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Q1.
(A) 0
(B) 100
(C) 200
(D) 400
Solution
= –200 (cos x)π0 = –200 (–1–1) = 400
Q2. ∫2–2 min (x–[x], – x – [–x])dx, where [.] is the greatest integer function =
(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)
= 1/4 + 1/4 + 1/4 + 1/4 = 1
Hence I = 1
Q3. (where [.] denotes greatest integer function) =
(A) 2-√2
(B) 2+√2
(C) 2√2
(D) √2
= (√2 – 1) + 2 (3/2 – √2) = √2 – 1 + 3 – 2 √2 = 2 + √2 – 2 √2 = 2 – √2
Q4. Qp/20 sin 2 x in tan x dx =
(A) 2
(B) 0
(C) 1
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.
Q5. O21/2 1/x sin e/e x – 1/x o/o dx =
Let I =
Put z = x - 1/x
Then I = is odd function
Q6. The value of ∫π/20 √sin x / √sin x + √cos x is
(A) p/2
(B) p/3
(C) π/4
(D) p
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
Q7. =
(A) 3/p log 2
(B) 2/p log 2
(C) 3/p log 3
(D) 3/p
r/n = x, 1/n = dx
Q8. 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
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
Q9. 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
By definition the lines max, (x, y) = 1 means.
x = 1 and y < 1 or y = 1 and x < 1
Required area
Q10. 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
= 9 / 4 + 1/ 4 + 1/ 4 = 11/4 sq units.
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