**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**.

All the schools affiliated with CBSE, follow the NCERT books for all subjects. You can check your syllabus from **NCERT Syllabus for Mathematics Class 12**.

If you want to recall **All Maths Formulas for Class 12**, you can find it by clicking this link.

If you want to recall **All** **Maths Formulas for Class 11**, you can find it by clicking this link.

**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, lim_{x→0} f(x)
= lim_{x→0} (5x – 3) = 5 × 0 – 3 = – 3 and f(0) = 5 × 0 – 3 = – 3

Since
lim_{x→0} f(x)
= f(x), therefore,
f is continuous
at x = 0.

**(ii)** At x = – 3, lim_{x→3} f(x)
= lim_{x→–3} (5x – 3) = 5 × (–3) – 3 = –18 and f(– 3) = –18

Since
lim_{x→–3} f(x)
= f(x), therefore,
f is continuous
at x = –3.

**(iii)** At x = 5, lim_{x→5} f(x)
= lim_{x→5} (5x – 3) = 5 × 5 – 3 = 22 and f(5) = 5 × 5 – 3 = 22

Since
lim_{x→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, lim_{x→3} f(x) = lim_{x→3} (2x² – 1) = 2 ×
(3)^{2} – 1 = 17 and f(3) = 2 × (3)^{2} – 1 = 17

Since
lim_{x→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) = (x

^{2}– 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, lim_{x→c} f(x) = lim_{x→c} (x
– 5) = c – 5 and f(c) = c – 5

Since
lim_{x→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) = x^{n}is continuous at x = n, where n is a positive integer.

**Solution:**

The given
function is f(x) = x^{n} which is a polynomial.

At x = n ∈ N, lim_{x→n} f(x) = lim_{x→n} x^{n} = x^{n} and f(x) = x^{n}

Since
lim_{x→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 bycontinuous at x =
0? At x = 1? At x = 2?

**Solution:**

**(i)** At x = 0, lim_{x→0-} f(x) = lim_{x→0-} x
= 0

And lim_{x→0+} f(x)
= lim_{x→0+} x = 0 ⇒ f(0) = 0

∴ f is continuous at x = 0.

**(ii)** At x = 1, lim_{x→1-} f(x) = lim_{x→1-} (x)
= 1

And lim_{x→1+} f(x)
= lim_{x→1+ }(x) = 5

∴ lim_{x→1-} f(x) ≠ lim_{x→1+} f(x)

∴ f is discontinuous at x = 1.

**(iii)** At x = 2, lim_{x→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**Solution:**

At x = 1, L.H.L. =
lim_{x→1-} f(x) = lim_{x→1-} (x + 5) = 6,

R.H.L. = lim_{x→1+} f(x) = lim_{x→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, lim_{x→c} (x + 5) = c + 5 = f(c)

At x = c > 1, lim_{x→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. = lim_{x→1+} f(x) = 4, therefore, f is discontinuous at x =
1.

At x = 3, L.H.L. = lim_{x→3-} f(x) = 4,

R.H.L. = lim_{x→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. = lim_{x→0-} (2x) = 0,

R.H.L. = lim_{x→0+} (0) = 0, f(0) = 0

Therefore, f is continuous at x = 0.

At x = 1, L.H.L. = lim_{x→1-} (0) = 0,

R.H.L. = lim_{x→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. = lim_{x→1-} f(x) = –2, f(–1) = –2,

R.H.L. = lim_{x→1+} f(x) = –2

Therefore, f is continuous at x = –1.

At x = 1, L.H.L. = lim_{x→1-} f(x) = 2, R.H.L. = lim_{x→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.
= lim

_{x→3-}(ax + 1) = 3a + 1,

R.H.L. = lim

_{x→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
Î» is the function defined bycontinuous at x = 0? What about continuity at x = 1?

**Solution:**

At x = 0, L.H.L.
= lim_{x→0-} Î»(x² – 2x) = 0,

R.H.L. = lim_{x→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 Î» ∈ R.

At x = 1, lim_{x→1} f(x) = lim_{x→1} (4x + l) = f(1),
therefore, f is continuous at x = 1.

Hence, f is not continuous at x = 0 for any value of Î» but f is continuous at x
= 1 for all values of Î».

**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. = lim_{x→c-} (x – [x]) = lim_{h→0} [(c
– h) – (c – 1)]

= lim_{h→0} (c – h – c + 1)) = 1 [∵ [c – h] = c – 1]

R.H.L. = lim_{x→c+} (x – [x]) = lim_{h→0} (c + h – [c
+ h])

= lim_{h→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 = Ï€?

**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

**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.**

**Maths Class
12 Ex 5.1 Question 28.**

**Maths Class
12 Ex 5.1 Question 29.**

**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.

**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**