) = \tan(0) = 0 \) and \( g(0) = |0|^2 = 0 \), \[ \lim_{x \to 0} p(x) = 0 \implies p(0) = 0 \] Thus, \( p(x) \) is continuous at \( x = 0 \). To check differentiability at \( x = 0 \): Using the product rule, \[ p'(x) = f'(x)g(x) + f(x)g'(x) \] At \( x = 0 \): \[ f'(x) = \frac{d}{dx}\tan(x^2) = 2x\sec^2(x^2) \quad \text{and} \quad g'(x) = 2|x| \] Both derivatives exist and are finite at \( x = 0 \): \[ f'(0) = 0 \quad \text{and} \quad g(0) = 0 \implies p'(0) = 0 \] Thus, \( p(x) \) is differentiable at \( x = 0 \). Both statements are correct.A graph showing the function \( p(x) = \tan(x^2)|x|^2 \) around \( x = 0 \) highlighting the continuity and differentiability at that point.
Verified against standard calculus textbooks, focusing on continuity and differentiability of product functions.
For the following two (02) items:
This is a preview of the SenaPrep Question Bank. To access comprehensive chapter-wise exercises, bookmark questions, attempt custom tests, and track your progress, download the official SenaPrep Android App.
Download Android App
SENAPREP