SATHEE: Limit Continuity and Differentiability 7 Question 2

Limit Continuity and Differentiability 7 Question 2

2. If $f : R \to R$ is a differentiable function and

$f (2) = 6$ , then $lim_{x \to 2} \int_{6}^{f (x)} \frac{2 t d t}{(x - 2)}$ is

(2019 Main, 9 April II)

(a) $12 f^{'} (2)$

(b) 0

(d) $2 f^{'} (2)$

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Answer:

Correct Answer: 2. (b)

Solution:

Let $f (x) = \tan^{- 1} \frac{\sin x - \cos x}{\sin x + \cos x} = \tan^{- 1} \frac{\tan x - 1}{\tan x + 1}$

[dividing numerator and denominator by $\cos x > 0, x \in 0, \frac{π}{2}$ ]

$\begin{aligned} = \tan^{- 1} \frac{\tan x - \tan \frac{π}{4}}{1 + \tan \frac{π}{4} (\tan x)} \\ = \tan^{- 1} \tan x - \frac{π}{4} \\ ∵ \frac{\tan A - \tan B}{1 + \tan A \tan B} = \tan (A - B) \end{aligned}$

Since, it is given that $x \in 0, \frac{π}{2}$ , so

$x - \frac{π}{4} \in - \frac{π}{4}, \frac{π}{4}$

Also, for $x - \frac{π}{4} \in - \frac{π}{4}, \frac{π}{4}$ ,

Then,

$\begin{aligned} f (x) = \tan^{- 1} \tan x & - \frac{π}{4} = x - \frac{π}{4} \\ ∵ \tan^{- 1} \tan θ = θ, for θ \in - \frac{π}{2}, \frac{π}{2} \end{aligned}$

Now, derivative of $f (x)$ w.r.t. $\frac{x}{2}$ is

$\begin{aligned} \frac{d (f (x))}{d (x / 2)} = 2 \frac{d f (x)}{d (x)} \\ = 2 \times \frac{d}{d x} x - \frac{π}{4} = 2 \end{aligned}$

Limit Continuity and Differentiability 7 Question 2

2. If $f : R \to R$ is a differentiable function and

Answer:

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Limit Continuity and Differentiability 7 Question 2

2. If f:R→R is a differentiable function and

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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2. If $f : R \to R$ is a differentiable function and