Here we provide RD Sharma Class 12 Ex 9.2 Solutions Chapter 9 Continuity for English medium students, Which will very helpful for every student in their exams. Students can download the RD Sharma Class 12 Ex 9.2 Solutions Chapter 9 Continuity book pdf download. Now you will get step-by-step solutions to each question.

Textbook | NCERT |

Class | Class 12th |

Subject | Maths |

Chapter | 9 |

Exercise | 9.2 |

Category | RD Sharma Solutions |

Table of Contents

**RD Sharma Class 12 Ex 9.2 Solutions Chapter 9 Continuity**

### Question 1. Prove that the function is continuous everywhere.

**Solution:**

We know sin x/ x is continuous everywhere since it is the composite function of the functions sin x and x which are continuous.

When x > 0, we have f(x) = x + 1.

Given that

Now, (LHL at x = 0) = lim

_{{x ⇢ 0–} }f(x)= lim

_{{h ⇢ 0}}f(0 – h)= lim

_{{h ⇢ 0}}f(-h)= lim{h ⇢ 0} (sin (-h)/(-h))

= lim

_{{h ⇢ 0}}(sin h/ h)= 1

(RHL at x = 0) = lim

_{{x ⇢ 0+} }f(x)= lim

_{{h ⇢ 0}}f(0 + h)= lim

_{{h ⇢ 0} }f(h)= lim

_{{h ⇢ 0}}(h + 1)= 1

and, f(0) = 0 + 1 = 1.

We observe that: lim

_{{x ⇢ 0–}}f(x) = lim_{{x ⇢ 0+}}f(x) = f(0).

Therefore, f(x) is everywhere continuous.

### Question 2. Discuss the continuity of the function .

**Solution:**

We have,

Now: (LHL at x = 0) = lim_{{x ⇢ 0–}} f(x)

= lim_{{h ⇢ 0}} f(0 – h)

= lim_{{h ⇢ 0}} f(–h)

= lim_{{h⇢ 0}} (–1)

= –1

(RHL at x = 0) = lim_{{x ⇢ 0+}} f(x)

= lim_{{h ⇢ 0}} f(0 + h)

= lim_{{h ⇢ 0}} (1)

= 1

We observe that, lim_{{x ⇢ 0–} }f(x) ≠ lim_{{x ⇢ 0+}} f(x).

**Therefore, f(x) is discontinuous at x = 0.**

### Question 3. Find the points of discontinuity, if any, of the following functions:

### (i)

**Solution:**

Since a polynomial function is everywhere continuous.

At x = 1, we have

(LHL at x = 1) = lim

_{{x ⇢ 1–}}f(x)= lim

_{{h ⇢ 0}}f(1 – h)= lim

_{{h ⇢ 0}}((1 – h)^{3 }– (1 – h )^{2}+ 2(1 – h) – 2)= 1 – 1 + 2 – 2

= 0

(RHL at x = 1) = lim

_{{x ⇢ 1+}}f(x)= lim

_{{h ⇢ 0}}f(1 + h)= lim{h ⇢ 0} ((1 + h)

^{3}– (1 + h )^{2}+ 2(1 + h) – 2)= 1 – 1 + 2 – 2

= 0

Also, f(1) = 4.

We observe that, lim

_{{x ⇢ 0–}}f(x) = lim_{{x ⇢ 0+}}f(x) ≠ f(1).

Therefore, f(x) is discontinuous only at x = 1.

### (ii)

**Solution:**

When x ≠ 2 then

f(x) =

=

=

= (x

^{2}+ 4)(x + 2)Since a polynomial function is everywhere continuous, (x

^{2}+ 4) and (x + 2) are continuous everywhere.So, the product function (x

^{2}+ 4)(x + 2) is continuous.Thus, f(x) is continuous at every x ≠ 2 .

We observe that lim

_{{x->2-}}f(x) = lim_{{x->2+}}f(x) = f(2)

Therefore, f(x) is discontinuous only at x = 2.

### (iii)

**Solution:**

When x < 0, then f(x) = sin x/ x.

Since sin x as well as the identity function x are everywhere continuous, the quotient function sin x/x is continuous at each x < 0.

For x > 0, f(x) becomes a polynomial function. Therefore, f(x) is continuous at each x > 0.

We have: (LHL at x = 0) = lim

_{{x->0-}}f(x)= lim

_{{h -> 0}}f(0 – h)= lim

_{{h -> 0}}f (-h)= lim

_{{h -> 0}}(sin(-h)/(-h))= lim

_{{h -> 0}}(sin h/h)= 1

(RHL at x = 0) = lim

_{x -> 0+}f(x)= lim

_{{h -> 0} }f(0 + h)= lim

_{{h -> 0}}f(h)= lim

_{{h -> 0}}(2h + 3)= 3

We observe that lim

_{{x -> 0-}}f(x) ≠ lim_{{x -> 0+}}f(x)

Therefore, f(x) is discontinuous only at x = 0.

### (iv)

**Solution:**

At x ≠ 0, then f(x) = sin 3x/ x.

Since the functions sin 3x and x are everywhere continuous. So, the quotient function sin 3x/x is continuous at each x ≠ 0.

We have: (LHL at x = 0) = lim

_{{x -> 0+}}f(x)= lim

_{{h -> 0}}f(0 + h)= lim

_{{h -> 0}}f(h)= lim

_{{h -> 0}}(sin 3h/h)= lim

_{{h -> 0}}3 (sin h/h) = 3(RHL at x = 0) = lim

_{{x -> 0+}}f(x)= lim

_{{h -> 0}}f(0 + h)= lim

_{{h -> 0}}f(h)= lim

_{{h -> 0}}(sin 3h/h)= lim

_{{h -> 0}}3 (sin 3h/h)= 3

Also, f(0) = 4.

We observe that lim

_{{x -> 0-}}f(x) = lim_{{x -> 0+}}f(x) = f(0)

Therefore, f(x) is discontinuous only at x = 0.

### (v)

**Solution:**

When x ≠ 0, then f(x) = sin x/ x + cos x.

We know that sin x as well as cos x are everywhere continuous. Thus, the given function is continuous at each x ≠ 0.

Let us consider the point x = 0.

Given:

We have: (LHL at x = 0) = lim

_{{x -> 0-}}f(x)= lim

_{{h -> 0}}f(0 – h)= lim

_{{h -> 0}}f(-h)= lim

_{{h -> 0}}[(sin (-h)/(-h)) + cos (-h)]= lim

_{{h -> 0}}sin(-h)/(-h) + lim_{{h -> 0}}cos(-h)= 1 + 1

= 2

(RHL at x = 0) = lim

_{{x -> 0+}}f(x)= lim

_{{h -> 0}}f(0 + h)= lim

_{{h -> 0}}f(h)= lim

_{{h -> 0}}[(sin h/h) + cos (-h)]= lim

_{{h -> 0}}sin h/h + lim_{{h -> 0}}cos(-h)= 1 + 1

= 2

Also, f(0) = 5.

We observe that lim

_{{x -> 0-}}f(x) = lim_{{x -> 0^+}}f(x) ≠ f(0)

Therefore, f(x) is discontinuous only at x = 0.

### (vi)

**Solution:**

When x ≠ 0, then

x

^{4}+ x^{3}+ 2x^{2}being a polynomial function is continuous everywhere.Also, tan

^{-1}x is everywhere continuous.Let us consider the point x = 0.

We have:

(LHL at x = 0) = lim

_{{x -> 0-} }f(x)= lim

_{{h -> 0}}f(0 – h)= lim

_{{h -> 0}}f(-h)=

=

= 0

(RHL at x = 0) = lim

_{{x -> 0+}}f(x)= lim

_{{h -> 0}}f(0 + h)= lim

_{{h -> 0}}f(h)=

=

= 0

Also, f(0) = 10.

We observe that lim

_{{x -> 0-}}f(x) = lim_{{x -> 0+}}f(x) ≠ f(0)

Therefore, f(x) is discontinuous only at x = 0.

### (vii)

**Solution:**

We have,

=

=

= 1/2

It is given that f(0) = 7.

We observe that lim_{{x -> 0}} f(x) ≠ f(0)

**Therefore, f(x) is discontinuous only at x = 0.**

### (viii)

**Solution:**

At x > 1, f(x) being a modulus function is continuous for each x > 1.

When x < 1, then f( x ) being a composite of polynomial and continuous functions would be continuous.

At x = 1, we have

(LHL at x = 1) = lim

_{{x -> 1-}}f(x)= lim

_{{h -> 0}}f(1 – h)=

= 1/4 – 3/2 + 13/4

= 2

(RHL at x = 1) = lim

_{{x -> 1+}}f(x)= lim

_{{h -> 0}}f(1 + h)= lim

_{{h -> 0}}|1 + h – 3|= |-2|

= 2

Also f(1) = |1 – 3| = |- 2| = 2

We observe that, lim

_{{x -> 1-}}f(x) = lim_{{x -> 1+}}f(x) = f(1)

Therefore, the given function is everywhere continuous.

### (ix)

**Solution:**

f(x) being a modulus function is continuous for each x ≤ – 3.

At – 3 < x < 3 f(x) being a polynomial function is continuous.

At x > 3, f(x) being a polynomial function is continuous.

At x = 3,

We have: (LHL at x = 3) = lim

_{{x -> 3-}}f(x)= lim

_{{h -> 0}}f(3 – h)= lim

_{{h -> 0}}-2(3 – h)= -6

(RHL at x = 3) = lim

_{{x -> 3+}}f(x)= lim

_{{h -> 0}}f(3 + h)= lim

_{{h -> 0}}6(3 + h) + 2= 20

We observe that lim

_{{x -> 3-}}f(x) ≠ lim_{{x -> 3+}}f(x)

Therefore, f(x) is discontinuous only at x = 3.

### (x)

**Solution:**

According to the question it is given that function f is defined at all the points of the real line.

Let us considered c be a point on the real line.

Case I: If c< 1, then f(c) = c

^{10}−1 andlim

_{{x-> c}}f(x) = lim_{{x->c} }(x^{10}– 1)= c

^{10}−1.∴ lim

_{{x->c} }f(x) = f(c)Hence, f is continuous for all x < 1.

Case II: If c = 1, then the left hand limit of f at x = 1.

The right hand limit of f at x = 1 is, lim

_{(x->1) }f(x) = lim_{(x->1) }(x^{2}) = 1^{2}= 1So we conclude that the left and right hand limit of f at x = 1 do not coincide. So, f is not continuous at x = 1.

Case III: If c>1, then f(c) = c

^{2}lim

_{(x->c) }f(x) = lim_{(x->c)}f(c) = c^{2}∴ lim

_{(x->c) }f(x) = f(c)

Therefore, f(x) is discontinuous only at x = 1.

### (xi)

**Solution:**

Let us considered a be a point on the real line.

Case I: if a < 0, then f(c) = 2a.

lim

_{{x->a}}(a) = 2a.∴ lim

_{{N -> 0}}f(x) = f(a)So, f is continuous at all points such that x < 0.

Case II: If 0 < a < 1 then f(x) and lim

_{{x->a}}f(x)=lim_{{x->a}}(0)=0 .∴ lim

_{{x->a}}f(x)=f(a)So, f is continuous at all points of the interval (0, 1).

Case III: If a =1 then f(a) = f(1) = 0.

The left hand limit of f at x = 1 is,

lim

_{{x->1}}f(x) = lim_{{x->1}}f(1)The right hand limit of f at x = 1 is,

lim

_{{x->1}}f(x) = lim_{{x->1}}(4x) = 4(1) = 4So we conclude that LHL ≠ RHL. Thus, f is not continuous at x = 1.

Case IV: If a > 1, then f(a) = 4a and lim

_{{x->a}}f(4x) = 4a .∴ lim

_{{x->a}}f(x) = f(a)

So, f(x) is discontinuous only at x = 1.

### (xii)

**Solution:**

It is evident that f is defined at all points of the real line. Let p be a real number.

Case I: if p ≠ 0 , then f (p) = sin p – cos p

lim

_{{x → p}}f(x) = lim_{{x→p}}( sin x – cos x ) = sin p – cos p∴ lim

_{{x →p}}f(x) = f(p)Therefore, f is continuous at all points x, such that x ≠ 0.

Case II: if p = 0 , then f (0) = – 1.

lim

_{{x →0-}}f(x) = lim_{{x →0^-}}(sin x – cos x) = sin 0 – cos 0 = 0 – 1 = -1lim

_{{x →0+}}f(x) = lim_{{x →0}}(sin x – cos x) = sin 0 – cos 0 = 0 – 1 = -1We observe that: lim

_{{x →0-}}f (x) = lim_{{x →0+}}f (x)= f(0)

Therefore, f is a continuous function everywhere.

### (xiii)

**Solution: **

The given function is defined at all points of the real line. Let us considered a be a point on the real line.

Case I: If a < -1 then f(a)= -2 and lim

_{{x->a}}(x) = lim_{{x->a}}(-2) = -2∴ lim

_{{x->a}}f(x) = f(a)f is continuous for all x < −1.

Case II: If a =1 then f(a) = f(-1) = -2

LHL = lim

_{{x->-1}}f(x) = lim_{{x->-1}}f(-2) = -2RHL = lim

_{{x->-1}}f(x) = lim_{{x->-1}}f(2x) = 2(-1) = -2We observe that lim

_{{x->-1}}f(x) = f(-1)Therefore, f is continuous at x = −1.

Case III: if -1 < a < 1,then f(a) = 2a

lim

_{{x->a}}f(x) = lim_{{x->a}}f(2x) = 2a∴ lim

_{{x->a}}f(x) = f(a)Therefore, f is continuous at all points of the interval (−1, 1).

Case IV: if a = 1, then f(c) = f(1) = 2(1) = 2.

LHL = lim

_{{x->1}}f(x) = lim_{{x->1}}2 = 2RHL = lim

_{{x->1}}f(x) = lim_{{x->1}}2 = 2We observe that: lim

_{{x->1}}f(x) = lim_{{x->1}}f(c)Therefore, f is continuous at x = 2.

Therefore, f is a continuous function everywhere.

### Question 4. In the following, determine the value of constant involved in the definition so that the given function is continuous:

### (i)

**Solution:**

If f( x ) is continuous at x = 0, then

⇒ lim{x -> 0} f(x) = f(0)

⇒ lim

_{{x -> 0}}sin 2x/5x = f(0)⇒ lim

_{{x -> 0}}2 sin 2x/10x = f(0)⇒ 2/5 lim

_{{x -> 0}}sin 2x/2x = f(0)⇒

k = 2/15

### (ii)

**Solution:**

If f(x) is continuous at x = 2, then

lim

_{{x -> 2-}}f(x) = lim_{{x -> 2+}}f(x)⇒ lim

_{{h -> 0}}(k (2 – h) + 5) = lim_{{h -> 0}}(2 + h -1)⇒ lim

_{{h -> 0}}f(2 – h) = lim_{{h -> 0}}f(2 + h)⇒ 2k + 5 = 1

⇒ 2k = – 4

⇒

k = – 2

### (iii)

**Solution:**

If f(x) is continuous at x = 0, then

lim

_{{x -> 0-}}f(x) = lim_{{x -> 0+}}f(x)⇒ lim

_{{h -> 0}}f(-h) = lim_{{h -> 0}}f(h)⇒ (iv)

Solution:If f(x) is continuous at x = 3 and 5, then

lim

_{{x -> 3-}}f(x) = lim_{{x -> 3+}}f(x)and lim

_{{x -> 5-}}f(x) = lim_{{x -> 5+}}f(x)⇒ lim

_{{h -> 0}}f(3 – h) = lim_{{h -> 0}}f(3 + h)and lim

_{{h -> 0}}f(5 – h) = lim_{{h -> 0}}f(5 + h)⇒ 2 = 3a + b and 5a + b = 9

⇒ 2 = 3a + b and 5a + b = 9

⇒

a = 7/2 and b = -17/2.## (v)

Solution:If f(x) is continuous at x = −1 and 0, then

lim

_{{x -> -1-}}f(x) = lim_{{x -> – 1+}}f(x) and lim_{{x -> 0-}}f(x) = lim_{{x -> 0+}}f(x)⇒ lim

_{{h -> 0}}f(-1 – h) = lim_{{h -> 0}}f(-1 + h) and lim_{{h -> 0}}f(-h) = lim_{{h -> 0}}f(h)⇒ lim

_{{h -> 0}}(4) = lim_{{h -> 0}}(a (-1 + h)^{2}+ b)Also, (vi)

Solution:If f(x) is continuous at x = 0, then

⇒

⇒

⇒

⇒

⇒

⇒

⇒ 2p/2 = -1/2

⇒

p = -1/2.## (vii)

Solution:If f(x) is continuous at x = 2 and x = 10, then

lim

_{{x -> 2-}}f(x) = lim_{{x -> 2+}}f(x) and lim_{{x -> {10}–}}f(x) = lim_{{x -> {10}+}}f(x)⇒ lim

_{{h -> 0}}f(2 – h) = lim_{{h -> 0}}f(2 + h) and lim_{{h -> 0}}f(10 – h) = lim_{{h -> 0}}f(10 + h)⇒ lim

_{{h -> 0}}(5) = lim_{{h -> 0}}(a (2 + h) + b)And lim

_{{h -> 0}}a (10 – h) + b = lim_{{h -> 0}}(21)On solving equations, we get,

⇒

a = 2 and b = 1.## (viii)

Solution:If f(x) is continuous at x = π/2, then

lim

_{{x -> π/2-}}f(x) = f(π/2)⇒ lim

_{{h -> 0}}f(π/2 – h) = f(π/2)⇒ lim

_{{h -> 0}}f(π/2 – h) = 3⇒

⇒

⇒ lim

_{{h -> 0}}(k sin h/2h) = 3⇒ k/2 lim

_{{h -> 0} }sin h/h =3⇒ k/2 = 3

⇒

k = 6## Question 5. The function is continuous on (0, ∞), then find the most suitable values of a and b.

Solution:Given that f is continuous on ( 0, ∞ ).

So, f is continuous at x = 1 and x = √2.

At x = 1, we have lim

_{{x -> 1-}}f(x)= lim

_{{h -> 0} }f(1 – h)= lim

_{{h -> 0}}[(1 – h)^{2}/a]= 1/a

At x = √2, we have

lim

_{{x -> √2-}}f(x) = lim_{{h -> 0}}f(√2 + h)= lim

_{{h -> 0}}(a)= a

f is continuous at x = 1 and √2.

⇒ 1/a = a and b

^{2}– 2b = a⇒ a

^{2 }= 1 and b^{2}– 2b = a⇒ a = ±1 and b

^{2}– 2b = a . . . (1)If a = 1, then b

^{2}– 2b = 1⇒ b

^{2}– 2b – 1 = 0⇒ b = = 1 ± √2

If a = −1, then b

^{2}– 2b = – 1⇒ b

^{2}– 2b + 1 = 0⇒ b = 1

Therefore, a = −1, b = 1 or a = 1, b =√2.are the most suitable values of a and b.## Question 6. Find the values of a and b so that the function f(x) defined by becomes continuous on [0, π].

Solution:f is continuous at x = π.

At x = π/4, we have

lim

_{{x -> π/4-}}f(x) = lim_{{h -> 0}}f(π/4 – h)= lim{h -> 0} [(π/4 – h) + √2a sin (π/4 – h)]

= π/4 + √2a sin π/4

= π/4 + a

= lim

_{{h -> 0}}[2 (π/4 + h) cot (π/4 + h) + b]= [2 π/4 cot π/4 + b]

= π/2 + b

⇒ – b – a = b and π/4 + a = π/2 + b

⇒

a = π/6 and b = -π/12## Question 7. The function f(x) is defined as follows: . If f is continuous on [0, 8], find the values of a and b.

Solution:Given that f is continuous on [0, 8].

So, f is continuous at x = 2 and x = 4

At x = 2,

lim

_{{x -> 2-}}f(x) = lim_{{h -> 0}}f(2 – h)= lim

_{{h -> 0}}(2 – h)^{2}+ a(2 – h) + b= 4 + 2a + b

lim

_{{x -> 2+} }f(x) = lim_{{h -> 0}}f(2 + h)= lim

_{{h -> 0}}[3(2 + h) + h]= 8

At x = 4,

lim

_{{x -> 4-}}f(x) = lim_{{h -> 0}}f(4 – h)= lim_{h -> 0} [3(4 – h) + 2]

= 14

lim

_{{x -> 4+}}f(x) = lim_{{h -> 0}}f(4 + h)= lim

_{{h -> 0}}[2a(4 + h) + 5b]= 8a + 5b

So, f is continuous at x = 2 and x = 4.

lim

_{{x -> 2-}}f(x) = lim_{{x -> 2+}}f(x)and, lim

_{{x -> 4-}}f(x) = lim_{{x -> 4+}}f(x)⇒ 4 + 2a + b = 8 and 8a + 5b = 14

⇒ 2a + b = 4 and 8a + 5b = 14

On solving, we get

a = 3 and b = -2## Question 8. If for x ≠ π/4, find the value which can be assigned to f(x) at x = π/4 so that the function f(x) becomes continuous every where in [0, π/2].

Solution:If x ≠ π/4, tan (π/4 – x) and cot2x are continuous in [0, π/2]. Then the function is continuous for each x ≠ π/4.

Now, let us assume that the point x = π/4.

We have,

(LHL at x = π/4) = lim

_{{x -> π/4-} }f(x)= lim

_{{h -> 0}}f(π/4 – h)=

= lim

_{{h -> 0}}tan h/tan 2h=

=

= 1/2

(RHL at x = π/4) = lim

_{{x -> π/4+}}f(x)= lim

_{{h -> 0}}f(π/4 + h)=

= lim{h -> 0} tan (-h)/tan (-2h)

= lim

_{{h -> 0} }tan h/tan 2h=

=

= 1/2

If f(x) is continuous at x = π/4 then

lim

_{{x -> π/4-}}f(x) = lim_{{x -> π/4+} }f(x) = f(π/4)∴ f(π/4) = 1/2

Hence, the function will be everywhere continuous.## Question 9. Discuss the continuity of the function .

Solution:When x < 2, f(x) being a polynomial function is continuous.

When x > 2, f(x) being a polynomial and continuous function is continuous.

At x = 2, we have:

(LHL at x = 2) = lim

_{{x -> 2-}}f(x)= lim

_{{h -> 0} }f(2 – h)= lim

_{{h -> 0} }(2(2 – h) – 1)= 4 – 1

= 3

(RHL at x = 2) = lim

_{{x -> 2+}}f(x)= lim

_{{h -> 0}}f(2 + h)= lim

_{{h -> 0}}3 (h + 2)/2= 3

Also, f(2) = 3(2)/2 = 3

∴

So, f(x) is continuous at x = 2.## Question 10. Discuss the continuity of f(x) = sin |x|.

Solution:f is clearly the composition of two functions, f = h o g, where g (x) = |x| and h (x) = sin x

Since, hog(x) = h(g(x)) = h(|x|) = \sin|x|

g(x)=|x| being a modulus function must be continuous for all real numbers.

Let us assume that a be a real number.

Case 1:

If a > 0 then g(a) = a

lim

_{{x -> c} }(g(x)) = lim_{{x -> c} }(x) = aSo, lim

_{{x -> c} }(g(x)) = g(a)So, g is the continuous on all the points, i.e., x > 0

Case 2:

If a < 0 then g(a) = -a

lim

_{{x -> c} }(g(x)) = lim_{{x -> c} }(-x) = -aSo, lim

_{{x -> c} }(g(x)) = g(a)So, g is the continuous on all the points x < 0

Case 3:

If a = 0 then g(a) = g(0) = 0

lim

_{{x -> 0–} }(g(x)) = lim_{{x -> 0–} }(-x) = 0lim

_{{x -> 0+} }(g(x)) = lim_{{x -> 0+} }(x) = 0So, lim

_{{x -> 0–} }(g(x)) = lim_{{x -> 0+}}(g(x)) = g(0)So, g is continus at point x = 0

So, lim

_{{x -> c} }(g(x)) = g(a)So we conclude that h(x) = sinx is defined for every real number.

Let us considered b be the real number. Now put x = b + k

If x->b , then k ->0

So, h(b) = sin b

lim

_{{x -> b} }(h(x)) = lim_{{x -> b} }sin x= lim

_{{k -> 0} }sin (b + k)= lim

_{{k -> 0} }(sinb cos k + cos b sink)= lim

_{{k -> 0} }(sinb cos k) + lim_{{k -> 0}}(cos b sink)= sinb cos 0 + cos b sin 0

= sin b + 0

= sin b

Hence, lim

_{{x -> c} }h(x) = g(c)So, h is continuous function

Hence, f(x) = hog(x) = h(g(x)) = h(|x|) = sin|x|

## Question 11. Prove that is everywhere continuous.

Solution:When x < 0, sin x/x being the composite of two continuous functions is continuous.

When x > 0, we have f(x) being a polynomial function is continuous.

At x = 0:

(LHL at x = 0) = lim

_{{x -> 0–}}f(x)= lim

_{{h -> 0}}f(0 – h)= lim

_{{h -> 0}}f(-h)= lim

_{{h -> 0}}(sin (-h)/(-h))= lim

_{{h -> 0}}(sin h/h)= 1

(RHL at x = 0) = lim

_{{x -> 0+}}f(x)= lim

_{{h -> 0}}f(0 + h)= lim

_{{h -> 0} }f(h)= lim

_{{h -> 0}}(h + 1)= 1

Also, f(x) = 0 + 1 = 1

So we conclude that lim

_{{x -> 0–}}f(x) = lim_{{x -> 0+}}f(x) = f(0)

So, f(x) is everywhere continuous.## Question 12. Show that the function g (x) = x − [x] is discontinuous at all integral points. Here [x] denotes the greatest integer function.

Solution:g is defined at all integral points. Let n be an integer. Then,

g(n) = n − [n]

= n − n

= 0

At x = n, we have:

LHL = lim

_{{x -> n–}}g(x) = lim_{{x->n–}}(x – [x])= lim

_{{x->n–}}(x) – lim_{x->n[x]^{–}}= n − (n − 1)

= 1

RHL = lim

_{{x->n+}}g(x) = lim_{{x->n+} }(x – [x])= lim

_{{x-> n+}}(x) – lim_{{x->n+}}[x]= n − n

= 0

So we conclude that lim

_{{x -> 0–}}f( x ) ≠ lim_{{x -> 0+} }f(x)

So, g is discontinuous at all integral points.## Question 13. Discuss the continuity of the following functions:

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

## (ii) f(x) = sin x − cos x

## (iii) f(x) = sin x cos x

Solution:We know that if g and h are two continuous functions, then g + h, g − h and g o h are also continuous.

Let g (x) = sin x, defined for every real number.

Let a be a real number. Put x = a + h

If x → a, then h → 0 g(a)=sin a.

lim

_{{x->a}}g(x) = lim_{{x->a}}sina= lim

_{{h->0}}sin (a+h)= lim

_{{h->0}}[sin a cos h + cos a sin h]= lim

_{{h->0}}(sin a cos h )+lim_{{h->0}}(cos a sin h)= sin a cos 0 + cos a sin 0

= sin (a + 0)

= sin a

∴ lim

_{{x->c}}g(x) = g(c)Similarly, cos x can be proved as a continuous function.

So we conclude that

(i) f (x) = g (x) + h (x) = sin x + cos x is a continuous function.

(ii) f (x) = g (x) − h (x) = sin x − cos x is a continuous function.

(iii) f (x) = g (x) h (x) = sin x cos x is a continuous function.## Question 14. Show that f (x) = cos x

^{2}is a continuous function.

Solution:f can be written as the composition of two functions as f = g o h, where g (x) = cos x and h (x) = x

^{2}∵ (g o h)(x) = g(h (x)) = g(x

^{2}) = cos(x^{2}) = f(x)Let c be a real number.

Then, g(c) = cos c

If x-> c , then h->0 and lim

_{{x->c}}g(x)= lim

_{{x->c}}cos c= cos c

∴ lim

_{{x->c}}g(x) = g(c)So, g(x) = cos x is a continuous function.

Now, h(x) = x

^{2}Let k be a real number, then h(k) = k

^{2}lim

_{x->h}(x) = lim_{{x->k}}x^{2 }= k^{2}∴ lim

_{{x->k}}h(x) = h(k)So, h is a continuous function.

So, f(x) being a composite of two continuous functions is a continuous function.## Question 15. Show that f (x) = |cos x| is a continuous function.

Solution:f is the composition of two functions as, f = g o h, where g(x) = |x| and h(x) = cos x

(g o h)(x) = g(h(x)) = g(cos x) = |cos x| = f(x)

Clearly, g(x) being a modulus function would be continuous at all points.

Now, h (x) = cos x. We know that h (x) = cos x is defined for every real number.

Let c be a real number.

Put x = c + h.

If x → c, then h → 0.

⇒ h(c) = cos c

So, h (x) = cos x is a continuous function.

Therefore, f(x) being a composite of two continuous functions is a continuous function.## Question 16. Find all the points of discontinuity of f defined by f (x) = |x| − |x + 1|.

Solution:f is the composition of two functions as f(x) = g(x) – h(x), where g(x) = |x| and h(x) = |x + 1|.

Let c be a real number.

Case I: If c < 0 , then g(c) = -c and lim

_{{x->c}}g(x) = lim_{{x->c}}= -c∴ lim

_{{x->c}}g(x) = g(c)So, g is continuous at all points x < 0.

Case II: If c < 0 , then g(c) = -c and lim

_{(x->c)}g(x) = lim_{(x->c)}(-x) = -c∴ lim

_{(x->c)}g(x) = g(c)So, g is continuous at all points x > 0.

Case III: if c = 0 , then g (c) = g(0) = 0

lim

_{{x->0-}}g(x) = lim_{{x->0-}}(- x) = 0lim

_{{x->0+}}g(x) = lim_{{x->0+}}(x) = 0∴ lim

_{{x->0+}}g(x) = lim_{{x->0+}}(x) = g(0)So, g is continuous everywhere.

Clearly, h is defined for every real number. Let c be a real number.

Case I: if c < – 1, then h (c) = – (c + 1)

lim

_{{x->c}}h (x) = lim_{{x->c}}[-(x + 1)]= -(c + 1)

∴ lim

_{{x-> c} }h (x) = h(c)

Therefore, f being a composite of two continuous functions is a continuous function.## Question 17. Determine if is a continuous function?

Solution:Let us assume that c be a real number.

Case I: If c ≠ 0 , then f(c)= c

^{2}sin (1/c)lim

_{{x->c}}f(x) = lim_{{x->c}}(x^{2}sin 1/x)= (lim

_{{x->c}}x^{2}) (lim_{{x->c} }sin 1/x)= c

^{2}sin (1/c)lim

_{{x->c}}f(x) = f(c)So, f is continuous at all points such that x ≠ 0

Case II: If c = 0 then f(0) = 0

lim

_{{x -> 0–}}f(x) = lim_{{x -> 0–}}(x^{2}sin 1/x) = lim_{{x -> 0}}(x^{2}sin 1/x)So, -1 ≤ sin 1/x ≤ 1, x ≠ 0

-x

^{2}≤ x^{2}sin 1/x ≤ x^{2}lim

_{{x -> 0}}(-x^{2}) ≤ lim_{{x -> 0}}(x^{2 }sin 1/x) ≤ lim_{{x -> 0}}x^{2}0 ≤ lim

_{{x -> 0}}(x^{2 }sin 1/x) ≤ 0lim

_{{x -> 0}}(x^{2 }sin 1/x) = 0lim

_{{x -> 0–}}f(x) = 0Similarly, lim

_{{x -> 0+}}f(x) = lim_{{x -> 0+}}(x^{2}sin 1/x) = lim_{{x -> 0}}(x^{2}sin 1/x) = 0

Thus, f is a continuous function.## Question 18. Given the function . Find the points of discontinuity of the function f(f(x)).

Solution:Here,

We observe that f(f( x )) is not defined at x + 2 = 0 and 2x + 5 = 0.

If x + 2 = 0, then x = – 2 and if 2x + 5 = 0, then x = -5/2

Hence, the function is discontinuous at x = -5/2 and – 2.## Question 19. Find all point of discontinuity of the function f(t) = , where t = 1/(x – 1).

Solution:Here,

Now, let u = 1/(x – 1)

Therefore f( u ) =

=

So, f (u ) is not defined at u = -2 and u = 1.

If u = – 2, then -2 = 1/(x – 1)

⇒ 2x = 1

⇒ x = 1/2

If u = 1, then 1 = 1/(x – 1)

⇒ x = 2

Hence, the function is discontinuous at x =1/2, 2.I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment in the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

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