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Class 12 NCERT Solutions- Mathematics Part I – Chapter 5 Continuity And Differentiability – Exercise 5.1 | Set 1
  • Last Updated : 30 Apr, 2021

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) f(x) = \frac{1}{x-5}, 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) f(x) = \frac{x^2-25}{x+5}, 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

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

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 = \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.

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

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

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. 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}

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. f(x)= \begin{cases} \frac{|x|}{x}, \hspace{0.2cm}x\neq0\\ 0,\hspace{0.2cm}x=0 \end{cases}

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. f(x)= \begin{cases} \frac{x}{|x|}, \hspace{0.2cm}x<0\\ -1,\hspace{0.2cm}x\geq0 \end{cases}

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

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

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. f(x)= \begin{cases} x^{10}-1, \hspace{0.2cm}x\leq1\\ x^2,\hspace{0.2cm}x>1 \end{cases}

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

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

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. 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}

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. 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}

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. 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}

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

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

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

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