Mathematics Test

Result

Mathematics Test - 30

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Question 1
3 / -1

Evaluate- $1^{2} + 2^{2} + 3^{2} + - - - - - - - + 20^{2}$ .

A
$3845$

B
$3500$

C
$2170$

D
$2870$

Solution

Given-

$1^{2} + 2^{2} + 3^{2} + - - - - - - - + 20^{2}$

The series shows the sum of the squares of first $20$ natural numbers.

According to the formula-

Sum of the squares of first $n$ natural numbers $S$ $= \frac{n (n + 1) (2 n + 1)}{6}$ where $n = 20$

$S = \frac{20 (20 + 1) (40 + 1)}{6}$

$S = \frac{20 \times 21 \times 41}{6}$

$S = 2870$
Question 2
3 / -1

A
g(x)

B
1 + g(x)

C
1 + xg(x)

D
None of these

Solution
Question 3
3 / -1

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 4
3 / -1

Three of six vertices of a regular hexagon are chosen at random. The probability that the triangle with these three vertices will be equilateral is

A
1/10

B
1/5

C
2/15

D
None of these

Solution

Let ABCDEF be a regular hexagon.

Three vertices out of its 6 vertices can be chosen in ⁶C₃ways. Therefore, 20 triangles can be made by joining three vertices at a time.

Out of these 20 triangles, only ACE and BDF are equilateral. Hence, the required probability = 1/10.
Question 5
3 / -1

What is the area of the figure bounded by the curves y² = 2x + 1 and x - y = 1?

A
32/3

B
4/3

C
8/3

D
16/3

Solution
Question 6
3 / -1

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 7
3 / -1

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
3 / -1

Find the value of the given integral.

$\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\tan x} \text{dx}$

A
$\frac{\pi}{4}$

B
$\frac{\pi}{2}$

C
$\frac{\pi}{6}$

D
None of these

Solution

The given integral is,

$I=\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\tan x} d x$ ----(1)

We know that, $\int_{0}^{a} \mathrm{f}(\mathrm{x}) \mathrm{dx}=\int_{0}^{\mathrm{a}} \mathrm{f}(\mathrm{a}-\mathrm{x}) \mathrm{d} \mathrm{x}$

By using the above property of definite integral, equation (1) becomes,

$\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\tan \left(\frac{\pi}{2}-\mathrm{x}\right)} \mathrm{dx}$

$\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\cot \mathrm{x}} \mathrm{dx}$ ----(2)

Adding equation (1) and (2), we get,

$\Rightarrow 2I=\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\tan x} d x+\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\cot x} d x$

$\Rightarrow 2I=\int_{0}^{\frac{\pi}{2}}\left[\frac{1}{1+\tan x}+\frac{1}{1+\cot x}\right] d x$

$\Rightarrow 2I=\int_{0}^{\frac{\pi}{2}}\left[\frac{1}{1+\tan x}+\frac{\tan x}{1+\tan x}\right] d x$

$\Rightarrow 2I=\int_{0}^{\frac{\pi}{2}}\left[\frac{1+\tan x}{1+\tan x}\right] d x$

$\Rightarrow 2I=\int_{0}^{\frac{\pi}{2}} 1 d x$

$\Rightarrow 2I=[x]_{0}^{\frac{\pi}{2}}$

$\Rightarrow 2I=\frac{\pi}{2}$

$\Rightarrow I=\frac{\pi}{4}$
Question 9
3 / -1

A triangle with vertices $(4,1),(1,1),(3,5)$ is a/an:

A
Isosceles and a right-angled triangle

B
Scalene triangle

C
Isosceles but not the right-angled triangle

D
Right-angled but not isosceles triangle

Solution

${A}(4,1), {B}(1,1)$ and ${C}(3,5)$ are the vertices of a triangle.

Then,

$\Rightarrow {AB}^{2}=(1-1)^{2}+(1-4)^{2}=9$

$\Rightarrow \mathrm{BC}^{2}=(5-1)^{2}+(3-1)^{2}=20$

$\Rightarrow A C^{2}=(5-1)^{2}+(3-4)^{2}=17$

Since, all 3 sides have different lengths, so it is a scalene triangle.
Question 10
3 / -1

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