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All questions of Continuity and Differentiability for Commerce Exam
Therefore, f(x) continuous everywhere.
If the function f (x) = x2– 8x + 12 satisfies the condition of Rolle’s Theorem on (2, 6), find the value of c such that f ‘(c) = 0- a)6
- b)4
- c)8
- d)2
Correct answer is option 'B'. Can you explain this answer?
If the function f (x) = x2– 8x + 12 satisfies the condition of Rolle’s Theorem on (2, 6), find the value of c such that f ‘(c) = 0
a)
6
b)
4
c)
8
d)
2
|
Dr Manju Sen answered |
f (x) = x2 - 8x + 12
Function satisfies the condition of Rolle's theorem for (2,6).
We need to find c for which f’(c) = 0
f’(x) = 2x – 8
f’(c) = 2c – 8 = 0
c = 4
Function satisfies the condition of Rolle's theorem for (2,6).
We need to find c for which f’(c) = 0
f’(x) = 2x – 8
f’(c) = 2c – 8 = 0
c = 4
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
When Rolle’s Theorem is verified for f(x) on [a, b] then there exists c such that- a)c ε [a, b] such that f'(c) = 0
- b)c ε (a, b) such that f'(c) = 0
- c)c ε (a, b] such that f'(c) = 0
- d)c ε [a, b) such that f'(c) = 0
Correct answer is option 'B'. Can you explain this answer?
When Rolle’s Theorem is verified for f(x) on [a, b] then there exists c such that
a)
c ε [a, b] such that f'(c) = 0
b)
c ε (a, b) such that f'(c) = 0
c)
c ε (a, b] such that f'(c) = 0
d)
c ε [a, b) such that f'(c) = 0
|
Abhinay Tripathi answered |
Answer is
B) c ∈ (a, b) such that f'(c) = 0.
Statement for Rolle’s Theorem :
Suppose that a function f(x) is continuous on the closed interval [a,b] and differentiable on the open interval (a,b). Then if f(a)=f(b), then there exists at least one point c in the open interval (a,b) for which f′(c)=0.
B) c ∈ (a, b) such that f'(c) = 0.
Statement for Rolle’s Theorem :
Suppose that a function f(x) is continuous on the closed interval [a,b] and differentiable on the open interval (a,b). Then if f(a)=f(b), then there exists at least one point c in the open interval (a,b) for which f′(c)=0.
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)
- a)– 1
- b)0
- c)1
- d)1/2
Correct answer is option 'D'. Can you explain this answer?
a)
– 1
b)
0
c)
1
d)
1/2
|
Angad Gupta answered |
We have to use L'Hopital Rule It is in the form 0/0 So first we have to differentiate it After differentiating we get sinx/2x Then again differentiate it We get cosx/2 and now we get the answer as 1/2
- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
a)
b)
c)
d)
|
Lavanya Menon answered |
On differentiating both sides with respect to x
- 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
Differentiate sin2(θ2 + 1) with respect to θ2- a)sin(2θ2 + 1)
- b)cos(2θ2 + 2)
- c)sin(2θ2 + 2)
- d)cos(2θ2 + 1)
Correct answer is option 'C'. Can you explain this answer?
Differentiate sin2(θ2 + 1) with respect to θ2
a)
sin(2θ2 + 1)
b)
cos(2θ2 + 2)
c)
sin(2θ2 + 2)
d)
cos(2θ2 + 1)
|
Gunjan Lakhani answered |
y = sin2(θ2+1)
v = θ2
dy/d(v) = dydθ/dvdθ
dy/dthη = sin2(V+1)
= 2sin(V+1)⋅cos(V+1)dv/dθ
= 2sin(θ2+1)cos(θ2+1)
= sin2(θ2+1).
v = θ2
dy/d(v) = dydθ/dvdθ
dy/dthη = sin2(V+1)
= 2sin(V+1)⋅cos(V+1)dv/dθ
= 2sin(θ2+1)cos(θ2+1)
= sin2(θ2+1).
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.
Geometrically the Mean Value theorem ensures that there is at least one point on the curve f(x) , whose abscissa lies in (a, b) at which the tangent is- a)Parallel to the x axis
- b)Parallel to the y axis
- c)Parallel to the line joining the end points of the curve
- d)Parallel to the line y = x
Correct answer is option 'C'. Can you explain this answer?
Geometrically the Mean Value theorem ensures that there is at least one point on the curve f(x) , whose abscissa lies in (a, b) at which the tangent is
a)
Parallel to the x axis
b)
Parallel to the y axis
c)
Parallel to the line joining the end points of the curve
d)
Parallel to the line y = x
|
Ujjwal Gupta answered |
No, option C is correct answer.
If 3 sin(xy) + 4 cos (xy) = 5, then 
= .....- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
If 3 sin(xy) + 4 cos (xy) = 5, then = .....
= .....a)
b)
c)
d)
|
|
Dr Manju Sen answered |
3sinxy + 4cosxy = 5
⇒ 5(3/5 sinxy + 4/5 cosxy) = 5
⇒ (3/5 sinxy + 4/5 cosxy) = 1
now (3/5)²+(4/5)² = 1
so let, 3/5 = cosA
⇒ 4/5 = sinA
So , (3/5 sinxy + 4/5 cosxy) = 1
⇒ (cosAsinxy + sinAcosxy) = 1
⇒ sin(A+xy) = 1
⇒ A + xy = 2πk + π/2 (k is any integer)
⇒ sin⁻¹(4/5) + xy = 2πk + π/2
differenciating both sides with respect to x
0 + xdy/dx + y = 0
dy/dx = -y/x
⇒ 5(3/5 sinxy + 4/5 cosxy) = 5
⇒ (3/5 sinxy + 4/5 cosxy) = 1
now (3/5)²+(4/5)² = 1
so let, 3/5 = cosA
⇒ 4/5 = sinA
So , (3/5 sinxy + 4/5 cosxy) = 1
⇒ (cosAsinxy + sinAcosxy) = 1
⇒ sin(A+xy) = 1
⇒ A + xy = 2πk + π/2 (k is any integer)
⇒ sin⁻¹(4/5) + xy = 2πk + π/2
differenciating both sides with respect to x
0 + xdy/dx + y = 0
dy/dx = -y/x
Can you explain the answer of this question below:The derivatve of f(x) = 
- A:
- B:
- C:
- D:
3x2
The answer is b.
The derivatve of f(x) =
3x2
|
.mie. answered |
Here in this function ....firstly... exponential and logarithmic fn are anti to each other...
and therefore cncl out ech other....
here..
in this e and log cncl out ech other... and we are left with ....
log x^3.....
acc to formula..
log m^n = n log m
therefore...
3 log x
so we reduced our eq to this
then taking derivative....
its 3/x
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
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
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
Find the derivate of y = sin4x + cos4x- a)– sin 2x
- b)4 sin3 x + 4 cos3 x
- c)– sin 4x
- d)4 sin x cos x cos 2x
Correct answer is option 'C'. Can you explain this answer?
Find the derivate of y = sin4x + cos4x
a)
– sin 2x
b)
4 sin3 x + 4 cos3 x
c)
– sin 4x
d)
4 sin x cos x cos 2x
|
Mohit Rajpoot answered |
y=sin4x, and z=cos4x
So by using chain rule
df(x)/dx = dsin4x/dx + dcos4x/dx
=dy4/dy * dy/dx + dz4/dz * dzdx
=dy4/dy * dsinxdx + dz4/dz * dcosx/dx
=4y(4−1)⋅cosx+4z(4−1)⋅(−sinx)
=4sin3xcosx − 4cos3xsinx
=4sinxcosx(sin2x − cos2x)
=2sin2x(−cos2x)
=−2sin2xcos2x
=−sin4x
So by using chain rule
df(x)/dx = dsin4x/dx + dcos4x/dx
=dy4/dy * dy/dx + dz4/dz * dzdx
=dy4/dy * dsinxdx + dz4/dz * dcosx/dx
=4y(4−1)⋅cosx+4z(4−1)⋅(−sinx)
=4sin3xcosx − 4cos3xsinx
=4sinxcosx(sin2x − cos2x)
=2sin2x(−cos2x)
=−2sin2xcos2x
=−sin4x
- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
a)
b)
c)
d)
|
Sushil Kumar answered |
now differentiate y with respect to x,
dy/dx = -{[x d/dx(1+x) - (1+x)dx/dx]}/(1+x2)
= -1/(1+x)2
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.
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
Let f and g be differentiable functions such that fog = I, the identity function. If g’ (a) = 2 and g (a) = b, then f ‘ (b) =- a)1/2
- b)2
- c)-2
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
Let f and g be differentiable functions such that fog = I, the identity function. If g’ (a) = 2 and g (a) = b, then f ‘ (b) =
a)
1/2
b)
2
c)
-2
d)
none of these
|
Divey Sethi answered |
f(g(x)) = x
f'(g(x)) g'(x) = 1
put x = a
f'(b) g'(a) = 1
2 f'(b) = 1
f'(b) = 1/2
f'(g(x)) g'(x) = 1
put x = a
f'(b) g'(a) = 1
2 f'(b) = 1
f'(b) = 1/2
The derivative of 2x tan x is- a)2x log 2 [ sec2 x + tan x]
- b)2x tan x [sec x + log 2]
- c)2x [sec2 x + log 2 tan x]
- d)2x [sec2 x + tan x]
Correct answer is option 'C'. Can you explain this answer?
The derivative of 2x tan x is
a)
2x log 2 [ sec2 x + tan x]
b)
2x tan x [sec x + log 2]
c)
2x [sec2 x + log 2 tan x]
d)
2x [sec2 x + tan x]
|
|
Sushil Kumar answered |
dy/dx = 2x(tanx)' + tanx (2x)'
dy/dx = 2x (secx)2 + 2x tanx log2 , {since , (ax)' = ax loga (x)'}
dy/dx = 2x[(Secx)2 + log2 tanx]
dy/dx = 2x (secx)2 + 2x tanx log2 , {since , (ax)' = ax loga (x)'}
dy/dx = 2x[(Secx)2 + log2 tanx]
- 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 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
|
Infinity Academy answered |
As, f (x) = ex
Similarly, f (-3) = e-3
Similarly, f (-3) = e-3
If f(x) = | x | ∀ x ∈ R, then- a)f is discontinuous at x = 0
- b)f is derivable at x = 0 and f ‘ (0) = 1
- c)f is derivable at x = 0 but f’ (0) ≠
- d)none of these
Correct answer is option 'D'. Can you explain this answer?
If f(x) = | x | ∀ x ∈ R, then
a)
f is discontinuous at x = 0
b)
f is derivable at x = 0 and f ‘ (0) = 1
c)
f is derivable at x = 0 but f’ (0) ≠
d)
none of these
|
Sakshi Jain answered |
|X| is a continuous function ,which is clear from its graph but at x=0,|X| is not differentiable since it has a sharp edge at x=0 . hence 'd' is 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....
The value of c for which Lagrange’s theorem f(x) = |x| in the interval [-1, 1] is- a)1/2
- b)1
- c)-1/2
- d)non-existent in the interval
Correct answer is option 'D'. Can you explain this answer?
The value of c for which Lagrange’s theorem f(x) = |x| in the interval [-1, 1] is
a)
1/2
b)
1
c)
-1/2
d)
non-existent in the interval
|
Rakhi Kumari answered |
Since Fx =|x| therefore it's graph will be considerd only in the x axis but not in y axis and also its been in the closed interval [-1, 1] so other options will be neglected
- 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
The differential coffcient 
of the equation yx = e(x - y) is :- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
The differential coffcient of the equation yx = e(x - y) is :
of the equation yx = e(x - y) is :a)
b)
c)
d)
|
Ambition Institute answered |
Taking log both the sides
xlogy = (x-y)loge
Differentiate it with respect to x, we get
x/y dy/dx = logy = loge - xloge dy/dx
dy/dx = (loge - logy)/(x/y + loge)
= y(1 - loge)/(x + y)
xlogy = (x-y)loge
Differentiate it with respect to x, we get
x/y dy/dx = logy = loge - xloge dy/dx
dy/dx = (loge - logy)/(x/y + loge)
= y(1 - loge)/(x + y)
- 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
- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
a)
b)
c)
d)
|
Knowledge Hub answered |
ey =1/ (x+5)
Taking log both side we get
log ey = log {1/(5+x)}
then, y = log {1/(5+x)}
Differentiating both side ,
dy/dx =( x+5) . {-1/(5+x)²}
dy/dx = -1/(5+x) ……..( 1)
Again Differentiating, d²y/dx²= 1/(5+x)²
d²y/dx²= {-1/(5+x)}²
From equation (1)
d²y/dx² = {dy/dx}²
Taking log both side we get
log ey = log {1/(5+x)}
then, y = log {1/(5+x)}
Differentiating both side ,
dy/dx =( x+5) . {-1/(5+x)²}
dy/dx = -1/(5+x) ……..( 1)
Again Differentiating, d²y/dx²= 1/(5+x)²
d²y/dx²= {-1/(5+x)}²
From equation (1)
d²y/dx² = {dy/dx}²
:
Direction: Read the following text and answer the following questions on the basis of the same:Ms. Remka of city school is teaching chain rule to her students with the help of a flow-chartThe chain rule says that if h and g are functions and f(x) = g(h(x)), then
Let f(x) = sin x and g(x) = x3d/dx sin x3 _______.- a)cos (x3)
- b)–cos (x3)
- c)3x2 sin (x3)
- d)3x2 cos (x3)
Correct answer is option 'D'. Can you explain this answer?
Direction: Read the following text and answer the following questions on the basis of the same:
Ms. Remka of city school is teaching chain rule to her students with the help of a flow-chart
The chain rule says that if h and g are functions and f(x) = g(h(x)), then
Let f(x) = sin x and g(x) = x3
d/dx sin x3 _______.
a)
cos (x3)
b)
–cos (x3)
c)
3x2 sin (x3)
d)
3x2 cos (x3)
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Varun Kapoor answered |
= 3x2 cosx3
If f (x) = [x sin p x] { where [x] denotes greatest integer function}, then f (x) is
- a) Continuous at x = 0
- b) Continuous in (-1, 0)
- c) Differentiable at x = 1
- d) Differentiable in (-1, 1)
Correct answer is option 'A'. Can you explain this answer?
If f (x) = [x sin p x] { where [x] denotes greatest integer function}, then f (x) is
a)
Continuous at x = 0b)
Continuous in (-1, 0)c)
Differentiable at x = 1d)
Differentiable in (-1, 1)
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Asha Choudhury answered |
Solution:
To determine the continuity and differentiability of f(x), we need to evaluate left-hand limit, right-hand limit and the derivative of f(x) at every point in the domain of f(x).
Left-hand limit of f(x):
As x approaches 0 from the left, [x] approaches -1 and sin(px) oscillates between -1 and 1. Therefore, f(x) oscillates between -x and x. Thus, the left-hand limit of f(x) as x approaches 0 is:
lim x→0- f(x) = lim x→0- [x sin(px)] = lim x→0- (-x) = 0.
Right-hand limit of f(x):
As x approaches 0 from the right, [x] approaches 0 and sin(px) oscillates between -1 and 1. Therefore, f(x) oscillates between 0 and x. Thus, the right-hand limit of f(x) as x approaches 0 is:
lim x→0+ f(x) = lim x→0+ [x sin(px)] = lim x→0+ (0) = 0.
Since the left-hand limit and right-hand limit are equal, f(x) is continuous at x=0.
Derivative of f(x):
For x≠0, f(x) is piecewise linear with slope sin(px) and jump discontinuity at every integer multiple of 1/p. Therefore, f(x) is not differentiable at these points.
Since f(x) is continuous at x=0 and not differentiable at any other point, the correct answer is option 'A'.
To determine the continuity and differentiability of f(x), we need to evaluate left-hand limit, right-hand limit and the derivative of f(x) at every point in the domain of f(x).
Left-hand limit of f(x):
As x approaches 0 from the left, [x] approaches -1 and sin(px) oscillates between -1 and 1. Therefore, f(x) oscillates between -x and x. Thus, the left-hand limit of f(x) as x approaches 0 is:
lim x→0- f(x) = lim x→0- [x sin(px)] = lim x→0- (-x) = 0.
Right-hand limit of f(x):
As x approaches 0 from the right, [x] approaches 0 and sin(px) oscillates between -1 and 1. Therefore, f(x) oscillates between 0 and x. Thus, the right-hand limit of f(x) as x approaches 0 is:
lim x→0+ f(x) = lim x→0+ [x sin(px)] = lim x→0+ (0) = 0.
Since the left-hand limit and right-hand limit are equal, f(x) is continuous at x=0.
Derivative of f(x):
For x≠0, f(x) is piecewise linear with slope sin(px) and jump discontinuity at every integer multiple of 1/p. Therefore, f(x) is not differentiable at these points.
Since f(x) is continuous at x=0 and not differentiable at any other point, the correct answer is option 'A'.
- 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
|
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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
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)
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
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