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USE CODE: CHEM20

```21:       Area bounded by y = g(x), x-axis and the lines x = -2, x = 3, where

and f(x) = x2 -  , is equal to

(A)  113/24 sq.units                       (B)         111/24 sq.units

(C)  117/24 sq.units                       (D)         121/24 sq,units

22:                  Area of the region which consists of all the points satisfying the conditions |x–y| + |x+y| < 8 and xy > 2, is equal to

(A)      4(7 – ln8)sq. units               (B)      4(9 – ln8)sq. units

(C)      2(7 – ln8)sq. units               (D)      2(9 – ln8)sq. units

Solution:       The expression |x–y| + |x+y| < 8, represents the interior region of the square formed by the lines x = ± 4, y = ± 4 and xy > 2. represents the region lying inside the hyperbola xy =2

Required area

= 4(7–3 In2) sq.units

= 4(7 – In8) sq.units

23:                  Area bounded by the parabola (y - 2)2 = x – 1, the tangent to it at the point P (2, 3) and the x-axis is equal to

(A)      9 sq. units                             (B)      6 sq. units

(C)      3 sq. units                             (D)      None of these

Solution:       (y - 2)2 = (x - 1) => 2(y - 2).  = 1

=> dy/dx = 1/2(y–2)

Thus equation of tangent at P(2, 3) is,

(y – 3) = 1/2 (x–2) i.e., x = 2y – 4

Required area

= ((y–2)3/3 – y2 + 5y)30 = 9 sq. units

24:                  Two lines draw through the point P(4, 0) divide the area bounded by the curves y = √2 πx/4 and x – axis, between the linea x = 2, and x = 4, in to three equal parts. Sum of the slopes of the drawn lines is equal to

(A) –2 2/π                                     (B)      –√2/π

(C) –√2/π                                     (D)      –4√2/π

Solution:       Area bounded by y = √2 .sin πx/4 and x-axis between the lines x = 2 and x = 4,

Let the drawn lines are L1: y – m1(x - 4) = 0 and L2: y – m2(x - 4) = 0, meeting the line x = 2 at the points A and B respectively Clearly A = (2, - 2m1); B= (2, -2m2)

25:                  If A =  is equal to

(A)   1/π+2 – A                               (B)      1/2 + 1/π+2 – A

(C)   1/2 – 1/π+2 – A                     (D)      1/2 + 1/π+2 + A

26.                   The value of the integral  is

(A) 1                                                                         (B) π/12

(C) π/6                                                                     (D) none of these

Solution:       Using the property  f (a + b – x) dx, the given integral

Hence (B) is the correct answer.

27.                   If I =   dx then

(A) 0                                                               (B) 2

(C)  π/2                                                         (D) 2 – π/2

Hence (D) is the correct answer.

28.                   If I =  , then

(A) 0 < I < 1                                                   (B) I > π/2

(C) I < √2π                                                     (D) I > 2 π

Solution:       Since x ∈ [0, π/2]  => 1 < 1 + sin3 x < 2

Hence (C) is the correct answer.

29.                   If f (a + b –x) = f (x) then ∫ba x f (x) dx is equal to

(A) a–b/2 ∫ba f(x) dx                                    (B) (a+b/2) ∫ba f(x) d x

(C) 0                                                              (D) none of these

Solution:        I = ∫ba x f (x) dx = ∫ba (a+b-x) f (a +b-x) dx

= (a + b) ∫ba  f(a +b –x) - ∫ba x f (a + b –x) dx

= ( a + b) ∫ba f (x) dx - ∫ba x f (x) d x

Hence I = (a+b/2) ∫ba f(x) dx.

Hence (B) is the correct answer.

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USE CODE: CHEM20