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Class 12 NCERT Solutions- Mathematics Part I – Chapter 5 Continuity And Differentiability – Exercise 5.1

Last Updated : 03 Apr, 2024
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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 = [Tex]\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (5x-3)[/Tex]

= (5(0) – 3) = -3

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

= (5(0) – 3)= -3

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

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

Hence, the function is continuous at x = 0.

Continuity at x = -3

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

= (5(-3) – 3) = -18

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

= (5(-3) – 3) = -18

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

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

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

Continuity at x = 5

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

= (5(5) – 3) = 22

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

= (5(5) – 3) = 22

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

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

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 = [Tex]\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (2x^2-1)[/Tex]

= (2(3)2 – 1) = 17

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

= (2(3)2 – 1) = 17

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

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

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 = [Tex]\lim_{x \to c^-} f(x) = \lim_{x \to c^-} (x-5)[/Tex]

= (c – 5) = c – 5

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

= (c – 5) = c – 5

Function value at x = c, f(c) = c – 5

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

Hence, the function is continuous at every real number.

(b) [Tex]f(x) = \frac{1}{x-5}[/Tex], x ≠ 5

Solution:

To prove the continuity of the function f(x) = [Tex]\frac{1}{x-5}     [/Tex], 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 = [Tex]\lim_{x \to c^-} f(x) = \lim_{x \to c^-} \frac{1}{(x-5)}\\= \frac{1}{(c-5)}[/Tex]

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

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

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

Hence, the function is continuous at every real number.

(c) [Tex]f(x) = \frac{x^2-25}{x+5}[/Tex], x ≠ -5

Solution:

To prove the continuity of the function f(x) = [Tex]\frac{x^2-25}{x+5}     [/Tex], 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 = [Tex]\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[/Tex]

= c – 5

Right limit = [Tex]\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[/Tex]

= c – 5

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

= c – 5

As, [Tex]\lim_{x \to c^-} f(x)=\lim_{x \to c^+} f(x) = f(c) = c – 5     [/Tex], 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 = [Tex]\lim_{x \to c^-} f(x) = \lim_{x \to c^-} |x-5|\\= \lim_{x \to c^-} -(x-5)[/Tex]

= -(c – 5) 

= 5 – c

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

= -(c – 5)

= 5 – c

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

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

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

When c > 5

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

= (c – 5)

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

= (c – 5)

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

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

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

When c = 5

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

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

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

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

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 = [Tex]\lim_{x \to n^-} f(x) = \lim_{x \to n^-} (x^n)[/Tex]

= nn

Right limit = [Tex]\lim_{x \to n^+} f(x) = \lim_{x \to n^+} (x^n)[/Tex]

= nn

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

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

Hence, the function is continuous at x = n.

Question 5. Is the function f defined by

[Tex]f(x)= \begin{cases} x, \hspace{0.2cm}x\leq1\\ 5,\hspace{0.2cm}x>1 \end{cases}[/Tex]

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

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 = [Tex]\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (x)\\= 0[/Tex]

Right limit = [Tex]\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x)\\= 0[/Tex]

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

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

Hence, the function is continuous at x = 0.

Continuity at x = 1

Left limit = [Tex]\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x)\\= 1[/Tex]

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

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

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

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

Continuity at x = 2

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

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

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

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

Therefore, the function is continuous at x = 2.

Find all points of discontinuity of f, where f is defined by

Question 6. [Tex]f(x)= \begin{cases} 2x+3, \hspace{0.2cm}x\leq2\\ 2x-3,\hspace{0.2cm}x>2 \end{cases}[/Tex]

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 = [Tex]\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (2x+3)[/Tex]

= (2(2) + 3)

= 7

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

= (2(2) – 3)

= 1

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

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

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

Question 7. [Tex]f(x)= \begin{cases} |x|+3, \hspace{0.2cm}x\leq-3\\ -2x,\hspace{0.2cm}-3<x<3\\ 6x+2,\hspace{0.2cm}x\geq3 \end{cases}[/Tex]

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 = [Tex]\lim_{x \to -3^-} f(x) = \lim_{x \to -3^-} (|x|+3)\\= \lim_{x \to -3^-} (-x+3)[/Tex]

= (-(-3) + 3)

= 6

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

= (-2(-3))

= 6

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

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

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

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

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

= (-2(3))

= -6

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

= (6(3) + 2)

= 20

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

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

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

Question 8. [Tex]f(x)= \begin{cases} \frac{|x|}{x}, \hspace{0.2cm}x\neq0\\ 0,\hspace{0.2cm}x=0 \end{cases}[/Tex]

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) = [Tex]\frac{-x}{x}    [/Tex]= -1, which is a constant

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

When x > 0, f(x) = [Tex]\frac{x}{x}    [/Tex]= 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 = [Tex]\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{|x|}{x}\\= \lim_{x \to 0^-} \frac{-x}{x}\\= -1[/Tex]

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

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

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

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

Question 9. [Tex]f(x)= \begin{cases} \frac{x}{|x|}, \hspace{0.2cm}x<0\\ -1,\hspace{0.2cm}x\geq0 \end{cases}[/Tex]

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) = [Tex]\frac{x}{-x}    [/Tex]= -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 = [Tex]\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{x}{|x|}\\= \lim_{x \to 0^-} \frac{x}{-x}\\= -1[/Tex]

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

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

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

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. [Tex]f(x)= \begin{cases} x+1, \hspace{0.2cm}x\geq1\\ x^2+1,\hspace{0.2cm}x<1 \end{cases}[/Tex]

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 = [Tex]\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^2+1)[/Tex]

= 1 + 1

= 2

Right limit = [Tex]\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x+1)[/Tex]

= 1 + 1

= 2

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

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

Hence, the function is continuous at x = 1.

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

Question 11. [Tex]f(x)= \begin{cases} x^3-3, \hspace{0.2cm}x\leq2\\ x^2+1,\hspace{0.2cm}x>2 \end{cases}[/Tex]

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 = [Tex]\lim_{x \to 2^-} f(x)= \lim_{x \to 2^-} (x^3-3)\\= 8-3\\=5[/Tex]

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

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

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

Hence, the function is continuous at x = 2.

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

Question 12. [Tex]f(x)= \begin{cases} x^{10}-1, \hspace{0.2cm}x\leq1\\ x^2,\hspace{0.2cm}x>1 \end{cases}[/Tex]

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 = [Tex]\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^{10}-1)[/Tex]

= 1 – 1

= 0

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

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

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

Hence, the function is discontinuous at x = 1.

Question 13. Is the function defined by

[Tex]f(x)= \begin{cases} x+5, \hspace{0.2cm}x\leq1\\ x-5,\hspace{0.2cm}x>1 \end{cases}[/Tex]

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 = [Tex]\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x+5)[/Tex]

= (1 + 5)

= 6

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

= (1 – 5)

= -4

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

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

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

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

Question 14. [Tex]f(x)= \begin{cases} 3, \hspace{0.2cm}0\leq x \leq1\\ 4,\hspace{0.2cm}1<x<3 \\ 5,\hspace{0.2cm}3\leq x \leq10 \end{cases}[/Tex]

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 = [Tex]\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} 3\\= 3[/Tex]

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

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

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

Hence, the function is discontinuous at x = 1.

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

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

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

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

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

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. [Tex]f(x)= \begin{cases} 2x, \hspace{0.2cm}x<0\\ 0,\hspace{0.2cm}0\leq x\leq1 \\ 4x,\hspace{0.2cm}x>1 \end{cases}[/Tex]

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 = [Tex]\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} 2x\\=2(0)\\ = 0[/Tex]

Right limit = [Tex]\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} 0\\= 0[/Tex]

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

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

Hence, the function is continuous at x = 0.

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

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

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

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

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

Hence, the function is discontinuous at x = 1.

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

Question 16. [Tex]f(x)= \begin{cases} -2, \hspace{0.2cm}x \leq-1\\ 2x,\hspace{0.2cm}-1<x<1 \\ 2,\hspace{0.2cm}x>1 \end{cases}[/Tex]

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 = [Tex]\lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} (-2)\\=-2[/Tex]

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

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

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

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

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

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

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

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

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

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

[Tex]f(x)= \begin{cases} ax+1, \hspace{0.2cm}x \leq3\\ bx+3,\hspace{0.2cm}x>3 \end{cases}[/Tex]

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:

[Tex]\lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) = f(3)[/Tex]

Continuity at x = 3,

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

= (a(3) + 1)

= 3a + 1

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

= (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

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

[Tex]f(x)= \begin{cases} \lambda(x^2-2x), \hspace{0.2cm}x \leq0\\ 4x+1,\hspace{0.2cm}x>0 \end{cases}[/Tex]

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

Solution:

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

[Tex]\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)[/Tex]

Continuity at x = 0,

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

= λ(02– 2(0)) = 0

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

= λ4(0) + 1 = 1

Function value at x = 0, f(0) = [Tex]\lambda(0^2-2(0)) = 0[/Tex]

As, 0 = 1 cannot be possible

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

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

Continuity at x = 1,

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

= (4(1) + 1) = 5

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

= 4(1) + 1 = 5

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

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

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 = [Tex]\lim_{x \to c^-} f(x) = \lim_{x \to c^-} (x-[x])[/Tex]

= (c – (c – 1)) = 1

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

= (c – c) = 0

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

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

Hence, the function is discontinuous at integral.

c be not an integer

Let’s check the continuity at x = c,

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

= (c – (c – 1)) = 1

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

= (c – (c – 1)) = 1

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

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

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 = [Tex]\lim_{x \to \pi^-} f(x) = \lim_{x \to \pi^-} (x^2 – sin \hspace{0.1cm}x + 5)[/Tex]

= (π2 – sinπ + 5) = π2 + 5

Right limit = [Tex]\lim_{x \to \pi^+} f(x) = \lim_{x \to \pi^+}(x^2 – sin \hspace{0.1cm}x + 5)[/Tex]

= (π2 – sinπ + 5) = π2 + 5

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

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

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

[Tex]\lim_{x \to c} f(x) = \lim_{h \to 0} f(c+h)[/Tex]

So, 

[Tex]\lim_{h \to 0} f(c+h) = \lim_{h \to 0} [/Tex](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

[Tex]\lim_{h \to 0} f(c+h) = \lim_{h \to 0}  [/Tex]((sinc cosh + cosc sinh) + (cosc cosh sinc sinh))

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

cos 0 = 1 and sin 0 = 0

[Tex]\lim_{h \to 0} f(c+h) [/Tex] = (sinc + cosc) = f(c)

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

As, [Tex]\lim_{x \to c} f(x)  [/Tex]= 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

[Tex]\lim_{x \to c} f(x) = \lim_{h \to 0} f(c+h)[/Tex]

So,

[Tex]\lim_{h \to 0} f(c+h) = \lim_{h \to 0}  [/Tex](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

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

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

cos 0 = 1 and sin 0 = 0

[Tex]\lim_{h \to 0} f(c+h) [/Tex] = (sinc cosc) = f(c)

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

As, [Tex]\lim_{x \to c} f(x) [/Tex] = 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

[Tex]\lim_{x \to c} f(x) = \lim_{h \to 0} f(c+h)[/Tex]

So,

[Tex]\lim_{h \to 0} f(c+h) = \lim_{h \to 0} [/Tex]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

[Tex]\lim_{h \to 0} f(c+h) = \lim_{h \to 0}  [/Tex]((sinc cosh + cosc sinh) × (cosc cosh sinc sinh))

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

cos 0 = 1 and sin 0 = 0

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

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

As, [Tex]\lim_{x \to c} f(x)  [/Tex]= 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

[Tex]\lim_{x \to c} f(x) = \lim_{h \to 0} f(c+h)[/Tex]

So,

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

Using the trigonometric identities, we get

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

[Tex]\lim_{h \to 0} f(c+h) = \lim_{h \to 0} [/Tex](cosc cosh sinc sinh)

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

cos 0 = 1 and sin 0 = 0

[Tex]\lim_{h \to 0} f(c+h) [/Tex] = (cosc) = f(c)

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

As, [Tex]\lim_{x \to c} f(x) [/Tex] = f(c) = (cosc)

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

Continuity of cosecant

Here,

f(x) = cosec x = [Tex]\frac{1}{sin \hspace{0.1cm}x}[/Tex]

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

Let’s take, x = c + h

When x⇢c then h⇢0

[Tex]\lim_{x \to c} f(x) = \lim_{h \to 0} f(c+h)[/Tex]

So,

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

Using the trigonometric identities, we get

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

[Tex]\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})[/Tex]

cos 0 = 1 and sin 0 = 0

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

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

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

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

Continuity of secant

Here,

f(x) = sec x = [Tex]\frac{1}{cos \hspace{0.1cm}x}[/Tex]

Let’s take, x = c + h

When x⇢c then h⇢0

[Tex]\lim_{x \to c} f(x) = \lim_{h \to 0} f(c+h)[/Tex]

So,

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

Using the trigonometric identities, we get

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

[Tex]\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})[/Tex]

[Tex]\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})[/Tex]

cos 0 = 1 and sin 0 = 0

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

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

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

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

Continuity of cotangent

Here,

f(x) = cot x = [Tex]\frac{cos \hspace{0.1cm}x}{sin \hspace{0.1cm}x}[/Tex]

Let’s take, x = c+h

When x⇢c then h⇢0

[Tex]\lim_{x \to c} f(x) = \lim_{h \to 0} f(c+h)[/Tex]

So,

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

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

[Tex]\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})[/Tex]

[Tex]\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})[/Tex]

cos 0 = 1 and sin 0 = 0

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

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

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

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

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

Question 23. Find all points of discontinuity of f, where

[Tex]f(x)= \begin{cases} \frac{sin \hspace{0.1cm}x}{x}, \hspace{0.2cm}x <0\\ x+1,\hspace{0.2cm}x\geq0 \end{cases}[/Tex]

Solution:

Here,

From the two continuous functions g and h, we get

[Tex]\frac{g(x)}{h(x)}     [/Tex]= continuous when h(x) ≠ 0

For x < 0, f(x) = [Tex]\frac{sin \hspace{0.1cm}x}{x}     [/Tex], 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 = [Tex]\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (\frac{sin \hspace{0.1cm}x}{x})\\= 1[/Tex]

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

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

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

Hence, the function is continuous at x = 0.

Hence, the function is continuous for any real number.

Question 24. Determine if f defined by

[Tex]f(x)= \begin{cases} x^2sin\frac{1}{x}, \hspace{0.2cm}x \neq0\\ 0,\hspace{0.2cm}x=0 \end{cases}[/Tex]

is a continuous function?

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 ≤ [Tex]sin(\frac{1}{x})     [/Tex]≤ 1 which is a finite number.

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

= (02 ×(finite number)) = 0

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

As, [Tex]\lim_{x \to 0} f(x) = f(0).[/Tex]

Hence, the function is continuous for any real number.

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

[Tex]f(x)= \begin{cases} sin\hspace{0.1cm}x-cos\hspace{0.1cm}x, \hspace{0.2cm}x\neq0\\ -1,\hspace{0.2cm}x=0 \end{cases}[/Tex]

Solution:

Continuity at x = 0,

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

= (sin0 − cos0) = 0 − 1 = −1

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

= (sin0 − cos0) = 0 − 1 = −1

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

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

Hence, the function is continuous at x = 0.

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

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

= (sinc cosc)

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

= (sinc cosc)

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

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

So concluding the results, we get

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

Find the values of k so that the function f is continuous at the indicated point in Exercises 26 to 29. 

Question 26. [Tex]f(x)= \begin{cases} \frac{k\hspace{0.1cm}cos\hspace{0.1cm}x}{\pi-2x}, \hspace{0.2cm}x\neq\frac{\pi}{2}\\ 3,x=\frac{\pi}{2} \end{cases} \hspace{0.1cm}\hspace{0.1cm} [/Tex] at x = π/2.

Solution:

Continuity at x = π/2

Let’s take x = [Tex]\frac{\pi}{2}+h[/Tex]

When x⇢π/2 then h⇢0

Substituting x = [Tex]\frac{\pi}{2}   [/Tex]+h, we get

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

Limit = [Tex]\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}[/Tex]

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

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

k/2 = 3

k = 6

Question 27. [Tex]f(x)= \begin{cases} kx^2,x\leq2\\ 3,x>2 \end{cases} \hspace{0.1cm}\hspace{0.1cm} [/Tex] at x = 2

Solution:

Continuity at x = 2

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

= k(2)2 = 4k

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

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

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

4k = 3

k = 3/4

Question 28. [Tex]f(x)= \begin{cases} kx+1,x\leq\pi\\ cos \hspace{0.2cm}x,x>\pi \end{cases} \hspace{0.1cm}\hspace{0.1cm} [/Tex] at x = π

Solution:

Continuity at x = π

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

= k(π) + 1

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

= cos(π) = -1

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

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

kπ + 1 = -1

k = -2/π

Question 29. [Tex]f(x)= \begin{cases} kx+1,x\leq5\\ 3x-5,x>5 \end{cases} \hspace{0.1cm}\hspace{0.1cm} [/Tex] at x = 5

Solution:

Continuity at x = 5

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

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

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

= 3(5) – 5 = 10

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

As, [Tex]\lim_{x \to 5^-} f(x) = \lim_{x \to 5^+} f(x)= f(5)   [/Tex]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

[Tex]f(x)= \begin{cases} 5,x\leq2\\ ax+b,2<x<10\\ 21,x\geq10 \end{cases}[/Tex]

is a continuous function

Solution:

Continuity at x = 2

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

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

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

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

2a + b = 5 ……………………(1)

Continuity at x = 10

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

= 10a + b

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

= 21

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

As, [Tex]\lim_{x \to 10^-} f(x) = \lim_{x \to 10^+} f(x)= f(10)   [/Tex]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 = [Tex]\lim_{h \to 0} g(c+h) = \lim_{h \to 0} (cos(c+h))\\ = \lim_{h \to 0}  [/Tex](cosc cosh sinc sinh)

= cosc cos0 − sinc sin0 = cosc

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

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

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

Continuity of h(x) = x2

Let’s check the continuity at x = c

Limit = [Tex]\lim_{x \to c} h(x) = \lim_{x \to c} (x^2)[/Tex]

= c2

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

As, [Tex]\lim_{x \to c} h(x) = h(c) = c^2[/Tex]

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 = [Tex]\lim_{x \to c} g(x) = \lim_{x \to c} (|x|)\\= \lim_{x \to c} (-x)\\ = -c[/Tex]

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

As, [Tex]\lim_{x \to c} g(x) = g(c) = -c[/Tex]

When c ≥ 0

Limit = [Tex]\lim_{x \to c} g(x) = \lim_{x \to c} (|x|)\\= \lim_{x \to c} (x)\\ = c[/Tex]

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

As, [Tex]\lim_{x \to c} g(x) = g(c) = c[/Tex]

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 = [Tex]\lim_{h \to 0} m(c+h) = \lim_{h \to 0} (cos(c+h))\\ = \lim_{h \to 0} [/Tex](cosc cosh sinc sinh)

= cosc cos0 − sinc sin0 = cosc

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

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

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 = [Tex]\lim_{x \to c} g(x) = \lim_{x \to c} (|x|)\\= \lim_{x \to c} (-x)\\ = -c[/Tex]

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

As, [Tex]\lim_{x \to c} g(x) = g(c) = -c[/Tex]

When c ≥ 0

Limit = [Tex]\lim_{x \to c} g(x) = \lim_{x \to c} (|x|)\\= \lim_{x \to c} (x)\\ = c[/Tex]

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

As, [Tex]\lim_{x \to c} g(x) = g(c) = c[/Tex]

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 = [Tex]\lim_{h \to 0} m(c+h) = \lim_{h \to 0} (sin(c+h))\\ = \lim_{h \to 0}  [/Tex](sinc cosh + cosc sinh

= sinc cos0 + cos csin0 = sinc

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

As, [Tex]\lim_{x \to c} m(x) = m(c) = sin c[/Tex]

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 = [Tex]\lim_{x \to c} g(x) = \lim_{x \to c} (|x|)\\= \lim_{x \to c} (-x)\\ = -c[/Tex]

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

As, [Tex]\lim_{x \to c} g(x) = g(c) = -c[/Tex]

When c ≥ 0

Limit = [Tex]\lim_{x \to c} g(x) = \lim_{x \to c} (|x|)\\= \lim_{x \to c} (x)\\ = c[/Tex]

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

As, [Tex]\lim_{x \to c} g(x) = g(c) = c[/Tex]

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 = [Tex]\lim_{x \to c} m(x) = \lim_{x \to c} (|x+1|)\\= \lim_{x \to c} -(x+1)[/Tex]

= -(c + 1)

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

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

When c ≥ -1

Limit = [Tex]\lim_{x \to c} m(x) = \lim_{x \to c} (|x|)\\= \lim_{x \to c} (x+1)[/Tex]

= c + 1

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

As, [Tex]\lim_{x \to c} m(x) [/Tex] = 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.



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