NCERT Solutions for Maths Class 12 Exercise 5.1

NCERT Solutions for Maths Class 12 Exercise 5.1

Hello Students! In this post, you will find the complete NCERT Solutions for Maths Class 12 Exercise 5.1.

You can download the PDF of NCERT Books Maths Chapter 5 for your easy reference while studying NCERT Solutions for Maths Class 12 Exercise 5.1.

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NCERT Solutions for Maths Class 12 Exercise 5.2

NCERT Solutions for Maths Class 12 Exercise 5.3

NCERT Solutions for Maths Class 12 Exercise 5.4

NCERT Solutions for Maths Class 12 Exercise 5.5

NCERT Solutions for Maths Class 12 Exercise 5.6

NCERT Solutions for Maths Class 12 Exercise 5.7

 

NCERT Solutions for Maths Class 12 Exercise 5.1

 

Maths Class 12 Ex 5.1 Question 1.

Prove that the function f(x) = 5x – 3 is continuous at x = 0, at x = – 3 and at x = 5.

Solution:

The given function is f(x) = 5x – 3.

(i) At x = 0, limx→0 f(x) = limx→0 (5x – 3) = 5 × 0 – 3 = – 3 and f(0) = 5 × 0 – 3 = – 3

Since limx→0 f(x) = f(x), therefore, f is continuous at x = 0.

(ii) At x = – 3, limx→3 f(x) = limx→–3 (5x – 3) = 5 × (–3) – 3 = –18 and f(– 3) = –18
Since limx→–3 f(x) = f(x), therefore, f is continuous at x = –3.

(iii) At x = 5, limx→5 f(x) = limx→5 (5x – 3) = 5 × 5 – 3 = 22 and f(5) = 5 × 5 – 3 = 22
Since limx→5 f(x) = f(x), therefore, f is continuous at x = 5.

Maths Class 12 Ex 5.1 Question 2.

Examine the continuity of the function f(x) = 2x² – 1 at x = 3.

Solution:

The given function is f(x) = 2x² – 1.
At x = 3, limx→3 f(x) = limx→3 (2x² – 1) = 2 × (3)2 – 1 = 17 and f(3) = 2 × (3)2 – 1 = 17
Since limx→3 f(x) = f(x), therefore, f is continuous at x = 3.

Maths Class 12 Ex 5.1 Question 3.

Examine the following functions for continuity.
(a) f(x) = x – 5
(b) f(x) = 1/(x – 5), x ≠ 5
(c) f(x) = (x2 – 25)/(x + 5), x ≠ 5
(d) f(x) = |x – 5|

Solution:
(a) f(x) = x – 5 => x – 5 is a polynomial.

At x = c R, limx→c f(x) = limx→c (x – 5) = c – 5 and f(c) = c – 5

Since limx→c f(x) = f(c), therefore, f is continuous at every c R.


Maths Class 12 Ex 5.1 Question 4.

Prove that the function f(x) = xn is continuous at x = n, where n is a positive integer.

Solution:
The given function is f(x) = xn which is a polynomial.

At x = n N, limx→n f(x) = limx→n xn  = xn and f(x) = xn

Since limx→n f(x) = f(x), therefore, f is continuous at x = n N.

Maths Class 12 Ex 5.1 Question 5.

Is the function f defined by

 

continuous at x = 0? At x = 1? At x = 2?

Solution:
(i) At x = 0, limx→0- f(x) = limx→0- x = 0

And limx→0+ f(x) = limx→0+ x = 0 f(0) = 0
f is continuous at x = 0.

(ii) At x = 1, limx→1- f(x) = limx→1- (x) = 1

And limx→1+ f(x) = limx→1+ (x) = 5
limx→1- f(x) ≠ limx→1+ f(x)
f is discontinuous at x = 1.

(iii) At x = 2, limx→2 f(x) = 5, f(2) = 5
f is continuous at x = 2.

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

Maths Class 12 Ex 5.1 Question 6.


Solution:



Maths Class 12 Ex 5.1 Question 7.



Solution:


Maths Class 12 Ex 5.1 Question 8.

Test the continuity of the function f (x) at x = 0


Solution: 


Maths Class 12 Ex 5.1 Question 9.

Solution:


Maths Class 12 Ex 5.1 Question 10.



Solution:


Maths Class 12 Ex 5.1 Question 11.

Solution:


 

Maths Class 12 Ex 5.1 Question 12.



Solution:



Maths Class 12 Ex 5.1 Question 13.

Is the function defined by

 
a continuous function?

Solution:

At x = 1, L.H.L. = limx→1- f(x) = limx→1- (x + 5) = 6,
R.H.L. = limx→1+ f(x) = limx→1+ (x – 5) = –4
f(1) = 1 + 5 = 6,
f(1) = L.H.L. ≠ R.H.L.
Therefore, f is not continuous at x = 1.
At x = c < 1, limx→c (x + 5) = c + 5 = f(c)
At x = c > 1, limx→c (x – 5) = c – 5 = f(c)
f is continuous at all points x R except x = 1.

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

Maths Class 12 Ex 5.1 Question 14.


Solution:

In the interval 0 ≤ x ≤ 1, f(x) = 3; therefore, f is continuous in this interval.
At x = 1, L.H.L. = lim x→1- f(x) = 3,
R.H.L. = limx→1+ f(x) = 4, therefore, f is discontinuous at x = 1.
At x = 3, L.H.L. = limx→3- f(x) = 4,
R.H.L. = limx→3+ f(x) = 5, therefore, f is discontinuous at x = 3.
Hence, f is not continuous at x = 1 and x = 3.

Maths Class 12 Ex 5.1 Question 15.


Solution:

At x = 0, L.H.L. = limx→0- (2x) = 0,
R.H.L. = limx→0+ (0) = 0, f(0) = 0
Therefore, f is continuous at x = 0.
At x = 1, L.H.L. = limx→1- (0) = 0,
R.H.L. = limx→1+ (4x) = 4 and f(1) = 0.
f(1) = L.H.L. ≠ R.H.L.
f is not continuous at x = 1.

When x < 0, f(x) = 2x is a polynomial.

It is continuous at all points x < 0.

When x > 1, f(x) = 4x is a polynomial.

It is continuous at all points x > 1.
When 0 ≤ x ≤ 1, f(x) = 0 is a continuous function.

Hence, the point of discontinuity is x = 1.

Maths Class 12 Ex 5.1 Question 16.


Solution:

At x = –1, L.H.L. = limx→1- f(x) = –2, f(–1) = –2,
R.H.L. = limx→1+ f(x) = –2
Therefore, f is continuous at x = –1.
At x = 1, L.H.L. = limx→1- f(x) = 2, R.H.L. = limx→1+ f(x) = 2 and f(1) = 2
Therefore, f is continuous at x = 1.
Hence, f is a continuous function.

Maths Class 12 Ex 5.1 Question 17.

Find the relationship between a and b so that the function f defined by

is continuous at x = 3.


Solution:


At x = 3, L.H.L. = limx→3- (ax + 1) = 3a + 1,
R.H.L. = limx→3+ (bx + 3) = 3b + 3 and f(3) = 3a + 1
f is continuous if L.H.L. = R.H.L. = f(3)
3a + 1 = 3b + 3 or 3(a – b) = 2
a – b = 2/3 or a = b + 2/3, for any arbitrary value of b.
Therefore, the value of a is corresponding to the value of b.

Maths Class 12 Ex 5.1 Question 18.

For what value of λ is the function defined by

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

Solution:

At x = 0, L.H.L. = limx→0- λ(x² – 2x) = 0,
R.H.L. = limx→0+ (4x + 1) = 1 and f(0) = 0
f(0) = L.H.L. ≠ R.H.L.
Therefore, f is not continuous at x = 0, for every value of λ
R.
At x = 1, limx→1 f(x) = limx→1 (4x + l) = f(1), therefore, f is continuous at x = 1.
Hence, f is not continuous at x = 0 for any value of λ but f is continuous at x = 1 for all values of λ.

Maths Class 12 Ex 5.1 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:
Let c be an integer, [c – h] = c – 1, [c + h] = c, [c] = c, g(x) = x – [x].
At x = c, L.H.L. = limx→c- (x – [x]) = limh→0 [(c – h) – (c – 1)]
= limh→0 (c – h – c + 1)) = 1          [
[c – h] = c – 1]
R.H.L. = limx→c+ (x – [x]) = limh→0 (c + h – [c + h])
= limh→0 (c + h – c) = 0 and f(c) = c – [c] = 0.
Thus, L.H.L. ≠ R.H.L. = f (c), therefore, f is not continuous at integral points.

Maths Class 12 Ex 5.1 Question 20.

Is the function defined by f(x) = x² – sin x + 5 continuous at x = π?

Solution:


Maths Class 12 Ex 5.1 Question 21.

Discuss the continuity of the following functions:
(a) f(x) = sin x + cos x
(b) f(x) = sin x – cos x
(c) f(x) = sin x · cos x

Solution:



Maths Class 12 Ex 5.1 Question 22.

Discuss the continuity of the cosine, cosecant, secant and cotangent functions.

Solution:


Maths Class 12 Ex 5.1 Question 23.

Find all points of discontinuity of f, where


Solution:


Maths Class 12 Ex 5.1 Question 24.

Determine if f defined by

 

is a continuous function?

Solution:

Maths Class 12 Ex 5.1 Question 25.

Examine the continuity of f, where f is defined by

 

Solution:

Find the values of k so that the function is continuous at the indicated point in Questions 26 to 29.

Maths Class 12 Ex 5.1 Question 26.



Solution:


Maths Class 12 Ex 5.1 Question 27.



Solution:


Maths Class 12 Ex 5.1 Question 28.



Solution:


Maths Class 12 Ex 5.1 Question 29.



Solution:


Maths Class 12 Ex 5.1 Question 30.

Find the values of a and b such that the function defined by

is a continuous function.

Solution:


Maths Class 12 Ex 5.1 Question 31.

Show that the function defined by f(x) = cos (x²) is a continuous function.

Solution:
We have, f(x) = cos x²,

Let g(x) = cos x and h(x) = x²
goh(x) = g(h(x)) = cos x²
Now, g and h both are continuous
x R.
Hence, f(x) = goh(x) = cos x² is also a continuous function at all x
R.

Maths Class 12 Ex 5.1 Question 32.

Show that the function defined by f(x) = |cos x| is a continuous function.

Solution:
We have, f(x) = |cos x|,

Let g(x) =|x| and h(x) = cos x,

goh(x) = g(h(x)) = g(cos x) = |cos x|
Now, g(x) = |x| and h(x) = cos x both are continuous for all values of x
R.
(goh)(x) is also continuous.
Hence, f(x) = goh(x) = |cos x| is continuous for all values of x
R.

Maths Class 12 Ex 5.1 Question 33.

Examine that sin |x| is a continuous function.

Solution:
We have, f(x) = sin |x|,

Let g(x) = sin x, h(x) = |x|, goh(x) = g(h(x)) = g(|x|) = sin |x| = f(x)
Now, g(x) = sin x and h(x) = |x| both are continuous for all x
R.
Hence, f(x) = goh(x) = sin |x| is continuous at all x R.

Maths Class 12 Ex 5.1 Question 34.

Find all the points of discontinuity of f defined by f(x) = |x| – |x + 1|.

Solution:
We have, f(x) = |x| – |x + 1|, when x < –1,
f(x) = –x – [– (x + 1)] = –x + x + 1 = 1
When –1 ≤ x < 0, f(x) = –x – (x + 1) = –2x – 1,
When x ≥ 0, f(x) = x – (x + 1) = –1


Related Links:

NCERT Solutions for Maths Class 12 Exercise 5.2

NCERT Solutions for Maths Class 12 Exercise 5.3

NCERT Solutions for Maths Class 12 Exercise 5.4

NCERT Solutions for Maths Class 12 Exercise 5.5

NCERT Solutions for Maths Class 12 Exercise 5.6

NCERT Solutions for Maths Class 12 Exercise 5.7

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