Limit Continuity and Differentiability 7 Question 2
2. If $f: R \rightarrow R$ is a differentiable function and
$f(2)=6$, then $\lim _{x \rightarrow 2} \int _6^{f(x)} \frac{2 t d t}{(x-2)}$ is
(2019 Main, 9 April II)
(a) $12 f^{\prime}(2)$
(b) 0
(c) $24 f^{\prime}(2)$
(d) $2 f^{\prime}(2)$
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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 $\cos x>0, x \in 0, \frac{\pi}{2}$ ]
$$ \begin{aligned} & =\tan ^{-1} \frac{\tan x-\tan \frac{\pi}{4}}{1+\tan \frac{\pi}{4}(\tan x)} \\ & =\tan ^{-1} \tan x-\frac{\pi}{4} \\ & \because \frac{\tan A-\tan B}{1+\tan A \tan B}=\tan (A-B) \end{aligned} $$
Since, it is given that $x \in 0, \frac{\pi}{2}$, so
$$ x-\frac{\pi}{4} \in-\frac{\pi}{4}, \frac{\pi}{4} $$
Also, for $x-\frac{\pi}{4} \in-\frac{\pi}{4}, \frac{\pi}{4}$,
Then,
$$ \begin{aligned} f(x)=\tan ^{-1} \tan x & -\frac{\pi}{4} \quad=x-\frac{\pi}{4} \\ & \because \tan ^{-1} \tan \theta=\theta, \text { for } \theta \in-\frac{\pi}{2}, \frac{\pi}{2} \end{aligned} $$
Now, derivative of $f(x)$ w.r.t. $\frac{x}{2}$ is
$$ \begin{aligned} & \frac{d(f(x))}{d(x / 2)}=2 \frac{d f(x)}{d(x)} \\ & \quad=2 \times \frac{d}{d x} \quad x-\frac{\pi}{4}=2 \end{aligned} $$
