Q: Let R denote the set of all real numbers. Then the area of the region \ (x, y) R R : x>0, y> 1 x , 5 x-4 y-1>0,4 x+4 y-17 <0 \ is

Area = 13 8 + 20 8 - _e 4= ( 33 8 - _e 4 )

Q: Let S= \ (x, y) R R : x 0, y 0, y^2 4 x, y^2 12-2 x . and .3 y+ 8 x 5 8 \ . If the area of the region S is 2 , then is equal to

Point of intersection of all curves is (2,2 2 ) Area = A _1+ A _2 2 = _0^2 2 x dx + 1 2 3 2 2 aligned & 2 =2 [ x^ 3 2 3 2 ]_0^2+3 2 \\ & 2 = 17 2 3 \\ & = 17 3 aligned

Q: Let f : 0 , 1 → 0 , 1 be the function defined by f x = x 3 3 - x 2 + 5 9 x + 17 36 . Consider the square region S = 0 , 1 × 0 , 1 . Let G = x , y ∈

Given, Function f x = x 3 3 - x 2 + 5 9 x + 17 36 Square region S = 0 , 1 × 0 , 1 Green region given by G = x , y ∈ S : y > f x Red region is given by R = x , y ∈ S : y < f x And L h = x , h ∈ S : x ∈ 0 , 1 be the horizontal line drawn at a height h ∈ 0 , 1 Now plotting the diagram of the above function we get, Now differentiating the function, f x = x 3 3 - x 2 + 5 9 x + 17 36 &#8

Q: Let n ≥ 2 be a natural number and f : 0 , 1 → ℝ be the function defined by f x = n 1 - 2 n x if 0 ≤ x ≤ 1 2 n 2 n 2 n x - 1 if 1 2 n ≤ x ≤ 3 4 n

Given, f x = n 1 - 2 n x , 0 ≤ x < 1 2 n 2 n 2 n x - 1 , 1 2 n ≤ x < 3 4 n 4 n 1 - n x , 3 4 n ≤ x ≤ 1 n n n - 1 n x - 1 , 1 n ≤ x ≤ 1 Here, x ∈ 0 , 1 Now f x = n - 2 n 2 x is decreasing in 0 , 1 2 n And f x = 4 n 2 x - 4 n increasing in 1 2 n , 3 4 n And f x = 4 n - 4 n 2 x decreasing in 3 4 n , 1 n And f x = n 2 x - n n - 1 increasing in 1 n , &

Q: Consider the functions f , g : ℝ → ℝ defined by f x = x 2 + 5 12 and g x = 2 1 - 4 x 3 , x ≤ 3 4 , 0 , x > 3 4 If α is the area of the region x , y ∈ ℝ

Plotting the diagram of the given function we have, Now by diagram we can see twice of A 1 and A 2 will be the required area, So, required area = 2 · A 1 + A 2 ⇒ α = ∫ 0 1 2 x 2 + 5 12 d x + 1 2 3 4 - 1 2 · 2 3 × 2 ⇒ α = x 3 3 + 5 x 12 0 1 2 + 1 2 ⋅ 1 4 ⋅ 2 3 × 2 ⇒ α = 1 4 + 1 12 × 2 ⇒ α = 2 3 ⇒ 9 α = 6

Q: The area of the region ( x , y ) : 0 ≤ x ≤ 9 4 , 0 ≤ y ≤ 1 , x ≥ 3 y , x + y ≥ 2 is

The region is a quadrilateral A B C D Now, area of the quadrilateral A B C D = area of triangle A B D + area of the triangle B C D So, area = 1 2 3 2 1 2 1 2 0 1 9 4 3 4 1 + 1 2 × 9 4 - 2 × 3 4 = 1 2 - 2 1 2 - 3 4 - 9 8 - 9 8 + 1 2 × 1 4 × 3 4 = 1 4 + 3 32 = 11 32

Q: Let the functions f : R → R and g : R → R be defined by f x = e x − 1 − e − x − 1 and g x = 1 2 e x − 1 + e 1 − x . Then the area of the region in the f

f x = e x − 1 − e − x − 1 = 0 , x ≤ 1 e x − 1 − e 1 - x , x > 1 and g x = 1 2 e x − 1 + e 1 − x At x = 0 , f 0 = 0 & g 0 = e + e - 1 2 At x = 1 , f 1 = 0 & g 1 = 1 2 e 1 - 1 + e 1 - 1 = 1 For point of intersection of the two curves for x > 1 , f x = g x ⇒ e x − 1 − e 1 - x = 1 2 e x − 1 + e 1 − x ⇒ 2 e x − 1 − 2 e 1 - x = e x − 1 + e 1 − x ⇒ e x − 1 = 3 e 1 − x ⇒ e x − 1 2 = 3 ⇒ x = 1 + ln 3 ∴ Required area = Area of the shaded region = ∫ 0 1 g x d x + ∫ 1 1 + ln 3 g x − f x d x = ∫ 0 1 1 2 e x

Question 1

Passage: Consider the curve C_1 given by y = e^ -x for x [0, 10 ], and the curve C_2 given by y = e^ -x ( x + x) for x [0, 10 ]. Let n be the total number of points of intersection of the curves C_1 a

Accepted Answer

To find the points of intersection of the curves C_1 and C_2, we equate their equations: e^ -x = e^ -x ( x + x) Since e^ -x 0, we have: x + x = 1 2 (x + 4 ) = 1 (x + 4 ) = 1 2 The general solution is x + 4 = 2k + 4 or x + 4 = 2k + 3 4 for any integer k. Thus, x = 2k or x = 2k + 2 . For x [0, 10 ], the first four points of intersection are: _1 = 0 _2 = 2 _3 = 2 _4 = 2 + 2 = 5 2 The area enclosed between the curves from x = _1 to x = _4 is given by: = _ 0 ^ 5 2 |e^ -x ( x + x - 1)| dx We analyze t

Question 2

Passage: Consider the ellipses given by x^2 + 4y^2 = 1 and 4x^2 + y^2 = 1. Question: If is the area of the common region that lies inside both the given ellipses, then the value of is ___________.

Accepted Answer

The given ellipses are E_1: x^2 + 4y^2 = 1 and E_2: 4x^2 + y^2 = 1. By symmetry, the common region is symmetric about both the coordinate axes and the lines y = x and y = -x. The total area is 8 times the area of the region in the first quadrant bounded by = 0 and = 4 . In polar coordinates, substituting x = r and y = r into the equation of E_2 (which is the inner boundary for 0 4 ), we get: 4r^2 ^2 + r^2 ^2 = 1 r^2 = 1 4 ^2 + ^2 The area of this sector is given by: A = 1 2 _ 0 ^ 4 r^2 d = 1 2 _

Question 3

Let R denote the set of all real numbers. Then the area of the region \ (x, y) R R : x>0, y> 1 x , 5 x-4 y-1>0,4 x+4 y-17 <0 \ is

Accepted Answer

Area = 13 8 + 20 8 - _e 4= ( 33 8 - _e 4 )

Question 4

Let S= \ (x, y) R R : x 0, y 0, y^2 4 x, y^2 12-2 x . and .3 y+ 8 x 5 8 \ . If the area of the region S is 2 , then is equal to

Accepted Answer

Point of intersection of all curves is (2,2 2 ) Area = A _1+ A _2 2 = _0^2 2 x dx + 1 2 3 2 2 aligned & 2 =2 [ x^ 3 2 3 2 ]_0^2+3 2 \ & 2 = 17 2 3 \ & = 17 3 aligned

Question 5

Let f : 0 , &#160; 1 &#8594; 0 , &#160; 1 be the function defined by f x = x 3 3 - x 2 + 5 9 x + 17 36 . Consider the square region S = 0 , &#160; 1 &#215; 0 , &#160; 1 . Let G = x , &#160; y &#8712;

Accepted Answer

Given, Function  f x = x 3 3 - x 2 + 5 9 x + 17 36 Square region S = 0 , &#160; 1 &#215; 0 , &#160; 1 Green region given by  G = x , &#160; y &#8712; S : y &#62; f x Red region is given by  R = x , &#160; y &#8712; S : y &#60; f x And  L h = x , &#160; h &#8712; S : x &#8712; 0 , &#160; 1 be the horizontal line drawn at a height h &#8712; 0 , &#160; 1 Now plotting the diagram of the above function we get, Now differentiating the function, f x = x 3 3 - x 2 + 5 9 x + 17 36 &#8

Question 6

Let n &#8805; 2  be a natural number and f : 0 , &#160; 1 &#8594; &#8477; be the function defined by f x = n 1 - 2 n x if 0 &#8804; x &#8804; 1 2 n 2 n 2 n x - 1 if 1 2 n &#8804; x &#8804; 3 4 n

Accepted Answer

Given, f x = n 1 - 2 n x , 0 &#8804; x &#60; 1 2 n 2 n 2 n x - 1 , 1 2 n &#8804; x &#60; 3 4 n 4 n 1 - n x , 3 4 n &#8804; x &#8804; 1 n n n - 1 n x - 1 , 1 n &#8804; x &#8804; 1 Here,  x &#8712; 0 , &#160; 1 Now  f x = n - 2 n 2 x  is decreasing in  0 , &#160; 1 2 n   And  f x = 4 n 2 x - 4 n  increasing in  1 2 n , &#160; 3 4 n And f x = 4 n - 4 n 2 x  decreasing in  3 4 n , &#160; 1 n And f x = n 2 x - n n - 1  increasing in  1 n , &

Question 7

Consider the functions f , g : &#8477; &#8594; &#8477; defined by f x = x 2 + 5 12 and g x = 2 1 - 4 x 3 , x &#8804; 3 4 , 0 , x &#62; 3 4 If &#945; is the area of the region x , y &#8712; &#8477;

Accepted Answer

Plotting the diagram of the given function we have, Now by diagram we can see twice of  A 1  and  A 2  will be the required area, So, required area = 2 &#183; A 1 + A 2 &#8658; &#945; = &#8747; 0 1 2 x 2 + 5 12 d x + 1 2 3 4 - 1 2 &#183; 2 3 &#215; 2 &#8658; &#945; = x 3 3 + 5 x 12 0 1 2 + 1 2 &#8901; 1 4 &#8901; 2 3 &#215; 2 &#8658; &#160;&#160; &#945; = 1 4 + 1 12 &#215; 2 &#8658; &#160;&#160; &#945; = 2 3 &#8658; 9 &#945; = 6

Question 8

The area of the region ( x , y ) : 0 &#8804; x &#8804; 9 4 , &#160;&#160; 0 &#8804; y &#8804; 1 , &#160;&#160; x &#8805; 3 y , &#160;&#160; x + y &#8805; 2  is

Accepted Answer

The region is a quadrilateral  A B C D Now, area of the quadrilateral  A B C D = area of triangle  A B D + area of the triangle  B C D   So, area  = 1 2 3 2 1 2 1 2 0 1 9 4 3 4 1 + 1 2 &#215; 9 4 - 2 &#215; 3 4 = 1 2 - 2 1 2 - 3 4 - 9 8 - 9 8 + 1 2 &#215; 1 4 &#215; 3 4 = 1 4 + 3 32 = 11 32

Question 9

Let the functions f : R &#8594; R and g : R &#8594; R be defined by f x = e x &#8722; 1 &#8722; e &#8722; x &#8722; 1 and g x = 1 2 e x &#8722; 1 + e 1 &#8722; x . Then the area of the region in the f

Accepted Answer

f x = e x − 1 − e − x − 1 = 0 , x ≤ 1 e x − 1 − e 1 - x , x > 1 and g x = 1 2 e x − 1 + e 1 − x At x = 0 , f 0 = 0 & g 0 = e + e - 1 2 At x = 1 , f 1 = 0 & g 1 = 1 2 e 1 - 1 + e 1 - 1 = 1 For point of intersection of the two curves for x > 1 , f x = g x ⇒ e x − 1 − e 1 - x = 1 2 e x − 1 + e 1 − x ⇒ 2 e x − 1 − 2 e 1 - x = e x − 1 + e 1 − x ⇒ e x − 1 = 3 e 1 − x ⇒ e x − 1 2 = 3 ⇒ x = 1 + ln 3 ∴ Required area = Area of the shaded region = ∫ 0 1 g x d x + ∫ 1 1 + ln 3 g x − f x d x = ∫ 0 1 1 2 e x

Question 10

The area of the region x , y : x y &#8804; 8 , &#160; 1 &#8804; y &#8804; x 2 is

Accepted Answer

To draw the inequality, let us draw the equation x y = 8 and y = 1 and y = x 2 For point of intersection (i) x y = 8 and y = 1 ⇒ A ( 8,1 ) (ii) x y = 8 and y = x 2 ⇒ x 3 = 8 ⇒ x = 2 ⇒ B ( 2 , 4 ) (iii) y = x 2 and y = 1 ⇒ C 1 , 1 and D 1 , 1 Now x y ≤ 8 ⇒ region which contains origin y ≥ 1 ⇒ region above line y = 1 y ≤ x 2 ⇒ region outside the parabola Now required area Method I: Using x-axis: A = ∫ 1 2 x 2 - 1 ⋅ d x + ∫ 2 8 8 x - 1 ⋅ d x A = x 3 3 - x 1 2 + 8 l n x - x 2 8 A = - 14 3 + 16 l n 2

Area Under Curves — Previous Year Questions & Solutions