Test: MCQs (One or More Correct Option): Limits, Continuity and Differentiability | JEE Advanced - JEE MCQ

Given that x + | y | = 2y
If y < 0 then x – y = 2y
⇒ y = x/3 ⇒ x < 0
If y = 0 then x = 0. If y > 0 then x + y = 2y
⇒ y = x ⇒ x > 0

Thus we can define 
Continuity at x = 0

As Lf ' ≠ Rf ' ⇒ f (x) is not differentiable at x = 0

We have 

Let us check differentiability of f (x) at x = 0

Since Lf ' (0) = Rf ' (0)
∴ f is differentiable at x = 0.

The graph of f (x) = 1 + | sin x | is as shown in the fig.

From graph it is clear that function is continuous every where but not differentiable at integral multiples of π (∴ at these points curve has sharp turnings)

We have, for  – 1 < x < 1 ⇒  0 < x sin π x < 1/2
∴ f (x) = [x sin π x] = 0
Also x sin p x becomes negative and numerically less than 1 when x is slightly greater than 1 and so by definition of [x],
f (x) = [x sin π x] = – 1 when 1 < x < 1 + h
Thus f (x) is constant and equal to 0 in the closed interval [– 1, 1] and so f (x) is continuous and differentiable in the open interval (– 1, 1).
At x = 1, f (x) is clearly discontinuous, since f (1– 0) = 0 and f (1 + 0) = – 1 and f (x) is non-differentiable at x = 1

The given function is,

⇒ f is differentiable at x = 0
Hence f is differentiable in (– ∞, ∞).

∴ f is differentiable at x = 1 and hence continuous at x = 1.
Again,  Lf ' (3) = – 1 and Rf ' (3) = 1
⇒ Lf ' (3) ≠ Rf ' (3)
⇒ f is not differentiable at x = 3

∴ The function tan [f (x)] is discontinuous at x = 2.

discontinuous at x = 2.
Thus both the given functions tan [f (x)] as well as
 are discontinuous on the interval [0, π].

∴ The given limit does not exist.

On (0, π)
(a) tan x = f (x)
We know that tan x is discontinuous at x = π/2

which exists on (0, π)
∴ f (x), being differentiable, is continuous on (0, π).

Clearly f (x) is continuous on (0, π) except



Here f (x) will be continuous on (0, π) if it is continuous at x = π/2.
At x =π/2,


∴ f (x) is not continuous on (0, π).

From graph of f ' (x), it is clear that f ' (x) is continuous but not differentiable at x = 0.

  is not continuous, therefore g' (x) is not continuous at x = 0.
At x = 0

which does not exist.
∴ f is not differentiable at x = 0.

From graph it is clear that f (x) is continuous every where and also differentiable everywhere except at  x = 1 and – 1.

From the figure it is clear that

From the graph it is clear that h is continuous for all x ∈ R , h' (x) = 1 for all x > 1 and h is not differentiable at x = 0 and 1.

From graph, f (x) is continuous everywhere but not differentiable at x = 1.

and L is finite.

∴ L is finite, limiting value of numerator zero which is so when should be 

i.e. a = 2 (∴ a > 0)

Applying L ‘Hospital’s rule again, we ge

Which is continuous and differentiable
∴ b and c are the correct options.

Also at x  = 0
Lf '(0) = sin 0 = 0; Rf '(0) = 1 – 0 = 1
∴ Lf '(0) ≠ Rf '(0)
⇒ f is not differentiable at x = 0
At x = 1
Lf '(1) = R'f (1) ⇒ f is differentiable at x = 1.
which is differentiable.
∴ All four options are correct.

As f is continuous for all n
∴At x = 2n, LHL = RHL = f (2n)

g(x) may be discontinuous at x = a or x = b.
Let us check the continuity of g(x) at x = a and x = b

∴ g (x) is continuous at x = a



∴ g is not differentiable at a and b

Let f and g be maximum at c₁ and c₂ respectively,

which shows that (a) and (d) are correct.

∴ hof is differentiable at x = 0.

 f(x) = a cos³(|x³ –x|) + b |x| sin (|x³ + x|)
(a) If a = 0, b = 1
⇒ f(x) = |x| sin |x³ + x|
= x sin (x³ + x), x ∈ R
∴ f is differentiable every where.
(b), (c) If a = 1, b = 0 ⇒ f(x) = cos³ (|x³ – x|)
= cos³(x³ – x)
which is differentiable every where.
(d) when a = 1, b = 1, f(x) = cos(x³ – x) + x sin (x³ + x) which is differentiable at x = 1
∴ Only a and b are the correct options.

f(x) = [x² – 3] is discontinuous at all integral points in 

∴ f is discontinuous exactly at four points in 

Also g (x) = (|x| + |4x- 7|)f (x)

Here f is not differentiable at   

and |x| + |4x – 7| is not differentiable at 0 and 7/4

∴ g(x) becomes differentiable at x = 7/4

Hence g(x) is non-differentiable at four points i.e.