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Objective Problems of Definite Problems

OBJECTIVE PROBLEMS

1:   1367_integration.JPG

(A)          0

(B)          100

(C)          200

(D)          400

Solution :          

                                  2141_integration.JPG

                                    = –200 (cos x)π0 = –200 (–1–1) = 400

2:                   2–2 min (x–[x], – x – [–x])dx,   where [.] is the greatest integer function =

                            (A)              0                                                              (B)          1

                           (C)               2                                                              (D)          None of these.

Solution :                           Let I = 2–2 min (x – [x], – x – x [–x])dx

                                                 = 2 2–2 min (x – [x], – x – [–x])dx (function is even)

                                  2382_integration.JPG

                               2410_integration.JPG

                                                = 1/4 + 1/4 + 1/4 + 1/4 = 1

                                                Hence I = 1

3:                               (where [.] denotes greatest integer function) =

                                (A)          2-√2                                                   (B)          2+√2

                                (C)          2√2                                                      (D)          √2

Solution :           1014_integration.JPG

                         56_integration.JPG 

                          = (√2 – 1) + 2 (3/2 – √2) = √2 – 1 + 3 – 2 √2 = 2 + √2 – 2 √2 = 2 – √2

4:                     Qp/20 sin 2 x in tan x dx = 

                                (A)          2                                                              (B)          0

                                (C)          1                                                              (D)          None of these.

Solution :           I = ∫π/20 sin 2x ln (tan x) dx                                                                             … (1)

                                                I = π/20 sin 2 (π/2 – x) ln tan (π/2 – x)dx

                                                I = π/20 sin 2x ln (cot x) dx                                                                              … (2)

                                                Adding (1) and (2)

                                                2I = π/20 sin 2x [ln (tan x) + ln (cot x)] dx = π/20 sin 2x ln 1 dx = 0.

  
5:                        O21/2 1/x sin e/e x – 1/x o/o dx =

                                (A)          0                                                              (B)          1

                                (C)          2                                                              (D)          √2

Solution :                           Let I = 2044_integration.JPG

                                                Put z = x - 1/x

                                                Then I = 2490_integration.JPG is odd function

  
6:                              The value of π/20 √sin x / √sin x + √cos x is

                                (A)          p/2                                                       (B)          p/3

                                (C)        π/4                                                       (D)          p

Solution :                           Let I = π/20 √sin x / √sin x + √cos x dx

                                                Here the lower limit is zero hence , we can replace   x by (a – x) i.e. by p/2 –x

                              1254_integration.JPG

7:                                     1052_integration.JPG  =

                                (A)       3/p log 2                                       (B)          2/p log 2

                                (C)      3/p log 3                                      (D)            3/p

Solution :           1356_integration.JPG

                                                r/n = x, 1/n = dx

                         2221_integration.JPG 
8:                                    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)          6 sq. units

                                (C)          4 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

                                 1070_integration.JPG

                                 209_ordinates of the maximum and minimum points.JPG 
9:                                    What is the area of a plane figure bounded by the points of the lines max (x, y) = 1 and x2 + y2 = 1 ?

                                (A)          p/2 sq. units                                        (B)          p/3 sq. units

                                (C)          p/4  sq. units                                        (D)          p sq. units

Solution :                   By definition the lines max, (x, y) = 1 means.

                                                x = 1 and y < 1 or y = 1 and x < 1

                                                Required area

                                         1160_definition the lines max.JPG

                                          1397_integration.JPG                                         

10:             The area bounded by the curve y = (x –1) (x – 2) (x – 3) lying between the ordinates x = 0 and x = 3 is

                            (A)              7/4 sq. units                                         (B)          4 sq. units

                           (C)             11/ 4 sq. units                                       (D)          3 sq. units

78_integration.JPG 

                   = 9 / 4 + 1/ 4 + 1/ 4 = 11/4  sq units.

                1741_lying between the ordinates.JPG

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