JEE Main Previous Year Questions (2016- 2024): Definite Integrals and Applications of Integrals | Mathematics for Airmen Group X - Airforce X Y / Indian Navy SSR PDF Download

Q.1. The area of the region, enclosed by the circle x² + y² = 2 which is not common to the region bounded by the parabola y² = x and the straight line y = x, is    (2020)
(1)
(2)
(3)
(4)
Ans. (4)
The given equation of circle is
x² + y² = 2....    (1)
Now, total area enclosed by the circle is πr² = π(√2)² = 2π sq. units
The area bounded by the parabola y² = x and y = x is


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

Q.2. The area (in sq. units) of the region is    (2020)
(1) 125/3
(2) 128/3
(3) 124/3
(4) 127/3
Ans. (2)
The given curves are  
Now, on solving y = 4x² and y = 8x + 12, we get
4x² - 8x - 12 = 0 ⇒ x² - 2x - 3 = 0
x = -1, 3 and y = 4, 36
Therefore

Hence, the area of shaded region is



Q.3. For a > 0, let the curves C₁ : y² = ax and C₂ : x² = ay intersect at origin O and at Point P. Let the line x = b (0 < b < a) intersect the chord OP and the x axis at points Q and R respectively. If the line x = b bisects the area bounded by the curves, C₁ and C₂, and the area of then ‘a’ satisfies the equation:    (2020)
(1) x⁶ - 6x³ + 4 = 0
(2) x⁶ - 12x³ + 4 = 0
(3) x⁶ + 6x³ - 4 = 0
(4) x⁶ - 12x³ - 4 = 0
Ans.(2)
We have


    (1)
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)
(1) 32/3
(2) 34/3
(3) 29/3
(4) 31/3
Ans. (1)
The equations of given curves are
x² < y...    (1)
2x + y < 3...    (2)
On solving y = x² = and 2x + y = 3, we get the intersection point x = − 1, 3.
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 =√3 is    (2020)
(1)
(2)
(3)
(4)
Ans. (2)
We have,

Required area, A = Area of trapezium ABCD
 sq. units

Q.6. The value of α for which is    (2020)
(1) log_e2
(2)
(3)
(4)
Ans. (1)
We have,





Thus,

Q.7. If then    (2020)
(1)
(2)
(3)
(4)
Ans. (2)
We have


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

Hence,

Q.8. If for all real triplets (a, b, c), f(x) = a + bx + cx² ; then is equal to    (2020)
(1)
(2)
(3)
(4)
Ans. (4)
We have
....    (1)

Now,

Q.9. Let z be a complex number such that  and Then, the value ofis    (2020)
(1) √10
(2) 7/2
(3) 15/4
(4) 2√3
Ans. (2)
We have


and


Now,

Q.10. If f(a + b + 1 - x) = f(x),  for all x, where a and b are fixed positive real numbers, then is equal to    (2020)
(1)
(2)
(3)
(4)
Ans. 
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 ⇒ dx = dt, so

Q.11. The value of is equal to    (2020)
(1) 2π
(2) 2π²
(3) π²
(4) 4π
Ans. (3)
We have
....    (1)

    (2)
From Eqs. (1) and (2), we get

....    (3)

    (4)
From Eqs. (3) and (4), we get


Q.12. If where c is a constant of integration, then is equal to    (2020)
(1)
(2) 2
(3) 9/8
(4) -2
Ans. (4)
Given,
... (1)
Now,


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

Hence,


Q.13.      (2019)
(1) 0
(2) 4/3
(3) 2/3
(4) -4/3
Ans. (2)
Solution.




Q.14. Let f be a differentiable function from R to R such that |f(x) - f(y)|≤ 2|x - y|^3/2, for all x, y, ∈ R. If f(0) = 1 then  is equal to:     (2019)
(1) 1 
(2) 2
(3) 1/2 
(4) 0
Ans. (1)
Solution.



∴  f(x) is a constant function.
Given f(0) = 1  ⇒   f(x) = 1
Hence, the integral


Q.15.  then the value of k is:     (2019)
(1) 4
(2) 1/2
(3) 1
(4) 2
Ans. (4)
Solution.


Hence, integral becomes.


∴ k = 2

Q.16. then f'(1/2) is:      (2019)
(1) 24/25
(2) 18/25
(3) 4/5
(4) 6/25
Ans. (1)
Solution.




Then


Q.17. Given:  and g (x) =  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 = √3 is
(1)
(2)
(3)
(4)
Ans. (2)
We have

Required area, A = Area of trapezium ABCD
sq. units

Q.18. The value of  where [t] denotes the greatest integer less than or equal to t, is:     (2019)




Ans. (3)
Solution.




Q.19. The value of the integral
(where [x] denotes the greatest integer less than or equal to x) is:       (2019)
(1) 0
(2) sin 4
(3) 4
(4) 4 -sin 4
Ans. (1)
Solution.

     

⇒ f(x) is odd function


Q.20.
        (2019)




Ans. (2)
Solution.


Let tan⁵x = t






Q.21. Let f and g be continuous functions on [0, a] such that f(x) =f(a - x) and g(x) + g(a - x) = 4,  equal to:      (2019)




Ans. (3)
Solution.


Let, the integral,









Q.22. The integral  is equal to:      (2019)




Ans. (4)
Solution.

Let 
⇒
⇒ x (ln x - 1) = ln t
On differentiating both sides w.r.t x we get

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



Q.23.
     (2019)
(1) π/4
(2) tan^-1(3)
(3) π/2
(4) tan^-1(2)
Ans. (4)
Solution.




Q.24. then the value of the 

       (2019)
(1) log_e3
(2) log_ee
(3) log_e2
(4) log_e1
Ans. (4)
Solution.


Then, equation (1) becomes,




Q.24.  , where g is a non-zero even function. If 

        (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 ∈ [0,1]   (2019)




Ans. (4)
Solution.




Q.27. The value of  where [t] denotes the greatest integer function, is:        (2019)
(1) π  
(2) -π 
(3) -2π
(4) 2π
Ans. (2)
Solution.



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


Q.28.         (2019)




Ans. (1)
Solution.



Q.29. is equal to:       (2019)
(1) 3^5/6 - 3^2/3
(2) 3^4/3 - 3^1/3
(3) 3^7/6 - 3^5/6
(4) 3^5/3 - 3^1/3
Ans. (3)
Solution.





Q.30.  then m.n is equal to:      (2019)

(2) 2
(3) 1/2
(4) -1
Ans. (4)
Solution.




Q.31. A value of α such that       (2019)
(1) -2
(2) 1/2
(3)
(4) 2
Ans. (1)
Solution.




Q.32. The area (in sq. units) bounded by the parabola y = x² - 1, the tangent at the point (2, 3) to it and the y-axis is:       (2019)
(1) 8/3
(2) 32/3
(3) 56/3
(4) 14/3
Ans. (1)
Solution.


equation of tangent at (2, 3)

Here the curve cuts Y-axis
∴ required area


Q.33. The area of the region A = {(x, y): 0 ≤ y ≤ x |x| + 1 and - 1 ≤ x ≤ 1} in sq. units is:        (2019)
(1) 2/3
(2) 2
(3) 4/3
(4) 1/3
Ans. (2)
Solution.


∴ Area of shaded region


Q.34. If the area enclosed between the curves y = kx² and x = ky², (k > 0), is 1 square unit. Then k is:       (2019)
(1) √3/2
(2) 1/√3
(3) √3
(4) 2/√3
Ans. (2)
Solution.

Two curves will intersect in the 1st quadrant at
∵ area of shaded region = 1.






Q.35. The area (in sq. units) of the region bounded by the curve x² = 4 y and the straight line x = 4y - 2 is:        (2019)
(1) 5/4
(2) 9/8
(3) 7/8
(4) 3/4
Ans. (2)
Solution.

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

x² - x - 2 = 0
x = 2, - 1



Q.36. The area (in sq. units) in the first quadrant bounded by the parabola,y = x² + 1, the tangent to it at the point (2, 5) and the coordinate axes is:        (2019)
(1) 8/3
(2) 37/24
(3) 187/24
(4) 14/3
Ans. (2)
Solution.

The equation of parabola x² = y - 1
The equation of tangent at (2, 5) to parabola is

y - 5 = 4(x - 2)
4x - y = 3
Then, the required area




Q.37. The area (in sq. units) of the region bounded by the parabola, y = x² + 2 and the lines, y = x + 1, x = 0 and x = 3, is:      (2019)
(1) 15/4
(2) 21/2
(3) 17/4
(4) 15/2
Ans. (4)
Solution.




Q.38. The area (in sq. units) of the region
     (2019)
(1) 53/6
(2) 8
(3) 59/6
(4) 26/3
Ans. (3)
Solution.
Since, the relation y ≤ x² + 3x represents the region below the parabola in the 1^st quadrant

∵ y = 4
⇒ x² + 3x = 4 ⇒ x = 1, -4
the required area = area of shaded region


Q.39. Let S(α) = {{x, y): y² ≤ x, 0 ≤ x ≤ α} and A(α) is area of the region S(α). If for a λ, 0 < λ < 4, A(λ) : A(4) = 2 : 5, then λ equals:     (2019)

Ans. (4)
Solution.





Q.40. The area (in sq. units) of the region A = {(x, y): x² ≤ y ≤ x + 2} is:     (2019)
(1) 10/3
(2) 9/2
(3) 31/6
(4) 13/6
Ans. (2)
Solution.

Required area is equal to the area under the curves y ≥ x² and y ≤ x + 2



Q.41. The area (in sq. units) of the region       (2019)
(1) 53/3
(2) 30
(3) 16
(4) 18
Ans. (4)
Solution.




Q.42. The region represented by |x - y| ≤ 2 and |x + y|≤ 2 is bounded by a:     (2019)
(1) square of side length 2√2 units
(2) rhombus of side length 2 units
(3) square of area 16 sq. units
(4) rhombus of area 8√2 sq. units
Ans. (1)
Solution.

By the diagram, region is square



Q.43. The area (in sq. units) of the region bounded by the curves y = 2^x and y = |x +1|, in the first quadrant is:     (2019)

(2) 3/2
(3) 1/2

Ans. (4)
Solution.




Q.44. If the area (in sq. units) of the region {(x, y) : y² ≤ 4x, x + y ≤ 1, x ≥ 0, y ≥ 0} is a √2 + b, then a - b is equal to:     (2019)
(1) 10/3
(2) 6
(3) 8/3
(4) -2/3
Ans. (2)
Solution. 
Consider y² = 4x and x + y = 1

Substituting x = 1 - y in the equation of parabola,
y² = 4(1 - y) ⇒ y² + 4y - 4 = 0
⇒ (y + 2)² = 8 ⇒ y + 2 = ±2√2
Hence, required area






Q.45. If the area (in sq. units) bounded by the parabola y² = 4λx and the line y = λx, λ > 0, is 1/9, then λ is equal to:       (2019)
(1) 2√6
(2) 48
(3) 24
(4) 4√3
Ans. (3)
Solution.
Given parabola y² = 4λx and the line y = λx

Q.46. The value of dx is:    (2018)
(1) π/8
(2) π/2
(3) 4π
(4) π/4
Ans. (4)
Solution.



Q.47. Let g(x) = cos x² ,f(x) = √x, and α ,β (α < β) be the roots of the quadratic equation 18x² - 9πx +π² = 0. Then the area (in sq. units) bounded by the curve  y = (gof)(x) and the lines x =α , x = β and y = 0 , is:    (2018)
(1)
(2)
(3)
(4)
Ans. (1)
Solution.
g(x) = cos x²
f(x) = √x
g(f (x)) = cos x


Q.48. The value of the integral    
      (2018)
(1) 3/4
(2) 0
(3)

(4)
Ans. (4)
Solution.



I = 3π/16

Q.49. The area (in sq. units) of the region {x ϵ R ∶ x ≥ 0, y ≥ x − 2 and y ≥ √x} is∶    (2018)
(1) 8/3
(2)
(3)
(4)
Ans. (2)
Solution.



Q.50. If the area of the region bounded by the curves, y = x², y = 1/x and the lines y = 0 and x = t (t > 1) is 1 sq. unit, then t is equal to:    (2018)
(1) e^2/3
(2) e^3/2
(3) 3/2
(4) 4/3
Ans. (1)
Solution.
The intersection point of y = x² and y = 1/x is (1, 1)
Area bounded by the curves is the region ABCDA is given as:




Q.51. The area (in sq. units) of the region {(x, y) : x ≥ 0, x + y ≤ 3, x² ≤ 4y and y ≤ 1 + √x} is     (2017)
(1) 5/2
(2) 59/12
(3) 3/2
(4) 7/3
Ans. (1)
Solution:

Area of shaded region

= 5/2 sq. unit

Q.52. The integral  is equal to    (2017)
(1) –1
(2) –2
(3) 2
(4) 4
Ans. (3)
Solution:







Q.53. 
      (2017)
(1) 13/256
(2) 15/64
(3) 13/32
(4) 15/128
Ans. (4)
Solution.


Q.54. The area (in sq. units) of the smaller portion enclosed between the curves, x² + y² = 4 and y² = 3x, is:    (2017)




Ans. (1)
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) : y² ≥ 2x and x² + y² ≤ 4x, x ≥ 0, y ≥ 0} is:     (2016)
(1)
(2)
(3)
(4)
Ans. (2)
Solution.
x² + y² ≤ 4x  &  y² ≥ 2x
To find point of intersection,
x² + y² = 4x ⇒ x² + 2x = 4x
⇒ x² - 2x = 0 ⇒ x (x - 2) = 0
⇒ x = 0  or  x = 2
∴ y = 0 or y = 2
Solve (x, y) = (0, 0) & (x, y) = (2, 2)

Area =


Q.57.
     (2016)
(1) log 2

(3) log 4
(4)
Ans. (1)
Solution.



Q.58. The area (in sq. units) of the region described by A = {(x, y) | y ≥ x² - 5x + 4, x + y ≥ 1, y ≤ 0} is:    (2016)
(1) 7/2
(2) 13/6
(3) 17/6
(4) 19/6
Ans. (4)
Solution.
A = {(x,y)|y ≥ x² - 5x + 4, x + y ≥ 1, y ≤ 0}
Here y ≥ x² - 5x + 4, x + y ≥ 1 , y ≤ 0



Q.59. For x ∈ R, x ≠ 0, if y(x) is a differentiable function such that  equals     (2016)
(where C is a constant)




Ans. (4)
Solution.





Q.60. The value of the integral   where [x] denotes the greatest integer less than or equal to x, is     (2016)
(1) 3
(2) 7
(3) 6
(4) 1/3
Ans. (1)
Solution.