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# Class 12 NCERT Solutions- Mathematics Part I – Chapter 5 Continuity And Differentiability – Exercise 5.1 | Set 2

• Last Updated : 28 Jul, 2022

### 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 =

= Î»(02– 2(0)) = 0

Right limit =

= Î»4(0) + 1 = 1

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

As, 0 = 1 cannot be possible

Hence, for no value of Î», 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 Î».

### Question 19. Show that the function defined by g (x) = x â€“ [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) = x2 â€“ sin x + 5 continuous at x = Ï€?

Solution:

Let’s check the continuity at x = Ï€,

f(x) = x2 â€“ sin x + 5

Let’s substitute, x = Ï€+h

When xâ‡¢Ï€, Continuity at x = Ï€

Left limit =

= (Ï€2 â€“ sinÏ€ + 5) = Ï€2 + 5

Right limit =

= (Ï€2 â€“ sinÏ€ + 5) = Ï€2 + 5

Function value at x = Ï€, f(Ï€) = Ï€2 â€“ sin Ï€ + 5 = Ï€2 + 5

As,

Hence, the function is continuous at x = Ï€ .

### (a) f(x) = sin x + cos x

Solution:

Here,

f(x) = sin x + cos x

Let’s take, x = c + h

When xâ‡¢c then hâ‡¢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 âˆ’ sinc sinh))

= ((sinc cos0 + cosc sin0) + (cosc cos0 âˆ’ 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 â€“ cos x

Solution:

Here,

f(x) = sin x – cos x

Let’s take, x = c+h

When xâ‡¢c then hâ‡¢0

So,

(sin(c + h) âˆ’ 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) âˆ’ (cosc cosh âˆ’ sinc sinh))

= ((sinc cos0 + cosc sin0) âˆ’ (cosc cos0 âˆ’ sinc sin0))

cos 0 = 1 and sin 0 = 0

= (sinc âˆ’ cosc) = f(c)

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

As,  = f(c) = sinc âˆ’ 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 xâ‡¢c then hâ‡¢0

So,

sin(c + h) Ã— 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) Ã— (cosc cosh âˆ’ sinc sinh))

= ((sinc cos0 + cosc sin0) Ã— (cosc cos0 âˆ’ sinc sin0))

cos 0 = 1 and sin 0 = 0

= (sinc Ã— cosc) = f(c)

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

As, = f(c) = sinc Ã— 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 xâ‡¢c then hâ‡¢0

So,

Using the trigonometric identities, we get

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

(cosc cosh âˆ’ sinc sinh)

= (cosc cos0 âˆ’ 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 – {nÏ€}, n âˆˆ Integer

Let’s take, x = c + h

When xâ‡¢c then hâ‡¢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 xâ‡¢c then hâ‡¢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 xâ‡¢c then hâ‡¢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) â‰  0

For x < 0, f(x) = , is continuous

Hence, f(x) is continuous x âˆˆ (-âˆž, 0)

Now, For x â‰¥ 0, f(x) = x + 1, which is a polynomial

As polynomial are continuous, hence f(x) is continuous x âˆˆ (0, âˆž)

So now, as f(x) is continuous in x âˆˆ (-âˆž, 0) U (0, âˆž)= 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.

### 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 âˆˆ = R – {0}

Let’s check the continuity at x = 0,

As, we know range of sin function is [-1,1]. So, -1 â‰¤ â‰¤ 1 which is a finite number.

Limit =

= (02 Ã—(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 âˆ’ cos0) = 0 âˆ’ 1 = âˆ’1

Right limit =

= (sin0 âˆ’ cos0) = 0 âˆ’ 1 = âˆ’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 câ‰ 0),

Left limit =

= (sinc âˆ’ cosc)

Right limit =

= (sinc âˆ’ 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 = Ï€/2.

Solution:

Continuity at x = Ï€/2

Let’s take x =

When xâ‡¢Ï€/2 then hâ‡¢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 = Ï€

Solution:

Continuity at x = Ï€

Left limit =

= k(Ï€) + 1

Right limit =

= cos(Ï€) = -1

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

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

kÏ€ + 1 = -1

k = -2/Ï€

### 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

### 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 (x2) 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 xâ‡¢c then hâ‡¢0

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

Limit = (cosc cosh âˆ’ sinc sinh)

= cosc cos0 âˆ’ 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 â‰¥ 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 â‰¥ 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 xâ‡¢c then hâ‡¢0

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

Limit = (cosc cosh âˆ’ sinc sinh)

= cosc cos0 âˆ’ 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 xâ‰¥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 â‰¥ 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 xâ‡¢c then hâ‡¢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 | â€“ | x + 1 |

Solution:

Let’s take

g(x) = |x|

m(x) = |x + 1|

g(x) – m(x) = | x | â€“ | 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 xâ‰¥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 â‰¥ 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 â‰¥ -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 â‰¥ -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| â€“ |x + 1| is also continuous.

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