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All questions of Chapter 6 - Application of Derivatives for JEE Exam
The function f(x) = ax, 0 < a < 1 is- a)increasing
- b)strictly decreasing on R
- c)neither increasing or decreasing
- d)decreasing
Correct answer is option 'D'. Can you explain this answer?
The function f(x) = ax, 0 < a < 1 is
a)
increasing
b)
strictly decreasing on R
c)
neither increasing or decreasing
d)
decreasing
|
Krishna Iyer answered |
f(x) = ax
Taking log bth the sides, log f(x) = xloga
f’(x)/ax = loga
f’(x) = ax loga {ax > 0 for all x implies R,
for loga e<a<1 that implies loga < 0}
Therefore, f’(x) < 0, for all x implies R
f(x) is a decreasing function.
Taking log bth the sides, log f(x) = xloga
f’(x)/ax = loga
f’(x) = ax loga {ax > 0 for all x implies R,
for loga e<a<1 that implies loga < 0}
Therefore, f’(x) < 0, for all x implies R
f(x) is a decreasing function.
Using approximation find the value of 
- a)2.025
- b)2.001
- c)2.01
- d)2.0025
Correct answer is option 'D'. Can you explain this answer?
Using approximation find the value of
a)
2.025
b)
2.001
c)
2.01
d)
2.0025
|
Gunjan Lakhani answered |
Let x=4, Δx=0.01
y=x^½ = 2
y+Δy = (x+ Δx)^½ = (4.01)^½
Δy = (dy/dx) * Δx
Δy = (x^(-1/2))/2 * Δx
Δy = (½)*(½) * 0.01
Δy = 0.25 * 0.01
Δy = 0.0025
So, (4.01)^½ = 2 + 0.0025 = 2.0025
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
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
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).
Find slope of normal to the curve y=5x2-10x + 7 at x=1- a)not defined
- b)-1
- c)1
- d)zero
Correct answer is option 'A'. Can you explain this answer?
Find slope of normal to the curve y=5x2-10x + 7 at x=1
a)
not defined
b)
-1
c)
1
d)
zero
|
Neha Sharma answered |
y = 5x2 - 10x + 7
dy/dx = 10x - 10
(At x = 1) 10(1) - 10
m1 = 0
As we know that slope, m1m2 = -1
=> 0(m2) = -1
m2 = -1/0 (which is not defined)
dy/dx = 10x - 10
(At x = 1) 10(1) - 10
m1 = 0
As we know that slope, m1m2 = -1
=> 0(m2) = -1
m2 = -1/0 (which is not defined)
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 |
The equation of the normal to the curve x2 = 4y which passes through the point (1, 2) is.- a)x + y – 3 = 0
- b)4x – y = 2
- c)4x – 2y = 0
- d)4x – 3y + 2= 0
Correct answer is option 'B'. Can you explain this answer?
The equation of the normal to the curve x2 = 4y which passes through the point (1, 2) is.
a)
x + y – 3 = 0
b)
4x – y = 2
c)
4x – 2y = 0
d)
4x – 3y + 2= 0
|
Sushil Kumar answered |
h2 = 4k
slope of normal=−1/(dy/dx) = −2h
equation of normal(y − k)= −2h(x−h)
k = 2 + 2/h(1 − h)
(h2) / 4 = 2 + 2/h (1 − h)
h = 2, k = 1
equation of line (y - 1)= -1(x - 2)
x + y = 3
slope of normal=−1/(dy/dx) = −2h
equation of normal(y − k)= −2h(x−h)
k = 2 + 2/h(1 − h)
(h2) / 4 = 2 + 2/h (1 − h)
h = 2, k = 1
equation of line (y - 1)= -1(x - 2)
x + y = 3
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.
Read the following text and answer the following questions, on the basis of the same:An architect designs a building for a multi-national company. The floor consists of a rectangular region with semicircular ends having a perimeter of 200 m as shown below:
If x and y represents the length and breadth of the rectangular region, then the relation between the variables is :- a)x + ?y = 100
- b)2x + ?y = 200
- c)?x + y = 50
- d)x + y = 100
Correct answer is option 'B'. Can you explain this answer?
Read the following text and answer the following questions, on the basis of the same:
An architect designs a building for a multi-national company. The floor consists of a rectangular region with semicircular ends having a perimeter of 200 m as shown below:
If x and y represents the length and breadth of the rectangular region, then the relation between the variables is :
a)
x + ?y = 100
b)
2x + ?y = 200
c)
?x + y = 50
d)
x + y = 100
|
|
Anjali Sharma answered |
Separate the interval 
open square brackets 0 comma straight pi over 2 close square brackets into sub-intervals in which f (x) = sin4 x + cos4 x is increasing or decreasing- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
Separate the interval open square brackets 0 comma straight pi over 2 close square brackets into sub-intervals in which f (x) = sin4 x + cos4 x is increasing or decreasing
open square brackets 0 comma straight pi over 2 close square brackets into sub-intervals in which f (x) = sin4 x + cos4 x is increasing or decreasinga)
b)
c)
d)
|
|
Nandini Iyer answered |
f’(x) = 4sin3x cosx - 4cos3x sinx
= - sin4x
As x is in the interval [0, ∏/2] hence 4x is in the interval [0, 2∏]
= - sin4x
As x is in the interval [0, ∏/2] hence 4x is in the interval [0, 2∏]
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.
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.
is monotonically decreasing in
The function 
increases in- a)(0, 1)
- b)(-∞,e)
- c)(e,∞)
- d)(1, e)
Correct answer is option 'C'. Can you explain this answer?
The function increases in
increases ina)
(0, 1)
b)
(-∞,e)
c)
(e,∞)
d)
(1, e)
|
Vikas Kapoor answered |
f'(x) = [(logx).2−2x.1/x](logx)2
= 2(logx−1)/(logx)^2
∴f'(x)>0
⇔logx−1>0
⇔logx>1
⇔logx>loge
⇔x>e
∴f(x) is increasing in (e,∞)
= 2(logx−1)/(logx)^2
∴f'(x)>0
⇔logx−1>0
⇔logx>1
⇔logx>loge
⇔x>e
∴f(x) is increasing in (e,∞)
The curve y = f(x) which satisfies the condition f'(x) > 0 and f"(x) < 0 for all real x, is- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
The curve y = f(x) which satisfies the condition f'(x) > 0 and f"(x) < 0 for all real x, is
a)
b)
c)
d)
|
|
Geetika Shah answered |
f’(x) > 0
=> f(x) is increasing
f’’(x) < 0 => f(x) is convex
=> f(x) is increasing
f’’(x) < 0 => f(x) is convex
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
The function f (x) = -3x + 12 on R.is- a)increasing
- b)strictly decreasing
- c)decreasing
- d)neither increasing or decreasing
Correct answer is option 'B'. Can you explain this answer?
The function f (x) = -3x + 12 on R.is
a)
increasing
b)
strictly decreasing
c)
decreasing
d)
neither increasing or decreasing
|
|
Aryan Khanna answered |
f(x) = -3x + 12
f(0) = -3(0) + 12 = 0 - 12 = 0
f(1) = 9
f(2)= 6
f(3) = 3
f(4) = 0
f(5) = -3
f(0) = -3(0) + 12 = 0 - 12 = 0
f(1) = 9
f(2)= 6
f(3) = 3
f(4) = 0
f(5) = -3
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.
The maximum value of 
is- a)(1/e)1/e
- b)(e)2/e
- c)(e)-1/e
- d)(e)1/e
Correct answer is option 'D'. Can you explain this answer?
The maximum value of is
isa)
(1/e)1/e
b)
(e)2/e
c)
(e)-1/e
d)
(e)1/e
|
Shiksha Academy answered |
For every real number (or) valued function f(x), the values of x which satisfies the equation f1(x)=0 are the point of it's local and global maxima or minima.
This occurs due to the fact that, at the point of maxima or minima, the curve of the function has a zero slope.
We have function f(x) = (1/x)x
We will be using the equation, y = (1/x)x
Taking in both sides we get
ln y = −xlnx
Differentiating both sides with respect to x.y.
dy/dx = −lnx−1
dy/dx =−y(lnx+1)
Equating dy/dx to 0, we get
−y(lnx+1)=0
Since y is an exponential function it can never be equal to zero, hence
lnx +1 = 0
lnx = −1
x = e(−1)
So, for the maximum value we put x = e^(−1)in f(x) to get the value of f(x) at the point.
f(e^−1) = e(1/e).
Hence the maximum value of the function is (e)1/e
This occurs due to the fact that, at the point of maxima or minima, the curve of the function has a zero slope.
We have function f(x) = (1/x)x
We will be using the equation, y = (1/x)x
Taking in both sides we get
ln y = −xlnx
Differentiating both sides with respect to x.y.
dy/dx = −lnx−1
dy/dx =−y(lnx+1)
Equating dy/dx to 0, we get
−y(lnx+1)=0
Since y is an exponential function it can never be equal to zero, hence
lnx +1 = 0
lnx = −1
x = e(−1)
So, for the maximum value we put x = e^(−1)in f(x) to get the value of f(x) at the point.
f(e^−1) = e(1/e).
Hence the maximum value of the function is (e)1/e
The equation of the tangent line to the curve y = 
which is parallel to the line 4x -2y + 3 = 0 is
- a)80x +40y – 193 = 0
- b)4x – 2y – 3 = 0
- c)80x – 40y + 193 = 0
- d)80x – 40y – 103 = 0
Correct answer is option 'D'. Can you explain this answer?
The equation of the tangent line to the curve y = which is parallel to the line 4x -2y + 3 = 0 is
which is parallel to the line 4x -2y + 3 = 0 isa)
80x +40y – 193 = 0
b)
4x – 2y – 3 = 0
c)
80x – 40y + 193 = 0
d)
80x – 40y – 103 = 0
|
Dabhi Bharat answered |
Given curve y=√5x-3 -2
y+2=√5x-3
dy/dx=5/2×√5x-3
m1=5/2(y+2)
given that line 4x-2y+3=0is parallel to tangent of curve
slope of line m2=2
m1=m2
5/2(y+2)=2
5/4=(y+2)
y=-3/4
from y+2=√5x-3
-3/4+2=√5x-3
5/4=√5x-3
5x=25/16+3
X=73/80
points are p(73/80,-3/4)
equ. of tangent :y-y1=m(x-x1)
y+3/4=2(x-73/80)
80x-73=40y+30
80x-40y-103=0
y+2=√5x-3
dy/dx=5/2×√5x-3
m1=5/2(y+2)
given that line 4x-2y+3=0is parallel to tangent of curve
slope of line m2=2
m1=m2
5/2(y+2)=2
5/4=(y+2)
y=-3/4
from y+2=√5x-3
-3/4+2=√5x-3
5/4=√5x-3
5x=25/16+3
X=73/80
points are p(73/80,-3/4)
equ. of tangent :y-y1=m(x-x1)
y+3/4=2(x-73/80)
80x-73=40y+30
80x-40y-103=0
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
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
Find the equation of tangent to 
which has slope 2.- a)2x – y = 1
- b)No tangent
- c)y – 2x = 0
- d)y – 2x = 3
Correct answer is option 'B'. Can you explain this answer?
Find the equation of tangent to which has slope 2.
which has slope 2.a)
2x – y = 1
b)
No tangent
c)
y – 2x = 0
d)
y – 2x = 3
|
|
Raghava Rao answered |
Y=1/(x-3)^2
dy/dx=(-1/(x-3)^2)
given slope=2
-1/(x-3)^2 =2
-1/2=(x-3)^2
negative number is not equal to square. so no tangent
dy/dx=(-1/(x-3)^2)
given slope=2
-1/(x-3)^2 =2
-1/2=(x-3)^2
negative number is not equal to square. so no tangent
In case of strict decreasing functions, slope of tangent and hence derivative is
- a)Negative
- b)either negative or zero.
- c)Zero
- d)Positive
Correct answer is option 'A'. Can you explain this answer?
In case of strict decreasing functions, slope of tangent and hence derivative is
a)
Negative
b)
either negative or zero.
c)
Zero
d)
Positive
|
|
Nishtha Mishra answered |
Slope of tangent and derivative of a function are closely related concepts in calculus. In the case of a strict decreasing function, the slope of the tangent line to the graph of the function is always negative. This implies that the derivative of the function is also negative.
Definition of a Strict Decreasing Function:
A function f(x) is said to be strictly decreasing on an interval if for any two values a and b in that interval, where a < b,="" the="" corresponding="" function="" values="" satisfy="" f(a)="" /> f(b).
Explanation:
1. Slope of the Tangent Line:
The slope of a tangent line to a curve at a particular point represents the rate at which the function is changing at that point. In the case of a strict decreasing function, the function values decrease as the input values increase. As a result, the tangent line to the graph of the function will have a negative slope.
2. Derivative of a Function:
The derivative of a function f(x) represents the rate of change of the function with respect to x. Mathematically, it is defined as the limit of the difference quotient as the change in x approaches zero:
f'(x) = lim(h -> 0) [(f(x+h) - f(x))/h]
For a strict decreasing function, as x increases, the function values decrease. This means that the numerator (f(x+h) - f(x)) will be negative for positive values of h. Dividing by a positive value of h will result in a negative difference quotient. Taking the limit as h approaches zero, the derivative of the function will be negative.
Conclusion:
In conclusion, for a strict decreasing function, the slope of the tangent line is always negative and the derivative of the function is also negative. This is because the function values decrease as the input values increase, leading to a negative rate of change.
Definition of a Strict Decreasing Function:
A function f(x) is said to be strictly decreasing on an interval if for any two values a and b in that interval, where a < b,="" the="" corresponding="" function="" values="" satisfy="" f(a)="" /> f(b).
Explanation:
1. Slope of the Tangent Line:
The slope of a tangent line to a curve at a particular point represents the rate at which the function is changing at that point. In the case of a strict decreasing function, the function values decrease as the input values increase. As a result, the tangent line to the graph of the function will have a negative slope.
2. Derivative of a Function:
The derivative of a function f(x) represents the rate of change of the function with respect to x. Mathematically, it is defined as the limit of the difference quotient as the change in x approaches zero:
f'(x) = lim(h -> 0) [(f(x+h) - f(x))/h]
For a strict decreasing function, as x increases, the function values decrease. This means that the numerator (f(x+h) - f(x)) will be negative for positive values of h. Dividing by a positive value of h will result in a negative difference quotient. Taking the limit as h approaches zero, the derivative of the function will be negative.
Conclusion:
In conclusion, for a strict decreasing function, the slope of the tangent line is always negative and the derivative of the function is also negative. This is because the function values decrease as the input values increase, leading to a negative rate of change.
Let f (x) be differentiable in (0, 4) and f (2) = f (3) and S = {c : 2 < c < 3, f’ (c) = 0} then- a)S has exactly one point
- b)S = { }
- c)S has atleast one point
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
Let f (x) be differentiable in (0, 4) and f (2) = f (3) and S = {c : 2 < c < 3, f’ (c) = 0} then
a)
S has exactly one point
b)
S = { }
c)
S has atleast one point
d)
none of these
|
|
Nandini Choudhury answered |
Conditions of Rolle’s Theorem are satisfied by f(x) in [2,3].Hence there exist atleast one real c in (2, 3) s.t. f ‘(c) = 0 . Therefore , the set S contains atleast one element
The function f (x) = x3 has a- a)point of inflexion at = 0
- b)local minima at x = 0
- c)local maxima at x = 0
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
The function f (x) = x3 has a
a)
point of inflexion at = 0
b)
local minima at x = 0
c)
local maxima at x = 0
d)
none of these
|
Amrita Sarkar answered |
f ‘(0) = 0 , f ‘’ (0) = 0 and f ‘’’(0) = 6 . So, f has a point of inflexion at 0.
The set of all x for which ln (1 + x) ≤ x is equal to- a) x > 0
- b)x>–1
- c)–1<x<0
- d) Null set
Correct answer is option 'B'. Can you explain this answer?
The set of all x for which ln (1 + x) ≤ x is equal to
a)
x > 0
b)
x>–1
c)
–1<x<0
d)
Null set
|
|
Amit Kumar answered |
f(x) = ln(1+x) - x ≤ 0
f(x) = ln(1+x) - x
f’(x) = 1/(1+x) - 1
= (1 - 1 - x)/(1+x)
= -x/(1+x)
f’(x) ≤ 0
-x/(1+x) ≤ 0
0 ≤ x /(1+x)
for(x = 2) (-2)/(1-2)
= 2 > 0, therefore x > - 1
f(x) = ln(1+x) - x
f’(x) = 1/(1+x) - 1
= (1 - 1 - x)/(1+x)
= -x/(1+x)
f’(x) ≤ 0
-x/(1+x) ≤ 0
0 ≤ x /(1+x)
for(x = 2) (-2)/(1-2)
= 2 > 0, therefore x > - 1
If the line y=x is a tangent to the parabola y=ax2+bx+c at the point (1,1) and the curve passes through (−1,0), then- a) a=b=−1, c=3
- b) a=b=1/2, c=0
- c) a=c=1/4, b=1/2
- d) a=0, b=c=1/2
Correct answer is option 'C'. Can you explain this answer?
If the line y=x is a tangent to the parabola y=ax2+bx+c at the point (1,1) and the curve passes through (−1,0), then
a)
a=b=−1, c=3
b)
a=b=1/2, c=0
c)
a=c=1/4, b=1/2
d)
a=0, b=c=1/2
|
|
Rajesh Gupta answered |
The correct option is C
a=c=1/4, b=1/2
y=x is a tangent
∴ slopes are equal.
dy/dx=2ax+b
⇒1=2a+b at (1,1)⋯(1)
Also, the parabola passes through (1,1)
⇒a+b+c=1⋯(2)
The parabola passes through (−1,0)
⇒0=a−b+c⋯(3)
Solving (1),(2),(3), we get -
∴a=c=1/4
and b=1/2
a=c=1/4, b=1/2
y=x is a tangent
∴ slopes are equal.
dy/dx=2ax+b
⇒1=2a+b at (1,1)⋯(1)
Also, the parabola passes through (1,1)
⇒a+b+c=1⋯(2)
The parabola passes through (−1,0)
⇒0=a−b+c⋯(3)
Solving (1),(2),(3), we get -
∴a=c=1/4
and b=1/2
The equation of the tangent to the curve y=(4−x2)2/3 at x = 2 is- a)x = 2
- b)x = – 2
- c)y = – 1.
- d)y = 2
Correct answer is option 'A'. Can you explain this answer?
The equation of the tangent to the curve y=(4−x2)2/3 at x = 2 is
a)
x = 2
b)
x = – 2
c)
y = – 1.
d)
y = 2
|
|
Siddharth Sengupta answered |
To find the equation of the tangent to the curve y = (4x^2)^(2/3) at x = 2, we need to follow these steps:
1. Find the derivative of the curve:
The derivative of y with respect to x can be found using the chain rule. Let's denote y as u^(2/3), where u = 4x^2.
dy/dx = (2/3) * u^(-1/3) * du/dx
= (2/3) * (4x^2)^(-1/3) * d(4x^2)/dx
= (2/3) * (4x^2)^(-1/3) * 8x
= (16/3) * (x^(-2/3)) * x
= (16/3) * x^(1/3)
2. Find the slope of the tangent line:
To find the slope of the tangent line at x = 2, substitute x = 2 into the derivative:
dy/dx = (16/3) * 2^(1/3)
= (16/3) * (∛2)
3. Find the y-coordinate of the point on the curve at x = 2:
Substitute x = 2 into the original equation y = (4x^2)^(2/3):
y = (4 * 2^2)^(2/3)
= (4 * 4)^(2/3)
= 16^(2/3)
= 4^2
= 16
4. Use the point-slope form of a line:
The equation of a line with slope m passing through the point (x1, y1) is given by:
y - y1 = m(x - x1)
Substituting the values we found:
y - 16 = (16/3) * (∛2)(x - 2)
5. Simplify the equation:
y - 16 = (16/3) * (∛2)x - (16/3) * (∛2) * 2
y - 16 = (16/3) * (∛2)x - (32/3) * (∛2)
y = (16/3) * (∛2)x - (32/3) * (∛2) + 16
Thus, the equation of the tangent to the curve y = (4x^2)^(2/3) at x = 2 is y = (16/3) * (∛2)x - (32/3) * (∛2) + 16, which can be simplified as y = (16/3) * (∛2)x - (32/3) * (∛2/3) + 16. The correct answer is option 'A', x = 2, which is not the correct equation of the tangent.
1. Find the derivative of the curve:
The derivative of y with respect to x can be found using the chain rule. Let's denote y as u^(2/3), where u = 4x^2.
dy/dx = (2/3) * u^(-1/3) * du/dx
= (2/3) * (4x^2)^(-1/3) * d(4x^2)/dx
= (2/3) * (4x^2)^(-1/3) * 8x
= (16/3) * (x^(-2/3)) * x
= (16/3) * x^(1/3)
2. Find the slope of the tangent line:
To find the slope of the tangent line at x = 2, substitute x = 2 into the derivative:
dy/dx = (16/3) * 2^(1/3)
= (16/3) * (∛2)
3. Find the y-coordinate of the point on the curve at x = 2:
Substitute x = 2 into the original equation y = (4x^2)^(2/3):
y = (4 * 2^2)^(2/3)
= (4 * 4)^(2/3)
= 16^(2/3)
= 4^2
= 16
4. Use the point-slope form of a line:
The equation of a line with slope m passing through the point (x1, y1) is given by:
y - y1 = m(x - x1)
Substituting the values we found:
y - 16 = (16/3) * (∛2)(x - 2)
5. Simplify the equation:
y - 16 = (16/3) * (∛2)x - (16/3) * (∛2) * 2
y - 16 = (16/3) * (∛2)x - (32/3) * (∛2)
y = (16/3) * (∛2)x - (32/3) * (∛2) + 16
Thus, the equation of the tangent to the curve y = (4x^2)^(2/3) at x = 2 is y = (16/3) * (∛2)x - (32/3) * (∛2) + 16, which can be simplified as y = (16/3) * (∛2)x - (32/3) * (∛2/3) + 16. The correct answer is option 'A', x = 2, which is not the correct equation of the tangent.
The true set of real values of x for which the function, f(x) = x ln x – x + 1 is positive is- a) (1, ∞)
- b)(1/e, ∞)
- c)[e, ∞)
- d)(0, 1) and (1, ∞)
Correct answer is option 'D'. Can you explain this answer?
The true set of real values of x for which the function, f(x) = x ln x – x + 1 is positive is
a)
(1, ∞)
b)
(1/e, ∞)
c)
[e, ∞)
d)
(0, 1) and (1, ∞)
|
|
Neha Joshi answered |
f(x)=xlnx−x+1
f(1)=0
On differentiating w.r.t x, we get
f′(x)=x*1/x+lnx−1
=lnx
Therefore,f′(x)>0 for allx∈(1,∞)
f′(x)<0 for allx∈(0,1)
lim x→0 f(x)=1
f(x)>0
for allx∈(0,1)∪(1,∞)
f(1)=0
On differentiating w.r.t x, we get
f′(x)=x*1/x+lnx−1
=lnx
Therefore,f′(x)>0 for allx∈(1,∞)
f′(x)<0 for allx∈(0,1)
lim x→0 f(x)=1
f(x)>0
for allx∈(0,1)∪(1,∞)
Find the approximate change in total surface area of a cube of side x metre caused by increase in side by 1%.- a)12 m2
- b)0.12x2 m2
- c)1.2x m2
- d)12x m2
Correct answer is option 'B'. Can you explain this answer?
Find the approximate change in total surface area of a cube of side x metre caused by increase in side by 1%.
a)
12 m2
b)
0.12x2 m2
c)
1.2x m2
d)
12x m2
|
Ayush Joshi answered |
Toolbox:
Let y=f(x)
Δx denote a small increment in x
Δy=f(x+Δx)−f(x)
dy=(dy/dx)Δx
Surface area of cube =6s^2
Step 1:
The side of the cube =x meters
Decrease in side =1%
= 0.01x
Increase in side =Δx
= −0.01x
Step 2:
Surface area of cube = 6s^2
= 6 * x^2
S = 6x^2
ds/dx = 12x [Differentiating with respect to x]
Approximate change in surface area of cube = ds/dx * Δx
=12x *(0.01x)
= 0.12x^2 m^2
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
|
Ameya Sengupta answered |
Given curve: 4y = |x-24|
To find the points on the curve at which tangents are parallel to the x-axis, we need to find the points where the slope of the tangent is zero (since the slope of the x-axis is zero).
First, let's find the derivative of the curve:
Differentiating both sides of the equation with respect to x, we get:
4(dy/dx) = d/dx |x-24|
To find the derivative of the absolute value function, we consider two cases: x-24 > 0 and x-24 < />
Case 1: x-24 > 0
In this case, the absolute value function simplifies to x-24. Taking the derivative, we get:
4(dy/dx) = d/dx (x-24)
4(dy/dx) = 1
Case 2: x-24 < />
In this case, the absolute value function simplifies to -(x-24). Taking the derivative, we get:
4(dy/dx) = d/dx (-(x-24))
4(dy/dx) = -1
Simplifying both cases, we get:
Case 1: dy/dx = 1/4
Case 2: dy/dx = -1/4
Now, let's find the points on the curve where the slope of the tangent is zero (parallel to the x-axis).
Setting dy/dx = 0, we get:
Case 1: 1/4 = 0 (No solution)
Case 2: -1/4 = 0 (No solution)
Therefore, there are no points on the curve where the tangents are parallel to the x-axis. Hence, the correct answer is option 'D' (none of these).
Note: It is important to carefully analyze the given curve and consider all possible cases when finding the points where tangents are parallel to a particular line. In this case, since the slope of the x-axis is zero, we need to find the points where the derivative is zero. However, after considering all cases, we find that there are no such points on the given curve.
To find the points on the curve at which tangents are parallel to the x-axis, we need to find the points where the slope of the tangent is zero (since the slope of the x-axis is zero).
First, let's find the derivative of the curve:
Differentiating both sides of the equation with respect to x, we get:
4(dy/dx) = d/dx |x-24|
To find the derivative of the absolute value function, we consider two cases: x-24 > 0 and x-24 < />
Case 1: x-24 > 0
In this case, the absolute value function simplifies to x-24. Taking the derivative, we get:
4(dy/dx) = d/dx (x-24)
4(dy/dx) = 1
Case 2: x-24 < />
In this case, the absolute value function simplifies to -(x-24). Taking the derivative, we get:
4(dy/dx) = d/dx (-(x-24))
4(dy/dx) = -1
Simplifying both cases, we get:
Case 1: dy/dx = 1/4
Case 2: dy/dx = -1/4
Now, let's find the points on the curve where the slope of the tangent is zero (parallel to the x-axis).
Setting dy/dx = 0, we get:
Case 1: 1/4 = 0 (No solution)
Case 2: -1/4 = 0 (No solution)
Therefore, there are no points on the curve where the tangents are parallel to the x-axis. Hence, the correct answer is option 'D' (none of these).
Note: It is important to carefully analyze the given curve and consider all possible cases when finding the points where tangents are parallel to a particular line. In this case, since the slope of the x-axis is zero, we need to find the points where the derivative is zero. However, after considering all cases, we find that there are no such points on the given curve.
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.
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].
Find the points of local maxima or minima for the function f(x) = x3.ex.- a)x=-3 is a point of local maxima
- b)x=-3 is a point of local minima
- c)x=0 is a point of local maxima
- d)x=0 is a point of local minima
Correct answer is option 'B'. Can you explain this answer?
Find the points of local maxima or minima for the function f(x) = x3.ex.
a)
x=-3 is a point of local maxima
b)
x=-3 is a point of local minima
c)
x=0 is a point of local maxima
d)
x=0 is a point of local minima
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Gitanjali Tiwari answered |
Solution:
The given function is f(x) = x3.ex.
To find the points of local maxima or minima, we need to find the critical points of the function.
Critical points: The points where the derivative of the function is either zero or does not exist.
f'(x) = 3x2.ex + x3.ex
Let f'(x) = 0, then
3x2.ex + x3.ex = 0
x2(ex + x) = 0
x = 0 or x = -ex
Now, we need to check the nature of critical points using the second derivative test.
f''(x) = 6x.ex + 6x2.ex + 2x3.ex
At x = 0,
f''(0) = 0
Thus, x = 0 is not a point of local maxima or minima.
At x = -ex,
f''(-ex) = 6(-ex).ex + 6(-ex)2.ex + 2(-ex)3.ex
f''(-ex) = -2ex3 < />
Thus, x = -ex is a point of local maxima.
Hence, option B is the correct answer.
Note: The second derivative test is used to determine the nature of critical points. If f''(x) > 0, then the critical point is a point of local minima. If f''(x) < 0,="" then="" the="" critical="" point="" is="" a="" point="" of="" local="" maxima.="" if="" f''(x)="0," then="" the="" test="" is="" inconclusive.="" 0,="" then="" the="" critical="" point="" is="" a="" point="" of="" local="" maxima.="" if="" f''(x)="0," then="" the="" test="" is="" />
The given function is f(x) = x3.ex.
To find the points of local maxima or minima, we need to find the critical points of the function.
Critical points: The points where the derivative of the function is either zero or does not exist.
f'(x) = 3x2.ex + x3.ex
Let f'(x) = 0, then
3x2.ex + x3.ex = 0
x2(ex + x) = 0
x = 0 or x = -ex
Now, we need to check the nature of critical points using the second derivative test.
f''(x) = 6x.ex + 6x2.ex + 2x3.ex
At x = 0,
f''(0) = 0
Thus, x = 0 is not a point of local maxima or minima.
At x = -ex,
f''(-ex) = 6(-ex).ex + 6(-ex)2.ex + 2(-ex)3.ex
f''(-ex) = -2ex3 < />
Thus, x = -ex is a point of local maxima.
Hence, option B is the correct answer.
Note: The second derivative test is used to determine the nature of critical points. If f''(x) > 0, then the critical point is a point of local minima. If f''(x) < 0,="" then="" the="" critical="" point="" is="" a="" point="" of="" local="" maxima.="" if="" f''(x)="0," then="" the="" test="" is="" inconclusive.="" 0,="" then="" the="" critical="" point="" is="" a="" point="" of="" local="" maxima.="" if="" f''(x)="0," then="" the="" test="" is="" />
If a differentiable function f (x) has a relative minimum at x = 0, then the function y = f (x) + a x + b has a relative minimum at x = 0 for- a)all b if a = 0
- b)all a > 0.
- c)all b > 0
- d)all a and all b
Correct answer is option 'A'. Can you explain this answer?
If a differentiable function f (x) has a relative minimum at x = 0, then the function y = f (x) + a x + b has a relative minimum at x = 0 for
a)
all b if a = 0
b)
all a > 0.
c)
all b > 0
d)
all a and all b
|
Manisha Choudhary answered |
And y has a relative minimum at x = 0 if f ‘(0)+a = 0 . a =0.
Read the following text and answer the following questions on the basis of the same: The shape of a toy is given as f(x) = 6(2x4 – x2). To make the toy beautiful 2 sticks which are perpendicular to each other were placed at a point (2, 3), above the toy.
Which value from the following may be abscissa of critical point?- a)± 1/4
- b)± 12
- c)± 1
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
Read the following text and answer the following questions on the basis of the same: The shape of a toy is given as f(x) = 6(2x4 – x2). To make the toy beautiful 2 sticks which are perpendicular to each other were placed at a point (2, 3), above the toy.
Which value from the following may be abscissa of critical point?
a)
± 1/4
b)
± 12
c)
± 1
d)
None of these
|
|
Hansa Sharma answered |
Critical point is point where f’(x) = 0
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