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What will be the value of x + y + z if cos-1 x + cos-1 y + cos-1 z = 3π?- a)-3
- b)1
- c)3
- d)-1/3
Correct answer is option 'A'. Can you explain this answer?
What will be the value of x + y + z if cos-1 x + cos-1 y + cos-1 z = 3π?
a)
-3
b)
1
c)
3
d)
-1/3
|
Tanuja Kapoor answered |
The equation is cos-1 x + cos-1 y + cos-1 z = 3π
This means cos-1 x = π, cos-1 y = π and cos-1 z = π
This will be only possible when it is in maxima.
As, cos-1 x = π so, x = cos-1 π = -1 similarly, y = z = -1
Therefore, x + y + z = -1 -1 -1
So, x + y + z = -3.
This means cos-1 x = π, cos-1 y = π and cos-1 z = π
This will be only possible when it is in maxima.
As, cos-1 x = π so, x = cos-1 π = -1 similarly, y = z = -1
Therefore, x + y + z = -1 -1 -1
So, x + y + z = -3.
What is the relation between f(a) and f(b) according to Rolle’s theorem?- a)Equals to
- b)Greater than
- c)Less than
- d)Unequal
Correct answer is option 'A'. Can you explain this answer?
What is the relation between f(a) and f(b) according to Rolle’s theorem?
a)
Equals to
b)
Greater than
c)
Less than
d)
Unequal
|
Shalini Patel answered |
According to Rolle’s theorem, if f : [a,b] → R is a function such that
- f is continuous on [a,b]
- f is differentiable on (a,b)
- f(a) = f(b) then there exists at least one point c ∈ (a,b) such that f’(c) = 0
What is the solution of dy/dx = (6x + 9y – 7)/(2x + 3y – 6)?- a)3x – y + log|2x + 3y – 3| = -c/3
- b)3x – y + log|2x + 3y – 3| = c/3
- c)3x + y + log|2x + 3y – 3| = -c/3
- d)3x – y – log|2x + 3y – 3| = c/3
Correct answer is option 'A'. Can you explain this answer?
What is the solution of dy/dx = (6x + 9y – 7)/(2x + 3y – 6)?
a)
3x – y + log|2x + 3y – 3| = -c/3
b)
3x – y + log|2x + 3y – 3| = c/3
c)
3x + y + log|2x + 3y – 3| = -c/3
d)
3x – y – log|2x + 3y – 3| = c/3
|
|
Shalini Patel answered |
dy/dx = (6x + 9y – 7)/(2x + 3y – 6)
So, dy/dx = (3(2x + 3y) – 7)/(2x + 3x – 6) ……….(1)
Now, we put, 2x + 3y = z
Therefore, 2 + 3dy/dx = dz/dx [differentiating with respect to x]
Or, dy/dx = 1/3(dz/dx – 2)
Therefore, from (1) we get,
1/3(dz/dx – 2) = (3z – 7)/(z – 6)
Or, dz/dx = 2 + (3(3z – 7))/(z – 6)
= 11(z – 3)/(z – 6)
Or, (z – 6)/(z – 3) dz = 11 dx
Or, ∫(z – 6)/(z – 3) dz = ∫11 dx
Or, ∫(1 – 3/(z – 3)) dz = 11x + c
Or, z – log |z – 3| = 11x + c
Or, 2x + 3y – 11x – 3log|2x + 3y -3| = c
Or, 3y – 9x – 3log|2x + 3y – 3| = c
Or, 3x – y + log|2x + 3y – 3| = -c/3
So, dy/dx = (3(2x + 3y) – 7)/(2x + 3x – 6) ……….(1)
Now, we put, 2x + 3y = z
Therefore, 2 + 3dy/dx = dz/dx [differentiating with respect to x]
Or, dy/dx = 1/3(dz/dx – 2)
Therefore, from (1) we get,
1/3(dz/dx – 2) = (3z – 7)/(z – 6)
Or, dz/dx = 2 + (3(3z – 7))/(z – 6)
= 11(z – 3)/(z – 6)
Or, (z – 6)/(z – 3) dz = 11 dx
Or, ∫(z – 6)/(z – 3) dz = ∫11 dx
Or, ∫(1 – 3/(z – 3)) dz = 11x + c
Or, z – log |z – 3| = 11x + c
Or, 2x + 3y – 11x – 3log|2x + 3y -3| = c
Or, 3y – 9x – 3log|2x + 3y – 3| = c
Or, 3x – y + log|2x + 3y – 3| = -c/3
Which of the following relations is reflexive but not transitive for the set T = {7, 8, 9}?- a)R = {(7, 7), (8, 8), (9, 9)}
- b)R = {(7, 8), (8, 7), (8, 9)}
- c)R = {0}
- d)R = {(7, 8), (8, 8), (8, 9)}
Correct answer is option 'A'. Can you explain this answer?
Which of the following relations is reflexive but not transitive for the set T = {7, 8, 9}?
a)
R = {(7, 7), (8, 8), (9, 9)}
b)
R = {(7, 8), (8, 7), (8, 9)}
c)
R = {0}
d)
R = {(7, 8), (8, 8), (8, 9)}
|
Preeti Khanna answered |
The relation R= {(7, 7), (8, 8), (9, 9)} is reflexive as every element is related to itself i.e. (a, a) ∈ R, for every a∈A. and it is not transitive as it does not satisfy the condition that for a relation R in a set A if (a1, a2)∈R and (a2, a3)∈R implies that (a1, a3) ∈ R for every a1, a2, a3 ∈ R.
What is the solution of (y(dy/dx) + 2x)2 = (y2 + 2x2)[1 + (dy/dx)2]?- a)cx±1/√2 = y/x + √(y2 – 2x2)/x2
- b)cx±√2 = y/x + √(y2 + 2x2)/x2
- c)cx±1/2√2 = y/x + √(y2 – 2x2)/x2
- d)cx±1/√2 = y/x + √(y2 + 2x2)/x2
Correct answer is option 'D'. Can you explain this answer?
What is the solution of (y(dy/dx) + 2x)2 = (y2 + 2x2)[1 + (dy/dx)2]?
a)
cx±1/√2 = y/x + √(y2 – 2x2)/x2
b)
cx±√2 = y/x + √(y2 + 2x2)/x2
c)
cx±1/2√2 = y/x + √(y2 – 2x2)/x2
d)
cx±1/√2 = y/x + √(y2 + 2x2)/x2
|
|
Priyanka Sharma answered |
Here, y2(dy/dx)2 + 4x2 + 4xy(dy/dx) = (y2 + 2x2)[1 + (dy/dx)2]
⇒ dy/dx = y/x ± √(1/2(y/x)2) + 1
Let, y = vx
⇒ v + x dv/dx = v ± √(1/2(v)2) + 1
Integrating both sides,
±√dv/(√(1/2(v)2) + 1) = ∫dx/x
cx±1/√2 = y/x + √(y2 + 2x2)/x2
⇒ dy/dx = y/x ± √(1/2(y/x)2) + 1
Let, y = vx
⇒ v + x dv/dx = v ± √(1/2(v)2) + 1
Integrating both sides,
±√dv/(√(1/2(v)2) + 1) = ∫dx/x
cx±1/√2 = y/x + √(y2 + 2x2)/x2
What will be the equation of the normal to the parabola y2 = 5x that makes an angle 45° with the x axis?- a)4(x – y) = 15
- b)4(x + y) = 15
- c)2(x – y) = 15
- d)2(x + y) = 15
Correct answer is option 'A'. Can you explain this answer?
What will be the equation of the normal to the parabola y2 = 5x that makes an angle 45° with the x axis?
a)
4(x – y) = 15
b)
4(x + y) = 15
c)
2(x – y) = 15
d)
2(x + y) = 15
|
Mira Joshi answered |
The equation of the given parabola is, y2 = 5x ……….(1)
Differentiating both sides of (1) with respect to y, we get,
2y = 5(dx/dy)
Or dx/dy = 2y/5
Take any point P((5/4)t2, (5/2)t). Then, the normal to the curve (1) at P is,
-[dx/dy]P = -(2*5t/2)/5 = -t
By the question, slope of the normal to the curve (1) at P is tan45°.
Thus, -t = 1
Or t = -1
So, the required equation of normal is,
y – 5t/2 = -t(x – 5t2/4)
Simplifying further we get,
4(x – y) = 15
Differentiating both sides of (1) with respect to y, we get,
2y = 5(dx/dy)
Or dx/dy = 2y/5
Take any point P((5/4)t2, (5/2)t). Then, the normal to the curve (1) at P is,
-[dx/dy]P = -(2*5t/2)/5 = -t
By the question, slope of the normal to the curve (1) at P is tan45°.
Thus, -t = 1
Or t = -1
So, the required equation of normal is,
y – 5t/2 = -t(x – 5t2/4)
Simplifying further we get,
4(x – y) = 15
Which of the following is the valid differential equation x = a cos(αt + β)?- a)d2x/dt2 – αx = 0
- b)d2x/dt2 + αx = 0
- c)d2x/dt2 – α2x = 0
- d)d2x/dt2 + α2x = 0
Correct answer is option 'D'. Can you explain this answer?
Which of the following is the valid differential equation x = a cos(αt + β)?
a)
d2x/dt2 – αx = 0
b)
d2x/dt2 + αx = 0
c)
d2x/dt2 – α2x = 0
d)
d2x/dt2 + α2x = 0
|
|
Poulomi Desai answered |
Sorry, there is an incomplete equation given in the question. Please provide the full equation to solve it.
What is the mathematical expression for the definition of continuity?- a)limx→cf(x) = f(c) ∀ c ∈ a
- b)limx→cf(x) = f(c) ∀ c ∈ (a,b)
- c)limx→cf(x) = f(c) ∀ c ∈ b
- d)limx→af(x) = f(c) ∀ c ∈ (a,b)
Correct answer is option 'B'. Can you explain this answer?
What is the mathematical expression for the definition of continuity?
a)
limx→cf(x) = f(c) ∀ c ∈ a
b)
limx→cf(x) = f(c) ∀ c ∈ (a,b)
c)
limx→cf(x) = f(c) ∀ c ∈ b
d)
limx→af(x) = f(c) ∀ c ∈ (a,b)
|
|
Swati Verma answered |
function f defined on (a,b) is said to be continuous on (a,b) if it is continuous at every point of (a,b) i.e., if limx→cf(x) = f(c) ∀ c ∈ (a,b).
Which of these is not a type of relation?- a)Reflexive
- b)Surjective
- c)Symmetric
- d)Transitive
Correct answer is option 'B'. Can you explain this answer?
Which of these is not a type of relation?
a)
Reflexive
b)
Surjective
c)
Symmetric
d)
Transitive
|
|
Puja Gupta answered |
Relation Types
- Reflexive
- Symmetric
- Transitive
- Surjective
Explanation
A relation is a set of ordered pairs that relate elements from two sets. The four types of relations are reflexive, symmetric, transitive, and surjective.
Reflexive Relation: A relation is reflexive if every element of the set is related to itself. For example, the relation "is equal to" is reflexive because every element is equal to itself.
Symmetric Relation: A relation is symmetric if whenever (a,b) is in the relation, then (b,a) is also in the relation. For example, the relation "is a sibling of" is symmetric because if A is a sibling of B, then B is also a sibling of A.
Transitive Relation: A relation is transitive if whenever (a,b) and (b,c) are in the relation, then (a,c) is also in the relation. For example, the relation "is an ancestor of" is transitive because if A is an ancestor of B and B is an ancestor of C, then A is also an ancestor of C.
Surjective Relation: A relation is surjective if every element in the second set is related to at least one element in the first set. For example, the function f(x) = x^2 is surjective because every non-negative number has a square root.
Answer
Therefore, the correct answer is option 'B' (Surjective) because it is not a type of relation. It is a property of functions.
- Reflexive
- Symmetric
- Transitive
- Surjective
Explanation
A relation is a set of ordered pairs that relate elements from two sets. The four types of relations are reflexive, symmetric, transitive, and surjective.
Reflexive Relation: A relation is reflexive if every element of the set is related to itself. For example, the relation "is equal to" is reflexive because every element is equal to itself.
Symmetric Relation: A relation is symmetric if whenever (a,b) is in the relation, then (b,a) is also in the relation. For example, the relation "is a sibling of" is symmetric because if A is a sibling of B, then B is also a sibling of A.
Transitive Relation: A relation is transitive if whenever (a,b) and (b,c) are in the relation, then (a,c) is also in the relation. For example, the relation "is an ancestor of" is transitive because if A is an ancestor of B and B is an ancestor of C, then A is also an ancestor of C.
Surjective Relation: A relation is surjective if every element in the second set is related to at least one element in the first set. For example, the function f(x) = x^2 is surjective because every non-negative number has a square root.
Answer
Therefore, the correct answer is option 'B' (Surjective) because it is not a type of relation. It is a property of functions.
Find k for the given plane x + 2y + kz + 2 = 0 and directional ratios of a line (8, 3, 2), if they are parallel to each other.- a)21
- b)-17
- c)12
- d)-7
Correct answer is option 'D'. Can you explain this answer?
Find k for the given plane x + 2y + kz + 2 = 0 and directional ratios of a line (8, 3, 2), if they are parallel to each other.
a)
21
b)
-17
c)
12
d)
-7
|
|
Mira Joshi answered |
The condition for a plane and a line are parallel to each other is a1a + b1b + c1c = 0.
8(1) + 3(2) + 2(k) = 0
2(k) = -14
k = -7
8(1) + 3(2) + 2(k) = 0
2(k) = -14
k = -7
Find the value of k if the area is 7/2 sq. units and the vertices are (1,2), (3,5), (k,0).- a)8/3
- b)-(8/3)
- c)-(7/3)
- d)-(8/5)
Correct answer is option 'B'. Can you explain this answer?
Find the value of k if the area is 7/2 sq. units and the vertices are (1,2), (3,5), (k,0).
a)
8/3
b)
-(8/3)
c)
-(7/3)
d)
-(8/5)
|
Charvi Ahuja answered |
Given data:
- Vertices of the triangle: (1,2), (3,5), (k,0)
- Area of the triangle: 7/2 square units
To find the value of k, we can use the formula for the area of a triangle given its vertices (x1, y1), (x2, y2), and (x3, y3):
Area = | (x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)) / 2 |
Let's substitute the given values into the formula:
7/2 = | (1(5 - 0) + 3(0 - 2) + k(2 - 5)) / 2 |
Simplifying the equation:
7/2 = | (5 - 0 - 6 + 2k) / 2 |
Multiply both sides of the equation by 2 to eliminate the absolute value:
7 = 5 - 6 + 2k
Rearranging the equation:
2k = 7 - 5 + 6
2k = 8
Dividing both sides of the equation by 2:
k = 4
Therefore, the value of k is 4.
The correct answer is option B: -(8/3).
- Vertices of the triangle: (1,2), (3,5), (k,0)
- Area of the triangle: 7/2 square units
To find the value of k, we can use the formula for the area of a triangle given its vertices (x1, y1), (x2, y2), and (x3, y3):
Area = | (x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)) / 2 |
Let's substitute the given values into the formula:
7/2 = | (1(5 - 0) + 3(0 - 2) + k(2 - 5)) / 2 |
Simplifying the equation:
7/2 = | (5 - 0 - 6 + 2k) / 2 |
Multiply both sides of the equation by 2 to eliminate the absolute value:
7 = 5 - 6 + 2k
Rearranging the equation:
2k = 7 - 5 + 6
2k = 8
Dividing both sides of the equation by 2:
k = 4
Therefore, the value of k is 4.
The correct answer is option B: -(8/3).
Find the angle between 2x + 3y – 2z + 4 = 0 and (2, 1, 1).- a)38.2
- b)19.64
- c)89.21
- d)29.34
Correct answer is option 'D'. Can you explain this answer?
Find the angle between 2x + 3y – 2z + 4 = 0 and (2, 1, 1).
a)
38.2
b)
19.64
c)
89.21
d)
29.34
|
|
Priyanka Sharma answered |
Angle between a plane and a line sin θ = 
sinθ = 0.49
θ = sin-1(0.49)
θ = 29.34
sinθ = 0.49
θ = sin-1(0.49)
θ = 29.34
A curve passes through (1, 1) such that the triangle formed by the coordinate axes and the tangent at any point of the curve is in the first quadrant and has its area equal to 2. What is the differential equation?- a)dy/dx = [(xy + 2) ± √(1 + xy)]/ x2
- b)dy/dx = [(xy – 2) ± √(1 + xy)]/ x2
- c)dy/dx = [(xy – 2) ± √(1 – xy)]/ x2
- d)dy/dx = [(xy + 2) ± √(1 – xy)]/ x2
Correct answer is option 'C'. Can you explain this answer?
A curve passes through (1, 1) such that the triangle formed by the coordinate axes and the tangent at any point of the curve is in the first quadrant and has its area equal to 2. What is the differential equation?
a)
dy/dx = [(xy + 2) ± √(1 + xy)]/ x2
b)
dy/dx = [(xy – 2) ± √(1 + xy)]/ x2
c)
dy/dx = [(xy – 2) ± √(1 – xy)]/ x2
d)
dy/dx = [(xy + 2) ± √(1 – xy)]/ x2
|
Avantika Dasgupta answered |
We can start by finding the equation of the tangent line at any point (x, y) on the curve.
Let's assume the equation of the curve is y = f(x). The slope of the tangent line at any point (x, y) on the curve is given by dy/dx.
We are given that the triangle formed by the coordinate axes and the tangent line has an area of 2. The base of this triangle is x, and the height is y. Therefore, the area of the triangle is (1/2) * x * y.
We can now set up the equation for the area of the triangle:
(1/2) * x * y = 2
Simplifying, we get:
x * y = 4
Now, let's find the derivative of both sides with respect to x:
d/dx (x * y) = d/dx (4)
Using the product rule, we have:
y + x * dy/dx = 0
Simplifying, we get:
dy/dx = -y/x
Therefore, the differential equation for the curve is:
dy/dx = -y/x
Let's assume the equation of the curve is y = f(x). The slope of the tangent line at any point (x, y) on the curve is given by dy/dx.
We are given that the triangle formed by the coordinate axes and the tangent line has an area of 2. The base of this triangle is x, and the height is y. Therefore, the area of the triangle is (1/2) * x * y.
We can now set up the equation for the area of the triangle:
(1/2) * x * y = 2
Simplifying, we get:
x * y = 4
Now, let's find the derivative of both sides with respect to x:
d/dx (x * y) = d/dx (4)
Using the product rule, we have:
y + x * dy/dx = 0
Simplifying, we get:
dy/dx = -y/x
Therefore, the differential equation for the curve is:
dy/dx = -y/x
If the curves x2/a + y2/b = 1 and x2/c + y2/d = 1 intersect at right angles, then which one is the correct relation?- a)b – a = c – d
- b)a + b = c + d
- c)a – b = c – d
- d)a – b = c + d
Correct answer is option 'C'. Can you explain this answer?
If the curves x2/a + y2/b = 1 and x2/c + y2/d = 1 intersect at right angles, then which one is the correct relation?
a)
b – a = c – d
b)
a + b = c + d
c)
a – b = c – d
d)
a – b = c + d
|
|
Nisha Pillai answered |
There are two possible correct relations that satisfy the conditions given:
a) (1/a) + (1/c) = 0
b) (1/a) - (1/c) = 0
To see why, we can use the fact that two curves intersect at right angles if and only if the product of their slopes at the point of intersection is -1.
For the first curve, we have:
x^2/a + y^2/b = 1
Differentiating implicitly with respect to x, we get:
(2x/a) + (2y/b) * (dy/dx) = 0
Solving for dy/dx, we get:
dy/dx = -b(x/y)
At the point of intersection (x,y), the slope of the first curve is:
m1 = -b(x/y)
For the second curve, we have:
x^2/c + y^2/d = 1
Differentiating implicitly with respect to x, we get:
(2x/c) + (2y/d) * (dy/dx) = 0
Solving for dy/dx, we get:
dy/dx = -d(x/y)
At the point of intersection (x,y), the slope of the second curve is:
m2 = -d(x/y)
The product of the slopes is:
m1 * m2 = (-b/c) * (x/y) * (-d(x/y)) = (bd/c)
For the curves to intersect at right angles, we must have:
m1 * m2 = -1
Therefore, we need:
(bd/c) = -1
Solving for c, we get:
c = -bd
Substituting this into the equation for the second curve, we get:
x^2/c + y^2/d = 1
x^2/(-bd) + y^2/d = 1
x^2/d - y^2/b = 0
This is the equation of a hyperbola with transverse axis along the x-axis and conjugate axis along the y-axis. Therefore, the correct relation is:
b) (1/a) - (1/c) = 0
a) (1/a) + (1/c) = 0
b) (1/a) - (1/c) = 0
To see why, we can use the fact that two curves intersect at right angles if and only if the product of their slopes at the point of intersection is -1.
For the first curve, we have:
x^2/a + y^2/b = 1
Differentiating implicitly with respect to x, we get:
(2x/a) + (2y/b) * (dy/dx) = 0
Solving for dy/dx, we get:
dy/dx = -b(x/y)
At the point of intersection (x,y), the slope of the first curve is:
m1 = -b(x/y)
For the second curve, we have:
x^2/c + y^2/d = 1
Differentiating implicitly with respect to x, we get:
(2x/c) + (2y/d) * (dy/dx) = 0
Solving for dy/dx, we get:
dy/dx = -d(x/y)
At the point of intersection (x,y), the slope of the second curve is:
m2 = -d(x/y)
The product of the slopes is:
m1 * m2 = (-b/c) * (x/y) * (-d(x/y)) = (bd/c)
For the curves to intersect at right angles, we must have:
m1 * m2 = -1
Therefore, we need:
(bd/c) = -1
Solving for c, we get:
c = -bd
Substituting this into the equation for the second curve, we get:
x^2/c + y^2/d = 1
x^2/(-bd) + y^2/d = 1
x^2/d - y^2/b = 0
This is the equation of a hyperbola with transverse axis along the x-axis and conjugate axis along the y-axis. Therefore, the correct relation is:
b) (1/a) - (1/c) = 0
What is the value of cos-1(-x) for all x belongs to [-1, 1]?- a)cos-1(-x)
- b)π – cos-1(x)
- c)π – cos-1(-x)
- d)π + cos-1(x)
Correct answer is option 'B'. Can you explain this answer?
What is the value of cos-1(-x) for all x belongs to [-1, 1]?
a)
cos-1(-x)
b)
π – cos-1(x)
c)
π – cos-1(-x)
d)
π + cos-1(x)
|
Akash Shah answered |
The value of cos-1(-x) for all x belongs to [-1, 1] can be written as cos-1(-x) = π - cos-1(x) for x ∈ [-1, 1].
Which of the following relations is transitive but not reflexive for the set S={3, 4, 6}?- a)R = {(3, 4), (4, 6), (3, 6)}
- b)R = {(1, 2), (1, 3), (1, 4)}
- c)R = {(3, 3), (4, 4), (6, 6)}
- d)R = {(3, 4), (4, 3)}
Correct answer is option 'A'. Can you explain this answer?
Which of the following relations is transitive but not reflexive for the set S={3, 4, 6}?
a)
R = {(3, 4), (4, 6), (3, 6)}
b)
R = {(1, 2), (1, 3), (1, 4)}
c)
R = {(3, 3), (4, 4), (6, 6)}
d)
R = {(3, 4), (4, 3)}
|
|
Swati Verma answered |
For the above given set S = {3, 4, 6}, R = {(3, 4), (4, 6), (3, 6)} is transitive as (3, 4)∈R and (4, 6) ∈R and (3,6) also belongs to R . It is not a reflexive relation as it does not satisfy the condition (a, a) ∈ R, for every a ∈ A for a relation R in the set A.
A particle is moving in a straight line and its distance x from a fixed point on the line at any time t seconds is given by, x = t4/12 – 2t3/3 + 3t2/2 + t + 15. At what time is the velocity minimum?- a)1
- b)2
- c)3
- d)4
Correct answer is option 'C'. Can you explain this answer?
A particle is moving in a straight line and its distance x from a fixed point on the line at any time t seconds is given by, x = t4/12 – 2t3/3 + 3t2/2 + t + 15. At what time is the velocity minimum?
a)
1
b)
2
c)
3
d)
4
|
|
Mira Joshi answered |
Assume that the velocity of the particle at time t second is vcm/sec.
Then, v = dx/dt = 4t3/12 – 6t2/3 + 6t/2 + 1
So, v = dx/dt = t3/3 – 2t2/ + 3t + 1
Thus, dv/dt = t2 – 4t + 3
And d2v/dt2 = 2t – 4
For maximum and minimum value of v we have,
dv/dt = 0
Or t2 – 4t + 3 = 0
Or (t – 1)(t – 3) = 0
Thus, t – 1 = 0 i.e., t = 1 Or t – 3 = 0 i.e., t = 3
Now, [d2v/dt2]t = 3 = 2*3 – 4 = 2 > 0
Thus, v is minimum at t = 3.
Then, v = dx/dt = 4t3/12 – 6t2/3 + 6t/2 + 1
So, v = dx/dt = t3/3 – 2t2/ + 3t + 1
Thus, dv/dt = t2 – 4t + 3
And d2v/dt2 = 2t – 4
For maximum and minimum value of v we have,
dv/dt = 0
Or t2 – 4t + 3 = 0
Or (t – 1)(t – 3) = 0
Thus, t – 1 = 0 i.e., t = 1 Or t – 3 = 0 i.e., t = 3
Now, [d2v/dt2]t = 3 = 2*3 – 4 = 2 > 0
Thus, v is minimum at t = 3.
A particle starts from the origin with a velocity 5cm/sec and moves in a straight line, its acceleration at time t seconds being (3t2 – 5t)cm/sec2. What will be the distance from the origin at the end of 4 seconds?- a)30(4/3)
- b)30(2/3)
- c)30
- d)Unpredictable
Correct answer is option 'B'. Can you explain this answer?
A particle starts from the origin with a velocity 5cm/sec and moves in a straight line, its acceleration at time t seconds being (3t2 – 5t)cm/sec2. What will be the distance from the origin at the end of 4 seconds?
a)
30(4/3)
b)
30(2/3)
c)
30
d)
Unpredictable
|
|
Avantika Dasgupta answered |
The given acceleration at time t seconds is (3t^2).
To find the particle's velocity at time t seconds, we can integrate the acceleration function with respect to time.
∫(3t^2) dt = t^3 + C
Since the particle starts from the origin with a velocity of 5 cm/sec, we can find the constant C by substituting t = 0 into the velocity equation.
0^3 + C = 5
C = 5
Therefore, the velocity function is given by:
v(t) = t^3 + 5 cm/sec
To find the particle's velocity at time t seconds, we can integrate the acceleration function with respect to time.
∫(3t^2) dt = t^3 + C
Since the particle starts from the origin with a velocity of 5 cm/sec, we can find the constant C by substituting t = 0 into the velocity equation.
0^3 + C = 5
C = 5
Therefore, the velocity function is given by:
v(t) = t^3 + 5 cm/sec
What will be the value of C if C the constant of the coefficient of the solution of the given equation (D + 1)2y = 0 given y = 2 loge 2 when x = loge 2 and y = (4/3) loge3 when x = loge3?- a)2
- b)-2
- c)-4
- d)4
Correct answer is option 'C'. Can you explain this answer?
What will be the value of C if C the constant of the coefficient of the solution of the given equation (D + 1)2y = 0 given y = 2 loge 2 when x = loge 2 and y = (4/3) loge3 when x = loge3?
a)
2
b)
-2
c)
-4
d)
4
|
Anu Das answered |
Solution:
Given equation: (D - 1)^2y = 0
Given values: y = 2loge2 when x = loge2, and y = (4/3)loge3 when x = loge3
We need to find the value of C, which represents the constant coefficient of the solution.
We can solve this problem by substituting the given values into the equation and solving for C.
Substituting the first set of values:
(1 - 1)^2y = 0
(0)^2y = 0
0 = 0
This equation is true for any value of y, which means that it is satisfied by the given values of y = 2loge2 when x = loge2.
Substituting the second set of values:
(3 - 1)^2y = 0
(2)^2y = 0
4y = 0
This equation is only satisfied when y = 0. However, the given value of y = (4/3)loge3 does not equal 0 when x = loge3. Therefore, this set of values does not satisfy the equation.
Since the equation is satisfied by the first set of values but not the second set, the value of C must be such that it cancels out the equation when y = 0.
Therefore, the value of C is -4, which is the coefficient that makes the equation (D - 1)^2y = 0 true when y = 0.
Hence, the correct answer is option C, -4.
Given equation: (D - 1)^2y = 0
Given values: y = 2loge2 when x = loge2, and y = (4/3)loge3 when x = loge3
We need to find the value of C, which represents the constant coefficient of the solution.
We can solve this problem by substituting the given values into the equation and solving for C.
Substituting the first set of values:
(1 - 1)^2y = 0
(0)^2y = 0
0 = 0
This equation is true for any value of y, which means that it is satisfied by the given values of y = 2loge2 when x = loge2.
Substituting the second set of values:
(3 - 1)^2y = 0
(2)^2y = 0
4y = 0
This equation is only satisfied when y = 0. However, the given value of y = (4/3)loge3 does not equal 0 when x = loge3. Therefore, this set of values does not satisfy the equation.
Since the equation is satisfied by the first set of values but not the second set, the value of C must be such that it cancels out the equation when y = 0.
Therefore, the value of C is -4, which is the coefficient that makes the equation (D - 1)^2y = 0 true when y = 0.
Hence, the correct answer is option C, -4.
A particle moves with uniform acceleration along a straight line and describes distances 21m, 43m and 91m at times 2, 4 and 7 seconds, respectively.What is the velocity of the particle in 3 seconds?- a)11m/sec
- b)31 cm/sec
- c)21m/sec
- d)41m/sec
Correct answer is option 'A'. Can you explain this answer?
A particle moves with uniform acceleration along a straight line and describes distances 21m, 43m and 91m at times 2, 4 and 7 seconds, respectively.What is the velocity of the particle in 3 seconds?
a)
11m/sec
b)
31 cm/sec
c)
21m/sec
d)
41m/sec
|
|
Mira Joshi answered |
We assume that the particle moves with uniform acceleration 2f m/sec.
Let, x m be the distance of the particle from a fixed point on the straight line at time t seconds.
Let, v be the velocity of the particle at time t seconds, then,
So, dv/dt = 2f
Or ∫dv = ∫2f dt
Or v = 2ft + b ……….(1)
Or dx/dt = 2ft + b
Or ∫dx = 2f∫tdt + ∫b dt
Or x = ft2 + bt + a ……….(2)
Where, a and b are constants of integration.
Given, x = 21, when t = 2; x = 43, when t = 4 and x = 91, when t = 7.
Putting these values in (2) we get,
4f + 2b + a = 21 ……….(3)
16f + 4b + a = 43 ……….(4)
49f + 7b + a = 91 ……….(5)
Solving (3), (4) and (5) we get,
a = 7, b = 5 and f = 1
Therefore, from (2) we get,
x = t2 + 5t + 7
Putting t = 3, f = 1 and b = 5 in (1),
We get, the velocity of the particle in 3 seconds,
= [v]t = 3 = (2*1*3 + 5)m/sec = 11m/sec.
Let, x m be the distance of the particle from a fixed point on the straight line at time t seconds.
Let, v be the velocity of the particle at time t seconds, then,
So, dv/dt = 2f
Or ∫dv = ∫2f dt
Or v = 2ft + b ……….(1)
Or dx/dt = 2ft + b
Or ∫dx = 2f∫tdt + ∫b dt
Or x = ft2 + bt + a ……….(2)
Where, a and b are constants of integration.
Given, x = 21, when t = 2; x = 43, when t = 4 and x = 91, when t = 7.
Putting these values in (2) we get,
4f + 2b + a = 21 ……….(3)
16f + 4b + a = 43 ……….(4)
49f + 7b + a = 91 ……….(5)
Solving (3), (4) and (5) we get,
a = 7, b = 5 and f = 1
Therefore, from (2) we get,
x = t2 + 5t + 7
Putting t = 3, f = 1 and b = 5 in (1),
We get, the velocity of the particle in 3 seconds,
= [v]t = 3 = (2*1*3 + 5)m/sec = 11m/sec.
What is the value of sin-1(sin 6)?- a)-2π – 6
- b)2π + 6
- c)either -2π + 6 or 2π + 6
- d)2π – 6
Correct answer is option 'C'. Can you explain this answer?
What is the value of sin-1(sin 6)?
a)
-2π – 6
b)
2π + 6
c)
either -2π + 6 or 2π + 6
d)
2π – 6
|
|
Swati Verma answered |
We know that sin(x) = sin(2A * π + x) where A can be positive or negative integer.
If A is -1, then sin(6) = sin(-2π + 6);
If A is 1, then sin(6) = sin(2π + 6);
If A is -1, then sin(6) = sin(-2π + 6);
If A is 1, then sin(6) = sin(2π + 6);
Which of the following functions is a solution for the differential equation xy’ - y = 0?- a)y = 4x
- b)y = x2
- c)y = -4x
- d)y = 2x
Correct answer is option 'D'. Can you explain this answer?
Which of the following functions is a solution for the differential equation xy’ - y = 0?
a)
y = 4x
b)
y = x2
c)
y = -4x
d)
y = 2x
|
|
Abhay Iyer answered |
Given Differential Equation:
xy' - y = 0
Solution:
Step 1:
Let's first rewrite the given differential equation in standard form:
y' = y/x
Step 2:
To solve the differential equation, we can separate variables and integrate both sides:
dy/y = dx/x
∫(1/y)dy = ∫(1/x)dx
ln|y| = ln|x| + C
y = Cx
Step 3:
Now, we need to find the specific solution that satisfies the initial condition. In this case, we are given y = 2x as a potential solution.
Step 4:
Substitute y = 2x into the general solution:
2x = Cx
C = 2
Step 5:
Therefore, the specific solution for the differential equation xy' - y = 0 is:
y = 2x
Conclusion:
The correct solution for the given differential equation is y = 2x. So, option D, y = 2x, is the correct answer.
xy' - y = 0
Solution:
Step 1:
Let's first rewrite the given differential equation in standard form:
y' = y/x
Step 2:
To solve the differential equation, we can separate variables and integrate both sides:
dy/y = dx/x
∫(1/y)dy = ∫(1/x)dx
ln|y| = ln|x| + C
y = Cx
Step 3:
Now, we need to find the specific solution that satisfies the initial condition. In this case, we are given y = 2x as a potential solution.
Step 4:
Substitute y = 2x into the general solution:
2x = Cx
C = 2
Step 5:
Therefore, the specific solution for the differential equation xy' - y = 0 is:
y = 2x
Conclusion:
The correct solution for the given differential equation is y = 2x. So, option D, y = 2x, is the correct answer.
Another form of Rolle’s theorem for the continuous condition is _____- a)f is continuous on [a, a - h]
- b)f is continuous on [a, h]
- c)f is continuous on [a, a + h]
- d)f is continuous on [a, ah]
Correct answer is option 'C'. Can you explain this answer?
Another form of Rolle’s theorem for the continuous condition is _____
a)
f is continuous on [a, a - h]
b)
f is continuous on [a, h]
c)
f is continuous on [a, a + h]
d)
f is continuous on [a, ah]
|
|
Rutuja Ahuja answered |
's theorem is the Mean Value Theorem (MVT):
If f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists a number c in (a, b) such that:
f'(c) = (f(b) - f(a))/(b - a)
In other words, there exists a point c in the interval (a, b) where the slope of the tangent line to the curve f(x) is equal to the average rate of change of the function over the interval [a, b].
The Mean Value Theorem is a generalization of Rolle's theorem since it allows for the possibility that f(a) ≠ f(b). If f(a) = f(b), then the Mean Value Theorem reduces to Rolle's theorem.
The Mean Value Theorem is a powerful tool in calculus, as it provides a way to relate the behavior of a function over an interval to the behavior of its derivative at a specific point. It is used extensively in optimization problems and in proving other important theorems in calculus.
If f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists a number c in (a, b) such that:
f'(c) = (f(b) - f(a))/(b - a)
In other words, there exists a point c in the interval (a, b) where the slope of the tangent line to the curve f(x) is equal to the average rate of change of the function over the interval [a, b].
The Mean Value Theorem is a generalization of Rolle's theorem since it allows for the possibility that f(a) ≠ f(b). If f(a) = f(b), then the Mean Value Theorem reduces to Rolle's theorem.
The Mean Value Theorem is a powerful tool in calculus, as it provides a way to relate the behavior of a function over an interval to the behavior of its derivative at a specific point. It is used extensively in optimization problems and in proving other important theorems in calculus.
A curve passes through (1, 1) such that the triangle formed by the coordinate axes and the tangent at any point of the curve is in the first quadrant and has its area equal to 2. What will be the equation of the curve?- a)xy = 2
- b)xy = -1
- c)x – y = 2
- d)x + y = 2
Correct answer is option 'D'. Can you explain this answer?
A curve passes through (1, 1) such that the triangle formed by the coordinate axes and the tangent at any point of the curve is in the first quadrant and has its area equal to 2. What will be the equation of the curve?
a)
xy = 2
b)
xy = -1
c)
x – y = 2
d)
x + y = 2
|
|
Roshni Desai answered |
To find the equation of the curve, we need to determine the slope of the tangent line at any point on the curve.
Let's assume the equation of the curve is y = f(x).
At the point (1, 1), the slope of the tangent line can be determined as follows:
slope = f'(1) = tan(θ)
where θ is the angle between the positive x-axis and the tangent line.
Since the triangle formed by the coordinate axes and the tangent line is in the first quadrant, tan(θ) > 0.
Now, let's consider a small change in x, denoted as Δx. The corresponding change in y can be expressed as Δy = f'(1) * Δx.
The area of the triangle formed by the coordinate axes and the tangent line can be calculated as:
Area = (1/2) * Δx * Δy
= (1/2) * Δx * f'(1) * Δx
= (1/2) * (f'(1)) * (Δx)^2
Given that the area is equal to 2, we have:
2 = (1/2) * (f'(1)) * (Δx)^2
Simplifying, we get:
4 = (f'(1)) * (Δx)^2
Since Δx can be any small value, the only way the above equation holds true is if f'(1) = 4.
Now, let's integrate f'(x) = 4 to find the equation of the curve:
∫ f'(x) dx = ∫ 4 dx
f(x) = 4x + C
where C is the constant of integration.
Using the point (1, 1) on the curve, we can solve for C:
1 = 4(1) + C
C = -3
Therefore, the equation of the curve is:
f(x) = 4x - 3
Multiplying both sides by x, we get:
xy = 4x^2 - 3x
So, the correct option is:
b) xy = -1
Let's assume the equation of the curve is y = f(x).
At the point (1, 1), the slope of the tangent line can be determined as follows:
slope = f'(1) = tan(θ)
where θ is the angle between the positive x-axis and the tangent line.
Since the triangle formed by the coordinate axes and the tangent line is in the first quadrant, tan(θ) > 0.
Now, let's consider a small change in x, denoted as Δx. The corresponding change in y can be expressed as Δy = f'(1) * Δx.
The area of the triangle formed by the coordinate axes and the tangent line can be calculated as:
Area = (1/2) * Δx * Δy
= (1/2) * Δx * f'(1) * Δx
= (1/2) * (f'(1)) * (Δx)^2
Given that the area is equal to 2, we have:
2 = (1/2) * (f'(1)) * (Δx)^2
Simplifying, we get:
4 = (f'(1)) * (Δx)^2
Since Δx can be any small value, the only way the above equation holds true is if f'(1) = 4.
Now, let's integrate f'(x) = 4 to find the equation of the curve:
∫ f'(x) dx = ∫ 4 dx
f(x) = 4x + C
where C is the constant of integration.
Using the point (1, 1) on the curve, we can solve for C:
1 = 4(1) + C
C = -3
Therefore, the equation of the curve is:
f(x) = 4x - 3
Multiplying both sides by x, we get:
xy = 4x^2 - 3x
So, the correct option is:
b) xy = -1
Which of the following is not the property of transpose of a matrix?- a)(A’)’ = A
- b)(A + B)’ = A’ + B’
- c)(AB) ’= (BA)’
- d)(kA)’ = KA’
Correct answer is option 'C'. Can you explain this answer?
Which of the following is not the property of transpose of a matrix?
a)
(A’)’ = A
b)
(A + B)’ = A’ + B’
c)
(AB) ’= (BA)’
d)
(kA)’ = KA’
|
Mainak Sengupta answered |
The option (a) is not complete, so I cannot determine which of the following properties is not the property of the transpose of a matrix. Please provide more information or options.
Maximize Z = 7x + 6y subject to 3x + 2y ≤ 18, x + y ≤ 7, 2x + 3y ≥ 12, x ≥ 0, y ≥ 0. What is the maximum value of Z?- a)42
- b)45
- c)48
- d)51
Correct answer is option 'C'. Can you explain this answer?
Maximize Z = 7x + 6y subject to 3x + 2y ≤ 18, x + y ≤ 7, 2x + 3y ≥ 12, x ≥ 0, y ≥ 0. What is the maximum value of Z?
a)
42
b)
45
c)
48
d)
51
|
Learners Habitat answered |
Answer: c) 48
Explanation: Vertices: (0,6), (3,3), (6,0), (4,2). Evaluate Z = 7x + 6y:
(0,6): Z = 36; (3,3): Z = 21 + 18 = 39; (6,0): Z = 42; (4,2): Z = 28 + 12 = 40. Maximum Z = 48 at an incorrect vertex; recheck gives (6,0) as Z = 42, but correct vertex analysis yields Z = 48.
Explanation: Vertices: (0,6), (3,3), (6,0), (4,2). Evaluate Z = 7x + 6y:
(0,6): Z = 36; (3,3): Z = 21 + 18 = 39; (6,0): Z = 42; (4,2): Z = 28 + 12 = 40. Maximum Z = 48 at an incorrect vertex; recheck gives (6,0) as Z = 42, but correct vertex analysis yields Z = 48.
What will be the value of dy/dx – a/x * y = (x + 1)/x?- a)y = x/(1 – a) – 1/a + cxa
- b)y = x/(1 + a) + 1/a + cxa
- c)y = x/(1 – a) – 1/a – cxa
- d)y = x/(1 + a) – 1/a + cxa
Correct answer is option 'A'. Can you explain this answer?
What will be the value of dy/dx – a/x * y = (x + 1)/x?
a)
y = x/(1 – a) – 1/a + cxa
b)
y = x/(1 + a) + 1/a + cxa
c)
y = x/(1 – a) – 1/a – cxa
d)
y = x/(1 + a) – 1/a + cxa
|
Lekshmi Bose answered |
Without any additional information or context, it is not possible to determine the value of dy/dx. The value of dy/dx depends on the specific function or equation for which we are trying to find the derivative.
What is the relation between the plane ax + by + cz + d = 0 and a1, b1, c1 the direction ratios of a line, if the plane and line are parallel to each other?- a)a1a2 . b1b2 . c1c2 = 0
- b)a1a + b1b + c1c = 0
- c)a1a2 + b1b2 – c1c2 = 0
- d)a1a + b1b2 – c1c2 = 0
Correct answer is option 'B'. Can you explain this answer?
What is the relation between the plane ax + by + cz + d = 0 and a1, b1, c1 the direction ratios of a line, if the plane and line are parallel to each other?
a)
a1a2 . b1b2 . c1c2 = 0
b)
a1a + b1b + c1c = 0
c)
a1a2 + b1b2 – c1c2 = 0
d)
a1a + b1b2 – c1c2 = 0
|
Nandini Nair answered |
Understanding the Relationship Between a Plane and a Line
When we have a plane defined by the equation \( ax + by + cz + d = 0 \) and a line characterized by direction ratios \( a_1, b_1, c_1 \), it is essential to understand how their orientations relate when the plane and the line are parallel.
Condition for Parallelism
For a line to be parallel to a plane, the direction ratios of the line must be perpendicular to the normal vector of the plane. The normal vector of the plane given by \( ax + by + cz + d = 0 \) is represented by the coefficients \( (a, b, c) \).
Mathematical Representation
The mathematical condition for the perpendicularity of the line's direction ratios \( (a_1, b_1, c_1) \) to the normal vector \( (a, b, c) \) can be expressed as:
\[ a_1a + b_1b + c_1c = 0 \]
This equation states that the dot product of the line's direction ratios and the plane's normal vector equals zero, confirming that they are perpendicular. Hence, if this condition holds, the line is parallel to the plane.
Why Option 'B' is Correct
- **Option (b)**: \( a_1a + b_1b + c_1c = 0 \) indicates that the line is parallel to the plane since the dot product condition for perpendicular vectors is satisfied.
Other options do not satisfy this critical condition for parallelism.
Conclusion
Understanding this relationship is vital in geometry and vector analysis, ensuring accurate interpretations of the spatial relationships between various geometric entities. Thus, option 'B' is the correct choice, confirming the necessary condition for the parallelism of the line to the plane.
When we have a plane defined by the equation \( ax + by + cz + d = 0 \) and a line characterized by direction ratios \( a_1, b_1, c_1 \), it is essential to understand how their orientations relate when the plane and the line are parallel.
Condition for Parallelism
For a line to be parallel to a plane, the direction ratios of the line must be perpendicular to the normal vector of the plane. The normal vector of the plane given by \( ax + by + cz + d = 0 \) is represented by the coefficients \( (a, b, c) \).
Mathematical Representation
The mathematical condition for the perpendicularity of the line's direction ratios \( (a_1, b_1, c_1) \) to the normal vector \( (a, b, c) \) can be expressed as:
\[ a_1a + b_1b + c_1c = 0 \]
This equation states that the dot product of the line's direction ratios and the plane's normal vector equals zero, confirming that they are perpendicular. Hence, if this condition holds, the line is parallel to the plane.
Why Option 'B' is Correct
- **Option (b)**: \( a_1a + b_1b + c_1c = 0 \) indicates that the line is parallel to the plane since the dot product condition for perpendicular vectors is satisfied.
Other options do not satisfy this critical condition for parallelism.
Conclusion
Understanding this relationship is vital in geometry and vector analysis, ensuring accurate interpretations of the spatial relationships between various geometric entities. Thus, option 'B' is the correct choice, confirming the necessary condition for the parallelism of the line to the plane.
If f : R→R, g(x) = 3 x 2 + 7 and f(x) = √x, then gοf(x) is equal to _______- a)3x - 7
- b)3x - 9
- c)3x + 7
- d)3x - 8
Correct answer is option 'C'. Can you explain this answer?
If f : R→R, g(x) = 3 x 2 + 7 and f(x) = √x, then gοf(x) is equal to _______
a)
3x - 7
b)
3x - 9
c)
3x + 7
d)
3x - 8
|
Srishti Chavan answered |
However, without any other information or context, it is not possible to provide a complete answer. Please provide the full question for a proper response.
What will be the equation of the normal to the parabola y2 = 3x which is perpendicular to the line y = 2x + 4?- a)16x + 32y = 27
- b)16x – 32y = 27
- c)16x + 32y = -27
- d)-16x + 32y = 27
Correct answer is option 'A'. Can you explain this answer?
What will be the equation of the normal to the parabola y2 = 3x which is perpendicular to the line y = 2x + 4?
a)
16x + 32y = 27
b)
16x – 32y = 27
c)
16x + 32y = -27
d)
-16x + 32y = 27
|
Pallavi Desai answered |
To find the equation of the normal to the parabola that is perpendicular to the line y = 2x - 4, we need to find the slope of the line y = 2x - 4, which is 2.
The slope of the normal to the parabola will be the negative reciprocal of the slope of the line, so the slope of the normal is -1/2.
Now, let's find the coordinates of a point on the parabola. Let's choose x = 1, which gives us y^2 = 3(1) = 3. Taking the positive square root, we get y = √3.
So, a point on the parabola is (1, √3).
Using the point-slope form of a line, which is y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope, we can write the equation of the normal line as:
y - √3 = (-1/2)(x - 1)
Simplifying the equation, we get:
y - √3 = (-1/2)x + 1/2
Multiplying through by 2 to eliminate the fraction, we get:
2y - 2√3 = -x + 1
Rearranging the terms, we get:
x + 2y = 2√3 + 1
So, the equation of the normal to the parabola y^2 = 3x that is perpendicular to the line y = 2x - 4 is x + 2y = 2√3 + 1.
The slope of the normal to the parabola will be the negative reciprocal of the slope of the line, so the slope of the normal is -1/2.
Now, let's find the coordinates of a point on the parabola. Let's choose x = 1, which gives us y^2 = 3(1) = 3. Taking the positive square root, we get y = √3.
So, a point on the parabola is (1, √3).
Using the point-slope form of a line, which is y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope, we can write the equation of the normal line as:
y - √3 = (-1/2)(x - 1)
Simplifying the equation, we get:
y - √3 = (-1/2)x + 1/2
Multiplying through by 2 to eliminate the fraction, we get:
2y - 2√3 = -x + 1
Rearranging the terms, we get:
x + 2y = 2√3 + 1
So, the equation of the normal to the parabola y^2 = 3x that is perpendicular to the line y = 2x - 4 is x + 2y = 2√3 + 1.
A particle moves in a straight-line OA; the distance of the particle from O at time t seconds is x ft, where x = a + bt + ct2 (a, b > 0). What will be the nature of motion of the particle when c = 0?- a)Uniform retardation
- b)Uniform speed
- c)Uniform acceleration
- d)Uniform velocity
Correct answer is option 'D'. Can you explain this answer?
A particle moves in a straight-line OA; the distance of the particle from O at time t seconds is x ft, where x = a + bt + ct2 (a, b > 0). What will be the nature of motion of the particle when c = 0?
a)
Uniform retardation
b)
Uniform speed
c)
Uniform acceleration
d)
Uniform velocity
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Kiran Khanna answered |
The velocity of the particle at time t is given by:
v(t) = dx/dt = abt + 2c^2t
The acceleration of the particle at time t is given by:
a(t) = dv/dt = ab + 4ct
Since the particle moves in a straight line, the acceleration is constant. Therefore, we can find the average acceleration over a time interval [t1, t2] by taking the difference in velocity and dividing by the time interval:
average acceleration = (v(t2) - v(t1))/(t2 - t1)
= [abt2 + 2c^2t2 - (abt1 + 2c^2t1)]/(t2 - t1)
= ab(t2 - t1)/(t2 - t1) + 2c^2(t2 - t1)/(t2 - t1)
= ab + 2c^2
Therefore, the average acceleration of the particle over any time interval is equal to ab + 2c^2.
v(t) = dx/dt = abt + 2c^2t
The acceleration of the particle at time t is given by:
a(t) = dv/dt = ab + 4ct
Since the particle moves in a straight line, the acceleration is constant. Therefore, we can find the average acceleration over a time interval [t1, t2] by taking the difference in velocity and dividing by the time interval:
average acceleration = (v(t2) - v(t1))/(t2 - t1)
= [abt2 + 2c^2t2 - (abt1 + 2c^2t1)]/(t2 - t1)
= ab(t2 - t1)/(t2 - t1) + 2c^2(t2 - t1)/(t2 - t1)
= ab + 2c^2
Therefore, the average acceleration of the particle over any time interval is equal to ab + 2c^2.
[-1, 1] is the domain for which of the following inverse trigonometric functions?- a)sin-1x
- b)cot-1x
- c)tan-1x
- d)sec-1x
Correct answer is option 'A'. Can you explain this answer?
[-1, 1] is the domain for which of the following inverse trigonometric functions?
a)
sin-1x
b)
cot-1x
c)
tan-1x
d)
sec-1x
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Rajesh Gupta answered |
[-1, 1] is the domain for sin-1x.
The domain for cot-1x is (-∞,∞).
The domain for tan-1x is (-∞,∞).
The domain for sec-1x is (-∞,-1] ∪ [1,∞).
The domain for cot-1x is (-∞,∞).
The domain for tan-1x is (-∞,∞).
The domain for sec-1x is (-∞,-1] ∪ [1,∞).
If, A and B are arbitrary constants then what will be the differential equation of y = Ax + B/x?- a)x2 d2 y/dx2 – xdy/dx + y = 0
- b)x2 d2 y/dx2 + xdy/dx + y = 0
- c)x2 d2 y/dx2 + xdy/dx – y = 0
- d)x2 d2 y/dx2 – xdy/dx – y = 0
Correct answer is option 'C'. Can you explain this answer?
If, A and B are arbitrary constants then what will be the differential equation of y = Ax + B/x?
a)
x2 d2 y/dx2 – xdy/dx + y = 0
b)
x2 d2 y/dx2 + xdy/dx + y = 0
c)
x2 d2 y/dx2 + xdy/dx – y = 0
d)
x2 d2 y/dx2 – xdy/dx – y = 0
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Avantika Dasgupta answered |
The differential equation for y = Ax + B/x is given by:
d^2y/dx^2 = 0
This is because the second derivative of y with respect to x is zero.
d^2y/dx^2 = 0
This is because the second derivative of y with respect to x is zero.
Bag 1 contains 4 white and 6 black balls while another Bag 2 contains 4 white and 3 black balls. One ball is drawn at random from one of the bags and it is found to be black. Find the probability that it was drawn from Bag 1.- a)12/13
- b)5/12
- c)7/11
- d)7/12
Correct answer is option 'D'. Can you explain this answer?
Bag 1 contains 4 white and 6 black balls while another Bag 2 contains 4 white and 3 black balls. One ball is drawn at random from one of the bags and it is found to be black. Find the probability that it was drawn from Bag 1.
a)
12/13
b)
5/12
c)
7/11
d)
7/12
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Sankar Banerjee answered |
Solution:
We are given that there are two bags, Bag 1 and Bag 2.
Bag 1 contains 4 white and 6 black balls.
Bag 2 contains 4 white and 3 black balls.
One ball is drawn at random from one of the bags and it is found to be black.
We need to find the probability that it was drawn from Bag 1.
Let A be the event that the ball is drawn from Bag 1.
Let B be the event that the ball is black.
We need to find P(A|B), the probability that the ball is drawn from Bag 1 given that it is black.
Using Bayes' theorem, we have:
P(A|B) = P(B|A) * P(A) / P(B)
P(B|A) is the probability of drawing a black ball given that Bag 1 is selected. This is equal to 6/10 = 3/5.
P(A) is the probability of selecting Bag 1. This is equal to 1/2.
P(B) is the probability of drawing a black ball. This can be calculated using the law of total probability.
P(B) = P(B|A) * P(A) + P(B|not A) * P(not A)
P(B|not A) is the probability of drawing a black ball given that Bag 2 is selected. This is equal to 3/7.
P(not A) is the probability of not selecting Bag 1. This is equal to 1/2.
Therefore,
P(B) = (3/5) * (1/2) + (3/7) * (1/2) = 39/70
Substituting the values in Bayes' theorem, we have:
P(A|B) = (3/5) * (1/2) / (39/70) = 7/12
Therefore, the probability that the black ball was drawn from Bag 1 is 7/12.
Hence, the correct option is (d) 7/12.
We are given that there are two bags, Bag 1 and Bag 2.
Bag 1 contains 4 white and 6 black balls.
Bag 2 contains 4 white and 3 black balls.
One ball is drawn at random from one of the bags and it is found to be black.
We need to find the probability that it was drawn from Bag 1.
Let A be the event that the ball is drawn from Bag 1.
Let B be the event that the ball is black.
We need to find P(A|B), the probability that the ball is drawn from Bag 1 given that it is black.
Using Bayes' theorem, we have:
P(A|B) = P(B|A) * P(A) / P(B)
P(B|A) is the probability of drawing a black ball given that Bag 1 is selected. This is equal to 6/10 = 3/5.
P(A) is the probability of selecting Bag 1. This is equal to 1/2.
P(B) is the probability of drawing a black ball. This can be calculated using the law of total probability.
P(B) = P(B|A) * P(A) + P(B|not A) * P(not A)
P(B|not A) is the probability of drawing a black ball given that Bag 2 is selected. This is equal to 3/7.
P(not A) is the probability of not selecting Bag 1. This is equal to 1/2.
Therefore,
P(B) = (3/5) * (1/2) + (3/7) * (1/2) = 39/70
Substituting the values in Bayes' theorem, we have:
P(A|B) = (3/5) * (1/2) / (39/70) = 7/12
Therefore, the probability that the black ball was drawn from Bag 1 is 7/12.
Hence, the correct option is (d) 7/12.
Which of the following relations is symmetric but neither reflexive nor transitive for a set A = {1, 2, 3}.- a)R = {(1, 2), (1, 3), (1, 4)}
- b)R = {(1, 2), (2, 1)}
- c)R = {(1, 1), (2, 2), (3, 3)}
- d)R = {(1, 1), (1, 2), (2, 3)}
Correct answer is option 'B'. Can you explain this answer?
Which of the following relations is symmetric but neither reflexive nor transitive for a set A = {1, 2, 3}.
a)
R = {(1, 2), (1, 3), (1, 4)}
b)
R = {(1, 2), (2, 1)}
c)
R = {(1, 1), (2, 2), (3, 3)}
d)
R = {(1, 1), (1, 2), (2, 3)}
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|
Swati Verma answered |
A relation in a set A is said to be symmetric if (a1, a2)∈R implies that (a1, a2)∈R,for every a1, a2∈R.
Hence, for the given set A={1, 2, 3}, R={(1, 2), (2, 1)} is symmetric. It is not reflexive since every element is not related to itself and neither transitive as it does not satisfy the condition that for a given relation R in a set A if (a1, a2)∈R and (a2, a3)∈R implies that (a1, a3)∈ R for every a1, a2, a3∈R.
Hence, for the given set A={1, 2, 3}, R={(1, 2), (2, 1)} is symmetric. It is not reflexive since every element is not related to itself and neither transitive as it does not satisfy the condition that for a given relation R in a set A if (a1, a2)∈R and (a2, a3)∈R implies that (a1, a3)∈ R for every a1, a2, a3∈R.
Which of the following functions is the solution of the differential equation dydx + 2y = 0?- a)y = -2e-x
- b)y = 2ex
- c)y = e-2x
- d)y = e2x
Correct answer is option 'C'. Can you explain this answer?
Which of the following functions is the solution of the differential equation dydx + 2y = 0?
a)
y = -2e-x
b)
y = 2ex
c)
y = e-2x
d)
y = e2x
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Nandini Nair answered |
To solve the given differential equation dy/dx + 2y = 0, we can use the method of separation of variables. This method involves separating the variables dy and dx and then integrating both sides of the equation.
Separation of Variables:
1. Start by isolating the dy term on one side of the equation:
dy = -2y dx
2. Now, divide both sides of the equation by y:
dy/y = -2 dx
Integration:
3. Integrate both sides of the equation with respect to their respective variables:
∫ (1/y) dy = ∫ (-2) dx
4. The integral of (1/y) with respect to y is ln|y| + C1, where C1 is the constant of integration.
The integral of -2 with respect to x is -2x + C2, where C2 is another constant of integration.
5. Therefore, we have the equation:
ln|y| + C1 = -2x + C2
6. Combining the constants of integration, we can write the equation as:
ln|y| = -2x + C, where C = C2 - C1
Exponential Function:
7. To eliminate the natural logarithm, we can take the exponential of both sides of the equation:
e^(ln|y|) = e^(-2x + C)
8. By the properties of logarithms, the exponential of ln|y| simplifies to |y|:
|y| = e^(-2x + C)
9. Since e^C is just another constant, we can rewrite the equation as:
|y| = Ce^(-2x)
10. Finally, we can express the absolute value as a positive constant:
y = Ce^(-2x), where C is a constant.
Conclusion:
The function y = Ce^(-2x) is the solution to the given differential equation dy/dx + 2y = 0. Therefore, option C is the correct answer.
Separation of Variables:
1. Start by isolating the dy term on one side of the equation:
dy = -2y dx
2. Now, divide both sides of the equation by y:
dy/y = -2 dx
Integration:
3. Integrate both sides of the equation with respect to their respective variables:
∫ (1/y) dy = ∫ (-2) dx
4. The integral of (1/y) with respect to y is ln|y| + C1, where C1 is the constant of integration.
The integral of -2 with respect to x is -2x + C2, where C2 is another constant of integration.
5. Therefore, we have the equation:
ln|y| + C1 = -2x + C2
6. Combining the constants of integration, we can write the equation as:
ln|y| = -2x + C, where C = C2 - C1
Exponential Function:
7. To eliminate the natural logarithm, we can take the exponential of both sides of the equation:
e^(ln|y|) = e^(-2x + C)
8. By the properties of logarithms, the exponential of ln|y| simplifies to |y|:
|y| = e^(-2x + C)
9. Since e^C is just another constant, we can rewrite the equation as:
|y| = Ce^(-2x)
10. Finally, we can express the absolute value as a positive constant:
y = Ce^(-2x), where C is a constant.
Conclusion:
The function y = Ce^(-2x) is the solution to the given differential equation dy/dx + 2y = 0. Therefore, option C is the correct answer.
What will be the required solution of d2y/dx2 – 3dy/dx + 4y = 0?- a)Ae-4x + Be-x
- b)Ae4x – Be-x
- c)Ae4x + Be-x
- d)Ae4x + Bex
Correct answer is option 'C'. Can you explain this answer?
What will be the required solution of d2y/dx2 – 3dy/dx + 4y = 0?
a)
Ae-4x + Be-x
b)
Ae4x – Be-x
c)
Ae4x + Be-x
d)
Ae4x + Bex
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Roshni Desai answered |
The required solution of d2y/dx2 will be a function that gives the second derivative of y with respect to x. It is a differential equation and the solution will involve integration and finding the constant of integration based on the given initial or boundary conditions. Without any specific equation or boundary conditions, it is not possible to provide a specific solution.
The matrix which follows the condition m>n is called as ____________- a)vertical matrix
- b)horizontal matrix
- c)diagonal matrix
- d)square matrix
Correct answer is option 'A'. Can you explain this answer?
The matrix which follows the condition m>n is called as ____________
a)
vertical matrix
b)
horizontal matrix
c)
diagonal matrix
d)
square matrix
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Janhavi Kaur answered |
It seems like you have not completed your sentence. Could you please clarify or provide more information about the condition m?
Value after differentiating cos (sinx) is _________- a)sin (sinx).cosx
- b)-sin (sinx).cosx
- c)sin (sinx)
- d)sin (cosx).cosx
Correct answer is option 'B'. Can you explain this answer?
Value after differentiating cos (sinx) is _________
a)
sin (sinx).cosx
b)
-sin (sinx).cosx
c)
sin (sinx)
d)
sin (cosx).cosx
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Kalyan Joshi answered |
To find the value after differentiating cos(sin(x)), we need to apply the chain rule. The chain rule states that if we have a composite function, F(g(x)), then the derivative of F with respect to x is given by F'(g(x)) * g'(x).
The given function cos(sin(x)) is a composite function where the outer function is cos(x) and the inner function is sin(x). So, we can apply the chain rule to differentiate it.
Let's break down the steps to differentiate cos(sin(x)):
Step 1: Identify the Inner and Outer Functions
The outer function is cos(x) and the inner function is sin(x).
Step 2: Find the Derivative of the Inner Function
The derivative of sin(x) with respect to x is cos(x). This is because the derivative of sin(x) is cos(x).
Step 3: Find the Derivative of the Outer Function
The derivative of cos(x) with respect to x is -sin(x). This is the basic derivative of cos(x).
Step 4: Apply the Chain Rule
According to the chain rule, we need to multiply the derivative of the outer function by the derivative of the inner function.
So, the derivative of cos(sin(x)) is:
- sin(sin(x)) * cos(x)
Answer: Option (b) -sin(sin(x)) * cos(x)
In summary, to find the value after differentiating cos(sin(x)), we applied the chain rule. The derivative of the outer function cos(x) is -sin(x), and the derivative of the inner function sin(x) is cos(x). Multiplying these two derivatives gives us -sin(sin(x)) * cos(x).
The given function cos(sin(x)) is a composite function where the outer function is cos(x) and the inner function is sin(x). So, we can apply the chain rule to differentiate it.
Let's break down the steps to differentiate cos(sin(x)):
Step 1: Identify the Inner and Outer Functions
The outer function is cos(x) and the inner function is sin(x).
Step 2: Find the Derivative of the Inner Function
The derivative of sin(x) with respect to x is cos(x). This is because the derivative of sin(x) is cos(x).
Step 3: Find the Derivative of the Outer Function
The derivative of cos(x) with respect to x is -sin(x). This is the basic derivative of cos(x).
Step 4: Apply the Chain Rule
According to the chain rule, we need to multiply the derivative of the outer function by the derivative of the inner function.
So, the derivative of cos(sin(x)) is:
- sin(sin(x)) * cos(x)
Answer: Option (b) -sin(sin(x)) * cos(x)
In summary, to find the value after differentiating cos(sin(x)), we applied the chain rule. The derivative of the outer function cos(x) is -sin(x), and the derivative of the inner function sin(x) is cos(x). Multiplying these two derivatives gives us -sin(sin(x)) * cos(x).
What will be the equation of the circle which touches the line x + 2y + 5 = 0 and passes through the point of intersection of the circle x2 + y2 = 1 and x2 + y2 + 2x + 4y + 1 = 0?- a)x2 + y2 + 2x + y = 0
- b)x2 + y2 + x + 2y = 1
- c)x2 + y2 + x + 2y = 0
- d)x2 + y2 + 2x + 2y = 1
Correct answer is option 'C'. Can you explain this answer?
What will be the equation of the circle which touches the line x + 2y + 5 = 0 and passes through the point of intersection of the circle x2 + y2 = 1 and x2 + y2 + 2x + 4y + 1 = 0?
a)
x2 + y2 + 2x + y = 0
b)
x2 + y2 + x + 2y = 1
c)
x2 + y2 + x + 2y = 0
d)
x2 + y2 + 2x + 2y = 1
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|
Pallavi Desai answered |
To find the equation of the circle, we need to determine its center and radius.
Step 1: Find the intersection point of the two given circles
The two given circles are:
1. x^2 + y^2 = 1
2. x^2 + y^2 - 2x - 4y + 1 = 0
By solving these two equations simultaneously, we can find the intersection points.
Subtracting equation 1 from equation 2, we get:
-2x - 4y + 1 = 0
Rearranging this equation, we get:
2x + 4y = 1
Dividing both sides by 2, we have:
x + 2y = 1/2
Now we have a system of equations:
x + 2y = 1/2
x - 2y = 0
By solving these equations simultaneously, we find the intersection point:
Adding the two equations, we get:
2x = 1/2
x = 1/4
Substituting x = 1/4 into either equation, we get:
1/4 + 2y = 1/2
2y = 1/2 - 1/4
2y = 1/4
y = 1/8
So the intersection point is (1/4, 1/8).
Step 2: Find the equation of the line tangent to the given line
The given line is:
x + 2y + 5 = 0
To find the equation of the line tangent to this line, we need to find the negative reciprocal of the slope of the given line. The given line can be rewritten as:
2y = -x - 5
y = -(1/2)x - 5/2
The slope of this line is -1/2, so the slope of the tangent line is 2.
The equation of the tangent line passing through the point (1/4, 1/8) is:
y - 1/8 = 2(x - 1/4)
Simplifying this equation, we get:
y - 1/8 = 2x - 1/2
y = 2x + 1/8 - 1/2
y = 2x - 3/8
Step 3: Find the equation of the circle
The equation of a circle with center (h, k) and radius r is given by:
(x - h)^2 + (y - k)^2 = r^2
Since the circle touches the line x + 2y + 5 = 0, the distance between the center of the circle and the line must be equal to the radius of the circle.
The distance from a point (x, y) to a line Ax + By + C = 0 is given by:
d = |Ax + By + C| / sqrt(A^2 + B^2)
For the line x + 2y + 5 = 0, A = 1, B = 2, and C = 5. Substituting the coordinates of the center of the circle (h, k) into the distance formula, we have:
|1h + 2k + 5| / sqrt(1^2 +
Step 1: Find the intersection point of the two given circles
The two given circles are:
1. x^2 + y^2 = 1
2. x^2 + y^2 - 2x - 4y + 1 = 0
By solving these two equations simultaneously, we can find the intersection points.
Subtracting equation 1 from equation 2, we get:
-2x - 4y + 1 = 0
Rearranging this equation, we get:
2x + 4y = 1
Dividing both sides by 2, we have:
x + 2y = 1/2
Now we have a system of equations:
x + 2y = 1/2
x - 2y = 0
By solving these equations simultaneously, we find the intersection point:
Adding the two equations, we get:
2x = 1/2
x = 1/4
Substituting x = 1/4 into either equation, we get:
1/4 + 2y = 1/2
2y = 1/2 - 1/4
2y = 1/4
y = 1/8
So the intersection point is (1/4, 1/8).
Step 2: Find the equation of the line tangent to the given line
The given line is:
x + 2y + 5 = 0
To find the equation of the line tangent to this line, we need to find the negative reciprocal of the slope of the given line. The given line can be rewritten as:
2y = -x - 5
y = -(1/2)x - 5/2
The slope of this line is -1/2, so the slope of the tangent line is 2.
The equation of the tangent line passing through the point (1/4, 1/8) is:
y - 1/8 = 2(x - 1/4)
Simplifying this equation, we get:
y - 1/8 = 2x - 1/2
y = 2x + 1/8 - 1/2
y = 2x - 3/8
Step 3: Find the equation of the circle
The equation of a circle with center (h, k) and radius r is given by:
(x - h)^2 + (y - k)^2 = r^2
Since the circle touches the line x + 2y + 5 = 0, the distance between the center of the circle and the line must be equal to the radius of the circle.
The distance from a point (x, y) to a line Ax + By + C = 0 is given by:
d = |Ax + By + C| / sqrt(A^2 + B^2)
For the line x + 2y + 5 = 0, A = 1, B = 2, and C = 5. Substituting the coordinates of the center of the circle (h, k) into the distance formula, we have:
|1h + 2k + 5| / sqrt(1^2 +
Which of the following functions is a solution for the differential equation y” + 6y = 0?- a)y = 5 cos3x
- b)y = 5 tan3x
- c)y = cos3x
- d)y = 6 cos3x
Correct answer is option 'A'. Can you explain this answer?
Which of the following functions is a solution for the differential equation y” + 6y = 0?
a)
y = 5 cos3x
b)
y = 5 tan3x
c)
y = cos3x
d)
y = 6 cos3x
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Shalini Patel answered |
Consider the function y = 5 cos3x
Differentiating w.r.t x, we get
y’ = dy/dx = -15 sin3x
Differentiating again w.r.t x, we get
y” = d2y/dx2 = -30 cos3x
⇒ y” + 6y=0.
Hence, the function y = 5 cos3x is a solution for the differential equation y” + 6y = 0.
Differentiating w.r.t x, we get
y’ = dy/dx = -15 sin3x
Differentiating again w.r.t x, we get
y” = d2y/dx2 = -30 cos3x
⇒ y” + 6y=0.
Hence, the function y = 5 cos3x is a solution for the differential equation y” + 6y = 0.
What is sec-1x in terms of tan-1?- a)tan-1√1 + x2
- b)tan-11+x2
- c)tan-1x
- d)tan-1√x2 - 1
Correct answer is option 'D'. Can you explain this answer?
What is sec-1x in terms of tan-1?
a)
tan-1√1 + x2
b)
tan-11+x2
c)
tan-1x
d)
tan-1√x2 - 1
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|
Aravind Rane answered |
(sec-1x) = cos-1(1/x) = tan-1(√(x^2-1)/x)
Find the angle between the planes x + 2y + 3z + 1 = 0 and (4, 1, -7).- a)- 29.34
- b)3.43
- c)11.23
- d)– 17.54
Correct answer is option 'A'. Can you explain this answer?
Find the angle between the planes x + 2y + 3z + 1 = 0 and (4, 1, -7).
a)
- 29.34
b)
3.43
c)
11.23
d)
– 17.54
|
|
Swara Sharma answered |
There are different methods to find the angle between two planes, but one common approach is to use the dot product. The normal vectors of two planes are perpendicular to them, so their dot product is zero. Therefore, if we can find the normal vectors of the two planes, we can use the dot product formula to find the cosine of the angle between them, and then take the inverse cosine to get the angle itself.
For the first plane, x + 2y + 3z + 1 = 0, we can read off the coefficients of x, y, and z to get the normal vector N1 = (1, 2, 3). Note that the magnitude of this vector is sqrt(1^2 + 2^2 + 3^2) = sqrt(14), but we don't need to normalize it for the dot product.
For the second plane, we need to use the given point (4, 1, -7) to find the normal vector. A normal vector is perpendicular to any vector lying in the plane, so we can subtract the point from a general point on the plane and then take the cross product of the resulting vectors. For example, if we let x, y, and z be arbitrary coordinates in the plane, then the vector from (4, 1, -7) to (x, y, z) is (x-4, y-1, z+7). We want this vector to be perpendicular to the given vector (4, 1, -7), so their dot product must be zero:
(4, 1, -7) dot (x-4, y-1, z+7) = 4(x-4) + (y-1) + (-7)(z+7) = 4x + y - 7z - 60 = 0
This is the equation of the plane, but we don't need it explicitly. Instead, we can read off the coefficients of x, y, and z to get the normal vector N2 = (4, 1, -7). Again, we don't need to normalize this vector for the dot product.
Now we can use the dot product formula:
N1 dot N2 = |N1| |N2| cos(theta)
where theta is the angle between the planes. Substituting the values we found, we get:
(1, 2, 3) dot (4, 1, -7) = sqrt(14) sqrt(66) cos(theta)
4 + 2 - 21 = sqrt(14) sqrt(66) cos(theta)
-15 = sqrt(924) cos(theta)
cos(theta) = -15/sqrt(924) = -5/22
Taking the inverse cosine (or arccosine) of this value, we get:
theta = 107.51 degrees
This is not one of the answer choices, but if we take the complementary angle (i.e. the angle between the normal vectors), we get:
theta' = 180 - 107.51 = 72.49 degrees
This is closest to answer choice (c) 11.23, but it is not exactly the same. However, it is possible that there is a mistake in the problem statement or the answer choices, or that we made an error in the calculation.
For the first plane, x + 2y + 3z + 1 = 0, we can read off the coefficients of x, y, and z to get the normal vector N1 = (1, 2, 3). Note that the magnitude of this vector is sqrt(1^2 + 2^2 + 3^2) = sqrt(14), but we don't need to normalize it for the dot product.
For the second plane, we need to use the given point (4, 1, -7) to find the normal vector. A normal vector is perpendicular to any vector lying in the plane, so we can subtract the point from a general point on the plane and then take the cross product of the resulting vectors. For example, if we let x, y, and z be arbitrary coordinates in the plane, then the vector from (4, 1, -7) to (x, y, z) is (x-4, y-1, z+7). We want this vector to be perpendicular to the given vector (4, 1, -7), so their dot product must be zero:
(4, 1, -7) dot (x-4, y-1, z+7) = 4(x-4) + (y-1) + (-7)(z+7) = 4x + y - 7z - 60 = 0
This is the equation of the plane, but we don't need it explicitly. Instead, we can read off the coefficients of x, y, and z to get the normal vector N2 = (4, 1, -7). Again, we don't need to normalize this vector for the dot product.
Now we can use the dot product formula:
N1 dot N2 = |N1| |N2| cos(theta)
where theta is the angle between the planes. Substituting the values we found, we get:
(1, 2, 3) dot (4, 1, -7) = sqrt(14) sqrt(66) cos(theta)
4 + 2 - 21 = sqrt(14) sqrt(66) cos(theta)
-15 = sqrt(924) cos(theta)
cos(theta) = -15/sqrt(924) = -5/22
Taking the inverse cosine (or arccosine) of this value, we get:
theta = 107.51 degrees
This is not one of the answer choices, but if we take the complementary angle (i.e. the angle between the normal vectors), we get:
theta' = 180 - 107.51 = 72.49 degrees
This is closest to answer choice (c) 11.23, but it is not exactly the same. However, it is possible that there is a mistake in the problem statement or the answer choices, or that we made an error in the calculation.
Another form of Rolle’s theorem for the differential condition is _____- a)f is differentiable on (a, ah)
- b)f is differentiable on (a, a - h)
- c)f is differentiable on (a, a/h)
- d)f is differentiable on (a, a + h)
Correct answer is option 'D'. Can you explain this answer?
Another form of Rolle’s theorem for the differential condition is _____
a)
f is differentiable on (a, ah)
b)
f is differentiable on (a, a - h)
c)
f is differentiable on (a, a/h)
d)
f is differentiable on (a, a + h)
|
Om Kumar answered |
Another form of Rolle's theorem is the generalized form of Rolle's theorem, also known as the Cauchy's mean value theorem.
The generalized form of Rolle's theorem states that if a function f(x) and its derivative g(x) are both continuous on the closed interval [a, b], and g(x) is non-zero at every point inside the open interval (a, b), then there exists at least one point c in (a, b) such that:
f'(c) = [f(b) - f(a)] / [g(b) - g(a)]
This form of Rolle's theorem is particularly useful when dealing with functions that have multiple roots or points where the derivative is zero. It allows us to find a point within the interval where the derivative of the function is equal to the average rate of change of the function over that interval.
The generalized form of Rolle's theorem states that if a function f(x) and its derivative g(x) are both continuous on the closed interval [a, b], and g(x) is non-zero at every point inside the open interval (a, b), then there exists at least one point c in (a, b) such that:
f'(c) = [f(b) - f(a)] / [g(b) - g(a)]
This form of Rolle's theorem is particularly useful when dealing with functions that have multiple roots or points where the derivative is zero. It allows us to find a point within the interval where the derivative of the function is equal to the average rate of change of the function over that interval.
Find dy/dx of y = sin (ax + b).- a)a.cos (ax + b)
- b)b.sin (ax + b)
- c)a.sin (ax + b)
- d)b.cos (ax + b)
Correct answer is option 'A'. Can you explain this answer?
Find dy/dx of y = sin (ax + b).
a)
a.cos (ax + b)
b)
b.sin (ax + b)
c)
a.sin (ax + b)
d)
b.cos (ax + b)
|
|
Swati Verma answered |
We differentiate the given function with the help of chain rule and hence
the outer function is differentiated into cos, and the inner function comes out to be a and the constant b becomes 0, which is multiplied to the whole function and the answer comes out to be
⇒ a.cos (ax + b).
the outer function is differentiated into cos, and the inner function comes out to be a and the constant b becomes 0, which is multiplied to the whole function and the answer comes out to be
⇒ a.cos (ax + b).
Rolle’s theorem is a special case of _____- a)Euclid’s theorem
- b)another form of Rolle’s theorem
- c)Lagrange’s mean value theorem
- d)Joule’s theorem
Correct answer is option 'C'. Can you explain this answer?
Rolle’s theorem is a special case of _____
a)
Euclid’s theorem
b)
another form of Rolle’s theorem
c)
Lagrange’s mean value theorem
d)
Joule’s theorem
|
Tejas Chawla answered |
Understanding Rolle's Theorem
Rolle's Theorem is a fundamental result in calculus, which states that if a function is continuous on a closed interval [a, b], differentiable on the open interval (a, b), and has equal values at the endpoints (f(a) = f(b)), then there exists at least one point c in (a, b) where the derivative f'(c) = 0.
Connection to Lagrange's Mean Value Theorem
Rolle's Theorem is a specific case of Lagrange's Mean Value Theorem (MVT). The MVT states that if a function is continuous on [a, b] and differentiable on (a, b), then there exists some c in (a, b) such that:
- f'(c) = (f(b) - f(a)) / (b - a)
Key Points of Lagrange's MVT
- It generalizes Rolle's Theorem by allowing for different function values at the endpoints.
- If f(a) = f(b), then (f(b) - f(a)) = 0, leading to f'(c) = 0, which is precisely the statement of Rolle's Theorem.
Why Option C is Correct
- Since Rolle's Theorem can be derived from Lagrange's MVT by applying the condition f(a) = f(b), it is accurate to say that it is a special case of the Mean Value Theorem.
- This establishes the hierarchy and relationship between these fundamental concepts in calculus.
In summary, understanding Rolle's theorem as a special case of Lagrange's Mean Value Theorem helps in grasping the broader implications of the behavior of functions in calculus.
Rolle's Theorem is a fundamental result in calculus, which states that if a function is continuous on a closed interval [a, b], differentiable on the open interval (a, b), and has equal values at the endpoints (f(a) = f(b)), then there exists at least one point c in (a, b) where the derivative f'(c) = 0.
Connection to Lagrange's Mean Value Theorem
Rolle's Theorem is a specific case of Lagrange's Mean Value Theorem (MVT). The MVT states that if a function is continuous on [a, b] and differentiable on (a, b), then there exists some c in (a, b) such that:
- f'(c) = (f(b) - f(a)) / (b - a)
Key Points of Lagrange's MVT
- It generalizes Rolle's Theorem by allowing for different function values at the endpoints.
- If f(a) = f(b), then (f(b) - f(a)) = 0, leading to f'(c) = 0, which is precisely the statement of Rolle's Theorem.
Why Option C is Correct
- Since Rolle's Theorem can be derived from Lagrange's MVT by applying the condition f(a) = f(b), it is accurate to say that it is a special case of the Mean Value Theorem.
- This establishes the hierarchy and relationship between these fundamental concepts in calculus.
In summary, understanding Rolle's theorem as a special case of Lagrange's Mean Value Theorem helps in grasping the broader implications of the behavior of functions in calculus.
Which of the following functions is a solution for the differential equation dy/dx -14x = 0?- a)y = 7x2
- b)y = 7x3
- c)y = x7
- d)y = 14x
Correct answer is option 'A'. Can you explain this answer?
Which of the following functions is a solution for the differential equation dy/dx -14x = 0?
a)
y = 7x2
b)
y = 7x3
c)
y = x7
d)
y = 14x
|
|
Nandini Nair answered |
Given: dy/dx -14x = 0
To find the solution for the given differential equation, we need to find a function y(x) that satisfies the equation.
To solve the differential equation, we can separate the variables and integrate both sides.
Separating the variables:
dy = 14x dx
Integrating both sides:
∫dy = ∫14x dx
Using the power rule of integration, we can integrate both sides:
y = 7x^2 + C
where C is the constant of integration.
Now, we have the general solution to the differential equation: y = 7x^2 + C.
To determine the specific solution, we need an initial condition or boundary condition. Since no initial condition is given in the question, we cannot determine the value of C.
Therefore, the solution to the differential equation dy/dx -14x = 0 is y = 7x^2 + C, where C is a constant.
Among the given options, option A (y = 7x^2) is the only function that satisfies the differential equation dy/dx -14x = 0, as it is the general solution without the constant of integration.
To find the solution for the given differential equation, we need to find a function y(x) that satisfies the equation.
To solve the differential equation, we can separate the variables and integrate both sides.
Separating the variables:
dy = 14x dx
Integrating both sides:
∫dy = ∫14x dx
Using the power rule of integration, we can integrate both sides:
y = 7x^2 + C
where C is the constant of integration.
Now, we have the general solution to the differential equation: y = 7x^2 + C.
To determine the specific solution, we need an initial condition or boundary condition. Since no initial condition is given in the question, we cannot determine the value of C.
Therefore, the solution to the differential equation dy/dx -14x = 0 is y = 7x^2 + C, where C is a constant.
Among the given options, option A (y = 7x^2) is the only function that satisfies the differential equation dy/dx -14x = 0, as it is the general solution without the constant of integration.
Which one of the following is correct if we differentiate the equation xy = aex + be-x two times?- a)x(d2y/ dx) + 2(dy/dx) = xy
- b)x(d2y/ dx) – 2(dy/dx) = xy
- c)3x(d2y/ dx) + 2(dy/dx) = xy
- d)x(d2y/ dx) + 2(dy/dx) = 2xy
Correct answer is option 'A'. Can you explain this answer?
Which one of the following is correct if we differentiate the equation xy = aex + be-x two times?
a)
x(d2y/ dx) + 2(dy/dx) = xy
b)
x(d2y/ dx) – 2(dy/dx) = xy
c)
3x(d2y/ dx) + 2(dy/dx) = xy
d)
x(d2y/ dx) + 2(dy/dx) = 2xy
|
|
Tanuja Kapoor answered |
We have xy = aex + be-x ……(1)
Differentiating (1) with respect to x, we get
x(dy/dx) + y = aex + be-x …..(2)
Differentiating (2) now, with respect to x, we get
x(d2y/ dx) + dy/dx + dy/dx = aex + be-x
From (1),
aex + be-x = xy, so that we get
x(d2y/ dx) + 2(dy/dx) = xy
which is the required differential equation.
Differentiating (1) with respect to x, we get
x(dy/dx) + y = aex + be-x …..(2)
Differentiating (2) now, with respect to x, we get
x(d2y/ dx) + dy/dx + dy/dx = aex + be-x
From (1),
aex + be-x = xy, so that we get
x(d2y/ dx) + 2(dy/dx) = xy
which is the required differential equation.
Value after differentiating cos (x2 + 5) is ________- a)5.sin (x2 + 5)
- b)-sin (x2 + 5).2x
- c)sin (x2 + 5).2x
- d)cos (x2 + 5).2x
Correct answer is option 'B'. Can you explain this answer?
Value after differentiating cos (x2 + 5) is ________
a)
5.sin (x2 + 5)
b)
-sin (x2 + 5).2x
c)
sin (x2 + 5).2x
d)
cos (x2 + 5).2x
|
|
Swati Verma answered |
We differentiate the given function with help of chain rule and hence the outer function becomes –sin and the inner function is differentiated into 2x, therefore the answer comes out to be -sin (x2 + 5).2x.
If, A normal is drawn at a point P(x, y) of a curve. It meets the x-axis at Q. If PQ is of constant length k. What kind of curve is passing through (0, k)?- a)Parabola
- b)Hyperbola
- c)Ellipse
- d)Circle
Correct answer is option 'D'. Can you explain this answer?
If, A normal is drawn at a point P(x, y) of a curve. It meets the x-axis at Q. If PQ is of constant length k. What kind of curve is passing through (0, k)?
a)
Parabola
b)
Hyperbola
c)
Ellipse
d)
Circle
|
|
Priyanka Sharma answered |
Equation of the normal at a point P(x, y) is given by
Y – y = -1/(dy/dx)(X – x) ….(1)
Let the point Q at the x-axis be (x1 , 0).
From (1), we get
y(dy/dx) = x1 – x ….(2)
Now, giving that PQ2 = k2
Or, x1 – x + y2 = k2
=>y(dy/dx) = ± √(k2 – y2) ….(3)
(3) is the required differential equation for such curves,
Now solving (3) we get,
∫-dy/√(k2 – y2) = ∫-dx
Or, x2 + y2 = k2 passes through (0, k)
Thus, it is a circle.
Y – y = -1/(dy/dx)(X – x) ….(1)
Let the point Q at the x-axis be (x1 , 0).
From (1), we get
y(dy/dx) = x1 – x ….(2)
Now, giving that PQ2 = k2
Or, x1 – x + y2 = k2
=>y(dy/dx) = ± √(k2 – y2) ….(3)
(3) is the required differential equation for such curves,
Now solving (3) we get,
∫-dy/√(k2 – y2) = ∫-dx
Or, x2 + y2 = k2 passes through (0, k)
Thus, it is a circle.
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