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All questions of Differentiation & Its Applications for JAMB Exam
- a)1
- b)1/3
- c)1/2
- d)0
Correct answer is option 'B'. Can you explain this answer?
a)
1
b)
1/3
c)
1/2
d)
0
|
Krishna Iyer answered |
lim(x → 0) (tanx-x)/x2 tanx
As we know that tan x = sinx/cosx
lim(x → 0) (sinx/cosx - x)/x2(sinx/cosx)
lim(x → 0) (sinx - xcosx)/(x2 sinx)
lim(x → 0) cosx - (-xsinx + cosx)/(x2cosx + sinx (2x))
lim(x → 0) (cosx + xsinx - cosx)/x2cosx + 2xsinx)
lim(x → 0) sinx/(xcosx + 2sinx)
Hence it is 0/0 form, apply L hospital rule
lim(x → 0) cosx/(-xsinx + cosx + 2cosx)
⇒ 1/(0+1+2)
= 1/3
As we know that tan x = sinx/cosx
lim(x → 0) (sinx/cosx - x)/x2(sinx/cosx)
lim(x → 0) (sinx - xcosx)/(x2 sinx)
lim(x → 0) cosx - (-xsinx + cosx)/(x2cosx + sinx (2x))
lim(x → 0) (cosx + xsinx - cosx)/x2cosx + 2xsinx)
lim(x → 0) sinx/(xcosx + 2sinx)
Hence it is 0/0 form, apply L hospital rule
lim(x → 0) cosx/(-xsinx + cosx + 2cosx)
⇒ 1/(0+1+2)
= 1/3
The maximum value of f (x) = sin x in the interval [π,2π] is
a) 6
b) 0
c) -2
d) -4
Correct answer is option 'B'. Can you explain this answer?
|
|
Kiran Mehta answered |
f(x) = sin x
f’(x) =cosx
f”(x) = -sin x
f”(3pi/2) = -sin(3pi/2)
= -(-1)
=> 1 > 0 (local minima)
f(pi) = sin(pi) = 0
f(2pi) = sin(2pi) = 0
Hence, 0 is the maxima.
f’(x) =cosx
f”(x) = -sin x
f”(3pi/2) = -sin(3pi/2)
= -(-1)
=> 1 > 0 (local minima)
f(pi) = sin(pi) = 0
f(2pi) = sin(2pi) = 0
Hence, 0 is the maxima.
Find the maximum and minimum values of f (x) = 2x3 – 24x + 107 in the interval [1, 3].- a)89, 69
- b)89, 75
- c)59, 56
- d)89, -9
Correct answer is option 'B'. Can you explain this answer?
Find the maximum and minimum values of f (x) = 2x3 – 24x + 107 in the interval [1, 3].
a)
89, 69
b)
89, 75
c)
59, 56
d)
89, -9
|
Anjana Sharma answered |
Toolbox:d/dx(x^n) = nx^n−1Maxima & Minima = f′(x) = 0Step 1:f(x) = 2x^3−24x+107Differentiating with
Let f be any function, such that 
exists, then 
- a)λℓ
- b)ℓ
- c)1
- d)λ
Correct answer is option 'A'. Can you explain this answer?
Let f be any function, such that exists, then
exists, then a)
λℓ
b)
ℓ
c)
1
d)
λ
|
Om Desai answered |
lim(x → a) f(x) = l
λf(x) = λl
λf(x) = λl
Find the approximate value of f(10.01) where f(x) = 5x2 +6x + 3- a)564.06
- b)564.01
- c)563.00
- d)563.01
Correct answer is option 'A'. Can you explain this answer?
Find the approximate value of f(10.01) where f(x) = 5x2 +6x + 3
a)
564.06
b)
564.01
c)
563.00
d)
563.01
|
Naina Sharma answered |
f(x) = 5x2 +6x + 3
f(10.01) = 5*(10.01)2 + 6*(10.01) + 3
To find (10.01)2
Let p=10, Δp=0.01
y=p2 = 100
y+Δy = (p+ Δp)2 = (10.01)2
Δy = (dy/dp) * Δp
Δy = 2*p* Δx
Δy = 2*10* 0.01
Δy = 20 * 0.01
Δy = 0.2
So, (10.01)2 = y + Δy
= 100.2
So,
f(10.01) = 5*(100.2) + 6*(10.01) + 3
= 501 + 60.06 + 3
= 564.06
f(10.01) = 5*(10.01)2 + 6*(10.01) + 3
To find (10.01)2
Let p=10, Δp=0.01
y=p2 = 100
y+Δy = (p+ Δp)2 = (10.01)2
Δy = (dy/dp) * Δp
Δy = 2*p* Δx
Δy = 2*10* 0.01
Δy = 20 * 0.01
Δy = 0.2
So, (10.01)2 = y + Δy
= 100.2
So,
f(10.01) = 5*(100.2) + 6*(10.01) + 3
= 501 + 60.06 + 3
= 564.06
The total revenue in Rupees received from the sale of x units of a product is given by R(x) = 5x2 + 22x + 35. Find the marginal revenue, when x = 7, where by marginal revenue we mean the rate of change of total revenue with respect to the number of items sold at an instant- a)Rs 7
- b)Rs 127
- c)Rs 92
- d)Rs 48
Correct answer is option 'C'. Can you explain this answer?
The total revenue in Rupees received from the sale of x units of a product is given by R(x) = 5x2 + 22x + 35. Find the marginal revenue, when x = 7, where by marginal revenue we mean the rate of change of total revenue with respect to the number of items sold at an instant
a)
Rs 7
b)
Rs 127
c)
Rs 92
d)
Rs 48
|
Gaurav Kumar answered |
A stone is dropped into a quiet lake and waves move in circles at a speed of 2cm per second. At the instant, when the radius of the circular wave is 12 cm, how fast is the enclosed area changing ?- a)Decreasing at the rate of 48π cm2 / sec
- b)Increasing at the rate of 24π cm2 / sec
- c)Increasing at the rate of 48π cm2 / sec
- d)Decreasing at the rate of 24π cm2 / sec
Correct answer is option 'C'. Can you explain this answer?
A stone is dropped into a quiet lake and waves move in circles at a speed of 2cm per second. At the instant, when the radius of the circular wave is 12 cm, how fast is the enclosed area changing ?
a)
Decreasing at the rate of 48π cm2 / sec
b)
Increasing at the rate of 24π cm2 / sec
c)
Increasing at the rate of 48π cm2 / sec
d)
Decreasing at the rate of 24π cm2 / sec
|
Hansa Sharma answered |
Rate of increase of radius dr/dt = 2 cm/s
Area of circle A = πr2
dA/dt = π*(2r)*(dr/dt)
= π*(24)*2
= 48π cm2/s
Rate of increase of area is 48π cm2/s (increasing as it is positive).
Area of circle A = πr2
dA/dt = π*(2r)*(dr/dt)
= π*(24)*2
= 48π cm2/s
Rate of increase of area is 48π cm2/s (increasing as it is positive).
Let (tan α) x + (sin α) y = α and (α cosec α) x + (cos α) y = 1 be two variable straight lines, α being the parameter. Let P be the point of intersection of the lines. In the limiting position when α→ 0, the coordinates of P are- a)(2 , 1)
- b) (–2 ,–1)
- c)(–2 , 1)
- d)(2 , –1)
Correct answer is option 'D'. Can you explain this answer?
Let (tan α) x + (sin α) y = α and (α cosec α) x + (cos α) y = 1 be two variable straight lines, α being the parameter. Let P be the point of intersection of the lines. In the limiting position when α→ 0, the coordinates of P are
a)
(2 , 1)
b)
(–2 ,–1)
c)
(–2 , 1)
d)
(2 , –1)
|
Maths Learner answered |
Can anyone explain the question and concept
The radius of air bubble is increasing at the rate of 0. 25 cm/s. At what rate the volume of the bubble is increasing when the radius is 1 cm.- a)4π cm3/s
- b)22π cm3/s
- c)2π cm3/s
- d)π cm3/s
Correct answer is option 'D'. Can you explain this answer?
The radius of air bubble is increasing at the rate of 0. 25 cm/s. At what rate the volume of the bubble is increasing when the radius is 1 cm.
a)
4π cm3/s
b)
22π cm3/s
c)
2π cm3/s
d)
π cm3/s
|
Rohan Yadav answered |
Given, the rate of increase of radius of the air bubble = 0.25 cm/s
We need to find the rate of increase of volume of the bubble when the radius is 1 cm.
Formula used:
Volume of a sphere = (4/3)πr^3
Differentiating both sides with respect to time t, we get:
dV/dt = 4πr^2(dr/dt)
where dV/dt is the rate of change of volume of the sphere with respect to time t and dr/dt is the rate of change of radius of the sphere with respect to time t.
Substituting the given values, we get:
dV/dt = 4π(1)^2(0.25) = π cm^3/s
Therefore, the rate of increase of volume of the bubble when the radius is 1 cm is π cm^3/s, which is the correct answer.
We need to find the rate of increase of volume of the bubble when the radius is 1 cm.
Formula used:
Volume of a sphere = (4/3)πr^3
Differentiating both sides with respect to time t, we get:
dV/dt = 4πr^2(dr/dt)
where dV/dt is the rate of change of volume of the sphere with respect to time t and dr/dt is the rate of change of radius of the sphere with respect to time t.
Substituting the given values, we get:
dV/dt = 4π(1)^2(0.25) = π cm^3/s
Therefore, the rate of increase of volume of the bubble when the radius is 1 cm is π cm^3/s, which is the correct answer.
Find the maximum profit that a company can make, if the profit function is given by P(x) = 41 + 24 x – 18x2- a)56
- b)49
- c)23
- d)89
Correct answer is option 'B'. Can you explain this answer?
Find the maximum profit that a company can make, if the profit function is given by P(x) = 41 + 24 x – 18x2
a)
56
b)
49
c)
23
d)
89
|
|
Aryan Khanna answered |
p’(x) = -24 - 36x
p”(x) = -36
Now, p’(x) = 0 ⇒ x = (-24)/36
x = -⅔
Also, p”(-⅔) = -36 < 0
By the second derivative test, x = -⅔
Therefore, maximum profit = p(-⅔)
= 41 - 24(-⅔) - 18(-⅔)^2
= 41 +16 - 8
⇒ 49
p”(x) = -36
Now, p’(x) = 0 ⇒ x = (-24)/36
x = -⅔
Also, p”(-⅔) = -36 < 0
By the second derivative test, x = -⅔
Therefore, maximum profit = p(-⅔)
= 41 - 24(-⅔) - 18(-⅔)^2
= 41 +16 - 8
⇒ 49
is equal to - a)1/6
- b)
- c)
- d)0
Correct answer is option 'C'. Can you explain this answer?
is equal to a)
1/6
b)
c)
d)
0
|
|
Stan answered |
Sinx -x/x^3 it is in 0/0 form
so use L hospital rule,
using it we get : cosX-1/3x^2
again using : -sinX/3*2X=-1/6 lim sinx/x=-1/6
A real number x when added to its reciprocal give minimum value to the sum when x is- a)1/2
- b)-1
- c)1
- d)2
Correct answer is option 'C'. Can you explain this answer?
A real number x when added to its reciprocal give minimum value to the sum when x is
a)
1/2
b)
-1
c)
1
d)
2
|
Krish Das answered |
Finding the Real Number that Gives Minimum Value to the Sum
Solution:
Let x be the real number. Then, its reciprocal is 1/x.
The sum of x and its reciprocal is x + 1/x.
To find the minimum value of this sum, we can use the concept of the arithmetic mean and geometric mean inequality.
We know that for any two positive numbers a and b, the arithmetic mean is (a+b)/2 and the geometric mean is √(ab).
The arithmetic mean is always greater than or equal to the geometric mean, i.e., (a+b)/2 ≥ √(ab).
Let's apply this inequality to x and 1/x.
The arithmetic mean of x and 1/x is (x + 1/x)/2.
The geometric mean of x and 1/x is √(x * 1/x) = √1 = 1.
By the arithmetic mean and geometric mean inequality, we have:
(x + 1/x)/2 ≥ √(x * 1/x) = 1
Multiplying both sides by 2 gives:
x + 1/x ≥ 2
Therefore, the minimum value of x + 1/x is 2, which is attained when x=1.
Hence, the real number x that gives minimum value to the sum x + 1/x is 1.
Solution:
Let x be the real number. Then, its reciprocal is 1/x.
The sum of x and its reciprocal is x + 1/x.
To find the minimum value of this sum, we can use the concept of the arithmetic mean and geometric mean inequality.
We know that for any two positive numbers a and b, the arithmetic mean is (a+b)/2 and the geometric mean is √(ab).
The arithmetic mean is always greater than or equal to the geometric mean, i.e., (a+b)/2 ≥ √(ab).
Let's apply this inequality to x and 1/x.
The arithmetic mean of x and 1/x is (x + 1/x)/2.
The geometric mean of x and 1/x is √(x * 1/x) = √1 = 1.
By the arithmetic mean and geometric mean inequality, we have:
(x + 1/x)/2 ≥ √(x * 1/x) = 1
Multiplying both sides by 2 gives:
x + 1/x ≥ 2
Therefore, the minimum value of x + 1/x is 2, which is attained when x=1.
Hence, the real number x that gives minimum value to the sum x + 1/x is 1.
A point c in the domain of a function f is called a critical point of f if- a)f’ (x) = 0 at x = c
- b)f is not differentiable at x = c
- c)Either f’ (c) = 0 or f is not differentiable
- d)f” (x) = 0, at x = c
Correct answer is option 'B'. Can you explain this answer?
A point c in the domain of a function f is called a critical point of f if
a)
f’ (x) = 0 at x = c
b)
f is not differentiable at x = c
c)
Either f’ (c) = 0 or f is not differentiable
d)
f” (x) = 0, at x = c
|
Ishan Choudhury answered |
A point C in the domain of a function f at which either f(c) = 0 or f is not differentiable.
The point f is called the critical point.
c is called the point of local maxima
If f ′(x) changes sign from positive to negative as x increases through c, that is, if f ′(x) > 0 at every point sufficiently close to and to the left of c, and f ′(x) < 0 at every point sufficiently close to and to the right of c.
c is called the point of local minima
If f ′(x) changes sign from negative to positive as x increases through c, that is, if f ′(x) < 0 at every point sufficiently close to and to the left of c, and f ′(x) > 0 at every point sufficiently close to and to the right of c.
c is called the point of inflexion
If f ′(x) does not change sign as x increases through c, then c is neither a point of local maxima nor a point of local minima.
The point f is called the critical point.
c is called the point of local maxima
If f ′(x) changes sign from positive to negative as x increases through c, that is, if f ′(x) > 0 at every point sufficiently close to and to the left of c, and f ′(x) < 0 at every point sufficiently close to and to the right of c.
c is called the point of local minima
If f ′(x) changes sign from negative to positive as x increases through c, that is, if f ′(x) < 0 at every point sufficiently close to and to the left of c, and f ′(x) > 0 at every point sufficiently close to and to the right of c.
c is called the point of inflexion
If f ′(x) does not change sign as x increases through c, then c is neither a point of local maxima nor a point of local minima.
The volume of cube is increasing at the constant rate of 3 cm3/s. Find the rate of change of edge of the cube when its edge is 5 cm.- a)25 cm3/sec
- b)25 cm/s
- c)1/25 cm/s
- d)1/25 cm3/s
Correct answer is option 'C'. Can you explain this answer?
The volume of cube is increasing at the constant rate of 3 cm3/s. Find the rate of change of edge of the cube when its edge is 5 cm.
a)
25 cm3/sec
b)
25 cm/s
c)
1/25 cm/s
d)
1/25 cm3/s
|
|
Aryan Khanna answered |
Let V be the instantaneous volume of the cube.
dV/dt = 3 cm3/s
Let x be the side of the cube.
V = x3
dV/dt = 3x2 * (dx/dt)
3 = 3*(52)*(dx/dt)
So, dx/dt = 1/25 cm/s
dV/dt = 3 cm3/s
Let x be the side of the cube.
V = x3
dV/dt = 3x2 * (dx/dt)
3 = 3*(52)*(dx/dt)
So, dx/dt = 1/25 cm/s
The maximum and minimum values of f(x) = 
are- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
The maximum and minimum values of f(x) = are
area)
b)
c)
d)
|
|
Aryan Khanna answered |
f(x) = sinx + 1/2cos2x
⇒ f'(x) = cos x – sin2x
Now, f'(x) = 0 gives cosx – sin2x = 0
⇒ cos x (1 – 2 sinx) = 0
⇒ cos x = 0, (1 – 2 sinx) = 0
⇒ cos x = 0, sinx = 1/2
⇒ x = π/6 , π/2
Now, f(0) = 1/2,
f(π/6) = 1/2 + 1/4 = 3/4,
f(π/2) = 1 – 1/2 = 1/2
Therefore, the absolute max value = 3/4 and absolute min = 1/2
⇒ f'(x) = cos x – sin2x
Now, f'(x) = 0 gives cosx – sin2x = 0
⇒ cos x (1 – 2 sinx) = 0
⇒ cos x = 0, (1 – 2 sinx) = 0
⇒ cos x = 0, sinx = 1/2
⇒ x = π/6 , π/2
Now, f(0) = 1/2,
f(π/6) = 1/2 + 1/4 = 3/4,
f(π/2) = 1 – 1/2 = 1/2
Therefore, the absolute max value = 3/4 and absolute min = 1/2
- a)π – 22/7
- b)π
- c)0
- d)(7x-22)/7
Correct answer is option 'A'. Can you explain this answer?
a)
π – 22/7
b)
π
c)
0
d)
(7x-22)/7
|
Lavanya Menon answered |
lim(x → π ) (x - 22/7)
Taking limit, we get
⇒ π - 22/7
Taking limit, we get
⇒ π - 22/7
f(x) = x5 – 5x4 + 5x3 – 1. The local maxima of the function f(x) is at x =- a)1
- b)5
- c)0
- d)3
Correct answer is option 'A'. Can you explain this answer?
f(x) = x5 – 5x4 + 5x3 – 1. The local maxima of the function f(x) is at x =
a)
1
b)
5
c)
0
d)
3
|
Rajat Patel answered |
f′(x)=5x4−20x3+15x2
f′(x)=5x2(x2−4x+3)
when f′(x)=0
⇒5x2(x2−4x+3)=0
⇒5x2(x−3)(x−1)=0
⇒x=0,x=3,x=1
is equal to 
is
Let f be a real valued function defined on (0, 1) ∪ (2, 4) such that f ‘ (x) = 0 for every x, then- a)f is constant function if f 1/2 = f (3)
- b)f is a constant function
- c)f is a constant function if f 1/2 = 0
- d)f is not a constant function
Correct answer is option 'A'. Can you explain this answer?
Let f be a real valued function defined on (0, 1) ∪ (2, 4) such that f ‘ (x) = 0 for every x, then
a)
f is constant function if f 1/2 = f (3)
b)
f is a constant function
c)
f is a constant function if f 1/2 = 0
d)
f is not a constant function
|
Nandini Choudhury answered |
f ‘ (x) = 0 ⇒ f (x)is constant in (0 , 1)and also in (2, 4). But this does not mean that f (x) has the same value in both the intervals . However , if f (c) = f (d) , where c ∈ (0 , 1) and d ∈ (2, 4) then f (x) assumes the same value at all x ∈ (0 ,1) U (2, 4) and hence f is a constant function.
As x → a, f(x) → l, then l is called the……..of the function f(X) which is symbolically written as…….- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
As x → a, f(x) → l, then l is called the……..of the function f(X) which is symbolically written as…….
a)
b)
c)
d)
|
|
Anand Kumar answered |
C is the answer
this is the definition of limit.
this is the definition of limit.
The derivate of the function 
- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
The derivate of the function
a)
b)
c)
d)
|
|
Gaurav Kumar answered |
Given, y = (sinx+cosx)/(sinx−cosx)
∴ dydx = [(sinx−cosx)(cosx−sinx)−(sinx+cosx)(cosx+sinx)]/(sinx−cosx)2
[by quotient rule]
= [(−sinx−cosx)2 − (sinx+cosx)2]/(sinx−cosx)2
− [(sinx−cosx)2 − (sinx+cosx)2]/(sinx−cosx)2
= − [(sin2x + cos2x − 2sinxcosx + sin2x + cos2x + 2sinxcosx)]/(sinx−cosx)2
= −2/(sinx−cosx)2
= -2/(sin2x + cos2x - 2sinxcosx)
= - 2/(1 - sin2x)
∴ dydx = [(sinx−cosx)(cosx−sinx)−(sinx+cosx)(cosx+sinx)]/(sinx−cosx)2
[by quotient rule]
= [(−sinx−cosx)2 − (sinx+cosx)2]/(sinx−cosx)2
− [(sinx−cosx)2 − (sinx+cosx)2]/(sinx−cosx)2
= − [(sin2x + cos2x − 2sinxcosx + sin2x + cos2x + 2sinxcosx)]/(sinx−cosx)2
= −2/(sinx−cosx)2
= -2/(sin2x + cos2x - 2sinxcosx)
= - 2/(1 - sin2x)
is equal to 
The value of 
- a)3/5
- b)3/2
- c)3/4
- d)2/5
Correct answer is option 'B'. Can you explain this answer?
The value of
a)
3/5
b)
3/2
c)
3/4
d)
2/5
|
Neha Joshi answered |
After applying L'Hôpital's Rule and taking the limit as �x approaches 0, the limit of the derivatives is 3223, which confirms our initial computation of the limit
Find two positive numbers x and y such that x + y = 60 and xy3 is maximum- a)x = 45, y = 15
- b)x = 15, y = 45
- c)x = 10, y = 50
- d)x = 30, y = 30
Correct answer is option 'B'. Can you explain this answer?
Find two positive numbers x and y such that x + y = 60 and xy3 is maximum
a)
x = 45, y = 15
b)
x = 15, y = 45
c)
x = 10, y = 50
d)
x = 30, y = 30
|
|
Gaurav Kumar answered |
two positive numbers x and y are such that x + y = 60.
x + y = 60
⇒ x = 60 – y ...(1)
Let P = xy3
∴ P =(60 – y)y3 = 60y3 – y4
Differentiating both sides with respect to y, we get
For maximum or minimum dP/dy = 0
⇒ 180y2 - 4y3 = 0
⇒ 4y2 (45 - y) = 0
⇒ y = 0 or 45 - y = 0
⇒ y = 0 or y = 45
⇒ y = 45 (∵ y = 0 is not possible)
Thus, the two positive numbers are 15 and 45.
x + y = 60
⇒ x = 60 – y ...(1)
Let P = xy3
∴ P =(60 – y)y3 = 60y3 – y4
Differentiating both sides with respect to y, we get
For maximum or minimum dP/dy = 0
⇒ 180y2 - 4y3 = 0
⇒ 4y2 (45 - y) = 0
⇒ y = 0 or 45 - y = 0
⇒ y = 0 or y = 45
⇒ y = 45 (∵ y = 0 is not possible)
Thus, the two positive numbers are 15 and 45.
- a)n
- b)n - 1
- c)(n + 1)
- d)-n
Correct answer is option 'D'. Can you explain this answer?
a)
n
b)
n - 1
c)
(n + 1)
d)
-n
|
Maulik Menon answered |
Consider the given function.
lim x→0 ((1-x)n −1)/x)
This is 0/0 form.
So, apply L-Hospital rule,
lim x→0 [nx-1x(1-x)n-1 −0)]/1
lim x→ 0(-n(1-x)n-1)
=-n
lim x→0 ((1-x)n −1)/x)
This is 0/0 form.
So, apply L-Hospital rule,
lim x→0 [nx-1x(1-x)n-1 −0)]/1
lim x→ 0(-n(1-x)n-1)
=-n
Every continuous function is- a)not differentiable
- b)not decreasing
- c)decreasing
- d)differentiable.
Correct answer is option 'A'. Can you explain this answer?
Every continuous function is
a)
not differentiable
b)
not decreasing
c)
decreasing
d)
differentiable.
|
Siddharth Rane answered |
Explanation:
Continuous functions are those functions which can be drawn without lifting a pen from the paper. In other words, a function is said to be continuous if it has no abrupt changes or jumps in its graph.
Not Necessory Differentiable:
A function is said to be differentiable at a point if its derivative exists at that point. However, a continuous function need not be differentiable at every point. There are many examples of continuous functions that are not differentiable at some points. One such example is the absolute value function, |x|. This function is continuous everywhere, but it is not differentiable at x = 0.
Not Decreasing:
A function is said to be decreasing if its value decreases as the input increases. However, a continuous function need not be decreasing. For example, the function f(x) = x^2 is continuous everywhere, but it is not decreasing.
Decreasing:
Some functions are decreasing. For example, the function f(x) = -x is a decreasing function. However, not all continuous functions are decreasing.
Differentiable:
A function is said to be differentiable if its derivative exists at every point in its domain. However, a continuous function need not be differentiable at every point.
Conclusion:
Therefore, the correct answer is option 'A' that every continuous function is not necessarily differentiable.
Continuous functions are those functions which can be drawn without lifting a pen from the paper. In other words, a function is said to be continuous if it has no abrupt changes or jumps in its graph.
Not Necessory Differentiable:
A function is said to be differentiable at a point if its derivative exists at that point. However, a continuous function need not be differentiable at every point. There are many examples of continuous functions that are not differentiable at some points. One such example is the absolute value function, |x|. This function is continuous everywhere, but it is not differentiable at x = 0.
Not Decreasing:
A function is said to be decreasing if its value decreases as the input increases. However, a continuous function need not be decreasing. For example, the function f(x) = x^2 is continuous everywhere, but it is not decreasing.
Decreasing:
Some functions are decreasing. For example, the function f(x) = -x is a decreasing function. However, not all continuous functions are decreasing.
Differentiable:
A function is said to be differentiable if its derivative exists at every point in its domain. However, a continuous function need not be differentiable at every point.
Conclusion:
Therefore, the correct answer is option 'A' that every continuous function is not necessarily differentiable.
The points on the curve 4 y = |x2−4| at which tangents are parallel to x – axis, are- a)(4, 3) and (– 4, – 3)
- b)(0, 1) only
- c)(2, 0) and (– 2, 0)
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
The points on the curve 4 y = |x2−4| at which tangents are parallel to x – axis, are
a)
(4, 3) and (– 4, – 3)
b)
(0, 1) only
c)
(2, 0) and (– 2, 0)
d)
none of these
|
Sinjini Datta answered |
Only when x = 0 and when x = 0 , y = 1. So, only at (0,1) tangent is parallel to x- axis.
The positive integer n so that limx→3 (xn – 3n)/(x – 3) = 108 is
- a)3
- b)4
- c)-2
- d)1
Correct answer is option 'B'. Can you explain this answer?
The positive integer n so that limx→3 (xn – 3n)/(x – 3) = 108 is
a)
3
b)
4
c)
-2
d)
1
|
Vaishnavi Iyer answered |
To find the value of the positive integer n that satisfies the given limit, we can start by evaluating the limit as x approaches 0 from the right side.
lim(x→0+) (sin(2nx))/(sin(x)) = 1/2
As x approaches 0, sin(x) approaches 0. Therefore, the denominator of the fraction approaches 0. For the limit to exist and be equal to 1/2, the numerator must also approach 0.
Since sin(2nx) is a periodic function, it will not approach 0 unless 2nx is a multiple of π. In other words, 2nx must be an integer multiple of π. Let's denote this integer multiple as kπ, where k is an integer.
2nx = kπ
Solving for x:
x = (kπ)/(2n)
Since we want to find a positive integer n, we need to find the smallest positive value of n that satisfies the equation above. This means we need to find the smallest positive integer k and n such that kπ/(2n) is equal to 0.
The smallest positive integer k that makes kπ/(2n) equal to 0 is k = 1, since any integer multiple of π divided by any positive integer n will not be equal to 0.
Therefore, the smallest positive integer n that satisfies the given limit is n = 1.
lim(x→0+) (sin(2nx))/(sin(x)) = 1/2
As x approaches 0, sin(x) approaches 0. Therefore, the denominator of the fraction approaches 0. For the limit to exist and be equal to 1/2, the numerator must also approach 0.
Since sin(2nx) is a periodic function, it will not approach 0 unless 2nx is a multiple of π. In other words, 2nx must be an integer multiple of π. Let's denote this integer multiple as kπ, where k is an integer.
2nx = kπ
Solving for x:
x = (kπ)/(2n)
Since we want to find a positive integer n, we need to find the smallest positive value of n that satisfies the equation above. This means we need to find the smallest positive integer k and n such that kπ/(2n) is equal to 0.
The smallest positive integer k that makes kπ/(2n) equal to 0 is k = 1, since any integer multiple of π divided by any positive integer n will not be equal to 0.
Therefore, the smallest positive integer n that satisfies the given limit is n = 1.
The 
n ∈ N is (where [*] denotes greatest integer function)- a)2n
- b)2n+1
- c)2n–1
- d)Does not exist
Correct answer is option 'C'. Can you explain this answer?
The n ∈ N is (where [*] denotes greatest integer function)
n ∈ N is (where [*] denotes greatest integer function)a)
2n
b)
2n+1
c)
2n–1
d)
Does not exist
|
|
Om Desai answered |
lim θ→0[nsinθ/θ]=n−1.....[from graph maximum value of sinx/x=1]
lim θ→0[ntanθ/θ]=n.....[from graph minimum value of tanx/x=1]
Hence, lim x→θ([[nsinθ/θ]+[ntanθ/θ])
= n−1+n =2n−1
lim θ→0[ntanθ/θ]=n.....[from graph minimum value of tanx/x=1]
Hence, lim x→θ([[nsinθ/θ]+[ntanθ/θ])
= n−1+n =2n−1
The equation of the tangent to the curve y = e2x at the point (0, 1) is- a)1 – y = 2 x
- b)y – 1 = 2 x
- c)y + 1 = 2 x
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
The equation of the tangent to the curve y = e2x at the point (0, 1) is
a)
1 – y = 2 x
b)
y – 1 = 2 x
c)
y + 1 = 2 x
d)
none of these
|
Nitya Yadav answered |
Finding the Derivative of y=e^(2x)
To find the equation of the tangent to the curve y = e^(2x) at the point (0,1), we first need to find the derivative of the function y = e^(2x) using the power rule of differentiation.
dy/dx = 2e^(2x)
Finding the Equation of the Tangent Line
Now that we have the derivative of the function, we can find the equation of the tangent line to the curve at the point (0,1) using the point-slope form of a line.
y - y1 = m(x - x1)
where m is the slope of the tangent line and (x1, y1) is the point on the curve where we want to find the tangent line.
Substituting x1 = 0 and y1 = 1, and m = dy/dx evaluated at (0,1), we get
y - 1 = 2e^(2*0)(x - 0)
y - 1 = 2x
Simplifying the equation, we get
y = 2x + 1
Therefore, the equation of the tangent to the curve y = e^(2x) at the point (0,1) is y - 1 = 2x. Option 'B' is the correct answer.
To find the equation of the tangent to the curve y = e^(2x) at the point (0,1), we first need to find the derivative of the function y = e^(2x) using the power rule of differentiation.
dy/dx = 2e^(2x)
Finding the Equation of the Tangent Line
Now that we have the derivative of the function, we can find the equation of the tangent line to the curve at the point (0,1) using the point-slope form of a line.
y - y1 = m(x - x1)
where m is the slope of the tangent line and (x1, y1) is the point on the curve where we want to find the tangent line.
Substituting x1 = 0 and y1 = 1, and m = dy/dx evaluated at (0,1), we get
y - 1 = 2e^(2*0)(x - 0)
y - 1 = 2x
Simplifying the equation, we get
y = 2x + 1
Therefore, the equation of the tangent to the curve y = e^(2x) at the point (0,1) is y - 1 = 2x. Option 'B' is the correct answer.
, x ≠ 0, then the value of the function at x = 0, so that f is continuous at x = 0, is
The function f (x) = x2, for all real x, is- a)neither decreasing nor increasing
- b)Decreasing
- c)Increasing
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
The function f (x) = x2, for all real x, is
a)
neither decreasing nor increasing
b)
Decreasing
c)
Increasing
d)
none of these
|
Pragati Patel answered |
Since f ‘(x) = 2x > 0 for x > 0,and f ‘ (x) = 2x < 0 for x < 0 ,therefore on R , f is neither increasing nor decreasing. Infact , f is strict increasing on [0 , ∞) and strict decreasing on (- ∞,0].
The function f(x) = log x- a)Has a local maximum but no local minimum value
- b)Has both ,a local minimum and a local maximum value
- c)Has neither a local minimum nor a local maximum value
- d)Has a local minimum but no local maximum value
Correct answer is option 'C'. Can you explain this answer?
The function f(x) = log x
a)
Has a local maximum but no local minimum value
b)
Has both ,a local minimum and a local maximum value
c)
Has neither a local minimum nor a local maximum value
d)
Has a local minimum but no local maximum value
|
|
Alok Mehta answered |
The domain is (0,∞).
Logarithm of zero is not defined. Any number raised to any power can’t be zero.
Logarithm of negative numbers is also not defined. The natural logarithm function ln(x) is defined only for x>0. The complex logarithmic function Log(z) is defined for negative values.
- a)1/3
- b)–(1/3)
- c)0
- d)not exist
Correct answer is option 'B'. Can you explain this answer?
a)
1/3
b)
–(1/3)
c)
0
d)
not exist
|
|
Lavanya Menon answered |
We know from trigonometry that
−1≤sin(1x)<−1 for all x≠0
Important: for limx→0 we don't care what happens when x=0
Since x<2>0 for all x≠0 , we can multiply through by x2 to get
−x2= x2sin(1/x)≤x2
Clearly limx→0(−x2)= 0 and limx→0 x2=0, so, by the squeeze theorem,
lim x→0 x2sin(1/x) = 0
−1≤sin(1x)<−1 for all x≠0
Important: for limx→0 we don't care what happens when x=0
Since x<2>0 for all x≠0 , we can multiply through by x2 to get
−x2= x2sin(1/x)≤x2
Clearly limx→0(−x2)= 0 and limx→0 x2=0, so, by the squeeze theorem,
lim x→0 x2sin(1/x) = 0
is the expected value of f at x = a given the values of f near x to the left of a. This value is called the……….of f at a.- a)Limit
- b)Left Hand limit
- c)Right Hand Limit
- d)Value of function
Correct answer is option 'B'. Can you explain this answer?
is the expected value of f at x = a given the values of f near x to the left of a. This value is called the……….of f at a.a)
Limit
b)
Left Hand limit
c)
Right Hand Limit
d)
Value of function
|
Learners Habitat answered |
When the x approaches a negative, the limit is left hand limit.
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