JEE Advanced Level Test: Area Under Curve With Solution - JEE MCQ

# JEE Advanced Level Test: Area Under Curve With Solution - JEE MCQ

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## 30 Questions MCQ Test - JEE Advanced Level Test: Area Under Curve With Solution

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JEE Advanced Level Test: Area Under Curve With Solution - Question 1

### The area bounded by [x] +[y] = 8 such that x, y > 0 is .... sq. units Where [.] is G.I.F.

Detailed Solution for JEE Advanced Level Test: Area Under Curve With Solution - Question 1

Area of 9 unit squares = 9 sq.units

JEE Advanced Level Test: Area Under Curve With Solution - Question 2

### Let f : [0, ∞)→ R be a continuous and strictly increasing function such that  The area enclosed by y = f (x), the x-axis and the ordinate at x = 3, is

Detailed Solution for JEE Advanced Level Test: Area Under Curve With Solution - Question 2

Given, f3(x) = ∫(0 to x)tf(t)dt
Differentiating throughout w.r.t x,
3f2(x) f′(x) = xf2(x)
⇒ f′(x)= x/3
​So, f(x)=∫(0 to x) x/3 dx = (x2)/6
Area = ∫f(x)dx=∫(0 to 3) (x2)/6 dx
= {(x3)/18}(0 to 3)
= 3/2

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JEE Advanced Level Test: Area Under Curve With Solution - Question 3

### The area bounded by x = x1, y = y1 and y = -(x + 1)2 where x1, y1 are the values of x, y satisfying the equation sin-1x + sin-1y = -π will be, (nearer to origin)

Detailed Solution for JEE Advanced Level Test: Area Under Curve With Solution - Question 3

JEE Advanced Level Test: Area Under Curve With Solution - Question 4

Area bounded between the curves

Detailed Solution for JEE Advanced Level Test: Area Under Curve With Solution - Question 4

JEE Advanced Level Test: Area Under Curve With Solution - Question 5

and  be two functions and let f1(x) = max {f(t), 0 < t < x, 0 < x <} and g1(x) = min {g(t), 0 < t < x, 0 < x < 1}. Then the area bounded by f1(x) < 0, g1(x) < 0 and x-axis is

JEE Advanced Level Test: Area Under Curve With Solution - Question 6

The values of the parameter a(a > 1) for which the area of the figure bounded by the pair of straight lines y2 – 3y + 2 = 0 and the curves  is greatest is.
(Here [.] denotes the greatest integer function).

Detailed Solution for JEE Advanced Level Test: Area Under Curve With Solution - Question 6

The curves  represent parabolas which are symmetric about yaxis. The equation y2 – 3y + 2 = 0 gives a pair of straight lines y = 1, y = 2 which are parallel to x-axis. The shaded region in figure determines the area bounded by the two parabolas and two lines. Let us slice this region into horizontal strips. For the approximating rectangle shown in figure. We have Length = x2 – x1, width = Δy, Area = (x2 - x1)dy
AT it can move vertically y = 1 and y = 2. So, Required area

JEE Advanced Level Test: Area Under Curve With Solution - Question 7

Area of region bounded by x2 + y2 < 4 and (|x| + |y|) < 2 is ____ square units.

Detailed Solution for JEE Advanced Level Test: Area Under Curve With Solution - Question 7

Using circles representation, required area = 8 sq. units

JEE Advanced Level Test: Area Under Curve With Solution - Question 8

The area of a circle is A1 and the area of a regular pentagon inscribed in the circle is A2 .Then A1 : A2 is

Detailed Solution for JEE Advanced Level Test: Area Under Curve With Solution - Question 8

JEE Advanced Level Test: Area Under Curve With Solution - Question 9

The area bounded by the curve  and x-axis is

Detailed Solution for JEE Advanced Level Test: Area Under Curve With Solution - Question 9

JEE Advanced Level Test: Area Under Curve With Solution - Question 10

the area of the region bounded by y = x and y = x+ sin x is

Detailed Solution for JEE Advanced Level Test: Area Under Curve With Solution - Question 10

JEE Advanced Level Test: Area Under Curve With Solution - Question 11

The area bounded by the curves y = x2, y = [x+1], x < 1 and the y - axis, where[.] denotes the greatest integer not exceeding x, is

Detailed Solution for JEE Advanced Level Test: Area Under Curve With Solution - Question 11

JEE Advanced Level Test: Area Under Curve With Solution - Question 12

The area of the smaller region in which the curve  denotes the greatest integer function, divides the circle (x – 2)2 + (y + 1)2 = 4, is equal to

Detailed Solution for JEE Advanced Level Test: Area Under Curve With Solution - Question 12

Circle has (2, -1) as its centre and radius of this circle is 2.
Thus if P(x, y) be any point on it, then x ∈ [0, 4].

The g(x) is increasing in [0, 4]

JEE Advanced Level Test: Area Under Curve With Solution - Question 13

The area bounded by the curves y = 2- |x -1| , y = sin x ; x = 0 and x =2 is

Detailed Solution for JEE Advanced Level Test: Area Under Curve With Solution - Question 13

JEE Advanced Level Test: Area Under Curve With Solution - Question 14

A curve passes through the point (0,1)and has the property that the slope of the curve at every point P is twice the y–coordinate of P . If the area bounded by the curve, the axes of coordinates and the line

Detailed Solution for JEE Advanced Level Test: Area Under Curve With Solution - Question 14

JEE Advanced Level Test: Area Under Curve With Solution - Question 15

The area bounded by the curve (y – arc sinx)2 = x – x2, is

Detailed Solution for JEE Advanced Level Test: Area Under Curve With Solution - Question 15

JEE Advanced Level Test: Area Under Curve With Solution - Question 16

The area of the region enclosed by the curve 5x2 + 6xy + 2y2 + 7x + 6y + 6 = 0 is

Detailed Solution for JEE Advanced Level Test: Area Under Curve With Solution - Question 16

Comparing the given equation with the general equation of second degree i.e.
ax2 + 2hxy + by2 + 2gx + 2fy + c = 0,
We have, abc + 2fgh – af2 – bg2 – ch2 = 60 – 45
And h2 – ab = 9 – 10 < 0.
So, the given equation represents an ellipse.
Rewriting the given equation as a quadratic in y, we obtain

Clearly, the values of y are real for x ∈ [1, 3]
When x = 1, we get y = -3 and x = 3 ⇒ y = -6.

JEE Advanced Level Test: Area Under Curve With Solution - Question 17

Let f(x) be a continuous function such that the area bounded by the curve y = f(x), the xaxis and the two ordinates x = 0 and

Detailed Solution for JEE Advanced Level Test: Area Under Curve With Solution - Question 17

Differentiating both sides w.r.t. a, we get

JEE Advanced Level Test: Area Under Curve With Solution - Question 18

The area enclosed between the curves y = sin4x cos3x, y = sin2xcos3x between x = 0 and

Detailed Solution for JEE Advanced Level Test: Area Under Curve With Solution - Question 18

Required area is equal to

JEE Advanced Level Test: Area Under Curve With Solution - Question 19

The area of the region bounded by the curves  which contains (1, 0) point in its interior is

Detailed Solution for JEE Advanced Level Test: Area Under Curve With Solution - Question 19

We observe that all powers of x in the above equation are even, so it is symmetric about y-axis. The curve intersects y-axis at (0, 1). Also,

Therefore, as x → ∞, y → -1 i.e. y = -1 is an asymptote to the curve.

This means that x is imaginary for y > 1 or y < -1 i.e. the curve lies between y = -1 and y = 1. At (0, 1) the tangent to the curve

JEE Advanced Level Test: Area Under Curve With Solution - Question 20

Area bounded by the curves y = ex , y= loge x and the lines x = 0, y = 0, y = 1 is

Detailed Solution for JEE Advanced Level Test: Area Under Curve With Solution - Question 20

Area = Area of  rectangle

JEE Advanced Level Test: Area Under Curve With Solution - Question 21

A circular arc of radius ‘1’ subtends an angle of ‘x’ radians, as shown in the figure. The point ‘R’ is the point of intersection of the two tangent lines at P & Q. Let T(x) be the area of triangle PQR and S(x) be area of the shaded region. Then

Detailed Solution for JEE Advanced Level Test: Area Under Curve With Solution - Question 21

JEE Advanced Level Test: Area Under Curve With Solution - Question 22

A point P moves inside a triangle formed by  such that  min {PA, PB, PC} = 1. The area formed by the curve traced by P is ....... sq. units

Detailed Solution for JEE Advanced Level Test: Area Under Curve With Solution - Question 22

Re quired area = Area of equilateral

JEE Advanced Level Test: Area Under Curve With Solution - Question 23

Area of the triangle formed by the tangent and normal at (1, 1) on the curve and the y-axis is

Detailed Solution for JEE Advanced Level Test: Area Under Curve With Solution - Question 23

Find equation of the tangent and normal and then put x=0 to evaluate vertices of triangle.
Then find area of triangle .

JEE Advanced Level Test: Area Under Curve With Solution - Question 24

Area bounded by the curve  and area bounded by latus rectum of

Detailed Solution for JEE Advanced Level Test: Area Under Curve With Solution - Question 24

Area bounded by C1 & C2 is λ1 = 16/3 sq.units
λ2 = Area bounded by C1 & its latus rectum = 8/3 sq.units

JEE Advanced Level Test: Area Under Curve With Solution - Question 25

The area of the part of circle x2 + y2 - 2x - 4y -1 = 0 above 2x - y = 0 is ...... sq. units

Detailed Solution for JEE Advanced Level Test: Area Under Curve With Solution - Question 25

Required area = Area of semi-circle

Since given line is diameter of circle

JEE Advanced Level Test: Area Under Curve With Solution - Question 26

The area bounded by the curves y = |x| – 1 and y – |x| 1is

JEE Advanced Level Test: Area Under Curve With Solution - Question 27

Area of closed curve 3(x -1)2 + 4(y2 - 3) = 0 is  where [.] is G.I.F

Detailed Solution for JEE Advanced Level Test: Area Under Curve With Solution - Question 27

JEE Advanced Level Test: Area Under Curve With Solution - Question 28

Let f(x) = x + sin x. The area bounded by y = f-1 (x), y = x, x ∈ [0, π] is

Detailed Solution for JEE Advanced Level Test: Area Under Curve With Solution - Question 28

The curves given by y = x + sin x and y = f-1(x) are images of each other in the line y = x.
Hence required area

JEE Advanced Level Test: Area Under Curve With Solution - Question 29

The area of the region bounded by the curves |x + y| < 2, |x – y| < 2 and 2x2 + 6y2 > 3 is

Detailed Solution for JEE Advanced Level Test: Area Under Curve With Solution - Question 29

2x2 + 6y2 > 3. . . . . . (1) area of ellipse

JEE Advanced Level Test: Area Under Curve With Solution - Question 30

Area enclosed by the closed curve 5x2 + 6xy + 2y2 + 7x + 6y + 6 = 0 is

Detailed Solution for JEE Advanced Level Test: Area Under Curve With Solution - Question 30

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