# Class 12 RD Sharma Solutions – Chapter 9 Continuity – Exercise 9.2 | Set 1

### 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)⇒

⇒ 0 = 1, which is not possible

Therefore, the given function is not continuous for any value of k.

### (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,

⇒ 4 = a + b and b = 1

⇒ a = 3 and b = 1.

### (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

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