SATHEE: Limit Continuity and Differentiability 4 Question 10

Limit Continuity and Differentiability 4 Question 10

10. A discontinuous function $y = f (x)$ satisfying $x^{2} + y^{2} = 4$ is given by $f (x) = \dots$ .

$(1982, 2 M)$

Analytical & Descriptive Questions

Determine $a$ and $b$ such that $f (x)$ is continuous at $x = 0$ .

(1994, 4M)

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Answer:

Correct Answer: 10. $a = \frac{2}{3}, b = e^{2 / 3} 12 . a = 8$

Solution:

$f (x) = [x] \sin \frac{π}{[x + 1]}$

We know that, $[x]$ is continuous on $R \sim I$ , where $I$ denotes the set of integers and $\sin \frac{π}{[x + 1]}$ is discontinuous for $[x + 1] = 0$ .

$\Rightarrow 0 \leq x + 1 < 1 \Rightarrow - 1 \leq x < 0$

Thus, the function is defined in the interval.

Clearly, RHL (at $x = 1) = 1 / 2$ and LHL $($ at $x = 1) = 1 / 2$ Also,

$f (x) = 1 / 2$

$∴ f (x)$ is continuous for all $x \in [0, 2]$ .

On differentiating Eq. (i), we get

Clearly, RHL (at $x = 1$ ) for $f^{'} (x) = 1$

and LHL (at $x = 1$ ) for $f^{'} (x) = 1$

Also,

$f (1) = 1$

Thus, $f^{'} (x)$ is continuous for all $x \in [0, 2]$ .

Again, differentiating Eq. (ii), we get

Clearly, RHL (at $x = 1) \neq LHL ($ at $x = 1$ )

Thus, $f^{''} (x)$ is not continuous at $x = 1$ .

or $f^{''} (x)$ is continuous for all $x \in [0, 2] - 1$ .

Limit Continuity and Differentiability 4 Question 10

10. A discontinuous function $y = f (x)$ satisfying $x^{2} + y^{2} = 4$ is given by $f (x) = \dots$ .

Analytical & Descriptive Questions

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

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Sample Mock Test

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NCERT Exercise Video Solution

Limit Continuity and Differentiability 4 Question 10

10. A discontinuous function y=f(x) satisfying x2+y2=4 is given by f(x)=….

Analytical & Descriptive Questions

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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10. A discontinuous function $y = f (x)$ satisfying $x^{2} + y^{2} = 4$ is given by $f (x) = \dots$ .