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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.
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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 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).
is
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 |
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:
The area of the rectangular region A expressed as a function of x is :- a)
- b)
- c)
- d)
Correct answer is option 'A'. 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:
The area of the rectangular region A expressed as a function of x is :
a)
b)
c)
d)
|
|
Krishna Iyer answered |
The abscissa of the point on the curve ay2 = x3, the normal at which cuts off equal intercepts from the coordinate axes is- a)
- b)
- c) –
- d)–
Correct answer is option 'B'. Can you explain this answer?
The abscissa of the point on the curve ay2 = x3, the normal at which cuts off equal intercepts from the coordinate axes is
a)
b)
c)
–
d)
–
|
|
Kavita Joshi answered |
(That should be "axes" -- plural.)
Find the normal; find the intercepts -- for a point (u, v) say.
To get there, differentiate to find the gradient - of the tangent, hence of the normal.
2 a y dy/dx = 3 x^2
So dy/dx = 3 x^2 / (2 a y) .
At the point (u, v), this is 3 u^2 / (2 a v)
So the gradient of the normal there is (-2 a v) / (3 u^2)
So the equation of the normal is
y - v = (-2 a v) / (3 u^2) * (x - u)
When x = 0,
y = v - 2 a v / (3 u^2) (-u)
= v (1 + 2 a / (3 u) )
When y = 0,
x = u + 3 u^2 / (2 a v) * v
= u (1 + 3 u / (2 a) )
Now, "equal intercepts" could mean just by size, or by size and sign. Take the latter:
Then
v (2a + 3u) / (3u) = u (2a + 3u) / (2a)
So either u = -3 a / 2,
or v / (3u) = u / (2a),
2 a v = 3 u^2 --
each to be taken with a v^2 = u^3
One possibility, which fits all three equations, is (u, v) = (0, 0).; and if either u = 0 or v = 0, the so is the other.
If u ≠ 0, v ≠ 0 then
first case:
u = -3 a / 2
so a v^2 = -27 a^3 / 8
so v^2 = -27 a^2 / 8 -- not possible
second case:
2 a v = 3 u^2
and a v^2 = u^3 divide
So v / 2 = u /3 substitute
so a 4u^2 / 9 = u^3,
(u = 0 or) u = 4 a / 9 substitute, again:
a v^2 = 64 a^3 / 729
v = (-/)+ 8 a / 27
The (only) point that cuts off equal intercepts on the axes is (4a/9, 8a/27).
(if we allow equal distances, we get one other point -- y changes sign.)
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 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 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
The length of the subtangent to the curve 
=3 at the point (4, 1) is- a)2
- b)1/2
- c)3
- d)4
Correct answer is option 'A'. Can you explain this answer?
The length of the subtangent to the curve =3 at the point (4, 1) is
=3 at the point (4, 1) isa)
2
b)
1/2
c)
3
d)
4
|
|
Anjali Sharma answered |
Length of subtangent =y(dx/dy)
Now given √x+√y=3
√y=3−√x
⇒1/2√ydy/dx = −1/(2√x) (On differentiating)
⇒dx/dy=−√x/√y
⇒ydx/dy=−√x/√y
⇒dx/dy/(4,1) = −√1√4
=−2
But the length can never be negative.
So length = 2.
Now given √x+√y=3
√y=3−√x
⇒1/2√ydy/dx = −1/(2√x) (On differentiating)
⇒dx/dy=−√x/√y
⇒ydx/dy=−√x/√y
⇒dx/dy/(4,1) = −√1√4
=−2
But the length can never be negative.
So length = 2.
The radius of a right circular cylinder of greatest curved surface which can be inscribed in a given right circular cone is- a) One third that of the cone
- b)1/
times that of the cone - c) 2/3 that of the cone
- d)1/2 that of the cone
Correct answer is option 'D'. Can you explain this answer?
The radius of a right circular cylinder of greatest curved surface which can be inscribed in a given right circular cone is
a)
One third that of the cone
b)
1/ times that of the cone
times that of the conec)
2/3 that of the cone
d)
1/2 that of the cone
|
Rajat Patel answered |
thenFind 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
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 minimum value of the function defined by f(x) = max (x, x + 1, 2 – x) is- a)0
- b)1/2
- c)1
- d)3/2
Correct answer is option 'D'. Can you explain this answer?
The minimum value of the function defined by f(x) = max (x, x + 1, 2 – x) is
a)
0
b)
1/2
c)
1
d)
3/2
|
|
Leelu Bhai answered |
Bhai dekh iss func ka graph bana ....mtlb ki f(x) = x, f(x) = x+ 1 aur f(x) = 2 - x , ye teeno func ek hi graph pe.now see, jo f(x) = x aur f(x) = x+1 hai woh to kahi bhi intersect nhi krege qki dono ||ᵉˡ line hai..aur jab tu f(x) = 2 - x ka graph banaega to inn dono ko intersect karte hue jyega.ab jo top most intersection point wohi sbse tagda hoga aur wohi f(x) = max. {x, x + 1, 2 - x} ki value hogi...so, f(x) = x + 1 and f(x) = 2 - x ka intersection iss graph ka top point hai ... to isko solve krke x = 1/2 aa rha hai aur iss x ki value ko f(x) me put krde to uski value 3/2 aayegi...and according to ques the value of f(x) = max. {x, x + 1, 2 - x} is the minimum value of function.i hope u will understand....
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
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.
increases in
is
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 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
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∏]
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
is
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.
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:
The extra area generated if the area of the whole floor is maximized is :- a)
- b)
- c)
- d)No change. Both areas are equal.
Correct answer is option 'D'. 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:
The extra area generated if the area of the whole floor is maximized is :
a)
b)
c)
d)
No change. Both areas are equal.
|
|
Raghav Bansal answered |
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
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.
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.
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:
The CEO of the multi-national company is interested in maximizing the area of the whole floor including the semi-circular ends. For this to happen the value of x should be- a)0 m
- b)30 m
- c)50 m
- d)80 m
Correct answer is option 'A'. 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:
The CEO of the multi-national company is interested in maximizing the area of the whole floor including the semi-circular ends. For this to happen the value of x should be
a)
0 m
b)
30 m
c)
50 m
d)
80 m
|
|
Geetika Shah answered |
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].
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
Chapter doubts & questions for Chapter 6 - Application of Derivatives - Mathematics Practice Tests: CUET Preparation 2024 is part of JEE exam preparation. The chapters have been prepared according to the JEE exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
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