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# 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.

### (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) =   = (x2 + 4)(x + 2)

Since a polynomial function is everywhere continuous, (x2 + 4) and (x + 2) are  continuous everywhere.

So, the product function (x2 + 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 x4 + x3 + 2x2 being a polynomial function is continuous everywhere.

Also, tan-1x 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) = c10 −1 and

lim{x-> c} f(x) = lim{x->c} (x10 – 1)

= c10 −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) (x2) = 12 = 1

So 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) = c2

lim(x->c) f(x) = lim(x->c) f(c) = c2

∴ 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) = 4

So 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 = -1

lim{x →0+}f(x) = lim{x →0}(sin x – cos x) = sin 0 –  cos 0 = 0 – 1 = -1

We 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) = -2

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

We 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 = 2

RHL = lim{x->1}f(x) = lim{x->1}2 = 2

We 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.

### (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 b2 – 2b = a

⇒ a2 = 1 and b2 – 2b = a

⇒ a = ±1  and  b2 – 2b = a          . . . (1)

If a = 1, then b2 – 2b = 1

⇒ b2 – 2b – 1 = 0

⇒ b = = 1 ± √2

If a = −1, then b2 – 2b = – 1

⇒ b2 – 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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