**Question 1.Examine the continuity of(i) f(x) = x**

^{3}+ 2 x

^{2}– x – 2 at x = -2Solution:

Given, f(x) = x

^{3}+ 2 x

^{2}– x – 2

f(x) is a polynomial function and hence it is continuous for all x ∈ R.

∴ f(x) is continuous at x = -2.

Solution:

= 8 for x = 3, at x = 3.

Solution:

f(3) = 8 ….(given)

∴ f(x) is discontinuous at x = 3.

**Question 2.Examine whether the function is continuous at the points indicated against them.(i) f(x) = x**

^{2}– 2x + 1, if x ≤ 2= 3x – 2, if x > 2, at x = 2.Solution:

= 20, for x = 1, at x = 1.

Solution:

Solution:

**Question 3.Find all the points of discontinuities of f(x) = [x] on the interval (-3, 2).Solution:**

f(x) = [x], x ∈ (-3, 2)

i.e., f(x) = -3, x ∈ (-3, -2)

= -2, x ∈ [-2, -1)

= -1, x ∈ [- 1, 0)

= 0, x ∈ [0, 1)

= 1, x ∈ [1, 2)

Similarly, f(x) is discontinuous at the points x = -1, x = 0, x = 1.

Thus all the integer values of x in the interval (-3, 2),

i.e., the points x = -2, x = -1, x = 0 and x = 1 are the required points of discontinuities.

**Question 4.Solution:**

**Question 5.Test the continuity of the following functions at the points or intervals indicated against them.Solution:**

**Question 6.Identify discontinuities for the following functions as either a jump or a removable discontinuity.**

It is a rational function and is discontinuous if

x – 7 = 0, i.e., x = 7

∴ f(x) is continuous for all x ∈ R, except at x = 7.

∴ f(7) is not defined.

Thus, limx→7f(x) exist but f(7) is not defined.

∴ f(x) has a removable discontinuity.

**(ii) f(x) = x ^{2} + 3x – 2, for x ≤ 4= 5x + 3, for x > 4.Solution:**

f(x) = x

^{2}+ 3x – 2, x ≤ 4

= 5x + 3, x > 4

f(x) is a polynomial function for both the intervals.

∴ f(x) is continuous for both the given intervals.

Let us test the continuity at x = 4.

∴ f(x) is discontinuous at x = 4.

∴ f(x) has a jump discontinuity at x = 4.

**(iii) f(x) = x ^{2} – 3x – 2, for x < -3 = 3 + 8x, for x > -3.Solution:**f(x) = x

^{2}– 3x – 2, x < -3 = 3 + 8x, x > -3

f(x) is a polynomial function for both the intervals.

∴ f(x) is continuous for both the given intervals.

Let us test the continuity at x = -3.

∴ f(x) is discontinuous at x = -3.

∴ f(x) has a jump discontinuity at x = -3

**(iv) f(x) = 4 + sin x, for x < π = 3 – cos x for x > π.Solution:**

f(x) = 4 + sin x, x < π = 3 – cos x, x > π

sin x and cos x are continuous for all x ∈ R.

4 and 3 are constant functions.

∴ 4 + sin x and 3 – cos x are continuous for all x ∈ R.

∴ f(x) is continuous for both the given intervals.

Let us test the continuity at x = π.

But f(π) is not defined.

∴ f(x) has a removable discontinuity at x = π.

**Question 7.Show that the following functions have a continuous extension to the point where f(x) is not defined. Also, find the extension.**

Here, f(0) is not defined.

Consider,

But f(0) is not defined.

∴ f(x) has a removable discontinuity at x = 0.

∴ The extension of the original function is

Consider,

But f(0) is not defined.

∴ f(x) has a removable discontinuity at x = 0.

∴ The extension of the original function is

∴ f(x) is continuous at x = 0.

Here, f(-1) is not defined.

Consider,

But f(-1) is not defined.

∴ f(x) has a removable discontinuity at x = -1.

∴ The extension of the original function is

**Question 8.Discuss the continuity of the following functions at the points indicated against them.Solution:**

**Question 9.Which of the following functions has a removable discontinuity? If it has a removable discontinuity, redefine the function so that it becomes continuous.Solution:**

**Question 10.**

Solution:

f(x) is continuous at x = 0, …..(given)

Solution:

f(x) is continuous at x = π, …..(given)

**Question 11.= k, for x = 0is continuous at x = 0, then find k.Solution:**

f(x) is continuous at x = 0 …..(given)

= k, for x = 0

is continuous at x = 0, then find k.

Solution:

f(x) is continuous at x = 0 …..(given)

= 4 for x = 0

= x^{2} + b – 3, for x < 0

is continuous at x = 0, find a and b.

Solution:

f(x) is continuous at x = 0 ……(given)

**(iv) For what values of a and b is the functionf(x) = ax + 2b + 18, for x ≤ 0= x**

^{2}+ 3a – b, for 0 < x ≤ 2 = 8x – 2, for x > 2,continuous for every x?Solution:

f(x) is continuous for every x …..(given)

∴ f(x) is continuous at x = 0 and x = 2.

As f(x) is continuous at x = 0,

limx→0−f(x)=limx→0+f(x)

∴ (2)

^{2}+ 3a – b = 8(2) – 2

∴ 4 + 3a – b = 14

∴ 3a – b = 10 …….(ii)

Subtracting (i) from (ii), we get

2a = 4

∴ a = 2

Substituting a = 2 in (i), we get

2 – b = 6

∴ b = -4

∴ a = 2 and b = -4

**(v) For what values of a and b is the function= 2x – a + b, for x ≥ 3continuous for every x on R?Solution:**

f(x) is continuous for every x on R …..(given)

∴ f(x) is continuous at x = 2 and x = 3.

As f(x) is continuous at x = 2,

∴ a(3)

^{2}– b(3) + 3 = 2(3) – a + b

∴ 9a – 3b + 3 = 6 – a + b

∴ 10a – 4b = 3 …..(ii)

Multiplying (i) by 2, we get

8a – 4b = 2 ….(iii)

Subtracting (ii) from (iii), we get

**Question 12.Discuss the continuity of f on its domain, wheref(x) = |x + 1|, for -3 ≤ x ≤ 2= |x – 5|, for 2 < x ≤ 7Solution:**

**Question 13.Solution:**

**Question 14.Determine the values of p and q such that the following function is continuous on the entire real number line.f(x) = x + 1, for 1 < x < 3= x**

^{2}+ px + q, for |x – 2| ≥ 1.Solution:

|x – 2| ≥ 1

∴ x – 2 ≥ 1 or x – 2 ≤ -1

∴ x ≥ 3 or x ≤ 1

∴ f(x) = x

^{2}+ px + q for x ≥ 3 as well as x ≤ 1

Thus, f(x) = x

^{2}+ px + q; x ≤ 1

= x + 1; 1 < x < 3 = x

^{2}+ px + q; x > 3

f(x) is continuous for all x ∈ R.

∴ f(x) is continuous at x = 1 and x = 3.

As f(x) is continuous at x = 1,

Subtracting (i) from (ii), we get

2p = -6

∴ p = -3

Substituting p = -3 in (i), we get

-3 + q = 1

∴ q = 4

∴ p = -3 and q = 4

**Question 15.Show that there is a root for the equation 2x ^{3} – x – 16 = 0 between 2 and 3.Solution:**

Let f(x) = 2x

^{3}– x – 16

f(x) is a polynomial function and hence it is continuous for all x ∈ R.

A root of f(x) exists, if f(x) = 0 for at least one value of x.

f(2) = 2(2)

^{3}– 2 – 16 = -2 < 0

f(3) = 2(3)

^{3}– 3 – 16 = 35 > 0

∴ f(2) < 0 and f(3) > 0

∴ By intermediate value theorem,

there has to be point ‘c’ between 2 and 3 such that f(c) = 0.

∴ There is a root of the given equation between 2 and 3.

**Question 16.Show that there is a root for the equation x ^{3} – 3x = 0 between 1 and 2.Solution:**

Let f(x) = x

^{3}– 3x

f(x) is a polynomial function and hence it is continuous for all x ∈ R.

A root of f(x) exists, if f(x) = 0 for at least one value of x.

f(1) = (1)

^{3}– 3(1) = -2 < 0

f(2) = (2)

^{3}– 3(2) = 2 > 0

∴ f(1) < 0 and f(2) > 0

∴ By intermediate value theorem,

there has to be point ‘c’ between 1 and 2 such that f(c) = 0.

There is a root of the given equation between 1 and 2.

**Question 17.Let f(x) = ax + b (where a and b are unknown)= x**

^{2}+ 5 for x ∈ RFind the values of a and b, so that f(x) is continuous at x = 1.Solution:

f(x) = x

^{2}+ 5, x ∈ R

∴ f(1) = 1 + 5 = 6

If f(x) = ax + b is continuous at x = 1, then

f(1) = limx→1(ax+b) = a + b

∴ 6 = a + b where, a, b ∈ R

∴ There are infinitely many values of a and b.

**Question 18.Activity: Suppose f(x) = px + 3 for a ≤ x ≤ b= 5 x**

^{2}– q for b < x ≤ cFind the condition on p, q, so that f(x) is continuous on [a, c], by filling in the boxes.Solution: