JEE Advanced (Single Correct MCQs): Limits, Continuity and Differentiability - JEE MCQ

By def. [x – π] is an integer whatever be the value of x.
And so p[x – π] is an integral multiple of p.
Consequently tan 
And since 1 + [x]2 ≠ 0 for any x, we conclude that f (x) = 0.
Thus f (x) is constant function and so, it is continuous and differentiable any no. of times, that is f '(x), f ''(x), f '''(x),... all exist for every x, their value being 0 at every pt. x. Hence, out of all the alternatives only (d) is correct.

f (x) = e^–x is one such function.

For f (x) to be continuous at x = 0

= a + b

The given function can be restated as

We have f : R → R, a differentiable function and f (1) = 4
NOTE THIS STEP

We have f (x)  = [tan² x]

tan x is an increasing function for 

When x is not an integer, both the functions [x] and cos   are continuous.
∴f (x) is continuous on all non integral points.

∴ f is continuous at all integral pts as well.
Thus, f is continuous everywhere.

We have f (x) = [x]² – [x²]

At x = 0,

f (1) = [1]² – [1²] = 1 – 1 = 0
∴ L.H.L. = R.H.L. = f (1)
∴ f (x) is continuous at x = 1.
Clearly at other integral pts f (x) is not continuous.

∴ Given function can be written as

This function is differentiable at all points except possibly at x = 1 and x = 2.


Lf  (2) ≠ Rf '(2)
∴ f is not differentiable at x = 2.

For x ∈ R 

KEY CONCEPT
A continuous function f (x) is not differentiable at x = a if graphically it takes a sharp turn at x = a.
Graph of f (x) = max {x, x³} is as shown with solid lines.

From graph of f (x) at x = – 1, 0, 1, we have sharp turns.
∴ f (x) is not differentiable at x = – 1, 0, 1.

Let us test each of four options.

The given function is

∴ f (x) is discontinuous at x = – 1
Also we can prove in the same way, that f (x) is discontinuous at x = 1
∴ f ' (x) can not be found for x = ± 1 or domain of f ' (x) = R – {– 1, 1}

Given that, 

For this limit to be finite, n – 3 = 0 ⇒ n = 3

Given that f ; R → R such that
f (1) = 3 and f ' (1) = 6

We are given that

∴ Applying L Hospital's rule, we get

Using L.H. Rule, we get

Graph of y = | | x | – 1| is as follows :

The graph has sharp turnings at x = – 1, 0 and 1; and hence not differentiable at x = – 1, 0, 1.

Given that f (x) is a contin uous and differ entiable function and 

Given that f (x) is differentiable on (0, ∞) with

[Linear differential equation]
Integrating factor
NOTE THIS STEP