Ans. Given:  

Here  is defined for  i.e., on  and also for  i.e., on 

 Domain of  is  = R

 For all  is a polynomial and hence continuous and for all is a continuous and hence it is also continuous on R – {2}.

Now Left Hand limit =  = 2 x 2 + 3 = 4 + 3 = 7

Right Hand limit =  = 2 x 2 – 3 = 4 – 3 = 1

Since   

Therefore,  does not exist and hence  is discontinuous at  (only)

Ans. Given:  

Here  is defined for  i.e., on  and for  i.e. on (−3,3)(−3,3) and also for  i.e., on 

 Domain of  is  = R

 For all  is a polynomial and hence continuous and for all  is a continuous and hence it is also continuous and also for all . Therefore,  is continuous on R – 

It is observed that  and  are partitioning points of domain R.

Now Left Hand limit = 

Right Hand limit = 

And 

Therefore,  is continuous at .

Again  Left Hand limit = 

Right Hand limit = 

Since 

Therefore,  does not exist and hence  is discontinuous at  (only).

Ans. Given:  i.e.,  if  and  if  

   if ,  if  and  if 

It is clear that domain of  is R as  is defined for ,  and .

For all ,  is a constant function and hence continuous.

For all ,  is a constant function and hence continuous.

Therefore  is continuous on R – {0}.

Now Left Hand limit = 

Right Hand limit = 

Since 

Therefore,  does not exist and hence  is discontinuous at  (only).

Ans. Given:  

At  L.H.L. =  And 

R.H.L. = 

Since   L.H.L. = R.H.L. = 

Therefore,  is a continuous function.

Now, for  

 

Therefore,  is a continuous at 

Now, for  

Therefore,  is a continuous at 

Hence, the function is continuous at all points of its domain.

Ans. Given:  

It is observed that  being polynomial is continuous for  and  for all  R.

Continuity at  R.H.L. = 

L.H.L. = 

And 

Since   L.H.L. = R.H.L. = 

Therefore,  is a continuous at  for all  R.

Hence,  has no point of discontinuity.

Ans. Given:  

At  L.H.L. = 

R.H.L. = 

Since   L.H.L. = R.H.L. = 

Therefore,  is a continuous at 

Now, for   and 

 

Therefore,  is a continuous for all  R.

Hence the function has no point of discontinuity.

Ans. Given:  

At  L.H.L. = 

R.H.L. = 

Since   L.H.L.  R.H.L.

Therefore,  is discontinuous at 

Now,   for   and for  limx→c(x2)=c2=f(c)limx→c⁡(x2)=c2=f(c)

Therefore,  is a continuous for all  R – {1}

Hence for all given function  is a point of discontinuity.

Ans. Given:  

At  L.H.L. = 

R.H.L. = 

Since   L.H.L.  R.H.L.

Therefore,  is discontinuous at 

Now,   for   and for  

Therefore,  is a continuous for all  R – {1}

Hence  is not a continuous function.

Ans. Given:  

In the interval  

  is continuous in this interval.

At  L.H.L. =  and R.H.L. = 

Since   L.H.L.  R.H.L.

Therefore,  is discontinuous at 

At  L.H.L. =  and R.H.L. = 

Since   L.H.L.  R.H.L.

Therefore,  is discontinuous at 

Hence,  is discontinuous at  and 

Ans. Given:  

At  L.H.L. =  and R.H.L. = 

Since   L.H.L. = R.H.L.

Therefore,  is continuous at 

At  L.H.L. =   and R.H.L. = 

Since   L.H.L.  R.H.L.

Therefore,  is discontinuous at 

When  being a polynomial function is continuous for all 

When . It is being a polynomial function is continuous for all 

Hence  is a point of discontinuity.

Ans. Given:  

At  L.H.L. =  and R.H.L. = 

Since   L.H.L. = R.H.L.

Therefore,  is continuous at 

At  L.H.L. =  and R.H.L. = 

Since   L.H.L. = R.H.L.

Therefore,  is continuous at 

Ans. Given:  

Continuity at 

Also

 

 

 

 

Ans. Since  is continuous at  

 

And 

Here, therefore should be L.H.L. = R.H.L.

 0 = 1, which is not possible.

∴∴ for no value of λλ, f(x) is continuous at x = 0

Again Since  is continuous at 

=>limx→1−f(x)=f(−1)=>limx→1−⁡f(x)=f(−1)

=>limx→1−4x+1=>limx→1−⁡4x+1

=>limh→04(1−h)+1=>limh→0⁡4(1−h)+1

=>4×1+1=>4×1+1

=> 5

And 

Here, L.H.L. = R.H.L.

∴∴ for any value of λλ, f(x) is continuous at x = 1.

Ans. Let n ∈∈ II
Then,limx→n−[x]=n−1limx→n−⁡[x]=n−1
∴[x]=n−1 ∀ x∈[n−1,n]∴[x]=n−1 ∀ x∈[n−1,n]
and g(n) = n – n = 0  [∴[n]=n because n∈I][∴[n]=n because n∈I]
Now, limx→n−g(x)=limx→n−(x−[x])=limx→n−x−limx→n−[x]limx→n−⁡g(x)=limx→n−⁡(x−[x])=limx→n−⁡x−limx→n−⁡[x]
=n−(n−1)=1=n−(n−1)=1
Also, limx→n+g(x)=limx→n+(x−[x])limx→n+⁡g(x)=limx→n+⁡(x−[x])
=limx→n+x−limx→n+[x]=n−n=0=limx→n+⁡x−limx→n+⁡[x]=n−n=0
Thus, limx→n−g(x)≠limx→n+g(x)limx→n−⁡g(x)≠limx→n+⁡g(x)
Hence, g(x) is discontinuous at all integral points. 

Ans. Given:  

L.H.L. = limx→π−(x2−sinx+5)limx→π−⁡(x2−sin⁡x+5) = limh→0[(π−h)2−sin(π−h)+5]=π2+5limh→0⁡[(π−h)2−sin⁡(π−h)+5]=π2+5

R.H.L. = limx→π+(x2−sinx+5)limx→π+⁡(x2−sin⁡x+5) = limh→0[(π+h)2−sin(π+h)+5]=π2+5limh→0⁡[(π+h)2−sin⁡(π+h)+5]=π2+5

And 

Since   L.H.L. = R.H.L. = 

Therefore,  is continuous at 

Ans. (a) Let  be an arbitrary real number then  

 

 

= 

= 

= 

Similarly, we have 

 

Therefore,  is continuous at 

Since,  is an arbitrary real number, therefore,  is continuous.

(b) Let  be an arbitrary real number then 

 

 

= 

= 

= 

Similarly, we have 

 

Therefore,  is continuous at 

Since,  is an arbitrary real number, therefore,  is continuous.

(c) Let  be an arbitrary real number then 

 

 

= 

= 

= 

Similarly, we have 

 

Therefore,  is continuous at 

Since,  is an arbitrary real number, therefore,  is continuous.

Ans. (a) Let  be an arbitrary real number then  

 

= 

=  = 

  for all  R

Therefore,  is continuous at 

Since,  is an arbitrary real number, therefore,  is continuous.

(b)  and domain  I

 

= 

= 

= 

= 

Therefore,  is continuous at 

Since,  is an arbitrary real number, therefore,  is continuous.

(c)  and domain  I

 

= 

= 

=  = 

Therefore,  is continuous at 

Since,  is an arbitrary real number, therefore,  is continuous.

(d)  and domain  I

 

=  = 

=  = 

Therefore,  is continuous at 

Since,  is an arbitrary real number, therefore,  is continuous.

Ans. Given:  

At  L.H.L. = 

R.H.L. = 

  is continuous at 

When  and  are continuous, then  is also continuous.

When  is a polynomial, then  is continuous.

Therefore,  is continuous at any point.

Ans. Here,  = 0 x a finite quantity = 0 

Also 

Since,   therefore, the function  is continuous at 

Ans.  The given function f is f(x) 

={sinx−cosx,ifx≠0−1ifx=0{sin⁡x−cos⁡x,ifx≠0−1ifx=0

It is evident that f is defined at all point of the real line.

Let c be a real number.

Case I:

If c≠≠0, then f(c) = sin c – cos c

limx→cf(x)=limx→c(sinx−cosx)=sinc−cosc∴limx→cf(x)=f(c)limx→c⁡f(x)=limx→c⁡(sin⁡x−cos⁡x)=sin⁡c−cosc∴limx→c⁡f(x)=f(c)

Therefore, f is continuous at all point x, such that x≠≠ 0

Case II:

If c = 0, then f(0)=-1

limx→0−f(x)=limx→0(sinx−cosx)=sin0−cos0=0−1=−1limx→0+f(x)=limx→0(sinx−cosx)=sin0−cos0=0−1=−1∴limx→0−f(x)=limx→0+f(x)=f(0)limx→0−⁡f(x)=limx→0⁡(sin⁡x−cos⁡x)=sin⁡0−cos0=0−1=−1limx→0+⁡f(x)=limx→0⁡(sin⁡x−cos⁡x)=sin⁡0−cos0=0−1=−1∴limx→0−⁡f(x)=limx→0+⁡f(x)=f(0)

Therefore, f is continuous at x=0

From the above observations, it can be concluded that f is continuous at every point of the real line.

Thus, f is a continuous function.

Ans. Here,       

 

 

Putting  where 

=  = 

=  = 

=  ……….(i)

And  ……….(ii)

  when  [Given]

Because  is continuous at 

 

 From eq. (i) and (ii),

 

Ans. Here, limx→2−f(x)=limx→2−kx2limx→2−⁡f(x)=limx→2−⁡kx2=  k×22k×22 

 limx→2+f(x)=3limx→2+⁡f(x)=3 and 

Since f(x)f(x) is continuous at x=2x=2 [given]

Therefore, limx→2−f(x)=limx→2+f(x)=f(2)limx→2−⁡f(x)=limx→2+⁡f(x)=f(2)

 

 

Ans. Here,  

And 

Also 

Since the given function is continuous at 

 

 

 

 

Ans. When  we have  which being a polynomial is continuous at each point  

And, when  we have  which being a polynomial is continuous at each point 

Now    

 ……….(i)

= 

limx→5−f(x)=limh→0f(5−h)=k(5−h)+1=5k−hk+1=5k+1limx→5−⁡f(x)=limh→0⁡f(5−h)=k(5−h)+1=5k−hk+1=5k+1  …….(i)

Since function is continuous, therefore, eq. (i) = eq. (ii)

 

 

 

Ans.  For  function is  constant, therefore it is continuous. 

For  function  polynomial, therefore, it is continuous.

For  function is  constant, therefore it is continuous.

For continuity at  

 

  ……….(i)

For continuity at  

 

 ……….(ii)

Solving eq. (i) and eq. (ii), we get

 and 

Ans. Let  and , then 

Now  and  being continuous it follows that their composite  is continuous.

Hence  is continuous function.

Ans. Given:  ….(i) 

 has a real and finite value for all  R.

 Domain of  is R.

Let  and 

Since  and  being cosine function and modulus function are continuous for all real 

Now,  being the composite function of two continuous functions is continuous, but not equal to 

Again,   [Using eq. (i)]

Therefore,  being the composite function of two continuous functions is continuous.

Ans. Let  and , then 

Now,  and  being continuous, it follows that their composite,  is continuous.

Therefore,  is continuous.

Ans. Given:  

When    = 

When   

When   

 

At  L.H.L. = 

R.H.L. = 

And f(−1)=−2×(−1)−1=1f(−1)=−2×(−1)−1=1

Therefore, at   is continuous.

At  L.H.L. =  

R.H.L. = 

And 

Therefore, at   is continuous.

Hence, there is no point of discontinuity.