SATHEE: Limit Continuity and Differentiability 4 Question 1

Limit Continuity and Differentiability 4 Question 1

1. If the function $f$ defined on $\frac{π}{6}, \frac{π}{3}$ by

$\begin{aligned} f (x) = \frac{\sqrt{2} \cos x - 1}{\cot x - 1}, x \neq \frac{π}{4} is continuous, \\ k, x = \frac{π}{4} \end{aligned}$

then $k$ is equal to

(2019 Main, 9 April I)

(a) $\frac{1}{2}$

(b) 2

(d) $\frac{1}{\sqrt{2}}$

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Answer:

Correct Answer: 1. (a)

Solution:

Given $\int_{6}^{f (x)} 4 t^{3} d t = (x - 2) g (x)$

$\Rightarrow g (x) = \frac{\int_{6}^{f (x)} 4 t^{3} d t}{(x - 2)}$

So, $lim_{x \to 2} g (x) = lim_{x \to 2} \frac{\int_{6}^{f (x)} 4 t^{3} d t}{x - 2}$

$∵ \frac{0}{0} form as x \to 2 \Rightarrow f (2) = 6$

$\begin{aligned} lim_{x \to 2} g (x) = lim_{x \to 2} \frac{4 (f (x))^{3} f^{'} (x)}{1} \\ ∵ \frac{d}{d x} \int_{φ_{1} (x)}^{φ_{2} (x)} f (t) d t = f (φ_{2} (x)), φ_{2}^{'} (x) - f (φ_{1} (x)) \cdot φ_{1}^{'} (x) \end{aligned}$

On applying limit, we get

$\begin{aligned} lim_{x \to 2} g (x) = & 4 (f (2))^{3} f^{'} (2) = 4 \times (6)^{3} \frac{1}{48}, \\ ∵ f (2) = 6 and f^{'} (2) = \frac{1}{48} \\ = & \frac{4 \times 216}{48} = 18 \end{aligned}$

Limit Continuity and Differentiability 4 Question 1

1. If the function $f$ defined on $\frac{π}{6}, \frac{π}{3}$ by

Answer:

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Limit Continuity and Differentiability 4 Question 1

1. If the function f defined on π6,π3 by

Answer:

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1. If the function $f$ defined on $\frac{π}{6}, \frac{π}{3}$ by