Mathematics Test

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Mathematics Test - 3

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Weekly Quiz Competition

Question 1
1 / -0

Find the value of the following integral.
\(\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x\)

A
\(\frac{\pi}{4}\)

B
\(\frac{\pi}{8}\)

C
\(\frac{\pi}{2}\)

D
\({\pi}\)

Solution

Let \(I=\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x\) ---- (1)
We know that,
\(\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\)
Using the above property of definite integrals in equation (1),
\(\Rightarrow I=\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\cos \left(\frac{\pi}{2}-x\right)}}{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}+\sqrt{\cos \left(\frac{\pi}{2}-x\right)}} d x\)
\(\Rightarrow I=\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x\) ---- (2)
Now adding equation (1) and (2),
\(\Rightarrow 2 I=\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\cos x}+\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x\)
\(\Rightarrow 2 I=\int_{0}^{\frac{\pi}{2}} 1 d x\)
\(\Rightarrow 2 I=[x]^{\frac{\pi}{2}}_{0}\)
\(\Rightarrow 2 I=\frac{\pi}{2}-0\)
\(\therefore I=\frac{\pi}{4}\)
Question 2
1 / -0

The area bounded by the curve y = 2x - x²and the straight line y = - x is given by

A
9/2

B
43/6

C
35/6

D
None of these

Solution
Question 3
1 / -0

The area of the region bounded by y = | x - 1 | and y = 1 is

A
1

B
2

C
1/2

D
None of these

Solution
Question 4
1 / -0

The area bounded by the curve y = x | x |, x-axis and the ordinates x = 1, x = -1 is given by

A
0

B
1/3

C
2/3

D
None of these

Solution
Question 5
1 / -0

The area lying in the first quadrant and bounded by the circle x² + y² = 4, the line x =√3 y and the x-axis is

A
π

B
π/2

C
π/3

D
None of these

Solution
Question 6
1 / -0

The area under the curve y = sin 2x + cos 2x between x = 0 and x = π/4 is

A
2 sq. units

B
1 sq. unit

C
3 sq. units

D
4 sq. units

Solution
Question 7
1 / -0

The area of the region bounded by the curve y² = 4x, y-axis and the line y = 3 is

A
2 sq. units

B
9/4 sq. units

C
6 √3sq. units

D
None of these

Solution
Question 8
1 / -0

The area of the figure bounded by y = sin x, y = cos x in the first quadrant is

A
2( √2- 1)

B
(2 √2+ 1) / 7

C
(2 √2- 1) / 7

D
None of these

Solution
Question 9
1 / -0

The area bounded by curve xy = c and x-axis between x = 1 and x = 4, is

A
c log 3 sq. units

B
2 log c sq. units

C
2c log 2 sq. units

D
2c log 5 sq. units

Solution
Question 10
1 / -0

If n(U) = 20, n(A) = 12, n(B) = 9, n(AB) = 4, where U is the universal set, A and B are subsets of U, then n | (A U B)^o| is equal to

A
17

B
9

C
11

D
3

Solution
Question 11
1 / -0

Period of cot 3 x - cos (4x + 3) is

A
π/3

B
π/4

C
π

D
π/2

Solution

Period of cot 3x is π/3 and period of cos (4x + 3) is π/2
⇒ L.C.M. is π, hence the answer.
Question 12
1 / -0

sin 30° + cos 60° + tan 45° is equal to

A
1

B
2

C
3

D
4

Solution

sin 30° + cos 60° + tan 45°
= (1/2) + (1/2) + 1
= 2
Question 13
1 / -0

The value of sin 10° + sin 20° + sin 30° + ........+ sin 360° is

A
1

B
0

C
-1

D
None of these

Solution

Since, sin 190° = - sin 10°, sin 200° = - sin 20°,
sin 210° = - sin 30°, sin 360° = sin 180° = 0 etc.
Hence all the terms in the expression cancels out, therefore answer is 0.
Question 14
1 / -0

α and β lie between 0 and π/4, cos(α + β) = 12/13 and sin(α - β) = 3/5.
sin 2α =

A
-16/65

B
0

C
56/65

D
64/65

Solution
Question 15
1 / -0

(tan 3x - tan 2x - tan x) is equal to

A
tan x tan 2x tan 3x

B
-tan x + tan 2x tan 3x

C
tan x tan x tan 3x - tan 2x tan 3x

D
None of these

Solution