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 =

= (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) = 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 =

= (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.

(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) = 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 =

= nn

Right limit =

= nn

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

As,

Hence, the function is continuous at x = n.

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 =

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.

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}.

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.

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

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