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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)
Ans. (4)
The given equation of circle is
x^{2} + y^{2} = 2.... (1)
Now, total area enclosed by the circle is πr^{2} = π(√2)^{2} = 2π sq. units
The area bounded by the parabola y^{2 }= 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 = 4x^{2} and y = 8x + 12, we get
4x^{2}  8x  12 = 0 ⇒ x^{2}  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 C_{1} : y^{2} = ax and C_{2 }: x^{2} = 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, C_{1} and C_{2}, and the area of then ‘a’ satisfies the equation: (2020)
(1) x^{6}  6x^{3} + 4 = 0
(2) x^{6}  12x^{3} + 4 = 0
(3) x^{6} + 6x^{3}  4 = 0
(4) x^{6}  12x^{3}  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
x^{2} < y... (1)
2x + y < 3... (2)
On solving y = x^{2} = 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_{e}2
(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 + cx^{2} ; 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) 2π^{2}
(3) π^{2}
(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)≤ 2x  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 tan^{5}x = 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_{e}3
(2) log_{e}e
(3) log_{e}2
(4) log_{e}1
Ans. (4)
Solution.
Then, equation (1) becomes,
Q.24. , where g is a nonzero 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 = x^{2}  1, the tangent at the point (2, 3) to it and the yaxis 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 Yaxis
∴ 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 = kx^{2} and x = ky^{2}, (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 x^{2} = 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
x^{2}  x  2 = 0
x = 2,  1
Q.36. The area (in sq. units) in the first quadrant bounded by the parabola,y = x^{2} + 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 x^{2} = 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 = x^{2} + 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 ≤ x^{2} + 3x represents the region below the parabola in the 1^{st} quadrant
∵ y = 4
⇒ x^{2} + 3x = 4 ⇒ x = 1, 4
the required area = area of shaded region
Q.39. Let S(α) = {{x, y): y^{2} ≤ 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): x^{2} ≤ 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 ≥ x^{2} 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) : y^{2} ≤ 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 y^{2} = 4x and x + y = 1
Substituting x = 1  y in the equation of parabola,
y^{2} = 4(1  y) ⇒ y^{2} + 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 y^{2} = 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 y^{2} = 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 x^{2} ,f(x) = √x, and α ,β (α < β) be the roots of the quadratic equation 18x^{2}  9πx +π^{2} = 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 x^{2}
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 = x^{2}, 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 = x^{2} 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, x^{2} ≤ 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, x^{2} + y^{2 }= 4 and y^{2} = 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) : y^{2} ≥ 2x and x^{2} + y^{2} ≤ 4x, x ≥ 0, y ≥ 0} is: (2016)
(1)
(2)
(3)
(4)
Ans. (2)
Solution.
x^{2} + y^{2} ≤ 4x & y^{2} ≥ 2x
To find point of intersection,
x^{2} + y^{2} = 4x ⇒ x^{2} + 2x = 4x
⇒ x^{2}  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 ≥ x^{2}  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 ≥ x^{2}  5x + 4, x + y ≥ 1, y ≤ 0}
Here y ≥ x^{2}  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.
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