A function f defined as follows: f x = 1 + 2 x , x ≤ 1 3 - x , x > 1 . Draw the graph of the function. Test the continuity at x = 1 .

A function is defined by f x = 1 + 2 x , x ≤ 1 3 - x , x > 1 . Now, when x ≤ 1 , the table values are: x 0 1 - 1 f x 1 3 - 1 Now, when x > 1 , the table values are: x 2 3 4 f x 1 0 - 1 Now, the graph of the given function is drawn below, Since the graph is broken. From the graph, we can say that, the function is discontinuous at x = 1

A function f defined as follows: f x = 2 x + 3 , x ≤ 0 3 - 2 x , x > 0 . Draw the graph of the function. Test the continuity at x = 0 .

A function is defined by f x = 2 x + 3 , x ≤ 0 3 - 2 x , x > 0 . Now, when x ≤ 0 , the table values are: x 0 - 1 - 2 f x 3 1 - 1 Now, when x > 0 , the table values are: x 1 2 3 f x 1 - 1 - 3 Now, the graph of the given function is drawn below, Since the graph is smooth i;e., neither broken nor have a hole. From the graph, we can say that, the function is continuous at x = 0

A function ϕ defined as follows: ϕ x = 2 x + 6 , - 3 ≤ x ≤ 0 6 , 0 < x < 2 2 x - 6 , 2 ≤ x ≤ 5 . Draw the graph of ϕ and find the point(s) of discontinuity of ϕ , if any.

A function is defined by ϕ x = 2 x + 6 , - 3 ≤ x ≤ 0 6 , 0 < x < 2 2 x - 6 , 2 ≤ x ≤ 5 . Case(i): When - 3 ≤ x ≤ 0 , then we have: ϕ ( x ) = 2 x + 6 , therefore, we have the table: x 0 - 1 - 2 ϕ ( x ) 6 4 2 Case (ii): When 0 < x < 2 , then we have ϕ ( x ) = 6 , so clearly, this a line which is parallel to x - axis. Case(iii): When 2 ≤ x ≤ 5 , then we have: ϕ ( x ) = 2 x - 6 , therefore, we have the table: x 2 3 4 ϕ ( x ) - 2 0 2 So, clearly the function is discontinuous at x = 2 .

If ϕ x = | x + 1 | - x , - 2 ≤ x ≤ 2 3 x - 5 , 2 < x < 4 , then draw the graph of ϕ .

A function is defined by ϕ x = | x + 1 | - x , - 2 ≤ x ≤ 2 3 x - 5 , 2 < x < 4 And it can be defined as ϕ x = - x - 1 - x , - 2 ≤ x < - 1 x + 1 - x - 1 ≤ x ≤ 2 3 x - 5 , 2 < x < 4 ⇒ ϕ x = - 2 x - 1 , - 2 ≤ x < - 1 1 - 1 ≤ x ≤ 2 3 x - 5 , 2 < x < 4 . Case (i): When - 2 ≤ x < - 1 , we have: ϕ x = - 2 x - 1 x - 2 - 1 - 1 . 5 ϕ x 3 1 2 Case (ii) When - 1 ≤ x ≤ 2 , we have: ϕ ( x ) = 1 So, clearly, for every value of x , the obtained line is parallel to the x - axis. Case(iii): When 2 < x < 4 , we have: ϕ ( x ) = 3 x - 5 x 2 . 5 3 3 . 5 ϕ x 2 . 5 4 5 . 5 Now, the graph of the given function is drawn below,

If ϕ x = | x + 1 | - x , - 2 ≤ x ≤ 2 3 x - 5 , 2 < x < 4 , find the point of discontinuity of ϕ , if any.

A function is defined by ϕ x = | x + 1 | - x , - 2 ≤ x ≤ 2 3 x - 5 , 2 < x < 4 And it can be defined as ϕ x = - x - 1 - x , - 2 ≤ x < - 1 x + 1 - x - 1 ≤ x ≤ 2 3 x - 5 , 2 < x < 4 ⇒ ϕ x = - 2 x - 1 , - 2 ≤ x < - 1 1 - 1 ≤ x ≤ 2 3 x - 5 , 2 < x < 4 . Case (i): When - 2 ≤ x ≤ - 1 , we have: ϕ x = - 2 x - 1 x - 2 - 1 - 1 . 5 ϕ x 3 1 2 Case (ii) When - 1 < x ≤ 2 , we have: ϕ ( x ) = 1 So, clearly, for every value of x , the obtained line is parallel to the x - axis. Case(iii): When 2 < x < 4 , we have: ϕ ( x ) = 3 x - 5 x 2 . 5 3 3 . 5 ϕ x 2 . 5 4 5 . 5 Now, the graph of the given function is drawn below, So, clearly the function is continuous at x = - 1 , 2

Show that f x = lim n → ∞ x + x 2 n sin x 1 + x 4 n is not continuous at the point x = 1 .

Given function: f x = lim n → ∞ x + x 2 n sin x 1 + x 4 n We know that, a function is continuous at x = a if lim x → a - f x = lim x → a + f x = f a i.e., LHL = RHL = f a We have, f 1 = 1 + 1 2 n sin 1 1 + 1 4 n = 1 + sin 1 2 We know that, lim n → ∞ x 2 n = 0 if 0 < x < 1 ∞ if x > 1 and lim n → ∞ x 4 n = 0 if 0 < x < 1 ∞ if x > 1 For 0 < x < 1 , we have f x = x + 0 · sin x 1 + 0 = x And for x > 1 , we have f x = lim n → ∞ x + x 2 n sin x 1 + x 4 n = lim n → ∞ x + x 2 n sin x x 4 n 1 + x 4 n x 4 n = 0 Thus, we have, f x = x if 0 < x < 1 1 + sin 1 2 if x = 1 0 if x > 1 Now, LHL at x = 1 , lim x → 1 - f x = lim x → 1 x = 1 Also, RHL at x = 1 , lim x → 1 + f x = lim x → 1 0 = 0 So, clearly LHL ≠ RHL Hence, the given function is discontinuous at x = 1 .

Find the value of a for which the function f x = ( 1 - cos 4 x ) tan x x 3 , x < 0 2 a 2 , x = 0 x 16 + x - 4 , x > 0 is continuous at the point x = 0 .

Given function: f x = ( 1 - cos 4 x ) tan x x 3 , x < 0 2 a 2 , x = 0 x 16 + x - 4 , x > 0 We know that, a function is continuous at x = a if lim x → a - f x = lim x → a + f x = f a i.e., LHL = RHL = f a → 1 Now, LHL at x = 0 , lim x → 0 - f x = lim x → 0 1 - cos 4 x tan x x 3 = lim x → 0 2 sin 2 2 x tan x x 3 = 2 lim x → 0 sin 2 2 x tan x x 2 x = 2 lim x → 0 sin 2 x x 2 × lim x → 0 tan x x = 2 × 2 2 × 1 = 8 Also, RHL at x = 0 , lim x → 0 + f x = lim x → 0 x 16 + x - 4 = lim x → 0 x 16 + x - 4 × 16 + x + 4 16 + x + 4 [Using Rationalization] = lim x → 0 x ( 16 + x + 4 ) 16 + x - 16 = lim x → 0 ( 16 + x + 4 ) [ ∵ x → 0 , so that x ≠ 0 ] = 16 + 0 + 4 = 4 + 4 = 8 and f 2 = 2 a 2 Since, f x is continuous at x = 0 , therefore, we have 2 a 2 = 8 ⇒ a 2 = 4 [Using 1 ] ⇒ a = ± 2 Thus, the value of a = ± 2 .

E M B I B E

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Calculus - A. Continuity and Differentiability>Chapter 7 - Continuity>EXERCISE

Parthasarathi Mukhopadhyay and Manabendra Nath Mukherjee Solutions for Exercise 1: EXERCISE

Author:Parthasarathi Mukhopadhyay & Manabendra Nath Mukherjee

Parthasarathi Mukhopadhyay Mathematics Solutions for Exercise - Parthasarathi Mukhopadhyay and Manabendra Nath Mukherjee Solutions for Exercise 1: EXERCISE

Attempt the practice questions from Exercise 1: EXERCISE with hints and solutions to strengthen your understanding. RUDIMENTS of MATHEMATICS For Class 12 of +2 Level solutions are prepared by Experienced Embibe Experts.

Questions from Parthasarathi Mukhopadhyay and Manabendra Nath Mukherjee Solutions for Exercise 1: EXERCISE with Hints & Solutions

MEDIUM

12th West Bengal Board

IMPORTANT

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Chapter 7 - Continuity>EXERCISE>Q 3

A function $f$ defined as follows: $f (x) = \{\begin{array}{cl} 1 + 2 x & , x \leq 1 \\ 3 - x & , x > 1 \end{array}$ . Draw the graph of the function. Test the continuity at $x = 1$ .

MEDIUM

12th West Bengal Board

IMPORTANT

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Chapter 7 - Continuity>EXERCISE>Q 4

A function $f$ defined as follows: $f (x) = \{\begin{cases} 2 x + 3, & x \leq 0 \\ 3 - 2 x, & x > 0 \end{cases}$ . Draw the graph of the function. Test the continuity at $x = 0$ .

MEDIUM

12th West Bengal Board

IMPORTANT

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Chapter 7 - Continuity>EXERCISE>Q 5

A function $ϕ$ defined as follows: $ϕ (x) = \{\begin{matrix} 2 x + 6 & , & - 3 \leq x \leq 0 \\ 6 & , & 0 < x < 2 \\ 2 x - 6 & , & 2 \leq x \leq 5 \end{matrix}$ . Draw the graph of $ϕ$ and find the point(s) of discontinuity of $ϕ$ , if any.

MEDIUM

12th West Bengal Board

IMPORTANT

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Chapter 7 - Continuity>EXERCISE>Q 6

If $ϕ (x) = \{\begin{matrix} | x + 1 | - x, & - 2 \leq x \leq 2 \\ 3 x - 5, & 2 < x < 4 \end{matrix}$ , then draw the graph of $ϕ$ .

MEDIUM

12th West Bengal Board

IMPORTANT

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Chapter 7 - Continuity>EXERCISE>Q 6

If $ϕ (x) = \{\begin{matrix} | x + 1 | - x, & - 2 \leq x \leq 2 \\ 3 x - 5, & 2 < x < 4 \end{matrix}$ , find the point of discontinuity of $ϕ$ , if any.

MEDIUM

12th West Bengal Board

IMPORTANT

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Chapter 7 - Continuity>EXERCISE>Q 7

A function $f (x) = | x - 1 | + | x | + | x + 1 |$ is defined in the interval $(0, 2)$ . Discuss its continuity at $x = 1$ .

HARD

12th West Bengal Board

IMPORTANT

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Chapter 7 - Continuity>EXERCISE>Q 20

Show that $f (x) = \lim_{n \to \infty} \frac{x + x^{2 n} \sin x}{1 + x^{4 n}}$ is not continuous at the point $x = 1$ .

HARD

12th West Bengal Board

IMPORTANT

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Chapter 7 - Continuity>EXERCISE>Q 21

Find the value of $a$ for which the function $f (x) = \{\begin{cases} \frac{(1 - \cos 4 x) \tan x}{x^{3}} & , x < 0 \\ 2 a^{2} & , x = 0 \\ \frac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4} & , x > 0 \end{cases}$ is continuous at the point $x = 0$ .

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