Discuss the continuity of the function f, where f is defined by: f(x)= ll 3, & if 0 x 1 4, & if 1<x<3 5, & if 3 x 10 .

The given function is f(x)= l 3, if 0 x 1 4, if 1<x<3 5, if 3 x 10 . The function f is defined at all points of interval [0, 10] Let k be the point in the interval [0, 10] Then, we have 5 cases i.e., 0 k < 1,k = 1, 1 < k < 3, k = 3 or 3 < k 10. Now, Case I: 0 k<1 Then, f(k) = 3 _ x k f(x) = _ x k (3) = 3 = f(k) Thus, _ x k f(x) = f(k) Hence, f is continuous in the interval [0, 10). Case II: k = 1 f(1) = 3 _ x 1^ - f(x) = _ x 1^ - (3) = 3 _ x 1^ + f(x) = _ x 1^ + (4) = 4 _ x 1 f ( x ) _ x, + f ( x ) Hence, f is not continuous at x = 1 Case III: 1 < k < 3 Then, f(k) = 4 _ x k f(x) = _ x k (4) = 4 = f(k) Thus, _ x k f ( x ) = f ( k ) Hence, f is continuous in (1, 3). Case IV: k = 3 _ x 3^ - f(x) = _ x 3^ - (4) = 4 _ x 3^ + f(x) = _ x 3^ + (5) = 5 _ x k ^ - f ( x ) _ x k ^ + f ( x ) Hence, f is not continuous at x = 3. Case V: 3 < k 10 Then, f(x) = 5 _ x k f(x) = _ x k (5) = 5 = f(k) Thus, _ x k f ( x ) = f ( k ) Hence, f is continuous at all points of the interval (3, 10]. Therefore, x = 1 and 3 are the points of discontinuity of f.

Discuss the continuity of the function f, where f is defined by: …

Download App

Question

Discuss the continuity of the function f, where f is defined by: $f(x)=\left\{\begin{array}{ll} {3,} & {\text { if } 0 \leq x \leq 1} \\ {4,} & {\text { if } 1<x<3} \\ {5,} & {\text { if } 3 \leq x \leq 10} \end{array}\right.$

✓

Answer

The given function is $f(x)=\left\{\begin{array}{l} {3, \text { if } 0 \leq x \leq 1} \\ {4, \text { if } 1<x<3} \\ {5, \text { if } 3 \leq x \leq 10} \end{array}\right.$
The function f is defined at all points of interval [0, 10]
Let k be the point in the interval [0, 10]
Then, we have 5 cases i.e., 0 $\le$ k < 1,k = 1, 1 < k < 3, k = 3 or 3 < k $\le$ 10.
Now, Case I: $0 \leq k<1$
Then, f(k) = 3
$\mathop {\lim }\limits_{x \to {{k}}} f(x) = \mathop {\lim }\limits_{x \to k} (3) = 3 = f(k)$
Thus, $\mathop {\lim }\limits_{x \to k} f(x) = f(k)$
Hence, f is continuous in the interval [0, 10).
Case II: k = 1
f(1) = 3
$\mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{x \to {1^ - }} (3) = 3$
$\mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} (4) = 4$
$ \Rightarrow \mathop {\lim }\limits_{x \to 1} {\text{f}}({\text{x}}) \ne \mathop {\lim }\limits_{x, + } {\text{f}}({\text{x}})$
Hence, f is not continuous at x = 1
Case III: 1 < k < 3
Then, f(k) = 4
$\mathop {\lim }\limits_{x \to {{k}}} f(x) = \mathop {\lim }\limits_{x \to k} (4) = 4 = f(k)$
Thus, $\mathop {\lim }\limits_{{{x}} \to {{k}}} {\text{f}}({\text{x}}) = {\text{f}}({\text{k}})$
Hence, f is continuous in (1, 3).
Case IV: k = 3
$\mathop {\lim }\limits_{x \to {3^ - }} f(x) = \mathop {\lim }\limits_{x \to {3^ - }} (4) = 4$
$\mathop {\lim }\limits_{x \to {3^ + }} f(x) = \mathop {\lim }\limits_{x \to {3^ + }} (5) = 5$
$\mathop {\lim }\limits_{x \to {{{k}}^ - }} {\text{f}}({\text{x}}) \ne \mathop {\lim }\limits_{{\text{x}} \to {{\text{k}}^ + }} {\text{f}}({\text{x}})$
Hence, f is not continuous at x = 3.
Case V: 3 < k $\le$ 10
Then, f(x) = 5
$\mathop {\lim }\limits_{x \to {{k}}} f(x) = \mathop {\lim }\limits_{x \to k} (5) = 5 = f(k)$
Thus, $\mathop {\lim }\limits_{{{x}} \to {{k}}} {{f}}({{x}}) = {{f}}({{k}})$
Hence, f is continuous at all points of the interval (3, 10].
Therefore, x = 1 and 3 are the points of discontinuity of f.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "1 Marks" from Continuity and Differentiability→Continuity and Differentiability — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Continuity and Differentiability — Maths STD 12 Science — Question

Answer

Need a full question paper?

Explore more

Similar questions