Chapter 5: Continuity and Differentiability

Finding the left hand and right hand limits for the given function, we have

So,

7 = 2k

k = 7/2 = 3.5

Therefore, the value of k is 3.5

The given function f(x) can be rewritten as,

So, k = ½

Therefore, the value of k is ½

Finding the left hand and right hand limits for the given function, we have

Therefore, the value of k is -1

Finding the left hand and right hand limits for the given function, we have

Therefore, the value of k is ± 1

Finding the left hand and right hand limit for the given function, we have

Now, as the left hand limit and the right hand limit are not equal and the value of both the limits are constant.

Hence, regardless the choice of k, the given function remains discontinuous at x = 0.

Finding the left hand and right hand limit for the given function, we have

So, -1 + a = a + b = 1 + b

-1 + a = a + b and 1 + b = a + b

We get, b = -1 and 1 + -1 = a + -1 ⇒ a = 1

Therefore, the value of a = 1 and b = -1

Given,

Now, the function will not be defined and continuous where

2x + 5 = 0 ⇒ x = -5/2

Therefore, x = -5/2 is the point of discontinuity.

Now,

if f(t) is discontinuous, then 2 – x = 0 ⇒ x = 2

And, 2x – 1 = 0 ⇒ x = ½

Therefore, the required points of discontinuity for the given function are 2 and ½.

Given,

f(x) = |sin x + cos x| at x = π

Now, put g(x) = sin x + cos x and h(x) = |x|

Hence, h[g(x)] = h(sin x + cos x) = |sin x + cos x|

Now,

g(x) = sin x + cos x is a continuous function since sin x and cos x are two continuous functions at x = π.

We know that, every modulus function is a common function is a continuous function everywhere.

Therefore, f(x) = |sin x + cos x| is continuous function at x = π.

We know that, a function f is differentiable at a point ‘a’ in its domain if

Lf’(c) = Rf’(c)

Therefore, f(x) is not differentiable at x = 2.