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All questions of Continuity and Differentiability for JEE Exam
Therefore, f(x) continuous everywhere.
- a)3 cos2 x. cos 4x
- b)cos 3x. cos3x
- c)-9 cos 3x. cos2x sinx
- d)3 sin2x.sin 4x
Correct answer is option 'D'. Can you explain this answer?
a)
3 cos2 x. cos 4x
b)
cos 3x. cos3x
c)
-9 cos 3x. cos2x sinx
d)
3 sin2x.sin 4x
|
Deepak Kapoor answered |
Given: y = sin 3x . sin3
= 3 sin2x.sin 4x
= 3 sin2x.sin 4x
Whta is the derivatve of y = log5 (x)- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
Whta is the derivatve of y = log5 (x)
a)
b)
c)
d)
|
Tanuja Kapoor answered |
y = log5 x = ln x/ln 5 → change of base
= ln x/ln 5
dy/dx = 1/ln5⋅1/x → 1/ln5 is a constant, so we don't change it
= 1/(x ln 5)
= ln x/ln 5
dy/dx = 1/ln5⋅1/x → 1/ln5 is a constant, so we don't change it
= 1/(x ln 5)
If xy = 2, then dy/dx is- a)-x2/2
- b)y2
- c)-2/x2
- d)-y/x
Correct answer is option 'D'. Can you explain this answer?
If xy = 2, then dy/dx is
a)
-x2/2
b)
y2
c)
-2/x2
d)
-y/x
|
Hansa Sharma answered |
xy = 2
x dy/dx + y = 0
x dy/dx = -y
dy/dx = -y/x
x dy/dx + y = 0
x dy/dx = -y
dy/dx = -y/x
A real function f is said to be continuous if it is continuous at every point in …… .- a)[-∞,∞]
- b)The range of f
- c)The domain of f
- d)Any interval of real numbers
Correct answer is option 'A'. Can you explain this answer?
A real function f is said to be continuous if it is continuous at every point in …… .
a)
[-∞,∞]
b)
The range of f
c)
The domain of f
d)
Any interval of real numbers
|
Saranya Choudhury answered |
Its domain. This means that for any point x in the domain of f, as x approaches a certain value a, the value of f(x) approaches f(a). In other words, there are no sudden jumps or gaps in the graph of f.
More formally, a function f is continuous at a point a if:
1. f(a) is defined (i.e. a is in the domain of f).
2. The limit of f(x) as x approaches a exists (i.e. the left and right-hand limits are equal).
3. The limit of f(x) as x approaches a is equal to f(a).
If a function is continuous at every point in its domain, it is called a continuous function. Continuous functions have many useful properties and are often used in mathematical models and real-world applications.
More formally, a function f is continuous at a point a if:
1. f(a) is defined (i.e. a is in the domain of f).
2. The limit of f(x) as x approaches a exists (i.e. the left and right-hand limits are equal).
3. The limit of f(x) as x approaches a is equal to f(a).
If a function is continuous at every point in its domain, it is called a continuous function. Continuous functions have many useful properties and are often used in mathematical models and real-world applications.
Derivatve of f(x) 
is given by
- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
Derivatve of f(x) is given by
is given bya)
b)
c)
d)
|
Neha Sharma answered |
y
The derivative of y = ef(x)is dy/dx = f'(x)ef(x)
In this case, f(x) = x2 , and the derivative of x2 = 2x
Therefore, f'(x)= 2x,
dy/dx = 2x
The derivative of y = ef(x)is dy/dx = f'(x)ef(x)
In this case, f(x) = x2 , and the derivative of x2 = 2x
Therefore, f'(x)= 2x,
dy/dx = 2x
Which of the following functions are not continuous.- a)
- b)
- c)ex
- d)
Correct answer is option 'A'. Can you explain this answer?
Which of the following functions are not continuous.
a)
b)
c)
ex
d)
|
Anu answered |
b,c,and d are continuous and [x] is discontinuous at integer points.
= .....
y = log(sec + tan x)
- a)sec x tan x – 1
- b)sec x
- c)sec x tan x + 1
- d)tan x
Correct answer is option 'B'. Can you explain this answer?
y = log(sec + tan x)
a)
sec x tan x – 1
b)
sec x
c)
sec x tan x + 1
d)
tan x
|
Lavanya Menon answered |
y = log(secx + tanx)
dy/dx = 1/(secx + tanx){(secxtanx) + sec2x}
= secx(secx + tanx)/(secx + tanx)
= secx
dy/dx = 1/(secx + tanx){(secxtanx) + sec2x}
= secx(secx + tanx)/(secx + tanx)
= secx
For what values of a and b, f is a continuous function.- a)a=2,b=0
- b)a=1,b=0
- c)a=0,b=2
- d)a=0,b=0
Correct answer is 'A'. Can you explain this answer?
For what values of a and b, f is a continuous function.a)
a=2,b=0
b)
a=1,b=0
c)
a=0,b=2
d)
a=0,b=0
|
|
Tejas Verma answered |
For continuity: LHL=RHL
at x=2,
LHL: x < 2 ⇒ f(x) = 2*a
RHL: x ≥ 2 ⇒ f(x) = 4
For continuity: LHL = RHL
⇒ 2a = 4 ⇒ a = 2
LHL: x < 2 ⇒ f(x) = 2*a
RHL: x ≥ 2 ⇒ f(x) = 4
For continuity: LHL = RHL
⇒ 2a = 4 ⇒ a = 2
at x = 0,
LHL: x < 0 ⇒ f(x) = b
RHL: x ≥ 0 ⇒ f(x) = 0 * a
For continuity: LHL = RHL
⇒ b = 0
LHL: x < 0 ⇒ f(x) = b
RHL: x ≥ 0 ⇒ f(x) = 0 * a
For continuity: LHL = RHL
⇒ b = 0
F(x) = tan (log x)
F'(x) =- a)sec2 (logx)
- b)
- c)
- d)-x sec2 (logx)
Correct answer is option 'B'. Can you explain this answer?
F(x) = tan (log x)
F'(x) =
F'(x) =
a)
sec2 (logx)
b)
c)
d)
-x sec2 (logx)
|
Virendra Singh answered |
I have chosen (b) but it is showing (d) is correct
Correct answer is option 'A'. Can you explain this answer?
|
Aryan Khanna answered |
y = tan-1(1-cosx)/sinx
y = tan-1{2sin2(x/2)/(2sin(x/2)cos(x/2)}
y = tan-1{tan x/2}
y = x/2 => dy/dx = 1/2
y = tan-1{2sin2(x/2)/(2sin(x/2)cos(x/2)}
y = tan-1{tan x/2}
y = x/2 => dy/dx = 1/2
Function f(x) = log x + 
is continuous at- a)(0,1)
- b)[-1,1]
- c)(0,∞)
- d)(0,1]
Correct answer is option 'D'. Can you explain this answer?
Function f(x) = log x + is continuous at
is continuous ata)
(0,1)
b)
[-1,1]
c)
(0,∞)
d)
(0,1]
|
Om Desai answered |
- [-1,1] cannot be continuous interval because log is not defined at 0.
- The value of x cannot be greater than 1 because then the function will become complex.
- (0,1) will not be considered because its continuous at 1 as well. Hence D is the correct option.
Can you explain the answer of this question below:
- A:
1/2
- B:
0
- C:
1
- D:
none of these
The answer is b.
1/2
0
1
none of these
|
Shivani answered |
Apply L-hospital rule.....u will get the ans....
- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
a)
b)
c)
d)
|
Poonam Reddy answered |
y + sin y = 5x
dy/dx + cos ydy/dx = 5
dy/dx = 5/(1+cos y)
dy/dx + cos ydy/dx = 5
dy/dx = 5/(1+cos y)
If x sin (a + y) = sin y, then 
is equal to- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
If x sin (a + y) = sin y, then is equal to
is equal toa)
b)
c)
d)
|
Swati Verma answered |
x sin(a+y) = sin y
⇒
⇒
Examine the continuity of function 
- a)Discontinuous at x=1,2
- b)Discontinuous at x=1
- c)Continuous everywhere.
- d)Discontinuous at x=2
Correct answer is option 'C'. Can you explain this answer?
Examine the continuity of function
a)
Discontinuous at x=1,2
b)
Discontinuous at x=1
c)
Continuous everywhere.
d)
Discontinuous at x=2
|
|
Om Desai answered |
Lim f (x) = lim (x-1)(x-2) at x tend to k
► So it get k2-3k+2
► Now f (k) = k2 -3k+2
► So f (x) =f (k) so continous at everywhere
- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
a)
b)
c)
d)
|
Suhani Dangarh answered |
Put x=tan thita. then you will get. 2 tan inverse x then differentiate
If f(x) = x + cot x, 
- a)-4
- b)2
- c)4
- d)-2
Correct answer is option 'C'. Can you explain this answer?
If f(x) = x + cot x,
a)
-4
b)
2
c)
4
d)
-2
|
|
Aryan Khanna answered |
f(x) = x + cot x
f’(x) = 1 + (-cosec2 x)
f”(x) = 0 - 2cosec x(-cosec x cot x)
= 2 cosec2 x cot x
f”(π/4) = 2 cosec2 (π/4) cot(π/4)
= 2 [(2)^½]2 (1)
= 4
f’(x) = 1 + (-cosec2 x)
f”(x) = 0 - 2cosec x(-cosec x cot x)
= 2 cosec2 x cot x
f”(π/4) = 2 cosec2 (π/4) cot(π/4)
= 2 [(2)^½]2 (1)
= 4
- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
a)
b)
c)
d)
|
Rocky Gupta answered |
X^a y^b = (x + y)^(a + b)
taking ln on both sides :-
alnx + b lny = (a + b) ln(x + y)
diff both sides w.r.t x :-
a/x + by'/y = ( (a + b)/(x + y) ) + (((a + b) y'))/(x + y)
or,
a/x - (a +b)/(x + y) = y'[((a + b) / (x + y)) - b/y]
or,
(ax + ay - ax - bx)/x = y' [ (ay + by - bx - by)/y] (cancel (x + y))
or,
y' = dy/dx = ( y (ay - bx) )/(x( ay - bx)) = y/x
therefore we can easily say that the option (D) is the correct answer
taking ln on both sides :-
alnx + b lny = (a + b) ln(x + y)
diff both sides w.r.t x :-
a/x + by'/y = ( (a + b)/(x + y) ) + (((a + b) y'))/(x + y)
or,
a/x - (a +b)/(x + y) = y'[((a + b) / (x + y)) - b/y]
or,
(ax + ay - ax - bx)/x = y' [ (ay + by - bx - by)/y] (cancel (x + y))
or,
y' = dy/dx = ( y (ay - bx) )/(x( ay - bx)) = y/x
therefore we can easily say that the option (D) is the correct answer
If f(x) = ex, then the value of f'(-3) is- a)log (3)
- b)e2
- c)log (-3)
- d)e-3
Correct answer is option 'D'. Can you explain this answer?
If f(x) = ex, then the value of f'(-3) is
a)
log (3)
b)
e
2
c)
log (-3)
d)
e-3
|
Naincy Tripathi answered |
As , f (x) = e^x now put (-3)in place of x as it is asking f (-3) f (-3) = e^-3
Differentiate 
with respect to x.- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
Differentiate with respect to x.
with respect to x.a)
b)
c)
d)
|
|
Tejas Verma answered |
y = e^(-x)2.................(1)
Put u = (-x)2
du/dx = -2x dx
Differentiating eq(1) y = eu
dy/du = e^u
⇒ dy/dx = (dy/du) * (du/dx)
= (eu) * (-2x)
⇒ - 2xe(-x2)
Put u = (-x)2
du/dx = -2x dx
Differentiating eq(1) y = eu
dy/du = e^u
⇒ dy/dx = (dy/du) * (du/dx)
= (eu) * (-2x)
⇒ - 2xe(-x2)
- a)2t
- b)1/2a
- c)-t/2a
- d)t/4a
Correct answer is option 'B'. Can you explain this answer?
a)
2t
b)
1/2a
c)
-t/2a
d)
t/4a
|
Knowledge Hub answered |
y = at4, x = 2at2
dy/dt = 4at3 dx/dt = 4at => dt/dx = 1/4at
Divide dy/dt by dx/dt, we get
dy/dx = t2
d2 y/dx2 = 2t dt/dx……………….(1)
Put the value of dt/dx in eq(1)
d2 y /dx2 = 2t(1/4at)
= 1/2a
dy/dt = 4at3 dx/dt = 4at => dt/dx = 1/4at
Divide dy/dt by dx/dt, we get
dy/dx = t2
d2 y/dx2 = 2t dt/dx……………….(1)
Put the value of dt/dx in eq(1)
d2 y /dx2 = 2t(1/4at)
= 1/2a
f (x) = max {x, x3},then the number of points where f (x) is not differentiable, are- a)1
- b) 2
- c)3
- d)4
Correct answer is option 'C'. Can you explain this answer?
f (x) = max {x, x3},then the number of points where f (x) is not differentiable, are
a)
1
b)
2
c)
3
d)
4
|
|
Gaurav Rane answered |
Solution:
To find the points where the given function is not differentiable, we need to find the points where the function is not continuous or where the derivative does not exist.
Step 1: Finding the points of non-differentiability where the function is not continuous.
Since the function is the maximum of two functions, we need to consider the points where these two functions are equal.
Let x = x³, then x³ - x = 0, which gives x = 0, 1.
Therefore, the function is not continuous at x = 0 and x = 1.
Step 2: Finding the points of non-differentiability where the derivative does not exist.
At x = 0, the left-hand derivative is 1 and the right-hand derivative is 0.
At x = 1, the left-hand derivative is 3 and the right-hand derivative is 0.
Therefore, the function is not differentiable at x = 0 and x = 1.
Hence, the function f(x) = max{x, x³} is not differentiable at three points, namely x = 0, x = 1, and wherever the function changes from x to x³ or vice versa.
Therefore, the correct option is (C) 3.
To find the points where the given function is not differentiable, we need to find the points where the function is not continuous or where the derivative does not exist.
Step 1: Finding the points of non-differentiability where the function is not continuous.
Since the function is the maximum of two functions, we need to consider the points where these two functions are equal.
Let x = x³, then x³ - x = 0, which gives x = 0, 1.
Therefore, the function is not continuous at x = 0 and x = 1.
Step 2: Finding the points of non-differentiability where the derivative does not exist.
At x = 0, the left-hand derivative is 1 and the right-hand derivative is 0.
At x = 1, the left-hand derivative is 3 and the right-hand derivative is 0.
Therefore, the function is not differentiable at x = 0 and x = 1.
Hence, the function f(x) = max{x, x³} is not differentiable at three points, namely x = 0, x = 1, and wherever the function changes from x to x³ or vice versa.
Therefore, the correct option is (C) 3.
What is the point of discontinuity for signum function?- a)x=1
- b)x=-1
- c)x=0
- d)function is continuous on R
Correct answer is option 'C'. Can you explain this answer?
What is the point of discontinuity for signum function?
a)
x=1
b)
x=-1
c)
x=0
d)
function is continuous on R
|
|
Arun Khanna answered |
It has a jumped discontinuity which means if the function is assigned some value at the point of discontinuity it cannot be made continuous. But the function is definitely discontinuous at x=0. the sgn function is a discontinuous function (isolated/jump discontinuity).
Let f (x) = [x], then f (x) is- a)continuous nowhere
- b)differentiable for all x∈ (R−I).
- c)differentiable for all x ∈ R
- d)continuous for all x ∈ R
Correct answer is option 'B'. Can you explain this answer?
Let f (x) = [x], then f (x) is
a)
continuous nowhere
b)
differentiable for all x∈ (R−I).
c)
differentiable for all x ∈ R
d)
continuous for all x ∈ R
|
Nishanth Verma answered |
f(x) = [x] is derivable at all x except at integral points i.e. on R – I .
Find the differential coefficient y = (sec 5x)5x - a)log ( sec 5x) + 5x log (tan 5x)
- b)5(sec 5x)5x [log (sec 5x) + 5x tan 5x ]
- c)(sec 5x)5x [log( sec 5x) + 5x log (tan 5x)]
- d)log (sec 5x) + 5x tan 5x
Correct answer is option 'B'. Can you explain this answer?
Find the differential coefficient y = (sec 5x)5x
a)
log ( sec 5x) + 5x log (tan 5x)
b)
5(sec 5x)5x [log (sec 5x) + 5x tan 5x ]
c)
(sec 5x)5x [log( sec 5x) + 5x log (tan 5x)]
d)
log (sec 5x) + 5x tan 5x
|
Angelica Das answered |
Take log of both sides then differentiate with respect to x. then replace the value of y in the equation
Find the second derivative of excosx- a)-2exsinx
- b)-exsinx
- c)ex(sinx + cosx)
- d)-2excosx
Correct answer is option 'A'. Can you explain this answer?
Find the second derivative of excosx
a)
-2exsinx
b)
-exsinx
c)
ex(sinx + cosx)
d)
-2excosx
|
|
Rounak Nair answered |
**Solution:**
To find the second derivative of the given function, we need to differentiate it twice with respect to x.
First, let's find the first derivative of the function:
f(x) = ex * cosx
Using the product rule, the derivative of f(x) is:
f'(x) = (ex * (-sinx)) + (cosx * ex)
= -ex * sinx + ex * cosx
= ex * (cosx - sinx)
Now, let's find the second derivative of the function. Taking the derivative of f'(x):
f''(x) = (ex * (-sinx)) + (ex * (-cosx))
= -ex * sinx - ex * cosx
= -ex * (sinx + cosx)
Therefore, the second derivative of excosx is -2exsinx, which is option A.
To find the second derivative of the given function, we need to differentiate it twice with respect to x.
First, let's find the first derivative of the function:
f(x) = ex * cosx
Using the product rule, the derivative of f(x) is:
f'(x) = (ex * (-sinx)) + (cosx * ex)
= -ex * sinx + ex * cosx
= ex * (cosx - sinx)
Now, let's find the second derivative of the function. Taking the derivative of f'(x):
f''(x) = (ex * (-sinx)) + (ex * (-cosx))
= -ex * sinx - ex * cosx
= -ex * (sinx + cosx)
Therefore, the second derivative of excosx is -2exsinx, which is option A.
What is/are conditions for a function to be continuous on (a,b)?- a)The function is continuous at each point of (a,b).
- b)The function is right continuous.
- c)The function is left continuous.
- d)Right continuous, left continuous, continuous at each point of (a,b).
Correct answer is option 'D'. Can you explain this answer?
What is/are conditions for a function to be continuous on (a,b)?
a)
The function is continuous at each point of (a,b).
b)
The function is right continuous.
c)
The function is left continuous.
d)
Right continuous, left continuous, continuous at each point of (a,b).
|
|
Nandini Iyer answered |
The three conditions required for a function f is said to be continuous on (a,b) if f is continuous at each point of (a,b), f is right continuous at x = a, f is left continuous at x = b.
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