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

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 continues 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

lim_x->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.

Summary

The practice problems in this set focus on testing students' understanding of continuity in functions. Problems typically involve determining the values of unknown constants to make functions continuous at specific points or over given intervals. Students are required to apply the concept of left-hand and right-hand limits, evaluate limits at critical points, and use the definition of continuity to solve equations. Some problems may also involve piecewise functions, requiring careful analysis of function behavior at the point where the function definition changes.