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and bounded by the lines x = 0 and 
be defined as f (t) = 
f(t )dt
Then the real roots of tlie equation x2 — f'(x) = 0 are
The area bounded by the curves y = 2√x, y = - x, x = 1 and x = 4 is given by- a)25
- b)33/2
- c)47/4
- d)101/6
Correct answer is option 'D'. Can you explain this answer?
The area bounded by the curves y = 2√x, y = - x, x = 1 and x = 4 is given by
a)
25
b)
33/2
c)
47/4
d)
101/6
|
Chirag Verma answered |
The given equations of the curves are
So, the required area is
So, the required area is
If the triple integral over the region bounded by the planes 2x + y + z = 4, x = 0, y = 0, z = 0 is given by 
then the function λ(x) – π(x, y) is - a)y
- b)x
- c)x + y
- d)x – y
Correct answer is option 'A'. Can you explain this answer?
If the triple integral over the region bounded by the planes 2x + y + z = 4, x = 0, y = 0, z = 0 is given by then the function λ(x) – π(x, y) is
a)
y
b)
x
c)
x + y
d)
x – y
|
Veda Institute answered |
From (1) & (2), we have
⇒ λ(x) = 4 – 2x.
µ(x, y) = 4 – 2x – y
⇒ λ – µ = 4 – 2x – 4 + 2x + y = y.
Consider the shaded triangular region P shown in the figure. What is 
xy dxdy?
- a)1/6
- b)2/9
- c)1/16
- d)1
Correct answer is option 'A'. Can you explain this answer?
Consider the shaded triangular region P shown in the figure. What is xy dxdy?
xy dxdy?a)
1/6
b)
2/9
c)
1/16
d)
1
|
|
Chirag Verma answered |
The intercept form of equation is
where, a is x intercept and b is y intercept.
So Given, equation is
implies x+2y=2
implies x=2(1 - y)
Limit of x is form 0 to 2(1 - y ) and limit of y is from 0 to 1.
Therefore,
= 1/6
where, a is x intercept and b is y intercept.
So Given, equation is
implies x+2y=2
implies x=2(1 - y)
Limit of x is form 0 to 2(1 - y ) and limit of y is from 0 to 1.
Therefore,
= 1/6
dydx is equal to
(x+y+z) dzdydx is 
The value of integral 
dxdy is
- a)
- b)√π
- c)π
- d)π/4
Correct answer is option 'D'. Can you explain this answer?
The value of integral dxdy is
dxdy isa)
b)
√π
c)
π
d)
π/4
|
|
Veda Institute answered |
I = 
dxdy
Putting x2 + y2 = r2, we get
Limits of r = 0 to ∞
and 0 = θ to π/2
Now, Putting r2 = t, we get
dxdyPutting x2 + y2 = r2, we get
Limits of r = 0 to ∞
and 0 = θ to π/2
Now, Putting r2 = t, we get
The area bounded by the curves y2 = 9x, x —y + 2 = 0 is given by- a)1
- b)1/2
- c)3/2
- d)5/4
Correct answer is option 'B'. Can you explain this answer?
The area bounded by the curves y2 = 9x, x —y + 2 = 0 is given by
a)
1
b)
1/2
c)
3/2
d)
5/4
|
|
Veda Institute answered |
The equations of the given curves are
y2 = 9x
x - y + 2 = 0
The curves (/') and (ii) intersect at
A(1, 3) and B(4, 6)
So, the required area
= 1/2
y2 = 9x
x - y + 2 = 0
The curves (/') and (ii) intersect at
A(1, 3) and B(4, 6)
So, the required area
= 1/2
If A is the region bounded by the parabolas y2 = 4x and x2 = 4y, then 
is equal to- a)48/5
- b)36/5
- c)32/5
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
If A is the region bounded by the parabolas y2 = 4x and x2 = 4y, then is equal to
is equal toa)
48/5
b)
36/5
c)
32/5
d)
None of these
|
|
Chirag Verma answered |
Let I = 
Solving the given equations
y2 =4x and x2 = 4y, we get x = 0, x = 4 The region of integration A is given by
Solving the given equations
y2 =4x and x2 = 4y, we get x = 0, x = 4 The region of integration A is given by
leads to
. What is q?
is equal to- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
a)
b)
c)
d)
|
|
Veda Institute answered |
We assume x + 2 = t2 dx = 2t dt
dydx by changing the order of integration is 
sin φ dr dφ dθ. The value of the integral is
dydxdz is equal to
If A is the region bounded by the parabolas y2 = 4x and x2 = 4y, then 
dxdy is equal to - a)48/5
- b)36/5
- c)32/5
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
If A is the region bounded by the parabolas y2 = 4x and x2 = 4y, then dxdy is equal to
dxdy is equal to a)
48/5
b)
36/5
c)
32/5
d)
None of these
|
|
Chirag Verma answered |
Let I = 
Solving the given equations
y2 =4x and x2 = 4y, we get x = 0, x = 4 The region of integration A is given by
Solving the given equations
y2 =4x and x2 = 4y, we get x = 0, x = 4 The region of integration A is given by
The area of the cardioid r = a (1 + cosθ) is given by- a)2
r drdθ - b)2
r drdθ - c)2
- d)2
r drdθ
Correct answer is option 'A'. Can you explain this answer?
The area of the cardioid r = a (1 + cosθ) is given by
a)
2 r drdθ
r drdθb)
2 r drdθ
r drdθc)
2
d)
2 r drdθ
r drdθ|
|
Chirag Verma answered |
The equation of the cardioid is
r = a (1 + cosθ)
Required area
r = a (1 + cosθ)
Required area
Let f(x) = 
then f decrease in the interval (x1, x2)- a)
- b)( - 2 , - 1 )
- c)( 1 ,2 )
- d)
Correct answer is option 'C'. Can you explain this answer?
Let f(x) = then f decrease in the interval (x1, x2)
then f decrease in the interval (x1, x2)a)
b)
( - 2 , - 1 )
c)
( 1 ,2 )
d)
|
Vikram Kapoor answered |
f(x) = ∫ex(x−1)(x−2)dx
For decreasing function ,f′(x)<0,f′(x)<0
⇒ex(x−1)(x−2) < 0
⇒(x−1)(x−2) < 0
⇒ ex(x−1)(x−2) < 0
⇒ (x−1)(x−2) < 0
⇒1 < x < 2,
⇒1 < x < 2,
∵ ex > 0 ∀ x ∈ R
dxdy
dydx
dydx
dxdy
is
dxdydz is equal to
is
abc cubic units
abc cubic units
,where D is bounded by y = x and y2 = 4x is equal to
The integral 
is equal to- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
The integral is equal to
a)
b)
c)
d)
|
|
Veda Institute answered |
The given integral is
From above triple integral, the projection in yz-plane is y = 0, y = 1 – z & z = 0, z = 1
So, the above integral can be rewrite as
From above triple integral, the projection in yz-plane is y = 0, y = 1 – z & z = 0, z = 1
So, the above integral can be rewrite as
The value of 
dxdy is- a)13.5
- b)27.0
- c)40.5
- d)54.0
Correct answer is option 'A'. Can you explain this answer?
The value of dxdy is
dxdy isa)
13.5
b)
27.0
c)
40.5
d)
54.0
|
|
Chirag Verma answered |
[change the order]
dx dy is equal to
dxdy is
- a)zero
- b)π
- c)π/2
- d)2
Correct answer is option 'D'. Can you explain this answer?
dxdy isa)
zero
b)
π
c)
π/2
d)
2
|
|
Veda Institute answered |
Step 1: Simplify by Changing Variables
To simplify this integral, note that we can use the property of separable integrals by changing variables. However, for clarity, let’s proceed with evaluating the inner integral directly.
Step 2: Evaluate the Inner Integral with Respect to x
Step 3: Integrate with Respect to y
Conclusion: The value of the integral is: 2.
So, the correct answer is: D: 2.
Step 3: Integrate with Respect to y
Conclusion: The value of the integral is: 2.
So, the correct answer is: D: 2.
The area bounded by the parabola y2 = 4ax and straight line x + y = 3a is- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
The area bounded by the parabola y2 = 4ax and straight line x + y = 3a is
a)
b)
c)
d)
|
|
Veda Institute answered |
We need to find the area bounded by the parabola y² = 4ax and the straight line x + y = 3a.
Step 1: Rewrite the Equations
The given equations are:
- Parabola: y² = 4ax
- Line: x + y = 3a, which can be rewritten as y = 3a - x
Step 2: Find Points of Intersection
To find the intersection points, substitute y = 3a - x into y² = 4ax:
(3a - x)² = 4ax.
Expanding and simplifying:
9a² - 6ax + x² = 4ax,
x² - 10ax + 9a² = 0.
This is a quadratic equation in x:
(x - 5a)² = 0.
So, x = 5a. Substituting x = 5a back into y = 3a - x:
y = 3a - 5a = -2a.
Thus, the point of intersection is (5a, -2a).
Step 3: Set Up the Integral for the Area
To find the area bounded by the parabola and the line, we integrate horizontally from x = 0 to x = 5a, finding the difference between the y-values of the line and the parabola at each x.
The area A is given by:
A = ∫05a ((3a - x) - √(4ax)) dx.
Step 4: Evaluate the Integral
We will split the integral into two parts:
1. Integral of (3a - x):
2. Integral of √(4ax):
Rewrite √(4ax) = 2√(ax), so we have:
Conclusion:
The correct answer, after calculating both integrals, is:
D: 10a² / 3.
Find the value of ∫∫xyex + y dxdy.- a)yey (xex-ex)
- b)(yey-ey)(xex-ex)
- c)(yey-ey)xex
- d)(yey-ey)(xex+ex)
Correct answer is option 'B'. Can you explain this answer?
Find the value of ∫∫xyex + y dxdy.
a)
yey (xex-ex)
b)
(yey-ey)(xex-ex)
c)
(yey-ey)xex
d)
(yey-ey)(xex+ex)
|
|
Chirag Verma answered |
Add constant automatically
Explanation: Given, ∫∫xyex + y dxdy
∫∫xyex ey dxdy= ∫yey dy∫xex dx=(yey-ey)(xex-ex).
By changing the order of integration in 
the value is- a)π/4
- b)πa/4
- c)πa
- d)a/4
Correct answer is option 'B'. Can you explain this answer?
By changing the order of integration in the value is
the value isa)
π/4
b)
πa/4
c)
πa
d)
a/4
|
|
Veda Institute answered |
To evaluate the integral:
∫0a ∫ya (x / (x² + y²)) dx dy
by changing the order of integration, we need to determine the region of integration in the x-y plane and then rewrite the integral accordingly.
Step 1: Determine the Region of Integration
The given limits indicate that:
- y ranges from 0 to a,
- For a fixed y, x ranges from y to a.
In the x-y plane, this describes the triangular region bounded by:
- y = 0,
- x = a,
- x = y.
Step 2: Change the Order of Integration
To change the order of integration, we need to describe the region in terms of x first:
- x ranges from 0 to a,
- For a fixed x, y ranges from 0 to x (since y ≤ x within this region).
Thus, we can rewrite the integral as:
Step 3: Evaluate the Inner Integral with Respect to y
Now, we evaluate the inner integral:
Since x is treated as a constant in the inner integral, we can factor it out:
Integrate with respect to y:
Since arctan(1) = π/4 and arctan(0) = 0, this becomes:
= (1/x) * (π/4) = π / (4x).
Step 4: Substitute Back and Integrate with Respect to x
Now our integral becomes:
This simplifies to:
Conclusion:
The value of the integral is:
B: πa / 4.
Let V be the region bounded by the planes x = 0, x = 2, y = 0, z = 0 and y + z = 1. Then the value of integral ∫∫V ∫ y dx dy dz is- a)1 / 2
- b)1 / 3
- c)4 / 3
- d)1
Correct answer is option 'B'. Can you explain this answer?
Let V be the region bounded by the planes x = 0, x = 2, y = 0, z = 0 and y + z = 1. Then the value of integral ∫∫V ∫ y dx dy dz is
a)
1 / 2
b)
1 / 3
c)
4 / 3
d)
1
|
|
Chirag Verma answered |
Let IV = ∫∫V ∫ y dx dy dz is
where V is the region bounded by the planes x = 0, x = 2, y = 0, z = 0 & y + z = 1
where V is the region bounded by the planes x = 0, x = 2, y = 0, z = 0 & y + z = 1
The value of 
- a)
- b)
- c)
- d)None o f these
Correct answer is option 'D'. Can you explain this answer?
The value of
a)
b)
c)
d)
None o f these
|
|
Chirag Verma answered |
The value of 
dydx
dydx
rdθdr is given by- a)
sinθdθ - b)
1/2 πrdr - c)
1/2 sin2θdθ - d)None of these
Correct answer is option 'C'. Can you explain this answer?
rdθdr is given bya)
sinθdθb)
1/2 πrdrc)
1/2 sin2θdθd)
None of these
|
|
Veda Institute answered |
The equations of given curves are
y(x2 + 2) = 3x
and 4y = x2
The curve (i) and (ii) intersect at A(2,1).
dxdy over the positive quadrant of the circle x2 + y2 = 1 is given by- a)1/6
- b)1/7
- c)1/14
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
dxdy over the positive quadrant of the circle x2 + y2 = 1 is given bya)
1/6
b)
1/7
c)
1/14
d)
None of these
|
|
Chirag Verma answered |
Let 
Changing to polar coordinates let x=r cosθ, y=r sinθ
Therefore,
= 1/14
Changing to polar coordinates let x=r cosθ, y=r sinθ
Therefore,
= 1/14The area bounded by the curve r = θ cosθ and the lines θ = 0 and θ = π/2 is given by
- a)
- b)
- c)
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
The area bounded by the curve r = θ cosθ and the lines θ = 0 and θ = π/2 is given by
a)
b)
c)
d)
None of these
|
|
Chirag Verma answered |
The equation of the given curve is
r =θcosθ -------(i)
So, the required area
r =θcosθ -------(i)
So, the required area
The volume of the tetrahedron bounded by the plane 
and the co-ordinate planes is equal to- a)1/3 πabc
- b)2/3 πabc
- c)1/6 πabc
- d)None of these
Correct answer is option 'D'. Can you explain this answer?
The volume of the tetrahedron bounded by the plane and the co-ordinate planes is equal to
and the co-ordinate planes is equal toa)
1/3 πabc
b)
2/3 πabc
c)
1/6 πabc
d)
None of these
|
|
Chirag Verma answered |
Here
Let u = x/a, v = y/b , w = z/c
Then dx = a du, dy = b dv, dz = c dw
So, Required volume
V =
abc du dv dw
where u + v + w ≤ 1, u ,v ,w ≥ 0
Thus
Let u = x/a, v = y/b , w = z/c
Then dx = a du, dy = b dv, dz = c dw
So, Required volume
V =
abc du dv dwwhere u + v + w ≤ 1, u ,v ,w ≥ 0
Thus
Hence, the correct answer is (c)
The area of the lemniscate r2 = a2 cos2θ is given by - a)4
- b)2
- c)4
- d)2
Correct answer is option 'A'. Can you explain this answer?
The area of the lemniscate r2 = a2 cos2θ is given by
a)
4
b)
2
c)
4
d)
2
|
|
Veda Institute answered |
The curve is r2 = a2 cos 2θ
The required area
The required area
The solution of the differential equation dy/dx = x log x is- a)y = x2 log x − x2/2 + c
- b)y = (x2/2) log x − x2/4 + c
- c)y = x2/2 + (x2/2) log x + c
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
The solution of the differential equation dy/dx = x log x is
a)
y = x2 log x − x2/2 + c
b)
y = (x2/2) log x − x2/4 + c
c)
y = x2/2 + (x2/2) log x + c
d)
None of these
|
|
Chirag Verma answered |
Given, dy = x log x dx
⇒ y = x2log x/2 − ∫ x/2 dx [integrating]
⇒ y = (x2/2) log x − x2/4 + c
⇒ y = x2log x/2 − ∫ x/2 dx [integrating]
⇒ y = (x2/2) log x − x2/4 + c
- a)cot x – x + C
- b)-cot x – x + C
- c)cot x + x + C
- d)-cot x + x + C
Correct answer is option 'B'. Can you explain this answer?
a)
cot x – x + C
b)
-cot x – x + C
c)
cot x + x + C
d)
-cot x + x + C
|
|
Chirag Verma answered |
We know that cot2 x = cosec2x – 1
∫cot²x dx = ∫ (cosec2x – 1) dx = -cot x -x + C. [Since, ∫cosec2x dx = -cot x + c]
Hence, the correct answer is option (b) -cot x – x + C.
is equal to
- a) esin^2y
- b)sin2y esin2y
- c)0
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
is equal toa)
esin^2y
b)
sin2y esin2y
c)
0
d)
none of these
|
|
Veda Institute answered |
ANSWER :- A
Solution :- limx→0 1/x[∫y→a e^sin2tdt− ∫x+y→a e^sin2tdt]
limx→0 [∫y→a e^sin2tdt− ∫x+y→a e^sin2tdt]/x
Hence it is 0/0 form, Apply L-hospital rule.
lim x-->0 0 - {e^(sin^2a).0 - e^(sin^2(x+y).1}/1
lim x-->0 e^(sin^2(x+y))
= e^(sin^2y)
The volume of the cylinder x2 + y2 = a2 bounded below by z = 0 and bounded above by z = h is given by - a)πah
- b)πa2h
- c)1/3πa3h
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
The volume of the cylinder x2 + y2 = a2 bounded below by z = 0 and bounded above by z = h is given by
a)
πah
b)
πa2h
c)
1/3πa3h
d)
None of these
|
|
Chirag Verma answered |
The equation of the cylinder is
x2 + y2 = a2
The equation of surface CDE is z = h
So, the required volume is
Let x = a sin θ
implies dx = a cosθ dθ
So, volume V
x2 + y2 = a2
The equation of surface CDE is z = h
So, the required volume is
Let x = a sin θ
implies dx = a cosθ dθ
So, volume V
is
?
Changing the order of integration in the double integral 
leads to
, then the value of q is- a)4y
- b)x
- c)16y2
- d)8
Correct answer is option 'A'. Can you explain this answer?
Changing the order of integration in the double integral leads to, then the value of q is
leads to, then the value of q isa)
4y
b)
x
c)
16y2
d)
8
|
|
Veda Institute answered |
Given, I = 
Equation of the line is x = Ay Starting val ue of x is zero and final value of x is 8 but we want the result of first integration in terms of y, so that we can integrate the result with respect to y to get the final result. So, limits are from 0 to 4y.
And integration limit for y is from 0 to 2.
So,
Hence, r=0, s=2, p=0, q=4y
Equation of the line is x = Ay Starting val ue of x is zero and final value of x is 8 but we want the result of first integration in terms of y, so that we can integrate the result with respect to y to get the final result. So, limits are from 0 to 4y.
And integration limit for y is from 0 to 2.
So,
Hence, r=0, s=2, p=0, q=4y
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