Skip to content
Related Articles

Related Articles

Class 12 RD Sharma Solutions – Chapter 9 Continuity – Exercise 9.2 | Set 2

View Discussion
Improve Article
Save Article
  • Last Updated : 01 Sep, 2021
View Discussion
Improve Article
Save Article

Question 8. If f\left( x \right) = \frac{\tan\left( \frac{\pi}{4} - x \right)}{\cot 2x}       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 \frac{\tan \left( \frac{\pi}{4} - x \right)}{\cot 2x}       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 \to 0} \left( \frac{\tan\left( \frac{\pi}{4} - \frac{\pi}{4} + h \right)}{\cot\left( \frac{\pi}{2} - 2h \right)} \right)

= lim{h -> 0} tan h/tan 2h

\lim_{h \to 0} \left( \frac{\frac{\tan \left( h \right)}{h}}{\frac{2 \tan \left( 2h \right)}{2h}} \right)

\frac{1}{2}\left( \frac{\lim_{h \to 0} \frac{\tan \left( h \right)}{h}}{\lim_{h \to 0} \frac{\tan \left( 2h \right)}{2h}} \right)

= 1/2

(RHL at x = π/4) = lim{x -> π/4+} f(x)

= lim{h -> 0} f(π/4 + h)

\lim_{h \to 0} \left( \frac{\tan \left( \frac{\pi}{4} - \frac{\pi}{4} - h \right)}{\cot \left( \frac{\pi}{2} + 2h \right)} \right)

= lim{h -> 0} tan (-h)/tan (-2h)

= lim{h -> 0} tan h/tan 2h

\lim_{h \to 0} \left( \frac{\frac{\tan \left( h \right)}{h}}{\frac{2 \tan \left( 2h \right)}{2h}} \right)

\frac{1}{2}\left( \frac{\lim_{h \to 0} \frac{\tan \left( h \right)}{h}}{\lim_{h \to 0} \frac{\tan \left( 2h \right)}{2h}} \right)

= 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 f\left( x \right) = \begin{cases}2x - 1 , & \text { if }  x < 2 \\ \frac{3x}{2} , & \text{ if  } x \geq 2\end{cases}      .

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

∴ \lim_{x \to 2^-} f\left( x \right) = \lim_{x \to 2^+} f\left( x \right) = f\left( 2 \right)

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) = a

So, 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) = -a

So, 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) = 0

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

So, 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 f\left( x \right) = \begin{cases}\frac{\sin x}{x} , & x < 0 \\ x + 1 , & x \geq 0\end{cases}    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 x2 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) = x2

∵  (g o h)(x) = g(h (x)) = g(x2) = cos(x2) = 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) = x2

Let k be a real number, then h(k) = k2

limx->h(x) = lim{x->k} x2 = k2

∴ 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) = 0

 lim{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 f\left( x \right) = \begin{cases}x^2 \sin\frac{1}{x} , & x \neq 0 \\ 0 , & x = 0\end{cases}    is a continuous function?

Solution:

Let us assume that c be a real number.

Case I: If  c ≠ 0 , then f(c)= c2 sin (1/c)

lim{x->c}f(x) = lim{x->c}(x2 sin 1/x) 

= (lim{x->c}x2) (lim{x->c} sin 1/x) 

= c2 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} (x2 sin 1/x) = lim{x -> 0} (x2 sin 1/x)

So, -1 ≤ sin 1/x ≤ 1, x ≠ 0

-x2 ≤ x2sin 1/x ≤ x2

lim{x -> 0} (-x2) ≤ lim{x -> 0} (x2 sin 1/x) ≤ lim{x -> 0} x2

0 ≤ lim{x -> 0} (x2 sin 1/x) ≤ 0

lim{x -> 0} (x2 sin 1/x) = 0

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

Similarly, lim{x -> 0+} f(x) = lim{x -> 0+} (x2 sin 1/x) = lim{x -> 0} (x2 sin 1/x) = 0

Thus, f is a continuous function.

Question 18. Given the function f\left( x \right) = \frac{1}{x + 2}+2   . Find the points of discontinuity of the function f(f(x)).

Solution:

Here, f[f(x)]=\frac{1}{\frac{1}{x+2}+2}=\frac{x+2}{2x+5}

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) = \frac{1}{t^2 + t - 2}   , where t = 1/(x – 1).

Solution:

Here, f\left( t \right) = \frac{1}{t^2 + t - 2}

 Now, let u = 1/(x – 1)

Therefore f( u ) = \frac{1}{u^2 + 2u - u - 2}

\frac{1}{\left( u + 2 \right)\left( u - 1 \right)}

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.


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!