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**Q.1. The area of the region, enclosed by the circle x ^{2} + y^{2} = 2 which is not common to the region bounded by the parabola y^{2} = x and the straight line y = x, is (2020)**

(1)

(2)

(3)

(4)

The given equation of circle is

x

Now, total area enclosed by the circle is Ï€r

The area bounded by the parabola y

Hence, the required area of the region = sq. units

(1) 125/3

(2) 128/3

(3) 124/3

(4) 127/3

The given curves are

Now, on solving y = 4x

4x

x = -1, 3 and y = 4, 36

Therefore

Hence, the area of shaded region is

(1) x

(2) x

(3) x

(4) x

We have

Now, the area of triangle PQR = 1/2

Substituting the value of b in Eq. (1), we get

**Q.4. The equations of given curves are (2020)**

The equations of given curves are

x

2x + y

On solving y = x

Now,

Hence, the area of shaded region is

**Q.5. Given: Then, the area (in sq. units) of the region bounded by the curves, y = f(x) and y = g(x) between the lines 2x = 1 and 2x =âˆš3 is (2020)**

We have,

Required area, A = Area of trapezium ABCD

sq. units

(1) log

(2)

(3)

(4)

We have,

Thus,

(1)

(2)

(3)

(4)

We have

Therefore, f (x) is decreasing in (1,2).

Now,

Hence,

(1)

(2)

(3)

(4)

We have

.... (1)

Now,

We have

and

Now,

Given,

f(x) = f(a + b + 1 - x)...(1)

f(x + 1) = f (a + b - x)...(2)

Now,

....(3)

....(4)

From Eqs. (2) and (3), we get

Let x + 1 = t â‡’ dx = dt, so

(1) 2Ï€

(3) Ï€

(4) 4Ï€

We have

.... (1)

(2)

From Eqs. (1) and (2), we get

.... (3)

(4)

From Eqs. (3) and (4), we get

Given,

... (1)

Now,

...(2)

From Eqs. (1) and (2), we get

Hence,

(1) 0

(2) 4/3

(3) 2/3

(4) -4/3

âˆ´ f(x) is a constant function.

Given f(0) = 1 â‡’ f(x) = 1

Hence, the integral

Hence, integral becomes.

âˆ´ k = 2

Then

We have

Required area, A = Area of trapezium ABCD

sq. units

(1) 0

(2) sin 4

(3) 4

(4) 4 -sin 4

â‡’ f(x) is odd function

Let tan

Let, the integral,

Let

â‡’

On differentiating both sides w.r.t x we get

When x = e then t = 1 and when x = 1 then t = 1/e.

** (2019)(1) log _{e}3(2) log_{e}e(3) log_{e}2(4) log_{e}1**

Then, equation (1) becomes,

** (2019)****Ans. **(1)

Solution.**Q.25. (2019)****Ans. **(2)

Solution.

Adding equation (1) and (2), we get**Q.26. The value of the integral **x** **âˆˆ [0,1]** (2019)****Ans. **(4)**Solution.****Q.27. The value of where [t] denotes ****the greatest integer function, is: (2019)****(1) Ï€ (2) -Ï€ (3) -2Ï€ (4) 2Ï€**

From (1) + (2), we get;

(2) 3

(3) 3

(4) 3

Ans.

Solution.

equation of tangent at (2, 3)

Here the curve cuts Y-axis

âˆ´ required area

(1) 2/3

(2) 2

(3) 4/3

(4) 1/3

âˆ´ Area of shaded region

Two curves will intersect in the 1st quadrant at

âˆµ area of shaded region = 1.

(1) 5/4

(2) 9/8

(3) 7/8

(4) 3/4

Ans.

Solution.

Let points of intersection of the curve and the line be P and Q

x

x = 2, - 1

The equation of parabola x

The equation of tangent at (2, 5) to parabola is

y - 5 = 4(x - 2)

4x - y = 3

Then, the required area

(2) 8

(3) 59/6

(4) 26/3

Since, the relation y â‰¤ x

âˆµ y = 4

â‡’ x

the required area = area of shaded region

**Ans. **(4)**Solution.****Q.40. The area (in sq. units) of the region A = {(x, y): x ^{2} â‰¤ y â‰¤ x + 2} is: (2019)**

Required area is equal to the area under the curves y â‰¥ x

(1) 53/3

(2) 30

(3) 16

(4) 18

(1) square of side length 2âˆš2 units

(2) rhombus of side length 2 units

(3) square of area 16 sq. units

(4) rhombus of area 8âˆš2 sq. units

Ans.

Solution.

By the diagram, region is square

(3) 1/2

Solution.

Consider y

Substituting x = 1 - y in the equation of parabola,

y

â‡’ (y + 2)

Hence, required area

Given parabola y

**Q.46. ****The value of dx is: (2018)(1) Ï€/8(2) Ï€/2(3) 4Ï€(4) Ï€/4Ans. **(4)

Solution.

(1)

(2)

(3)

(4)

Ans. (1)

Solution.

g(x) = cos x

f(x) = âˆšx

g(f (x)) = cos x

(2018)

(1) 3/4

(2) 0

(3)

**(4)Ans. (4)Solution.**

I = 3Ï€/16

(1) 8/3

(2)

(3)

(4)

Ans.

Solution.

(1) e

(2) e

(3) 3/2

(4) 4/3

Ans.

Solution.

The intersection point of y = x

Area bounded by the curves is the region ABCDA is given as:

(1) 5/2

(2) 59/12

(3) 3/2

(4) 7/3

Ans.

Solution:

Area of shaded region

= 5/2 sq. unit

(1) â€“1

(2) â€“2

(3) 2

(4) 4

Ans.

Solution:

**Q.55. (2017)****(1) 4****(2) 2****(3) 3****(4) 1****Ans. **(4)**Solution.****Q.56. ****The area (in sq. units) of the region {(x, y) : y ^{2} â‰¥ 2x and x^{2} + y^{2} â‰¤ 4x, x â‰¥ 0, y â‰¥ 0} is: (2016)**(2)

(1)

(2)

(3)

(4)

Ans.

Solution.

x

To find point of intersection,

x

â‡’ x

â‡’ x = 0 or x = 2

âˆ´ y = 0 or y = 2

Solve (x, y) = (0, 0) & (x, y) = (2, 2)

Area =

(1) log 2

(3) log 4

(4)

Ans.

Solution.

(1) 7/2

(2) 13/6

(3) 17/6

(4) 19/6

A = {(x,y)|y

Here y

(where C is a constant)

Ans.

Solution.

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