NCERT Solutions Class 12 maths Chapter-5 (Continuity And Differentiability)Exercise 5.1

Jolly September 06, 2022

NCERT Solutions Class 12 maths Chapter-5 (Continuity And Differentiability)Exercise 5.1

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-5 (Continuity And Differentiability)Exercise 5.1 This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.
Solutions Class 12 maths Chapter-5 (Continuity And Differentiability)Exercise 5.1
NCERT Question-Answer

Class 12 Mathematics

Chapter-5 (Continuity And Differentiability)

Questions and answers given in practice

Chapter-5 (Continuity And Differentiability)

Exercise 5.1

Set 1

Question 1. Prove that the function f(x) = 5x – 3 is continuous at x = 0, at x = – 3 and at x = 5.

Solution:

To prove the continuity of the function f(x) = 5x – 3, first we have to calculate limits and function value at that point.

Continuity at x = 0

Left limit = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (5x-3)

= (5(0) – 3) = -3

Right limit = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (5x-3)

= (5(0) – 3)= -3

Function value at x = 0, f(0) = 5(0) – 3 = -3

As, \lim_{x \to 0^-} f(x)=\lim_{x \to 0^+} f(x) = f(0) = -3, 

Hence, the function is continuous at x = 0.

Continuity at x = -3

Left limit = \lim_{x \to -3^-} f(x) = \lim_{x \to -3^-} (5x-3)

= (5(-3) – 3) = -18

Right limit = \lim_{x \to -3^+} f(x) = \lim_{x \to -3^+} (5x-3)

= (5(-3) – 3) = -18

Function value at x = -3, f(-3) = 5(-3) – 3 = -18

As, \lim_{x \to -3^-} f(x)=\lim_{x \to -3^+} f(x) = f(-3) = -18

Hence, the function is continuous at x = -3.

Continuity at x = 5

Left limit = \lim_{x \to 5^-} f(x) = \lim_{x \to 5^-} (5x-3)

= (5(5) – 3) = 22

Right limit = \lim_{x \to 5^+} f(x) = \lim_{x \to 5^+} (5x-3)

= (5(5) – 3) = 22

Function value at x = 5, f(5) = 5(5) – 3 = 22

As, \lim_{x \to 5^-} f(x)=\lim_{x \to 5^+} f(x) = f(5) = 22

Hence, the function is continuous at x = 5.

Question 2. Examine the continuity of the function f(x) = 2x2 – 1 at x = 3.

Solution:

To prove the continuity of the function f(x) = 2x2 – 1, first we have to calculate limits and function value at that point.

Continuity at x = 3

Left limit = \lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (2x^2-1)

= (2(3)2 – 1) = 17

Right limit = \lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (2x^2-1)

= (2(3)2 – 1) = 17

Function value at x = 3, f(3) = 2(3)2 – 1 = 17

As, \lim_{x \to 3^-} f(x)=\lim_{x \to 3^+} f(x) = f(3) = 17,

Hence, the function is continuous at x = 3.

Question 3. Examine the following functions for continuity.

(a) f(x) = x – 5

Solution:

To prove the continuity of the function f(x) = x – 5, first we have to calculate limits and function value at that point.

Let’s take a real number, c

Continuity at x = c

Left limit = \lim_{x \to c^-} f(x) = \lim_{x \to c^-} (x-5)

= (c – 5) = c – 5

Right limit = \lim_{x \to c^+} f(x) = \lim_{x \to c^+} (x-5)

= (c – 5) = c – 5

Function value at x = c, f(c) = c – 5

As, \lim_{x \to c^-} f(x)=\lim_{x \to c^+} f(x) = f(c) = c-5,for any real number c

Hence, the function is continuous at every real number.

(b) Solutions Class 12 maths Chapter-5 (Continuity And Differentiability), x ≠ 5

Solution:

To prove the continuity of the function f(x) = \frac{1}{x-5}, first we have to calculate limits and function value at that point.

Let’s take a real number, c

Continuity at x = c and c ≠ 5

Left limit = \lim_{x \to c^-} f(x) = \lim_{x \to c^-} \frac{1}{(x-5)}\\= \frac{1}{(c-5)}

Right limit = \lim_{x \to c^+} f(x) = \lim_{x \to c^+} \frac{1}{(x-5)}\\= \frac{1}{(c-5)}

Function value at x = c, f(c) = \frac{1}{c-5}

As, \lim_{x \to c^-} f(x)=\lim_{x \to c^+} f(x) = f(c) = \frac{1}{c-5},for any real number c

Hence, the function is continuous at every real number.

(c)Solutions Class 12 maths Chapter-5 (Continuity And Differentiability), x ≠ -5

Solution:

To prove the continuity of the function f(x) = \frac{x^2-25}{x+5}, first we have to calculate limits and function value at that point.

Let’s take a real number, c

Continuity at x = c and c ≠ -5

Left limit = \lim_{x \to c^-} f(x) = \lim_{x \to c^-} \frac{x^2-25}{(x+5)}\\= \lim_{x \to c^-} \frac{(x-5)(x+5)}{(x+5)}\\= \lim_{x \to c^-}x-5

= c – 5

Right limit = \lim_{x \to c^+} f(x) = \lim_{x \to c^+} \frac{x^2-25}{(x+5)}\\= \lim_{x \to c^+} \frac{(x-5)(x+5)}{(x+5)}\\= \lim_{x \to c^+}x-5

= c – 5

Function value at x = c, f(c) = \frac{c^2-25}{(c+5)}\\= \frac{(c-5)(c+5)}{(c+5)}

= c – 5

As, \lim_{x \to c^-} f(x)=\lim_{x \to c^+} f(x) = f(c) = c - 5, for any real number c

Hence, the function is continuous at every real number.

(d) f(x) = |x – 5|

Solution:

To prove the continuity of the function f(x) = |x – 5|, first we have to calculate limits and function value at that point.

Here,

As, we know that modulus function works differently.

In |x – 5|, |x – 5| = x – 5 when x>5 and |x – 5| = -(x – 5) when x < 5

Let’s take a real number, c and check for three cases of c:

Continuity at x = c

When c < 5

Left limit = \lim_{x \to c^-} f(x) = \lim_{x \to c^-} |x-5|\\= \lim_{x \to c^-} -(x-5)

= -(c – 5) 

= 5 – c

Right limit = \lim_{x \to c^+} f(x) = \lim_{x \to c^+} |x-5|\\= \lim_{x \to c^+} -(x-5)

= -(c – 5)

= 5 – c

Function value at x = c, f(c) = |c – 5| = 5 – c

As, \lim_{x \to c^-} f(x)=\lim_{x \to c^+} f(x) = f(c) = 5-c,

Hence, the function is continuous at every real number c, where c<5.

When c > 5

Left limit = \lim_{x \to c^-} f(x) = \lim_{x \to c^-} |x-5|\\= \lim_{x \to c^-} (x-5)

= (c – 5)

Right limit = \lim_{x \to c^+} f(x) = \lim_{x \to c^+} |x-5|\\= \lim_{x \to c^+} (x-5)

= (c – 5)

Function value at x = c, f(c) = |c – 5| = c – 5

As, \lim_{x \to c^-} f(x)=\lim_{x \to c^+} f(x) = f(c) = c-5,

Hence, the function is continuous at every real number c, where c > 5.

When c = 5

Left limit = \lim_{x \to 5^-} f(x) = \lim_{x \to 5^-} |x-5|\\= \lim_{x \to 5^-} |5-5|\\= 0

Right limit = \lim_{x \to 5^+} f(x) = \lim_{x \to 5^+} |x-5|\\= \lim_{x \to 5^+} |5-5|\\= 0

Function value at x = c, f(c) = |5 – 5| = 0

As, \lim_{x \to 5^-} f(x)=\lim_{x \to 5^+} f(x) = f(5) = 0,

Hence, the function is continuous at every real number c, where c = 5.

Hence, we can conclude that, the modulus function is continuous at every real number.

Question 4. Prove that the function f(x) = xn is continuous at x = n, where n is a positive integer.

Solution:

To prove the continuity of the function f(x) = xn, first we have to calculate limits and function value at that point.

Continuity at x = n

Left limit = \lim_{x \to n^-} f(x) = \lim_{x \to n^-} (x^n)

= nn

Right limit = \lim_{x \to n^+} f(x) = \lim_{x \to n^+} (x^n)

= nn

Function value at x = n, f(n) = nn

As, \lim_{x \to n^-} f(x)=\lim_{x \to n^+} f(x) = f(n) = n^n,

Hence, the function is continuous at x = n.

Question 5. Is the function f defined by

Solutions Class 12 maths Chapter-5 (Continuity And Differentiability)

continuous at x = 0? At x = 1? At x = 2?

Solution:

To prove the continuity of the function f(x), first we have to calculate limits and function value at that point.

Continuity at x = 0

Left limit = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (x)\\= 0

Right limit = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x)\\= 0

Function value at x = 0, f(0) = 0

As, \lim_{x \to 0^-} f(x)=\lim_{x \to 0^+} f(x) = f(0) = 0,

Hence, the function is continuous at x = 0.

Continuity at x = 1

Left limit = \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x)\\= 1

Right limit = \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (5)\\= 5

Function value at x = 1, f(1) = 1

As, \lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x),

Hence, the function is not continuous at x = 1.

Continuity at x = 2

Left limit = \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (5)\\= 5

Right limit = \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (5)\\= 5

Function value at x = 2, f(2) = 5

As, \lim_{x \to 2^-} f(x)=\lim_{x \to 2^+} f(x) = f(2) = 5,

Therefore, the function is continuous at x = 2.

Question 6. Solutions Class 12 maths Chapter-5 (Continuity And Differentiability)

Solution:

Here, as it is given that

For x ≤ 2, f(x) = 2x + 3, which is a polynomial

As polynomial are continuous, hence f(x) is continuous x ∈ (-∞, 2)

Now, For x > 2, f(x) = 2x – 3, which is a polynomial

As polynomial are continuous, hence f(x) is continuous x ∈ (2, ∞)

So now, as f(x) is continuous in x ∈ (-∞, 2) U (2, ∞) = R – {2}

Let’s check the continuity at x = 2,

Left limit = \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (2x+3)

= (2(2) + 3)

= 7

Right limit = \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (2x-3)

= (2(2) – 3)

= 1

Function value at x = 2, f(2) = 2(3) + 3 = 7

As, \lim_{x \to 2^-} f(x) \neq \lim_{x \to 2^+} f(x)

Therefore, the function is discontinuous at only x = 2.

Question 7. Solutions Class 12 maths Chapter-5 (Continuity And Differentiability)

Solution:

Here, as it is given that

For x ≤ -3, f(x) = |x| + 3, 

As, we know that modulus function works differently.

In |x|, |x – 0| = x when x > 0 and |x – 0| = -x when x < 0

f(x) = -x + 3, which is a polynomial

As polynomial are continuous, hence f(x) is continuous x ∈ (-∞, -3)

For -3 < x < 3, f(x) = -2x, which is a polynomial

As polynomial are continuous, hence f(x) is continuous x ∈ (-3, 3)

Now, for x ≥ 3, f(x) = 6x + 2, which is a polynomial

As polynomial are continuous, hence f(x) is continuous x ∈ (3, ∞)

So now, as f(x) is continuous in x ∈ (-∞, -3) U(-3, 3) U (3, ∞) = R – {-3, 3}

Let’s check the continuity at x = -3,

Left limit = \lim_{x \to -3^-} f(x) = \lim_{x \to -3^-} (|x|+3)\\= \lim_{x \to -3^-} (-x+3)

= (-(-3) + 3)

= 6

Right limit = \lim_{x \to -3^+} f(x) = \lim_{x \to -3^+} (-2x)

= (-2(-3))

= 6

Function value at x = -3, f(-3) = |-3| + 3 = 3 + 3 = 6

As, \lim_{x \to -3^-} f(x)=\lim_{x \to -3^+} f(x) = f(-3) = 6,

Hence, the function is continuous at x = -3.

Now, let’s check the continuity at x = 3,

Left limit = \lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (-2x)

= (-2(3))

= -6

Right limit = \lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (6x+2)

= (6(3) + 2)

= 20

Function value at x = 3, f(3) = 6(3) + 2 = 20

As, \lim_{x \to 3^-} f(x) \neq \lim_{x \to 3^+} f(x),

Therefore, the function is discontinuous only at x = 3.

Question 8. Solutions Class 12 maths Chapter-5 (Continuity And Differentiability)

Solution:

As, we know that modulus function works differently.

In |x|, |x – 0| = x when x > 0 and |x – 0|= -x when x < 0

When x < 0, f(x) = \frac{-x}{x}= -1, which is a constant

As constant functions are continuous, hence f(x) is continuous x ∈ (-∞, 0).

When x > 0, f(x) = \frac{x}{x}= 1, which is a constant

As constant functions are continuous, hence f(x) is continuous x ∈ (0, ∞).

So now, as f(x) is continuous in x ∈ (-∞, 0) U(0, ∞) = R – {0}

Let’s check the continuity at x = 0,

Left limit = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{|x|}{x}\\= \lim_{x \to 0^-} \frac{-x}{x}\\= -1

Right limit = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{|x|}{x}\\= \lim_{x \to 0^+} \frac{x}{x}\\= 1

Function value at x = 0, f(0) = 0

As, \lim_{x \to 0^-} f(x) \neq \lim_{x \to 0^+} f(x)

Hence, the function is discontinuous at only x = 0.

Question 9. Solutions Class 12 maths Chapter-5 (Continuity And Differentiability)

Solution:

As, we know that modulus function works differently.

In |x|, |x – 0| = x when x > 0 and |x – 0| = -x when x < 0

When x < 0, f(x) = \frac{x}{-x}= -1, which is a constant

As constant functions are continuous, hence f(x) is continuous x ∈ (-∞, 0).

When x > 0, f(x) = -1, which is a constant

As constant functions are continuous, hence f(x) is continuous x ∈ (0, ∞).

So now, as f(x) is continuous in x ∈ (-∞, 0) U(0, ∞) = R – {0}

Let’s check the continuity at x = 0,

Left limit = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{x}{|x|}\\= \lim_{x \to 0^-} \frac{x}{-x}\\= -1

Right limit = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (-1)\\= -1

Function value at x = 0, f(0) = -1

As, \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = -1

Hence, the function is continuous at x = 0.

So, we conclude that the f(x) is continuous at any real number. Hence, no point of discontinuity.

Question 10. Solutions Class 12 maths Chapter-5 (Continuity And Differentiability)

Solution:

Here,

When x ≥1, f(x) = x + 1, which is a polynomial

As polynomial functions are continuous, hence f(x) is continuous x ∈ (1, ∞)

When x < 1, f(x) = x2 + 1, which is a polynomial

As polynomial functions are continuous, hence f(x) is continuous x ∈ (-∞, 1)

So now, as f(x) is continuous in x ∈ (-∞, 1) U (1, ∞) = R – {1}

Let’s check the continuity at x = 1,

Left limit = \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^2+1)

= 1 + 1

= 2

Right limit = \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x+1)

= 1 + 1

= 2

Function value at x = 1, f(1) = 1 + 1 = 2

As, \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) = 2

Hence, the function is continuous at x = 1.

So, we conclude that the f(x) is continuous at any real number.

Question 11. Solutions Class 12 maths Chapter-5 (Continuity And Differentiability)

Solution:

Here,

When x ≤ 2, f(x) = x3 + 3, which is a polynomial

As polynomial functions are continuous, hence f(x) is continuous x ∈ (-∞, 2)

When x > 2, f(x) = x2 + 1, which is a polynomial

As polynomial functions are continuous, hence f(x) is continuous x ∈ (2, ∞)

So now, as f(x) is continuous in x ∈ (-∞, 2) U(2, ∞) = R – {2}

Let’s check the continuity at x = 2,

Left limit = \lim_{x \to 2^-} f(x)= \lim_{x \to 2^-} (x^3-3)\\= 8-3\\=5

Right limit = \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (x^2+1)\\= 4+1\\=5

Function value at x = 2, f(2) = 8 – 3 = 5

As, \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2) = 5

Hence, the function is continuous at x = 2.

So, we conclude that the f(x) is continuous at any real number.

Question 12. Solutions Class 12 maths Chapter-5 (Continuity And Differentiability)

Solution:

Here,

When x ≤ 1, f(x) = x10 – 1, which is a polynomial

As polynomial functions are continuous, hence f(x) is continuous x ∈ (-∞, 1)

When x >1, f(x) = x2, which is a polynomial

As polynomial functions are continuous, hence f(x) is continuous x ∈ (1, ∞)

So now, as f(x) is continuous in x ∈ (-∞, 1) U (1, ∞) = R – {1}

Let’s check the continuity at x = 1,

Left limit = \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^{10}-1)

= 1 – 1

= 0

Right limit = \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x^2)\\= 1

Function value at x = 1, f(1) = 1 – 1 = 0

As, \lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x)

Hence, the function is discontinuous at x = 1.

Question 13. Is the function defined by

Solutions Class 12 maths Chapter-5 (Continuity And Differentiability)

a continuous function?

Solution:

Here, as it is given that

For x ≤ 1, f(x) = x + 5, which is a polynomial

As polynomial are continuous, hence f(x) is continuous x ∈ (-∞, 1)

Now, For x > 1, f(x) = x – 5, which is a polynomial

As polynomial are continuous, hence f(x) is continuous x ∈ (1, ∞)

So now, as f(x) is continuous in x ∈ (-∞, 1) U (1, ∞) = R – {1}

Let’s check the continuity at x = 1,

Left limit = \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x+5)

= (1 + 5)

= 6

Right limit = \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x-5)

= (1 – 5)

= -4

Function value at x = 1, f(1) = 5 + 1 = 6

As, \lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x)

Hence, the function is continuous for only R – {1}.

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

Question 14. Solutions Class 12 maths Chapter-5 (Continuity And Differentiability)

Solution:

Here, as it is given that

For 0 ≤ x ≤ 1, f(x) = 3, which is a constant

As constants are continuous, hence f(x) is continuous x ∈ (0, 1)

Now, For 1 < x < 3, f(x) = 4, which is a constant 

As constants are continuous, hence f(x) is continuous x ∈ (1, 3)

For 3 ≤ x ≤ 10, f(x) = 5, which is a constant

As constants are continuous, hence f(x) is continuous x ∈ (3, 10)

So now, as f(x) is continuous in x ∈ (0, 1) U (1, 3) U (3, 10) = (0, 10) – {1, 3}

Let’s check the continuity at x = 1,

Left limit = \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} 3\\= 3

Right limit = \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} 4\\= 4

Function value at x = 1, f(1) = 3

As, \lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x)

Hence, the function is discontinuous at x = 1.

Now, let’s check the continuity at x = 3,

Left limit = \lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} 4\\= 4

Right limit = \lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} 5\\= 5

Function value at x = 3, f(3) = 4

As, \lim_{x \to 3^-} f(x) \neq \lim_{x \to 3^+} f(x)

Hence, the function is discontinuous at x = 3.

So concluding the results, we get

Therefore, the function f(x) is discontinuous at x = 1 and x = 3.

Question 15. Solutions Class 12 maths Chapter-5 (Continuity And Differentiability)

Solution:

Here, as it is given that

For x < 0, f(x) = 2x, which is a polynomial

As polynomial are continuous, hence f(x) is continuous x ∈ (-∞, 0)

Now, For 0 ≤ x ≤ 1, f(x) = 0, which is a constant

As constant are continuous, hence f(x) is continuous x ∈ (0, 1)

For x > 1, f(x) = 4x, which is a polynomial

As polynomial are continuous, hence f(x) is continuous x ∈ (1, ∞)

So now, as f(x) is continuous in x ∈ (-∞, 0) U (0, 1) U (1, ∞)= R – {0, 1}

Let’s check the continuity at x = 0,

Left limit = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} 2x\\=2(0)\\ = 0

Right limit = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} 0\\= 0

Function value at x = 0, f(0) = 0

As, \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x)

Hence, the function is continuous at x = 0.

Now, let’s check the continuity at x = 1,

Left limit = \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} 0\\= 0

Right limit = \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} 4x\\= \lim_{x \to 1^+} 4(1)\\= 4

Function value at x = 1, f(1) = 0

As, \lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x)

Hence, the function is discontinuous at x = 1.

Therefore, the function is continuous for only R – {1}

Question 16. Solutions Class 12 maths Chapter-5 (Continuity And Differentiability)

Solution:

Here, as it is given that

For x ≤ -1, f(x) = -2, which is a constant

As constant are continuous, hence f(x) is continuous x ∈ (-∞, -1)

Now, For -1 ≤ x ≤ 1, f(x) = 2x, which is a polynomial

As polynomial are continuous, hence f(x) is continuous x ∈ (-1, 1)

For x > 1, f(x) = 2, which is a constant

As constant are continuous, hence f(x) is continuous x ∈ (1, ∞)

So now, as f(x) is continuous in x ∈ (-∞, -1) U (-1, 1) U (1, ∞)= R – {-1, 1}

Let’s check the continuity at x = -1,

Left limit = \lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} (-2)\\=-2

Right limit = \lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} (2x)\\= (2(-1))\\= -2

Function value at x = -1, f(-1) = -2

As, \lim_{x \to -1^-} f(x) = \lim_{x \to -1^+} f(x) = f(-1) = -2

Hence, the function is continuous at x = -1.

Now, let’s check the continuity at x = 1,

Left limit = \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (2x)\\= (2(1))\\= 2

Right limit = \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2)\\= 2

Function value at x = 1, f(1) = 2(1) = 2

As, \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) = 2

Hence, the function is continuous at x = 1.

Therefore, the function is continuous for any real number.

Question 17. Find the relationship between a and b so that the function f defined by

Solutions Class 12 maths Chapter-5 (Continuity And Differentiability)

is continuous at x = 3.

Solution:

As, it is given that the function is continuous at x = 3.

It should satisfy the following at x = 3:

\lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) = f(3)

Continuity at x = 3,

Left limit = \lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (ax+1)

= (a(3) + 1)

= 3a + 1

Right limit = \lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (bx+3)

= (b(3) + 3)

= 3b + 3

Function value at x = 3, f(3) = a(3) + 1 = 3a + 1

So equating both the limits, we get

3a + 1 = 3b + 3

3(a – b) = 2

a – b = 2/3

Exercise 5.1

Set 2

Question 18. For what value of λ is the function defined 

bySolutions Class 12 maths Chapter-5 (Continuity And Differentiability)

continuous at x = 0? What about continuity at x = 1?

Solution:

To be continuous function, f(x) should satisfy the following at x = 0:

\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)

Continuity at x = 0,

Left limit = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \lambda(x^2-2x)

= λ(02– 2(0)) = 0

Right limit = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (4x+1)

= λ4(0) + 1 = 1

Function value at x = 0, f(0) = \lambda(0^2-2(0)) = 0

As, 0 = 1 cannot be possible

Hence, for no value of λ, f(x) is continuous.

But here, \lim_{x \to 0^-} f(x) \neq \lim_{x \to 0^+} f(x)

Continuity at x = 1,

Left limit = \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (4x+1)

= (4(1) + 1) = 5

Right limit = \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (4x+1)

= 4(1) + 1 = 5

Function value at x = 1, f(1) = 4(1) + 1 = 5

As, \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) = 5

Hence, the function is continuous at x = 1 for any value of λ.

Question 19. Show that the function defined by g (x) = x – [x] is discontinuous at all integral points. Here [x] denotes the greatest integer less than or equal to x. 

Solution:

[x] is greatest integer function which is defined in all integral points, e.g.

[2.5] = 2

[-1.96] = -2

x-[x] gives the fractional part of x.

e.g: 2.5 – 2 = 0.5

c be an integer

Let’s check the continuity at x = c,

Left limit = \lim_{x \to c^-} f(x) = \lim_{x \to c^-} (x-[x])

= (c – (c – 1)) = 1

Right limit = \lim_{x \to c^+} f(x) = \lim_{x \to c^+} (x-[x])

= (c – c) = 0

Function value at x = c, f(c) = c – = c – c = 0

As, \lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x)

Hence, the function is discontinuous at integral.

c be not an integer

Let’s check the continuity at x = c,

Left limit = \lim_{x \to c^-} f(x) = \lim_{x \to c^-} (x-[x])

= (c – (c – 1)) = 1

Right limit = \lim_{x \to c^+} f(x) = \lim_{x \to c^+} (x-[x])

= (c – (c – 1)) = 1

Function value at x = c, f(c) = c – = c – (c – 1) = 1

As, \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x)=f(1)=1

Hence, the function is continuous at non-integrals part.

Question 20. Is the function defined by f(x) = x2 – sin x + 5 continuous at x = π?

Solution:

Let’s check the continuity at x = π,

f(x) = x2 – sin x + 5

Let’s substitute, x = π+h

When x⇢π, Continuity at x = π

Left limit = \lim_{x \to \pi^-} f(x) = \lim_{x \to \pi^-} (x^2 - sin \hspace{0.1cm}x + 5)

= (π2 – sinπ + 5) = π2 + 5

Right limit = \lim_{x \to \pi^+} f(x) = \lim_{x \to \pi^+}(x^2 - sin \hspace{0.1cm}x + 5)

= (π2 – sinπ + 5) = π2 + 5

Function value at x = π, f(π) = π2 – sin π + 5 = π2 + 5

As, \lim_{x \to \pi^-} f(x) = \lim_{x \to \pi^+} f(x) = f(\pi)

Hence, the function is continuous at x = π .

Question 21. Discuss the continuity of the following functions:

(a) f(x) = sin x + cos x 

Solution:

Here, 

f(x) = sin x + cos x

Let’s take, x = c + h

When x⇢c then h⇢0

\lim_{x \to c} f(x) = \lim_{h \to 0} f(c+h)

So, 

\lim_{h \to 0} f(c+h) = \lim_{h \to 0}(sin(c + h) + cos(c + h))

Using the trigonometric identities, we get

sin(A + B) = sin A cos B + cos A sin B

cos(A + B) =  cos A cos B – sin A sin B

\lim_{h \to 0} f(c+h) = \lim_{h \to 0}((sinc cosh + cosc sinh) + (cosc cosh − sinc sinh))

\lim_{h \to 0} f(c+h)= ((sinc cos0 + cosc sin0) + (cosc cos0 − sinc sin0))

cos 0 = 1 and sin 0 = 0

\lim_{h \to 0} f(c+h) = (sinc + cosc) = f(c)

Function value at x = c, f(c) = sinc + cosc

As, \lim_{x \to c} f(x)= f(c) = sinc + cosc

Hence, the function is continuous at x = c.

(b) f(x) = sin x – cos x

Solution:

Here,

f(x) = sin x – cos x

Let’s take, x = c+h

When x⇢c then h⇢0

\lim_{x \to c} f(x) = \lim_{h \to 0} f(c+h)

So,

\lim_{h \to 0} f(c+h) = \lim_{h \to 0}(sin(c + h) − cos(c + h))

Using the trigonometric identities, we get

sin(A + B) = sin A cos B + cos A sin B

cos(A + B) =  cos A cos B – sin A sin B

\lim_{h \to 0} f(c+h) = \lim_{h \to 0}((sinc cosh + cosc sinh) − (cosc cosh − sinc sinh))

\lim_{h \to 0} f(c+h) = ((sinc cos0 + cosc sin0) − (cosc cos0 − sinc sin0))

cos 0 = 1 and sin 0 = 0

\lim_{h \to 0} f(c+h) = (sinc − cosc) = f(c)

Function value at x = c, f(c) = sinc − cosc

As, \lim_{x \to c} f(x) = f(c) = sinc − cosc

Hence, the function is continuous at x = c.

(c) f(x) = sin x . cos x

Solution:

Here,

f(x) = sin x + cos x

Let’s take, x = c+h

When x⇢c then h⇢0

\lim_{x \to c} f(x) = \lim_{h \to 0} f(c+h)

So,

\lim_{h \to 0} f(c+h) = \lim_{h \to 0}sin(c + h) × cos(c + h))

Using the trigonometric identities, we get

sin(A + B) = sin A cos B + cos A sin B

cos(A + B) =  cos A cos B – sin A sin B

\lim_{h \to 0} f(c+h) = \lim_{h \to 0}((sinc cosh + cosc sinh) × (cosc cosh − sinc sinh))

\lim_{h \to 0} f(c+h)= ((sinc cos0 + cosc sin0) × (cosc cos0 − sinc sin0))

cos 0 = 1 and sin 0 = 0

\lim_{h \to 0} f(c+h) = (sinc × cosc) = f(c)

Function value at x = c, f(c) = sinc × cosc

As, \lim_{x \to c} f(x)= f(c) = sinc × cosc

Hence, the function is continuous at x = c.

Question 22. Discuss the continuity of the cosine, cosecant, secant and cotangent functions.

Solution:

Continuity of cosine

Here,

f(x) = cos x

Let’s take, x = c+h

When x⇢c then h⇢0

\lim_{x \to c} f(x) = \lim_{h \to 0} f(c+h)

So,

\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (cos\hspace{0.1cm} (c+h))

Using the trigonometric identities, we get

cos(A + B) =  cos A cos B – sin A sin B

\lim_{h \to 0} f(c+h) = \lim_{h \to 0}(cosc cosh − sinc sinh)

\lim_{h \to 0} f(c+h) = (cosc cos0 − sinc sin0)

cos 0 = 1 and sin 0 = 0

\lim_{h \to 0} f(c+h) = (cosc) = f(c)

Function value at x = c, f(c) = (cosc)

As, \lim_{x \to c} f(x) = f(c) = (cosc)

Hence, the cosine function is continuous at x = c.

Continuity of cosecant

Here,

f(x) = cosec x = \frac{1}{sin \hspace{0.1cm}x}

Domain of cosec is R – {nπ}, n ∈ Integer

Let’s take, x = c + h

When x⇢c then h⇢0

\lim_{x \to c} f(x) = \lim_{h \to 0} f(c+h)

So,

\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{1}{sin \hspace{0.1cm}(c+h)})

Using the trigonometric identities, we get

sin(A + B) = sin A cos B + cos A sin B

\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{1}{sin\hspace{0.1cm} c\hspace{0.1cm} cos\hspace{0.1cm} h+cos\hspace{0.1cm} c\hspace{0.1cm} sin\hspace{0.1cm} h})\\ \lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{1}{sin\hspace{0.1cm} c\hspace{0.1cm} cos\hspace{0.1cm} 0+cos\hspace{0.1cm} c\hspace{0.1cm} sin\hspace{0.1cm} 0})

cos 0 = 1 and sin 0 = 0

\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{1}{sin\hspace{0.1cm} c})

Function value at x = c, f(c) = \frac{1}{sin\hspace{0.1cm} c}

As, \lim_{x \to c} f(x) = f(c) = \frac{1}{sin\hspace{0.1cm} c}

Hence, the cosecant function is continuous at x = c.

Continuity of secant

Here,

f(x) = sec x = \frac{1}{cos \hspace{0.1cm}x}

Let’s take, x = c + h

When x⇢c then h⇢0

\lim_{x \to c} f(x) = \lim_{h \to 0} f(c+h)

So,

\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{1}{cos \hspace{0.1cm}(c+h)})

Using the trigonometric identities, we get

cos(A + B) =  cos A cos B – sin A sin B

\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{1}{cos\hspace{0.1cm} c\hspace{0.1cm} cos\hspace{0.1cm} h-sin\hspace{0.1cm} c\hspace{0.1cm} sin\hspace{0.1cm} h})

\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{1}{cos\hspace{0.1cm} c\hspace{0.1cm} cos\hspace{0.1cm} 0-sin\hspace{0.1cm} c\hspace{0.1cm} sin\hspace{0.1cm} 0})

cos 0 = 1 and sin 0 = 0

\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{1}{cos\hspace{0.1cm} c})

Function value at x = c, f(c) = \frac{1}{cos\hspace{0.1cm} c}

As, \lim_{x \to c} f(x) = f(c) = \frac{1}{cos\hspace{0.1cm} c}

Hence, the secant function is continuous at x = c.

Continuity of cotangent

Here,

f(x) = cot x = \frac{cos \hspace{0.1cm}x}{sin \hspace{0.1cm}x}

Let’s take, x = c+h

When x⇢c then h⇢0

\lim_{x \to c} f(x) = \lim_{h \to 0} f(c+h)

So,

\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{cos \hspace{0.1cm}(c+h)}{sin \hspace{0.1cm}(c+h)})

Using the trigonometric identities, we get

sin(A + B) = sin A cos B + cos A sin B

cos(A + B) =  cos A cos B – sin A sin B

\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{cos\hspace{0.1cm} c\hspace{0.1cm} cos\hspace{0.1cm} h-sin\hspace{0.1cm} c\hspace{0.1cm} sin\hspace{0.1cm} h}{sin\hspace{0.1cm} c\hspace{0.1cm} cos\hspace{0.1cm} h+cos\hspace{0.1cm} c\hspace{0.1cm} sin\hspace{0.1cm} h})

\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{cos\hspace{0.1cm} c\hspace{0.1cm} cos\hspace{0.1cm} 0-sin\hspace{0.1cm} c\hspace{0.1cm} sin\hspace{0.1cm} 0}{sin\hspace{0.1cm} c\hspace{0.1cm} cos\hspace{0.1cm} 0+cos\hspace{0.1cm} c\hspace{0.1cm} sin\hspace{0.1cm} 0})

cos 0 = 1 and sin 0 = 0

\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{cos\hspace{0.1cm} c}{sin\hspace{0.1cm} c})

\lim_{h \to 0} f(c+h) = (\frac{cos\hspace{0.1cm} c}{sin\hspace{0.1cm} c})

Function value at x = c, f(c) = \frac{cos\hspace{0.1cm} c}{sin\hspace{0.1cm} c}

As, \lim_{x \to c} f(x) = f(c) = \frac{cos\hspace{0.1cm} c}{sin\hspace{0.1cm} c}

Hence, the cotangent function is continuous at x = c.

Question 23. Find all points of discontinuity of f, whereSolutions Class 12 maths Chapter-5 (Continuity And Differentiability)

Solution:

Here,

From the two continuous functions g and h, we get

\frac{g(x)}{h(x)}= continuous when h(x) ≠ 0

For x < 0, f(x) = \frac{sin \hspace{0.1cm}x}{x}, is continuous

Hence, f(x) is continuous x ∈ (-∞, 0)

Now, For x ≥ 0, f(x) = x + 1, which is a polynomial

As polynomial are continuous, hence f(x) is continuous x ∈ (0, ∞)

So now, as f(x) is continuous in x ∈ (-∞, 0) U (0, ∞)= R – {0}

Let’s check the continuity at x = 0,

Left limit = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (\frac{sin \hspace{0.1cm}x}{x})\\= 1

Right limit = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x+1)\\= (1+0)\\= 1

Function value at x = 0, f(0) = 0 + 1 = 1

As, \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = 1

Hence, the function is continuous at x = 0.

Hence, the function is continuous for any real number.

Question 24. Determine if f defined bySolutions Class 12 maths Chapter-5 (Continuity And Differentiability)

Solution:

Here, as it is given that

For x = 0, f(x) = 0, which is a constant

As constant are continuous, hence f(x) is continuous x ∈ = R – {0}

Let’s check the continuity at x = 0,

As, we know range of sin function is [-1,1]. So, -1 ≤ sin(\frac{1}{x})≤ 1 which is a finite number.

Limit = \lim_{x \to 0} f(x) = \lim_{x \to 0} (x^2 sin(\frac{1}{x}))

= (02 ×(finite number)) = 0

Function value at x = 0, f(0) = 0

As, \lim_{x \to 0} f(x) = f(0).

Hence, the function is continuous for any real number.

Question 25. Examine the continuity of f, where f is defined by

Solutions Class 12 maths Chapter-5 (Continuity And Differentiability)

Solution:

Continuity at x = 0,

Left limit = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (sin\hspace{0.1cm}x-cos\hspace{0.1cm}x)

= (sin0 − cos0) = 0 − 1 = −1

Right limit = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (sin\hspace{0.1cm}x-cos\hspace{0.1cm}x)

= (sin0 − cos0) = 0 − 1 = −1

Function value at x = 0, f(0) = sin 0 – cos 0 = 0 – 1 = -1

As, \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = -1

Hence, the function is continuous at x = 0.

Continuity at x = c (real number c≠0),

Left limit = \lim_{x \to c^-} f(x) = \lim_{x \to c^-} (sin\hspace{0.1cm}x-cos\hspace{0.1cm}x)

= (sinc − cosc)

Right limit = \lim_{x \to c^+} f(x) = \lim_{x \to c^+} (sin\hspace{0.1cm}x-cos\hspace{0.1cm}x)

= (sinc − cosc)

Function value at x = c, f(c) = sin c – cos c

As, \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c) = (sin\hspace{0.1cm}c-cos\hspace{0.1cm}c)

So concluding the results, we get

The function f(x) is continuous at any real number.

Question 26. Solutions Class 12 maths Chapter-5 (Continuity And Differentiability) at x = π/2.

Solution:

Continuity at x = π/2

Let’s take x = \frac{\pi}{2}+h

When x⇢π/2 then h⇢0

Substituting x = \frac{\pi}{2}+h, we get

cos(A + B) = cos A cos B – sin A sin B

Limit = \lim_{h \to 0} f(\frac{\pi}{2}+h) = \lim_{h \to 0} (\frac{k\hspace{0.1cm}cos(\frac{\pi}{2}+h)}{\pi-2(\frac{\pi}{2}+h)}\\= \lim_{h \to 0} (\frac{k(cos(\frac{\pi}{2})cos h-sin(\frac{\pi}{2})sinh)}{\pi-\pi-2h)}\\= \lim_{h \to 0} (\frac{k(0 \times cos h-1\times sinh)}{-2h)}\\= \lim_{h \to 0} (\frac{k(-sinh)}{-2h)}\\ = \frac{k}{2} \lim_{h \to 0} (\frac{(sinh)}{h)}\\ = \frac{k}{2}

Function value at x = \frac{\pi}{2}, f(\frac{\pi}{2})= 3

As, \lim_{x \to \frac{\pi}{2}} f(x) = f(\frac{\pi}{2})should satisfy, for f(x) being continuous

k/2 = 3

k = 6

Question 27. Solutions Class 12 maths Chapter-5 (Continuity And Differentiability) at x = 2

Solution:

Continuity at x = 2

Left limit = \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (kx^2)

= k(2)2 = 4k

Right limit = \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (3)\\= 3

Function value at x = 2, f(2) = k(2)2 = 4k

As, \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x)= f(2) should satisfy, for f(x) being continuous

4k = 3

k = 3/4

Question 28. Solutions Class 12 maths Chapter-5 (Continuity And Differentiability) at x = π

Solution:

Continuity at x = π

Left limit = \lim_{x \to \pi^-} f(x) = \lim_{x \to \pi^-} (kx+1)

= k(π) + 1

Right limit = \lim_{x \to \pi^+} f(x) = \lim_{x \to \pi^+} (cos x)

= cos(π) = -1

Function value at x = π, f(π) = k(π) + 1

As, \lim_{x \to \pi^-}f(x) = \lim_{x \to \pi^+} f(x)= f(\pi) should satisfy, for f(x) being continuous

kπ + 1 = -1

k = -2/π

Question 29. Solutions Class 12 maths Chapter-5 (Continuity And Differentiability) at x = 5

Solution:

Continuity at x = 5

Left limit = \lim_{x \to 5^-} f(x) = \lim_{x \to 5^-} (kx+1)

= k(5) + 1 = 5k + 1

Right limit = \lim_{x \to 5^+} f(x) = \lim_{x \to 5^+} (3x-5)

= 3(5) – 5 = 10

Function value at x = 5, f(5) = k(5) + 1 = 5k + 1

As, \lim_{x \to 5^-} f(x) = \lim_{x \to 5^+} f(x)= f(5)should satisfy, for f(x) being continuous

5k + 1 = 10

k = 9/5 

Question 30. Find the values of a and b such that the function defined by

Solutions Class 12 maths Chapter-5 (Continuity And Differentiability)

is a continuous function

Solution:

Continuity at x = 2

Left limit = \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (5)\\= 5

Right limit = \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (ax+b)\\= 2a+b

Function value at x = 2, f(2) = 5

As, \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x)= f(2)should satisfy, for f(x) being continuous at x = 2

2a + b = 5 ……………………(1)

Continuity at x = 10

Left limit = \lim_{x \to 10^-} f(x) = \lim_{x \to 10^-} (ax+b)

= 10a + b

Right limit = \lim_{x \to 10^+} f(x) = \lim_{x \to 10^+} (21)

= 21

Function value at x = 10, f(10) = 21

As, \lim_{x \to 10^-} f(x) = \lim_{x \to 10^+} f(x)= f(10)should satisfy, for f(x) being continuous at x = 10

10a + b = 21 ……………………(2)

Solving the eq(1) and eq(2), we get

a = 2

b = 1

Question 31. Show that the function defined by f(x) = cos (x2) is a continuous function

Solution:

Let’s take

g(x) = cos x

h(x) = x2

g(h(x)) = cos (x2)

To prove g(h(x)) continuous, g(x) and h(x) should be continuous.

Continuity of g(x) = cos x

Let’s check the continuity at x = c

x = c + h

g(c + h) = cos (c + h)

When x⇢c then h⇢0

cos(A + B) =  cos A cos B – sin A sin B

Limit = \lim_{h \to 0} g(c+h) = \lim_{h \to 0} (cos(c+h))\\ = \lim_{h \to 0}(cosc cosh − sinc sinh)

= cosc cos0 − sinc sin0 = cosc

Function value at x = c, g(c) = cos c

As, \lim_{x \to c} g(x) = g(c) = cos\hspace{0.1cm} c

The function g(x) is continuous at any real number.

Continuity of h(x) = x2

Let’s check the continuity at x = c

Limit = \lim_{x \to c} h(x) = \lim_{x \to c} (x^2)

= c2

Function value at x = c, h(c) = c2

As, \lim_{x \to c} h(x) = h(c) = c^2

The function h(x) is continuous at any real number.

As, g(x) and h(x) is continuous then g(h(x)) = cos(x2) is also continuous.

Question 32. Show that the function defined by f(x) = | cos x | is a continuous function. 

Solution:

Let’s take

g(x) = |x|

m(x) = cos x

g(m(x)) = |cos x|

To prove g(m(x)) continuous, g(x) and m(x) should be continuous.

Continuity of g(x) = |x|

As, we know that modulus function works differently.

In |x – 0|, |x| = x when x ≥ 0 and |x| = -x when x < 0

Let’s check the continuity at x = c

When c < 0

Limit = \lim_{x \to c} g(x) = \lim_{x \to c} (|x|)\\= \lim_{x \to c} (-x)\\ = -c

Function value at x = c, g(c) = |c| = -c

As, \lim_{x \to c} g(x) = g(c) = -c

When c ≥ 0

Limit = \lim_{x \to c} g(x) = \lim_{x \to c} (|x|)\\= \lim_{x \to c} (x)\\ = c

Function value at x = c, g(c) = |c| = c

As, \lim_{x \to c} g(x) = g(c) = c

The function g(x) is continuous at any real number.

Continuity of m(x) = cos x

Let’s check the continuity at x = c

x = c + h

m(c + h) = cos (c + h)

When x⇢c then h⇢0

cos(A + B) =  cos A cos B – sin A sin B

Limit = \lim_{h \to 0} m(c+h) = \lim_{h \to 0} (cos(c+h))\\ = \lim_{h \to 0}(cosc cosh − sinc sinh)

= cosc cos0 − sinc sin0 = cosc

Function value at x = c, m(c) = cos c

As, \lim_{x \to c} m(x) = m(c) = cos \hspace{0.1cm}c

The function m(x) is continuous at any real number.

As, g(x) and m(x) is continuous then g(m(x)) = |cos x| is also continuous.

Question 33. Examine that sin | x | is a continuous function.

Solution:

Let’s take

g(x) = |x|

m(x) = sin x

m(g(x)) = sin |x|

To prove m(g(x)) continuous, g(x) and m(x) should be continuous.

Continuity of g(x) = |x|

As, we know that modulus function works differently.

In |x-0|, |x|=x when x≥0 and |x|=-x when x<0

Let’s check the continuity at x = c

When c < 0

Limit = \lim_{x \to c} g(x) = \lim_{x \to c} (|x|)\\= \lim_{x \to c} (-x)\\ = -c

Function value at x = c, g(c) = |c| = -c

As, \lim_{x \to c} g(x) = g(c) = -c

When c ≥ 0

Limit = \lim_{x \to c} g(x) = \lim_{x \to c} (|x|)\\= \lim_{x \to c} (x)\\ = c

Function value at x = c, g(c) = |c| = c

As, \lim_{x \to c} g(x) = g(c) = c

The function g(x) is continuous at any real number.

Continuity of m(x) = sin x

Let’s check the continuity at x = c

x = c + h

m(c + h) = sin (c + h)

When x⇢c then h⇢0

sin(A + B) = sin A cos B + cos A sin B

Limit = \lim_{h \to 0} m(c+h) = \lim_{h \to 0} (sin(c+h))\\ = \lim_{h \to 0}(sinc cosh + cosc sinh) 

= sinc cos0 + cos csin0 = sinc

Function value at x = c, m(c) = sin c

As, \lim_{x \to c} m(x) = m(c) = sin c

The function m(x) is continuous at any real number.

As, g(x) and m(x) is continuous then m(g(x)) = sin |x| is also continuous.

Question 34. Find all the points of discontinuity of f defined by f(x) = | x | – | x + 1 |

Solution:

Let’s take

g(x) = |x|

m(x) = |x + 1|

g(x) – m(x) = | x | – | x + 1 |

To prove g(x) – m(x) continuous, g(x) and m(x) should be continuous.

Continuity of g(x) = |x|

As, we know that modulus function works differently.

In |x – 0|, |x| = x when x≥0 and |x| = -x when x < 0

Let’s check the continuity at x = c

When c < 0

Limit = \lim_{x \to c} g(x) = \lim_{x \to c} (|x|)\\= \lim_{x \to c} (-x)\\ = -c

Function value at x = c, g(c) = |c| = -c

As, \lim_{x \to c} g(x) = g(c) = -c

When c ≥ 0

Limit = \lim_{x \to c} g(x) = \lim_{x \to c} (|x|)\\= \lim_{x \to c} (x)\\ = c

Function value at x = c, g(c) = |c| = c

As, \lim_{x \to c} g(x) = g(c) = c

The function g(x) is continuous at any real number.

Continuity of m(x) = |x + 1|

As, we know that modulus function works differently.

In |x + 1|, |x + 1| = x + 1 when x ≥ -1 and |x + 1| = -(x + 1) when x < -1

Let’s check the continuity at x = c

When c < -1

Limit = \lim_{x \to c} m(x) = \lim_{x \to c} (|x+1|)\\= \lim_{x \to c} -(x+1)

= -(c + 1)

Function value at x = c, m(c) = |c + 1| = -(c + 1)

As, \lim_{x \to c} m(x) = m(c) = -(c+1)

When c ≥ -1

Limit = \lim_{x \to c} m(x) = \lim_{x \to c} (|x|)\\= \lim_{x \to c} (x+1)

= c + 1

Function value at x = c, m(c) = |c| = c + 1

As, \lim_{x \to c} m(x) = m(c) = c + 1

The function m(x) is continuous at any real number.

As, g(x) and m(x) is continuous then g(x) – m(x) = |x| – |x + 1| is also continuous.

Chapter-5 (Continuity And Differentiability)

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