11: The area bounded by the curve y = x (3 – x)2 , the x – axis and the ordinates of the maximum and minimum points of the curve is
(A) 2 sq. units
(B) 4 sq. units
(C) 3 sq. units
(D) 8 sq. units
Solution: y = x (3 – x)2 After solving , we get x =1 and x = 3 are points of maximum and minimum respectively. Now the shaded region is the required region
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12: The area enclosed by the curve |x + 1| + |y + 1| = 2 is
(A) 3 sq. units (B) 4 sq. units
(C) 5 sq. units (D) 8 sq. units
Solution: Shift the origin at the point (-1, -1) So |x + 1| + |y + 1| = 2 becomes |x| + | y | = 2 Hence required area is = 4 × 1/2 × 2 × 2 = 8sq. units |
13: The area common to the curves y = x3 and y = √x is
(A) 2 (B) 4
(C) 8 (D) None of these
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14: The area of the region consisting of points (x, y) satisfying |x ± y | < 2 and x2 + y2 > 2 is
(A) 8 – 2π sq. units
(B) 4 – 2π sq. units
(C) 1 – 2π sq. units
(D) 2π sq. units
Solution: Shaded region is the required one.
Required Area = 4 × 1/2 × 2 × 2 – π.2 = 8 – 2π sq. unit
15: ∫xo [sint] dt where x ∈ and [2nπ, (4n+1)]π, n ∈ N and [.] denotes the greatest integer function is equal to.
(A) -nπ (B) -(n +1)π
(C) -2nπ (D) -2(n +1)π
16: If f(π) = 2 and ∫πo [f(x) + f"(x)] sin x dx = 5 then f(0) is equal to, (it given that f(x) is continuous in [0, π])
(A) 7 (B) 3 (C) 5 (D) 1
=> f(π) + f(0) = 5 (given)
=> f(0) = 5 – f(π) = 5 – 2 = 3
17: Let f(x) is a continuous function for all real values of x and satisfies + a then value of ‘a’ is equal to
(A) –1/24 (B) 17/168 (C) 1/7 (D) None of these
Diff. both sides of (i) w. r. t. x we get;
f(x) = 0 – x2 f(x) + 2x15 + 2x5
18: . Then complete set of values of x for which f(x) <lnx is
(A) (0, 1] (B) [1, ∞)
(C) (0, ∞) (D) None of these
Solution:
Let g(x) = f(x) - ln(x). x ∈ R +
=> g ’(x) = f ’(x) – 1/x = ex – 1 / x > 0 ∀ x ∈ R+
=> g'(x) is increasing for ∀ x ∈ R+
g(1) = f(1) - ln1 = 0 – 0 = 0
=> g(x) > 0 ∀ x > 1 and g(x) > 0 ∀ x ∈ (0, 1]
=> ln x > f(x) ∀ x ∈ (0, 1]
19: , where [.] denotes the greatest integer function, and x ∈ R+, is equal to
(A) 1/In2 ([x] + 2[x]–1) (B) 1/In2 ([x] + 2[x])
(C) 1/In2 ([x] – 2[x]) (D) 1/In2 ([x] + 2[x]+1)
Solution: Let n < x < n + 1 where n ∈ I, I > 0
20: Let I1 = then
(A) I1 > I2 > I3 > I4 (B) I2 > I3 > I4 > I1
(C) I3 > I4 > I1 > I2 (D) I2 > I1 > I3 > I4
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