JEE Advanced Previous Year Questions (2018 - 2023): Limits, Continuity and Differentiability

# JEE Advanced Previous Year Questions (2018 - 2023): Limits, Continuity and Differentiability | Mathematics (Maths) for JEE Main & Advanced PDF Download

 Table of contents 2023 2022 2021 2020 2019 2018

## 2023

Q1: Let f : (0, 1)→ R be the function defined as , where [x] denotes the greatest integer less than or equal to x. Then which of the following statements is(are) true?(a) The function f is discontinuous exactly at one point in (0, 1)
(b) There is exactly one point in (0, 1) at which the function f is continuous but NOT differentiable
(c) The function f is NOT differentiable at more than three points in (0, 1)
(d) The minimum value of the function f is -1/512                  [JEE Advanced 2023 Paper 2]
Ans:
(a) & (b)

f (x) is discontinuous at x = 3/4 only

(x) is non-differentiable at x = 1/2 and 3/4
Minimum values of f(x) occur at x = 5/12 whose value is  -1/432

## 2022

Q1: For positive integer n, define

Then, the value of  is equal to :
(a)

(b)
(c)
(d)                  [JEE Advanced 2022 Paper 2]
Ans: (b)

Q2: If  then the value of  is ___________.     [JEE Advanced 2022 Paper 2]
Ans:
5
Given,

[Neglecting higher power of x]

Q3: Let α be a positive real number. Let  be the functions defined by

Then the value of  is_______.      [JEE Advanced 2022 Paper 1]
Ans:  0.49 to 0.51
[As f(x) is continuous function so we can write this]
Now,

From graph you can see

= 2

## 2021

Q1: Let f : R  R be defined by Then which of the following statements is (are) TRUE?
(a) f is decreasing in the interval (−2, −1)
(b) f is increasing in the interval (1, 2)
(c) f is onto
(d) Range of f is [−3/2, 2]                [JEE Advanced 2021 Paper 1]
Ans:
(a) & (b)
Given,
---- (i)

Sign scheme for f'(x)
Here, f is decreasing in the interval (2, 1) and f is increasing in the interval (1, 2).
Now, and

∴ Range =

Hence, f(x) is into.f(x) has local maxima at x = 4
and local minima at x = 0.

## 2020

Q1: Let f : R → R and g : R → R be functions satisfying f(x + y) = f(x) + f(y) + f(x)f(y)
and f(x) = xg(x) for all x, y ∈ R.
If , then which of the following statements is/are TRUE?

(a) f is differentiable at every x∈R
(b) If g(0) = 1, then g is differentiable at every x ∈ R
(c) The derivative f'(1) is equal to 1
(d) The derivative f'(0) is equal to 1          [JEE Advanced 2020 Paper 2]
Ans:
(a), (b) & (d)
The given function f : R  R is satisfying as

Therefore, f(x) = e− 1 is differentiable at every x ∈ R.
and Now,

LHD (at x = 0) of

and, RHD (at x = 0) of

So, if g(0) = 1, then g is differentiable at every x ∈ R.

Q2: Let the function f : R  R be defined by f(x) = x3  x2 + (x  1)sin x and let g : R  R be an arbitrary function. Let fg : R  R be the product function defined by (fg)(x) = f(x)g(x). Then which of the following statements is/are TRUE?
(a) If g is continuous at x = 1, then fg is differentiable at x = 1
(b) If f g is differentiable at x = 1, then g is continuous at x = 1
(c) If g is differentiable at x = 1, then fg is differentiable at x = 1
(d) If f g is differentiable at x = 1, then g is differentiable at x = 1     [JEE Advanced 2020 Paper 1]
Ans:
(a) & (c)
Given functions f : R  R be defined
by f(x) = x3  x2 + (x + 1) sin x and g : R  R be an arbitrary function.
Now, let g is continuous at x = 1, then

{ f(1) = 0 and g is continuous at x = 1, so g(1  h) = g(1)}

Similarly,

RHD and LHD of function fg at x = 1 is finitely exists and equal, so fg is differentiable at x = 1
Now, let (fg)(x) is differentiable at x = 1, so

It does not mean that g(x) is continuous or differentiable at x = 1.
But if g is differentiable at x = 1, then it must be continuous at x = 1 and so fg is differentiable at x = 1.

Q3: The value of the limit  is ________.                    [JEE Advanced 2020 Paper 2]Ans: 8
The Limit

= 8

Q4: let e denote the base of the natural logarithm. The value of the real number a for which the right hand limit is equal to a non-zero real number, is ____ .    [JEE Advanced 2020 Paper 1]
Ans: 1
The right hand limit

The above limit will be non-zero, if a = 1. And at a = 1, the value of the limit is

## 2019

Q1: For a ∈ R, |a| > 1, let

(a) -6
(b) -7
(c) 8
(d) -9                      [JEE Advanced 2019 Paper 2]
Ans:
(c) & (d)

Hence, options (c) and (d) are correct.

Q2: Let f : R  R be given by

Then which of the following options is/are correct?
(a) f is increasing on (−∞, 0)
(b) f' is not differentiable at x = 1
(c) f is onto
(d) f' has a local maximum at x = 1             [JEE Advanced 2019 Paper 1]
Ans:  (
b), (c) & (d)
Given function f : R  R is

So,

At x = 1, f"(1-) = 2 > 0 and f"(1+) = 48 = 4 < 0
f'(x) is not differentiable at x = 1 and f'(x) has a local maximum at x = 1.
For x  (−∞, 0)
f'(x) = 5x4 + 20x3 + 30x2 + 20x + 3
and since
f'(1) = 520 + 20 + 30  20 + 3 = 2 < 0
So, f(x) is not increasing on x (−∞, 0).
Now, as the range of function f(x) is R, so f is onto function.
Hence, options (b), (c) and (d) are correct.

## 2018

Q1: For every twice differentiable function f : R → [−2, 2] with (f(0))2 + (f′(0))2 = 85, which of the following statement(s) is(are) TRUE?
(a) There exist r, s ∈ R, where r < s, such that f is one-one on the open interval (r, s)
(b) There exists x0 ∈ (−4, 0) such that |f'(x0)| ≤ 1
(c) (d) There exists α ∈ (−4, 4) such that f(α) + f"(α) = 0 and f'(α) ≠ 0      [JEE Advanced 2018 Paper 1]Ans:
(a), (b) & (d)
We have,
(f(0))2 + (f′(0))2 = 85
and f : R → [−2, 2]
(a) Since, f is twice differentiable function, so f is continuous function.
This is true for every continuous function.
Hence, we can always find x  (r,  s), where f(x) is one-one.
This statement is true.
(b) By L.M.V.T.

Range of f is [−2, 2]

Hence, |f'(x0)| = 1.
Hence, statement is true.
(c) As no function is given, then we assume

Now,
and  does not exists.Hence, statement is false.
(d) From option b,

hence,

Now, let p  (4, 0) for which g(p) = 5
Similarly, let q be smallest positive number q  (0, 4)
such that g(q) = 5
Hence, by Rolle's theorem is (p, q)
g'(c) = 0 for α  (4, 4) and since g(x) is greater than 5 as we move from x = p to x = q
and f(x))2  4
(f'(x))2  1 in (p, q)
Thus, g'(c) = 0
f'f + f'f" = 0
So, f(α) + f"(α) = 0 and f'(α 0
Hence, statement is true.

Q2: Let f : R  R and g : R  R be two non-constant differentiable functions. If f'(x) = (e(f(x) − g(x))) g'(x) for all x ∈ R and f(1) = g(2) = 1, then which of the following statement(s) is (are) TRUE?
(a) f(2) < 1 − loge2
(b) f(2) > 1 − loge2
(c) g(1) > 1 − loge2
(d) g(1) < 1 − loge2                 [JEE Advanced 2018 Paper 1]
Ans:
(b) & (c)
We have,

On integrating both side, we get

At x = 1

At x = 2

From Eqs. (i) and (ii)

We know that, e−x is decreasing
f(2) < loge 21
f(2) > 1  loge 2

g(1) > loge 2

Q3: Let  and  be functions defined by
(i)
(ii) the inverse trigonometric function tan−1x assumes values in (-π/2, π/2)
(iii) , where for t ∈ R, [t] denotes the greatest integer less than or equal to t,
(iv)

(a) P → 2 ; Q → 3 ; R → 1 ; S → 4
(b) P → 4 ; Q → 1 ; R → 2 ; S → 3
(c) P → 4 ; Q → 2 ; R → 1 ; S → 3
(d) P → 2 ; Q → 1 ; R → 4 ; S → 3                   [JEE Advanced 2018 Paper 2 ]
Ans:
(d)

(i) Given,
f1 : R  R and f1(x) =
f1(x) is continuous at x = 0
Now,

At x = 0
f1'(x) does not exists.
f1(x) is not differential at x = 0
Hence, option (2) for P.

(ii) Given,

Clearly, f2(x) is not continuous at x = 0.
Option (1) for Q.

(iii) Given, f3(x) = [sin(loge(x + 2))], where [ ] is G.I.F.
and f3 : (1, eπ/2  2)  R
It is given,

It is differentiable and continuous at x = 0.
Option (4) for R
(iv) Given,

Now,

thus

Again,

does not exists.
Since, does not exists.
Hence, f'(x) is not continuous at x = 0.
Option (3) for S.

Q4: The value of  is ________.   [JEE Advanced 2018 Paper 1]
Ans:
8

= 22  x 2
= 8

The document JEE Advanced Previous Year Questions (2018 - 2023): Limits, Continuity and Differentiability | Mathematics (Maths) for JEE Main & Advanced is a part of the JEE Course Mathematics (Maths) for JEE Main & Advanced.
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