# 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.

**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 =

= (5(0) – 3) = -3

Right limit =

= (5(0) – 3)= -3

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

As, ,

Hence, the function is continuous at x = 0.

**Continuity at x = -3**

Left limit =

= (5(-3) – 3) = -18

Right limit =

= (5(-3) – 3) = -18

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

As,

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

**Continuity at x = 5**

Left limit =

= (5(5) – 3) = 22

Right limit =

= (5(5) – 3) = 22

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

As,

Hence, the function is continuous at x = 5.

**Question
2. Examine the continuity of the function f(x) = 2x**^{2 }**–
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 =

= (2(3)2 – 1) = 17

Right limit =

= (2(3)2 – 1) = 17

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

As,

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 =

= (c – 5) = c – 5

Right limit =

= (c – 5) = c – 5

Function value at x = c, f(c) = c – 5

As, for any real number c

Hence, the function is continuous at every real number.

**(b) **, x ≠ 5

**Solution:**

To prove the continuity of the function f(x) = , 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 =

Right limit =

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

As, for any real number c

Hence, the function is continuous at every real number.

**(c), x ≠ -5**

**Solution:**

To prove the continuity of the function f(x) = , 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 =

= c – 5

Right limit =

= c – 5

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

= c – 5

As, , 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 =

= -(c – 5)

= 5 – c

Right limit =

= -(c – 5)

= 5 – c

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

As,

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

**When c > 5**

Left limit =

= (c – 5)

Right limit =

= (c – 5)

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

As, ,

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

**When c = 5**

Left limit =

Right limit =

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

As,

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) = x**^{n}** 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 =

= nn

Right limit =

= nn

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

As,

Hence, the function is continuous at x = n.

**Question 5. Is the function f defined by**

**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 =

Right limit =

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

As,

Hence, the function is continuous at x = 0.

**Continuity at x = 1**

Left limit =

Right limit =

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

As, ,

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

**Continuity at x = 2**

Left limit =

Right limit =

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

As, ,

**Therefore, the function is continuous at x = 2.**

**Question 6. **

**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 =

= (2(2) + 3)

= 7

Right limit =

= (2(2) – 3)

= 1

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

As,

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

**Question 7. **

**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 =

= (-(-3) + 3)

= 6

Right limit =

= (-2(-3))

= 6

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

As,

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

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

Left limit =

= (-2(3))

= -6

Right limit =

= (6(3) + 2)

= 20

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

As,

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

**Question 8. **

**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) = = -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 =

Right limit =

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

As,

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

**Question 9. **

**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) = = -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 =

Right limit =

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

As,

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. **

**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 =

= 1 + 1

= 2

Right limit =

= 1 + 1

= 2

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

As,

Hence, the function is continuous at x = 1.

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

**Question 11. **

**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 =

Right limit =

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

As,

Hence, the function is continuous at x = 2.

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

**Question 12. **

**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 =

= 1 – 1

= 0

Right limit =

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

As,

**Hence, the function is discontinuous at x = 1.**

**Question 13. Is the function defined by**

**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 =

= (1 + 5)

= 6

Right limit =

= (1 – 5)

= -4

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

As,

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

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

**Question 14. **

**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 =

Right limit =

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

As,

Hence, the function is discontinuous at x = 1.

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

Left limit =

Right limit =

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

As,

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. **

**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 =

Right limit =

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

As,

Hence, the function is continuous at x = 0.

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

Left limit =

Right limit =

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

As,

Hence, the function is discontinuous at x = 1.

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

**Question 16. **

**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 =

Right limit =

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

As,

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

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

Left limit =

Right limit =

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

As,

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**

**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:

Continuity at x = 3,

Left limit =

= (a(3) + 1)

= 3a + 1

Right limit =

= (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 **

**by**

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

**Solution:**

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

**Continuity at x = 0,**

Left limit =

= λ(02– 2(0)) = 0

Right limit =

= λ4(0) + 1 = 1

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

As, 0 = 1 cannot be possible

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

But here,

**Continuity at x = 1,**

Left limit =

= (4(1) + 1) = 5

Right limit =

= 4(1) + 1 = 5

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

As,

**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 =

= (c – (c – 1)) = 1

Right limit =

= (c – c) = 0

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

As,

**Hence, the function is discontinuous at integral.**

**c be not an integer**

Let’s check the continuity at x = c,

Left limit =

= (c – (c – 1)) = 1

Right limit =

= (c – (c – 1)) = 1

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

As,

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

**Question
20. Is the function defined by f(x) = x**^{2}** –
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 =

= (π2 – sinπ + 5) = π2 + 5

Right limit =

= (π2 – sinπ + 5) = π2 + 5

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

As,

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

So,

(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

((sinc cosh + cosc sinh) + (cosc cosh − sinc sinh))

= ((sinc cos0 + cosc sin0) + (cosc cos0 − sinc sin0))

cos 0 = 1 and sin 0 = 0

= (sinc + cosc) = f(c)

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

As, = 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

So,

(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

((sinc cosh + cosc sinh) − (cosc cosh − sinc sinh))

= ((sinc cos0 + cosc sin0) − (cosc cos0 − sinc sin0))

cos 0 = 1 and sin 0 = 0

= (sinc − cosc) = f(c)

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

As, = 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

So,

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

((sinc cosh + cosc sinh) × (cosc cosh − sinc sinh))

= ((sinc cos0 + cosc sin0) × (cosc cos0 − sinc sin0))

cos 0 = 1 and sin 0 = 0

= (sinc × cosc) = f(c)

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

As, = 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

So,

Using the trigonometric identities, we get

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

(cosc cosh − sinc sinh)

= (cosc cos0 − sinc sin0)

cos 0 = 1 and sin 0 = 0

= (cosc) = f(c)

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

As, = f(c) = (cosc)

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

**Continuity of cosecant**

Here,

f(x) = cosec x =

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

Let’s take, x = c + h

When x⇢c then h⇢0

So,

Using the trigonometric identities, we get

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

cos 0 = 1 and sin 0 = 0

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

As,

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

**Continuity of secant**

Here,

f(x) = sec x =

Let’s take, x = c + h

When x⇢c then h⇢0

So,

Using the trigonometric identities, we get

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

cos 0 = 1 and sin 0 = 0

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

As,

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

**Continuity of cotangent**

Here,

f(x) = cot x =

Let’s take, x = c+h

When x⇢c then h⇢0

So,

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

cos 0 = 1 and sin 0 = 0

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

As,

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

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

**Solution:**

Here,

From the two continuous functions g and h, we get

= continuous when h(x) ≠ 0

For x < 0, f(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 =

Right limit =

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

As,

Hence, the function is continuous at x = 0.

**Hence, the function is continuous for any real number.**

**Question 24. Determine if f defined by**

**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 ≤ ≤ 1 which is a finite number.

Limit =

= (02 ×(finite number)) = 0

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

As,

Hence, the function is continuous for any real number.

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

**Solution:**

**Continuity at x = 0,**

Left limit =

= (sin0 − cos0) = 0 − 1 = −1

Right limit =

= (sin0 − cos0) = 0 − 1 = −1

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

As,

Hence, the function is continuous at x = 0.

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

Left limit =

= (sinc − cosc)

Right limit =

= (sinc − cosc)

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

As,

So concluding the results, we get

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

**Question 26. at x = π/2.**

**Solution:**

Continuity at x = π/2

Let’s take x =

When x⇢π/2 then h⇢0

Substituting x = +h, we get

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

Limit =

Function value at x = = 3

As, should satisfy, for f(x) being continuous

k/2 = 3

k = 6

**Question 27. at x = 2**

**Solution:**

Continuity at x = 2

Left limit =

= k(2)2 = 4k

Right limit =

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

As, should satisfy, for f(x) being continuous

4k = 3

k = 3/4

**Question 28. at x = π**

**Solution:**

Continuity at x = π

Left limit =

= k(π) + 1

Right limit =

= cos(π) = -1

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

As, should satisfy, for f(x) being continuous

kπ + 1 = -1

k = -2/π

**Question 29. at x = 5**

**Solution:**

Continuity at x = 5

Left limit =

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

Right limit =

= 3(5) – 5 = 10

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

As, 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**

**is a continuous function**

**Solution:**

**Continuity at x = 2**

Left limit =

Right limit =

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

As, should satisfy, for f(x) being continuous at x = 2

2a + b = 5 ……………………(1)

**Continuity at x = 10**

Left limit =

= 10a + b

Right limit =

= 21

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

As, 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 (x**^{2}**)
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 = (cosc cosh − sinc sinh)

= cosc cos0 − sinc sin0 = cosc

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

As,

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

**Continuity of h(x) = x2**

Let’s check the continuity at x = c

Limit =

= c2

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

As,

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 =

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

As,

**When c ≥ 0**

Limit =

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

As,

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 = (cosc cosh − sinc sinh)

= cosc cos0 − sinc sin0 = cosc

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

As,

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 =

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

As,

**When c ≥ 0**

Limit =

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

As,

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 = (sinc cosh + cosc sinh)

= sinc cos0 + cos csin0 = sinc

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

As,

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 =

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

As,

**When c ≥ 0**

Limit =

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

As,

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 =

= -(c + 1)

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

As,

**When c ≥ -1**

Limit =

= c + 1

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

As, = 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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