SATHEE: Limit Continuity and Differentiability 6 Question 1

Limit Continuity and Differentiability 6 Question 1

1. For the function $f (x) = x \cos \frac{1}{x}, x \geq 1$ ,

(a) for atleast one $x$ in the interval

(2009)

$[1, \infty), f (x + 2) - f (x) < 2$

(b) $lim_{x \to \infty} f^{'} (x) = 1$

(d) $f^{'} (x)$ is strictly decreasing in the interval $[1, \infty)$

Analytical & Descriptive Questions

$\begin{aligned} 2. Let f (x) = \begin{array}{cc} x + a, & if x < 0 \\ | x - 1 |, & if x \geq 0 \end{array} \\ g (x) = \begin{array}{cl} x + 1, & if x < 0 \\ (x - 1)^{2} + b, & if x \geq 0 \end{array} \end{aligned}$

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Answer:

Correct Answer: 1. (b, c, d)

Solution:

Given function, $g (x) = | f (x) |$

where $f : R \to R$ be differentiable at $c \in R$ and $f (c) = 0$ , then for function ’ $g$ ’ at $x = c$

$\begin{aligned} g^{'} (c) & = lim_{h \to 0} \frac{g (c + h) - g (c)}{h} [where h > 0] \\ = lim_{h \to 0} \frac{| f (c + h) | - | f (c) |}{h} = lim_{h \to 0} \frac{| f (c + h) |}{h} [as f (c) = 0 (given)] \\ = lim_{h \to 0} | \frac{f (c + h) - f (c)}{h} | \\ = | lim_{h \to 0} \frac{f (c + h) - f (c)}{h} | \\ = | f^{'} (c) | [∵ f is differentiable at x = c] \end{aligned}$

Now, if $f^{'} (c) = 0$ , then $g (x)$ is differentiable at $x = c$ , otherwise LHD (at $x = c$ ) and RHD (at $x = c$ ) is different.

Limit Continuity and Differentiability 6 Question 1

1. For the function $f (x) = x \cos \frac{1}{x}, x \geq 1$ ,

Analytical & Descriptive Questions

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Limit Continuity and Differentiability 6 Question 1

1. For the function f(x)=xcos⁡1x,x≥1,

Analytical & Descriptive Questions

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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1. For the function $f (x) = x \cos \frac{1}{x}, x \geq 1$ ,