IIT JAM Mathematics Practice Test- 14 - Mathematics MCQ

if -1 to 2 integration of f(x) is 5 then 2 to -1 it will be -5.

Let

Put x = rcos θ, y = rsinθ

and dxdy = rdθdr

The smaller area enclosed by the circle, x²+y²=4 and the line, x = 1 is represented by the shaded area ACBA as shown in the diagram.

It can be observed that,

2∫1 to 2 (4-x²)^1/2dx

= 2 [x/2(4-x²) + 2sin^-1(x/2)]1 to 2

= 2[2*pi/2 - (3)^1/2 /2 +2pi/6]

= 2[pi - (3)^1/2 /2 +pi/3]

8pi/3 - (3)^1/2

Given y = x² and y = 2 - x² from these eq. we get

Required area

= 2-1 = 1

Region D is 2x ≤ x² + y² ≤ 4x

i.e. x² + y² = 2x and x² + y² = 4x (x - 1)² + y² = 1 and (x - 2)² + y² = 4 polar form

x → rcos θ

y → rsin θ

dxdy → rdθdr

then

from equation (1) and equation (2)

Area of the smallest part is 

so, area of the largest part is area of the circle -area of the smallest part

Given curve is xy - 3x - 2y -10 = 0 

=>(x - 2)y = 3x +10

= 3 (1 ) + 16 [log(2) - log (1 )] = 3 + 16 log 2

Limits are y = x, y = ∞

X = 0, X = ∞

Changing the order of the integration

By the sum of the series

R = [0, 1 : 0, 3] i.e. 0 < x < 1 , 0 < y < 3 , 0 < x + y < 4

Required Area

The equations of the given curve are 

x sinθ  + y cosθ  = f'(θ )

x cosθ  - y sinθ  = f"(θ  ) 

Solving these equations, we get 

x = sinθ  f'(θ ) + cosθ  f"(θ) 

y = cosθ  f'(θ ) - sinθ  f"(θ ) 

Now differentiate these equation w.r. to θ we get

Hence required length of arc of the curve is

The given equation of the cuives are

Hence the required area is

The intersecting points of the graphs are

 and r(x) = 2

therefore, the volume is

Given double integral is

 this integral is not solvable.

Limits are y = 0, y = x

and x = 0, x = a

then by changing the order of integration we get

Differentiating the equation of the curve

Therefore the required length

Now equation becomes

⇒ equation (1) has 3 real distinct roots.

When the equation of the curve of integration is taken by expressing x as a single valued function of y, 

we put x = y²

so that

Here the limits of y are given by the circle and the straight line y = x +2a. The limits of x are given by the straight lines x = 0 and x = a

Therefore, the integral extends to all points in the space bounded by the axis of y, the circle AB, the straight line AL and the straight line LM. Draw BH and M perpendiculars to AL. Now when the order of integration is changed, we take strips paralles to axis of x instead of that is y, thus the integral breaks up into three parts: First corresponding to the area BAH, second to the rectangle BHKM and the third to the triangle MKL, hence

so