Limit Continuity and Differentiability 4 Question 1
1. If the function $f$ defined on $\frac{\pi}{6}, \frac{\pi}{3}$ by
$$ \begin{aligned} & f(x)=\frac{\sqrt{2} \cos x-1}{\cot x-1}, \quad x \neq \frac{\pi}{4} \text { is continuous, } \\ & k, \quad x=\frac{\pi}{4} \end{aligned} $$
then $k$ is equal to
(2019 Main, 9 April I)
(a) $\frac{1}{2}$
(b) 2
(c) 1
(d) $\frac{1}{\sqrt{2}}$
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Answer:
Correct Answer: 1. (a)
Solution:
- Given $\int _6^{f(x)} 4 t^{3} d t=(x-2) g(x)$
$$ \Rightarrow \quad g(x)=\frac{\int _6^{f(x)} 4 t^{3} d t}{(x-2)} $$
So, $\lim _{x \rightarrow 2} g(x)=\lim _{x \rightarrow 2} \frac{\int _6^{f(x)} 4 t^{3} d t}{x-2}$
$$ \because \frac{0}{0} \text { form as } x \rightarrow 2 \Rightarrow f(2)=6 $$
$$ \begin{aligned} & \lim _{x \rightarrow 2} g(x)=\lim _{x \rightarrow 2} \frac{4(f(x))^{3} f^{\prime}(x)}{1} \\ & \quad \because \frac{d}{d x} \int _{\varphi _1(x)}^{\varphi _2(x)} f(t) d t=f\left(\varphi _2(x)\right), \varphi _2^{\prime}(x)-f\left(\varphi _1(x)\right) \cdot \varphi _1^{\prime}(x) \end{aligned} $$
On applying limit, we get
$$ \begin{aligned} \lim _{x \rightarrow 2} g(x)= & 4(f(2))^{3} f^{\prime}(2)=4 \times(6)^{3} \frac{1}{48}, \\ & \because f(2)=6 \text { and } f^{\prime}(2)=\frac{1}{48} \\ = & \frac{4 \times 216}{48}=18 \end{aligned} $$
