Continuity question and solution

 Q1 For the function f(x)= x^2.   if 0<=x<2

                                              3x-2.   if 2<=x<=4

                                             x^3.    if.  x>4

Check continuity at x=0. and x=2.

Solution

Limit  at x<0

Limit x<0 not exist so function is not continuous at x=0.

Limit x<2

x^2

2^2=4

Limit at x=2

3x-2

3*2-2=4

Limit  at x>2

3x-2

3*2-2=4

Left hand limit x<2 is equal to right hand limit x>2 and x=2

So function is continuous at x=2.

Q2. The given function f(x)=k(x^2+2) if x<=0

                                                   4x+6.       if x>0

If f(x) is continuous at x=0 find the value of k.

Solution

Left hand limit

At x<0

k(x^2+2)

=2k

Right hand limit

At x>0

4x+6

=6

If function is continuous at x=0 so left hand limit equal to the right hand limit.

2k=6

k=3

Q3. f(x)=kx+1 if x<=π

                 cosx.  If x>π

Is continuous at x=π .find the value of k.

Solution

Left hand limit

x<π

kx+1

kπ+1

Right hand limit 

x>π

cosx

=cosπ

=-1

If function is continuous at x=π .so left hand limit equal to the right hand limit.

kπ+1=-1

kπ=-2

K=-2/π

Q4. f(x)=k(x^2+2).     if x<=0

                3x+1.            if.  x>0

Function is continuous at x=0 find the value of k.

Solution 

Left hand limit 

x<0

k(x^2+2)

k(2)

=2k

Right hand limit

x>0

3x+1

=1

If function is continuous at x=0 so left hand limit equal to the right hand limit.

2k=1

k=1/2