Q. 115.0( 2 Votes )

Find all points of discontinuity of f, where f is defined as follows:



OR

Find , if y = (cos x)x + (sin x)1/x.

Answer :

The given function is defined at all points of real line. So, let d be any point on the number line.

Case I: When d < -3


For this case, we can define function as,


f(x) = -x + 3, x ≤ -3


So,


At d < -3, f(x) = -x + 3


And f(d) = -d + 3


That is,




Therefore, the function is continuous at all points x < -3.


Case II: When d = -3


We have the function f(x) = -x + 3, x = -3


So,


At d = -3, f(x) = -x + 3


And f(-3) = -(-3) + 3


f(-3) = 3 + 3


f(-3) = 6


For the function to be continuous at x = -3,


Left Hand Limit (LHL) = Right Hand Limit (RHL) = f(-3)


That is,


For LHL,


We have the function f(x) = -x + 3, x < -3


So,






For RHL,


We have the function f(x) = -2x, x > -3


So,





Since,


Therefore, the function is continuous at x = -3.


Case III: When -3 < d < 3


For this case, we have the function f(x) = -2x, -3 < x < 3


At -3 < d < 3, f(x) = -2x


And f(d) = -2d


That is,




Therefore, the function is continuous at all points -3 < x < 3.


Case IV: When d = 3


We have the function f(x) = 6x + 2, x = 3


So,


At d = 3, f(x) = 6x + 2


And f(3) = 6(3) + 2


f(3) = 18 + 2


f(3) = 20


For the function to be continuous at x = 3,


Left Hand Limit (LHL) = Right Hand Limit (RHL) = f(3)


That is,


For LHL,


We have the function f(x) = -2x, x < 3


So,





For RHL,


We have the function f(x) = 6x + 2, x > 3


So,






Since,


Therefore, the function is discontinuous at x = 3.


Case V: When d > 3


For this case, we have the function f(x) = 6x + 2, x > 3


So,


At d > 3, f(x) = 6x + 2


And f(d) = 6d + 2


That is,




Therefore, the function is continuous at all points x > 3.


Thus, the function is discontinuous at x = 3.


OR


We need to find the differentiation of y with respect to x.


Let u = (cos x)x …(i)


Let v = (sin x)1/x …(ii)


For u = (cos x)x


Taking log on both sides, we get


log u = log (cos x)x


log u = x log (cos x)


Differentiate u with respect to x,







Substitute value of u from eq (i),


…(iii)


For v = (sin x)1/x


Taking log on both sides, we get


log v = log (sin x)1/x



Differentiate v with respect to x,








Substitute value of v from eq (ii),


…(iv)


We know,


y = u + v



Substituting the value of and from eq (iii) and (iv),



Rearranging it,



Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses