SATHEE: Limit Continuity and Differentiability 7 Question 7

Limit Continuity and Differentiability 7 Question 7

7. Let $f : (- 1, 1) \to R$ be a function defined by $f (x) = max - x ∣, - \sqrt{1 - x^{2}}$ . If $K$ be the set of all points at which $f$ is not differentiable, then $K$ has exactly

(2019 Main, 10 Jan II)

(a) three elements

(b) five elements

(d) one element

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Answer:

Correct Answer: 7. (b)

Solution:

We have, $x \log_{e} (\log_{e} x) - x^{2} + y^{2} = 4$ , which can be written as

$y^{2} = 4 + x^{2} - x \log_{e} (\log_{e} x)$

Now, differentiating Eq. (i) w.r.t. $x$ , we get

$2 y \frac{d y}{d x} = 2 x - x \frac{1}{\log_{e} x} \cdot \frac{1}{x} - 1 \cdot \log_{e} (\log_{e} x)$

[by using product rule of derivative]

$\Rightarrow \frac{d y}{d x} = \frac{2 x - \frac{1}{\log_{e} x} - \log_{e} (\log_{e} x)}{2 y}$

Now, at $x = e, y^{2} = 4 + e^{2} - e \log_{e} (\log_{e} e)$

$\begin{aligned} = 4 + e^{2} - e \log_{e} (1) = 4 + e^{2} - 0 \\ = e^{2} + 4 \\ \Rightarrow y = \sqrt{e^{2} + 4} \\ \frac{d y}{d x} = \frac{2 e - 1 - 0}{2 \sqrt{e^{2} + 4}} = \frac{2 e - 1}{2 \sqrt{e^{2} + 4}} [using Eq. (ii)] \end{aligned}$

Limit Continuity and Differentiability 7 Question 7

7. Let $f : (- 1, 1) \to R$ be a function defined by $f (x) = max - x ∣, - \sqrt{1 - x^{2}}$ . If $K$ be the set of all points at which $f$ is not differentiable, then $K$ has exactly

Answer:

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Limit Continuity and Differentiability 7 Question 7

7. Let f:(−1,1)→R be a function defined by f(x)=max−x∣,−1−x2. If K be the set of all points at which f is not differentiable, then K has exactly

Answer:

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Medical Preparation

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NCERT Exercise Video Solution

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7. Let $f : (- 1, 1) \to R$ be a function defined by $f (x) = max - x ∣, - \sqrt{1 - x^{2}}$ . If $K$ be the set of all points at which $f$ is not differentiable, then $K$ has exactly