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