# Class 12 NCERT Solutions- Mathematics Part I – Chapter 5 Continuity And Differentiability – 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(0

^{2}– 2(0)) = 0Right 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 integerLet’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 integerLet’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) = x

^{2}– sin x + 5Let’s substitute, x = π+h

When x⇢π, Continuity at x = π

Left limit =

= (π

^{2}– sinπ + 5) = π^{2 }+ 5Right limit =

= (π

^{2}– sinπ + 5) = π^{2 }+ 5Function value at x = π, f(π) = π

^{2}– sin π + 5 = π^{2}+ 5As,

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+coscAs, = 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−coscAs, = 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×coscAs, = 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 cosineHere,

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 cosecantHere,

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 secantHere,

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 cotangentHere,

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

### is a continuous function?

**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 =

= (0

^{2}×(finite number)) = 0Function 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 = −1Right limit =

= (

sin0 −cos0) = 0 − 1 = −1Function 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.

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

### 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 }= 4kRight limit =

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

^{2}= 4kAs, 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 = 2Left 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 = 10Left 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) = x

^{2}g(h(x)) = cos (x

^{2})To prove g(h(x)) continuous, g(x) and h(x) should be continuous.

Continuity of g(x) = cos xLet’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 =coscFunction value at x = c, g(c) = cos c

As,

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

Continuity of h(x) = x^{2}Let’s check the continuity at x = c

Limit =

= c

^{2}Function value at x = c, h(c) = c

^{2}As,

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

As, g(x) and h(x) is continuous then g(h(x)) = cos(x

^{2}) 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 < 0Limit =

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

As,

When c ≥ 0Limit =

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 xLet’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 =coscFunction 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 < 0Limit =

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

As,

When c ≥ 0Limit =

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 xLet’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 =sincFunction 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 < 0Limit =

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

As,

When c ≥ 0Limit =

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 < -1Limit =

= -(c + 1)

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

As,

When c ≥ -1Limit =

= 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.