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All questions of Continuity and Differentiability for JEE Exam
Therefore, f(x) continuous everywhere.
is continuous at 
then k = 
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
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.
- 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
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)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)– 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
If x2 + y2 = 1 then- a)yy'' – 2(y')2 + 1 = 0
- b)yy'' + (y')2 + 1 = 0
- c)yy'' + (y')2 – 1 = 0
- d)yy'' + 2(y')2 + 1 = 0
Correct answer is option 'B'. Can you explain this answer?
If x2 + y2 = 1 then
a)
yy'' – 2(y')2 + 1 = 0
b)
yy'' + (y')2 + 1 = 0
c)
yy'' + (y')2 – 1 = 0
d)
yy'' + 2(y')2 + 1 = 0
|
Sarthak Khanna answered |
x2 + y2 = 1 ⇒ 2x + 2yy' = 0 ⇒ x + yy'= 0
⇒ 1+ yy'' + (y')2 = 0 ⇒ yy'' + (y')2+1 = 0
⇒ 1+ yy'' + (y')2 = 0 ⇒ yy'' + (y')2+1 = 0
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.
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)
|
Krishna Iyer 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
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
The value of a for which the sum of the squares of the roots of the equation x2 – (a – 2) x – a – 1 = 0 assume the least value is- a)1
- b)0
- c)3
- d)2
Correct answer is option 'A'. Can you explain this answer?
The value of a for which the sum of the squares of the roots of the equation x2 – (a – 2) x – a – 1 = 0 assume the least value is
a)
1
b)
0
c)
3
d)
2
|
Jaya Das answered |
x2 - (a - 2) x -a-1 = 0
⇒ α + b = a- 2 ; α β = -(a+ 1)
α2 + β2 = (α + β)2 - 2αβ = a2 - 2a +6 = (a -1)2+5
For min. value of α2 + β2 where α is an integer
⇒ a = 1.
⇒ α + b = a- 2 ; α β = -(a+ 1)
α2 + β2 = (α + β)2 - 2αβ = a2 - 2a +6 = (a -1)2+5
For min. value of α2 + β2 where α is an integer
⇒ a = 1.
The equation 2 tan x + 5x – 2 = 0 has- a) No solution
- b)At least one real solution in [0, π/4]
- c) Two real solution in [0, π/4]
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
The equation 2 tan x + 5x – 2 = 0 has
a)
No solution
b)
At least one real solution in [0, π/4]
c)
Two real solution in [0, π/4]
d)
None of these
|
|
Krishna Iyer answered |
Let f(x) = 2tan x + 5x – 2
Now, f(0) = – 2 < 0
and f(π/4 ) = 2 + 5π/4 – 2
= 5π/5 > 0,
the line intersects the x-axis at x = ⅖
since 0 < ⅖ < π/4
Thus, f(x) = 0 has atleast real solution root in ( 0, π/4)
Now, f(0) = – 2 < 0
and f(π/4 ) = 2 + 5π/4 – 2
= 5π/5 > 0,
the line intersects the x-axis at x = ⅖
since 0 < ⅖ < π/4
Thus, f(x) = 0 has atleast real solution root in ( 0, π/4)
, -1 ≤ x < 0 is continuous in the interval [–1, 1], then `p' is equal to:- a)–1
- b)– 1/2
- c) 1/2
- d)1
Correct answer is option 'B'. Can you explain this answer?
, -1 ≤ x < 0 is continuous in the interval [–1, 1], then `p' is equal to:a)
–1
b)
– 1/2
c)
1/2
d)
1
|
Preeti Iyer answered |
lim f(x) = -½
Apply L hospital
lim(x → 0) p/2(1+px)½ + p/2(1-px)½ = -½
⇒ 2p = -1
p = -1/2
Apply L hospital
lim(x → 0) p/2(1+px)½ + p/2(1-px)½ = -½
⇒ 2p = -1
p = -1/2
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.
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.
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
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]
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
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
⇒
⇒
- a)1−y2
- b)1+y2
- c)y2+1
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
a)
1−y2
b)
1+y2
c)
y2+1
d)
none of these
|
My Vengeance Ends Now answered |
The correct answer is a)1-y^2
:
The function 
is continuous at exactly two points then the possible values of ' a ' are- a)( 2, ∞)
- b)(-∞,3)
- c)(-∞, - 1) ∪ (3, ∞)
- d)R
Correct answer is option 'C'. Can you explain this answer?
The function is continuous at exactly two points then the possible values of ' a ' are
is continuous at exactly two points then the possible values of ' a ' area)
( 2, ∞)
b)
(-∞,3)
c)
(-∞, - 1) ∪ (3, ∞)
d)
R
|
|
Tejas Verma answered |
f (x) is continuous when x2 - ax + 3 =2 - x
⇒ x2 - a -1 x + 1 = 0. This must have two distinct roots ⇒ Δ > 0 ⇒ (a -1)2 - 4 > 0
⇒ x2 - a -1 x + 1 = 0. This must have two distinct roots ⇒ Δ > 0 ⇒ (a -1)2 - 4 > 0
Let f : R → R be a function defined by f(x) = Min {x + 1, |x| + 1}. Then which of the following is true ?- a)f(x) ³ 1 for all x Î R
- b) f(x) is not differentiable at x = 1
- c)f(x) is differentiable everywhere
- d) f(x) is not differentiable at x = 0
Correct answer is option 'C'. Can you explain this answer?
Let f : R → R be a function defined by f(x) = Min {x + 1, |x| + 1}. Then which of the following is true ?
a)
f(x) ³ 1 for all x Î R
b)
f(x) is not differentiable at x = 1
c)
f(x) is differentiable everywhere
d)
f(x) is not differentiable at x = 0
|
|
Nikita Singh answered |
f(x) = Min {x + 1,∣x∣ + 1}
x = 0 ⇒ ∣x∣ = x
f(x) = min {x + 1,x + 1}
= x + 1
x < 0,∣x∣ = −x
f(x) = min {x + 1,−x + 1}
−x + 1> x + 1
x < 0
⇒x < 0, f(x) = x + 1
x > 0,f (x) = x + 1
⇒ f(x) = x + 1 for all x
∴f(x) is differential everywhere (continuous and constant slope)
x = 0 ⇒ ∣x∣ = x
f(x) = min {x + 1,x + 1}
= x + 1
x < 0,∣x∣ = −x
f(x) = min {x + 1,−x + 1}
−x + 1> x + 1
x < 0
⇒x < 0, f(x) = x + 1
x > 0,f (x) = x + 1
⇒ f(x) = x + 1 for all x
∴f(x) is differential everywhere (continuous and constant slope)
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)none of these
Correct answer is option 'B'. Can you explain this answer?
a)
b)
c)
d)
none of these
|
Seblewongel Girma answered |
F"(x)=(2logx-(3/ln10))÷x^3
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