Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 8 Continuity Ex 8.1 Questions and Answers.

## Maharashtra State Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1

Question 1.

Examine the continuity of

(i) f(x) = x^{3} + 2x^{2} – x – 2 at x = -2

Solution:

Given, f(x) = x^{3} + 2x^{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.

(ii) f(x) = sin x, for x ≤ \(\frac{\pi}{4}\)

= cos x, for x > \(\frac{\pi}{4}\), at x = \(\frac{\pi}{4}\)

Solution:

(iii) f(x) = \(\frac{x^{2}-9}{x-3}\), for x ≠ 3

= 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^{3} – 2x + 1, if x ≤ 2

= 3x – 2, if x > 2, at x = 2.

Solution:

(ii) f(x) = \(\frac{x^{2}+18 x-19}{x-1}\), for x ≠ 1

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

Solution:

(iii) f(x) = \(\frac{x}{\tan 3 x}+2\), for x < 0

= \(\frac{7}{3}\), for x ≥ 0, at x = 0.

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.

Discuss the continuity of the function f(x) = |2x + 3|, at x = \(\frac{-3}{2}\).

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.

(i) f(x) = \(\frac{x^{2}-10 x+21}{x-7}\)

Solution:

Given, f(x) = \(\frac{x^{2}-10 x+21}{x-7}\)

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, \(\lim _{x \rightarrow 7} \mathrm{f}(x)\) 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.

(i) f(x) = \(\frac{1-\cos 2 x}{\sin x}\), for x ≠ 0.

Solution:

f(x) = \(\frac{1-\cos 2 x}{\sin x}\), for x ≠ 0

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

f(x) = \(\frac{1-\cos 2 x}{\sin x}\) for x ≠ 0

= 0 for x = 0

∴ f(x) is continuous at x = 0.

(ii) f(x) = \(\frac{3 \sin ^{2} x+2 \cos x(1-\cos 2 x)}{2\left(1-\cos ^{2} x\right)}\), for x ≠ 0.

Solution:

f(x) = \(\frac{3 \sin ^{2} x+2 \cos x(1-\cos 2 x)}{2\left(1-\cos ^{2} x\right)}\), for x ≠ 0

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

f(x) = \(\frac{3 \sin ^{2} x+2 \cos x(1-\cos 2 x)}{2\left(1-\cos ^{2} x\right)}\), x ≠ 0

= \(\frac{7}{2}\), x = 0

∴ f(x) is continuous at x = 0.

(iii) f(x) = \(\frac{x^{2}-1}{x^{3}+1}\), for x ≠ -1

Solution:

f(x) = \(\frac{x^{2}-1}{x^{3}+1}\), for x ≠ -1

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

f(x) = \(\frac{x^{2}-1}{x^{3}+1}\), x ≠ -1

= \(-\frac{2}{3}\), x = -1

∴ f(x) is continuous at x = \(-\frac{2}{3}\)

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.

(i) If f(x) = \(\frac{\sqrt{2+\sin x}-\sqrt{3}}{\cos ^{2} x}\), for x ≠ \(\frac{\pi}{2}\), is continuous at x = \(\frac{\pi}{2}\) then find f(\(\frac{\pi}{2}\)).

Solution:

f(x) is continuous at x = \(\frac{\pi}{2}\), …..(given)

(ii) If f(x) = \(\frac{\cos ^{2} x-\sin ^{2} x-1}{\sqrt{3 x^{2}+1}-1}\) for x ≠ 0, is continuous at x = 0 then find f(0).

Solution:

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

(iii) If f(x) = \(\frac{4^{x-\pi}+4^{\pi-x}-2}{(x-\pi)^{2}}\) for x ≠ π, is continuous at x = π, then find f(π).

Solution:

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

Question 11.

(i) If f(x) = \(\frac{24^{x}-8^{x}-3^{x}+1}{12^{x}-4^{x}-3^{x}+1}\), for x ≠ 0

= k, for x = 0

is continuous at x = 0, then find k.

Solution:

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

(ii) If f(x) = \(\frac{5^{x}+5^{-x}-2}{x^{2}}\), for x ≠ 0

= k, for x = 0

is continuous at x = 0, then find k.

Solution:

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

(iii) If f(x) = \(\frac{\sin 2 x}{5 x}\) – a, for x > 0

= 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 function

f(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,

\(\lim _{x \rightarrow 0^{-}} \mathrm{f}(x)=\lim _{x \rightarrow 0^{+}} \mathrm{f}(x)\)

∴ (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

f(x) = \(\frac{x^{2}-4}{x-2}\), for x < 2

= ax^{2} – bx + 3, for 2 ≤ x < 3

= 2x – a + b, for x ≥ 3

continuous 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

-2a = -1

∴ a = \(\frac{1}{2}\)

Substituting a = \(\frac{1}{2}\) in (i), we get

4(\(\frac{1}{2}\)) – 2b = 1

∴ 2 – 2b = 1

∴ 1 = 2b

∴ b = \(\frac{1}{2}\)

∴ a = \(\frac{1}{2}\) and b = \(\frac{1}{2}\)

Question 12.

Discuss the continuity of f on its domain, where

f(x) = |x + 1|, for -3 ≤ x ≤ 2

= |x – 5|, for 2 < x ≤ 7

Solution:

Question 13.

Discuss the continuity of f(x) at x = \(\frac{\pi}{4}\) where,

f(x) = \(\frac{(\sin x+\cos x)^{3}-2 \sqrt{2}}{\sin 2 x-1}\), for x ≠ \(\frac{\pi}{4}\)

= \(\frac{3}{\sqrt{2}}\), for x = \(\frac{\pi}{4}\)

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 ∈ R

Find 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) = \(\lim _{x \rightarrow 1}(a x+b)\) = 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

= 5x^{2} – q for b < x ≤ c

Find the condition on p, q, so that f(x) is continuous on [a, c], by filling in the boxes.

Solution: