Find ‘a’ and ‘b’, if the function given by is differentiable at x = 1.

OR


Determine the values of ‘a’ and ‘b’ such that the following function is continuous at x = 0:



As the function f(x) is differentiable at x = 1 hence it is also continuous at x = 1 because differentiable function is also continuous.


As f(x) is continuous at x = 1


left hand limit of 1 = right hand limit of 1 = f(1)


As f(x) = 2x + 1 for x ≥ 1


f(1) = 2(1) + 1 = 3 …(p)


Now left-hand limit



For x < 1 f(x) = ax2 + b



left hand limit = a (1)2 + b


left hand limit = a + b …(q)


Now right-hand limit



For x > 1 f(x) = 2x + 1


right hand limit = 2(1) + 1


right hand limit = 3 …(r)


Equating (p), (q) and (r) as f(x) is continuous


a + b = 3 …(i)


As f(x) is differentiable at x = 1 hence the left-hand derivative and the right hand derivative should be equal


Let us find the left-hand derivative



We have to check differentiability at 1



For x = 1 f(x) = 2x + 1 and left-hand limit refers to the left hand side of 1 which means less than 1, for x < 1 f(x) = ax2 + b






Using (i)





left hand derivative = 2a …(j)


Now right hand derivative



We have to check differentiability at 1



For x = 1 f(x) = 2x + 1 and right hand limit refers to the right hand side of 1 which means greater than 1, for x > 1 f(x) = 2x + 1




right hand derivative = 2 …(k)


Equate (j) and (k) as f(x) is differentiable


2a = 2


a = 1


Pute a = 1 in (i)


1 + b = 3


b = 2


Hence if f(x) is differentiable at x = 1 then a = 1 and b = 2


OR


For f(x) to be continuous at x = 0 left hand limit of 0 = right hand limit of 0 = f(0)


f(0) = 2 …given …(i)


Now left-hand limit


left hand limit


For x < 0



left hand limit


Divide numerator and denominator by x



Now for multiply divide by (a + 1)



We know that




Now let us find the right-hand limit



For x > 0



right hand limit


Multiply divide by sin bx




We know that



right hand limit = 2 …(iii)


Note: Verify using L'Hopital's rule that


Equate (i), (ii) and (iii) for f(x) to be continuous at x = 0





a + 1 = 1


a = 1 - 1


a = 0


As there is no equation for b.


So, b can be any number other than zero because the continuity of f(x) is not dependent on b.


(Why can’t b = 0? Observe that when x > 0 , if here b = 0 then f(x) would be undefined because denominator will become 0 hence b ≠ 0)


Hence for f(x) to be continuous a = 0 and b can be any number but b ≠ 0.


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