SATHEE: Limit Continuity and Differentiability 5 Question 1

Limit Continuity and Differentiability 5 Question 1

1. Let $f : R \to R$ be a continuously differentiable function such that $f (2) = 6$ and $f^{'} (2) = \frac{1}{48}$ . If $\int_{6}^{f (x)} 4 t^{3} d t = (x - 2) g (x)$ , then $lim_{x \to 2} g (x)$ is equal to (2019 Main, 12 April I)

(a) 18

(b) 24

(d) 36

$\frac{\sin (p + 1) x + \sin x}{x}, x < 0$

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Answer:

Correct Answer: 1. (a)

Solution:

Given, $f (x) = x \cos \frac{1}{x}, x \geq 1 \Rightarrow f^{'} (x) = \frac{1}{x} \sin \frac{1}{x} + \cos \frac{1}{x}$

$\Rightarrow f^{''} (x) = - \frac{1}{x^{3}} \cos \frac{1}{x}$

Now, $lim_{x \to \infty} f^{'} (x) = 0 + 1 = 1 \Rightarrow$ Option (b) is correct.

Option (d) is correct.

As $f^{'} (1) = \sin 1 + \cos 1 > 1$

$f^{'} (x)$ is strictly decreasing and $lim_{x \to \infty} f^{'} (x) = 1$

So, graph of $f^{'} (x)$ is shown as below.

Now, in $[x, x + 2], x \in [1, \infty), f (x)$ is continuous and differentiable so by LMVT,

$f^{'} (x) = \frac{f (x + 2) - f (x)}{2}$

As, $f^{'} (x) > 1$

For all $x \in [1, \infty)$

$\Rightarrow \frac{f (x + 2) - f (x)}{2} > 1 \Rightarrow f (x + 2) - f (x) > 2$

For all $x \in [1, \infty)$

Limit Continuity and Differentiability 5 Question 1

1. Let $f : R \to R$ be a continuously differentiable function such that $f (2) = 6$ and $f^{'} (2) = \frac{1}{48}$ . If $\int_{6}^{f (x)} 4 t^{3} d t = (x - 2) g (x)$ , then $lim_{x \to 2} g (x)$ is equal to (2019 Main, 12 April I)

Answer:

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Limit Continuity and Differentiability 5 Question 1

1. Let f:R→R be a continuously differentiable function such that f(2)=6 and f′(2)=148. If ∫6f(x)4t3dt=(x−2)g(x), then limx→2g(x) is equal to (2019 Main, 12 April I)

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. Let $f : R \to R$ be a continuously differentiable function such that $f (2) = 6$ and $f^{'} (2) = \frac{1}{48}$ . If $\int_{6}^{f (x)} 4 t^{3} d t = (x - 2) g (x)$ , then $lim_{x \to 2} g (x)$ is equal to (2019 Main, 12 April I)