Mathematics Test

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Mathematics Test - 22

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Weekly Quiz Competition

Question 1
1 / -0

Find the coordinates of the centre of the circle passing through the points (0, 0), (–2, 1) and (–3, 2).

A
$(2, \frac{1}{3})$

B
$(3, \frac{11}{2})$

C
$(\frac{9}{7}, \frac{12}{7})$

D
$(\frac{3}{2}, \frac{11}{2})$

Solution
Question 2
1 / -0

Find the unit digit in $7^{65} \times 6^{41} \times 3^{57}$ =?

A
6

B
7

C
5

D
8

Solution

Unit digit in $(7)^{65}$
$=$ Unit digit in $7^{4 \times 16+1}$
$=$ Unit digit in $7^{1}=7$ Unit digit in $6^{41}=6\left[\because \text { unit digit in }(6)^{n}=6\right]$
Unit digit in $3^{57}$
$=$ Unit digit in $3^{4 \times 14+1}$
$=$ Unit digit is $3^{1}=3$ Thus, unit digit in $7^{65} \times 6^{41} \times 3^{57}$
$=$ Unit digit in $7 \times 6 \times 3$
$=$ Unit digit in $126=6$

Unit digit in $(7)^{65}$
$=$ Unit digit in $7^{4 \times 16+1}$
$=$ Unit digit in $7^{1}=7$ Unit digit in $6^{41}=6\left[\because \text { unit digit in }(6)^{n}=6\right]$
Unit digit in $3^{57}$
$=$ Unit digit in $3^{4 \times 14+1}$
$=$ Unit digit is $3^{1}=3$ Thus, unit digit in $7^{65} \times 6^{41} \times 3^{57}$
$=$ Unit digit in $7 \times 6 \times 3$
$=$ Unit digit in $126=6$
Question 3
1 / -0

$What is \int \frac{\sin \sqrt{x}}{\sqrt{x}} d x is equal to?$

A
$\frac{\cos \sqrt{x}}{2} + c$

B
$2 \cos \sqrt{x} + c$

C
$\frac{- \cos \sqrt{x}}{2} + c$

D
$- 2 \cos \sqrt{x} + c$

Solution

$\begin{array}{l} Let I = \int \frac{\sin \sqrt{x}}{\sqrt{x}} d x \\ Put t = \sqrt{x} \Rightarrow d t = \frac{1}{2 \sqrt{x}} d x \\ I = \int 2 \sin t d t = - 2 \cos t + c \\ = - 2 \cos \sqrt{x} + c \end{array}$
Question 4
1 / -0

For a frequency distribution, mean, median and mode are connected by the relation?

A
Mode = 3 Mean – 2 Median

B
Mode = 2 Median – 3 Mean

C
Mode = 3 Median –2 Mean

D
Mode = 3 median + 2 Mean

Solution

Mode = 3 Median –2 Mean

Hence correct answer is option C
Question 5
1 / -0

What is $\int_{0}^{1} x e^{x} d x$ equal to?

A
1

B
-1

C
0

D
E

Solution

Let $I=\int_{0}^{1} \begin{array}{rl}x & e^{x} \\ \text { I } & \text { II }\end{array} d x$
Using integration by parts
$I=\left[x . e^{x}\right]_{0}^{1}-\int_{0}^{1} 1 . e^{x} d x$
$=(1 . e-0)-\left[e^{x}\right]_{0}^{1}$
$=1$
Let $I=\int_{0}^{1} \begin{array}{cc}x & e^{x} \\ \text { I } & \text { II }\end{array} d x$
Using integration by parts
$I=\left[x . e^{x}\right]_{0}^{1}-\int_{0}^{1} 1 . e^{x} d x$
$=(1 . e-0)-\left[e^{x}\right]_{0}^{1}$
$=1$
Question 6
1 / -0

If ∆ABC ~ ∆DEF such that AB = 1.2 cm & DE = 1.4cm. Find the ratio of areas of ∆ABC & ∆DEF.

A
6 : 7

B
16 : 81

C
36 : 49

D
4 : 9

Solution

We know that the ratio of areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.

$\begin{array}{l} ∴ \frac{area (Δ A B C)}{area (Δ D E F)} = \frac{A B^{2}}{D E^{2}} \\ \Rightarrow \frac{area (Δ A B C)}{area (Δ D E F)} = \frac{(1.2)^{2}}{(1.4)^{2}} = {(\frac{12}{14})}^{2} = \frac{36}{49} \end{array}$
Question 7
1 / -0

A circle is inscribed in a ∆ABC having sides 8 cm, 10cm and 12cm as shown in figure. Find AD, BE & CF.

A
4cm, 2cm, 3cm

B
7cm, 5cm, 3cm

C
3cm, 2cm, 1cm.

D
7cm, 6cm, 3cm

Solution
Question 8
1 / -0

The ratio in which the line joining (2, 4, 5), (3, 5, –4) is divided by the yz plane is?

A
2 : 3

B
3 : 2

C
–2 : 3

D
4 : –3

Solution
Question 9
1 / -0

The sides of a triangle are 6.5 cm, 10 cm & x cm, where x is a positive number. What is the smallest possible value of x among the following?

A
3.5

B
4

C
4.5

D
2.8

Solution

A triangle can be formed only when sum of two sides of triangle is greater than the third side. Hence the smallest possible value of x will be 4.

i.e. 6.5 + 10 > 4

6.5 + 4 > 10

4 + 10 > 6.5
Question 10
1 / -0

If $\alpha \& \beta$ are the zeros of the polynomial $f(x)=x^{2}-5 x+k$ such that $\alpha-\beta=1,$ Find the value of $k$.

A
5

B
4

C
6

D
7

Solution

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Mathematics Test - 22

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Mathematics Test - 22

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Question 1

2,13

3,112

97,127

32,112

Solution

Question 2

6

7

5

8

Solution

Question 3

cosx2+c

2cosx+c

-cosx2+c

-2cosx+c

Solution

Question 4

Mode = 3 Mean – 2 Median

Mode = 2 Median – 3 Mean

Mode = 3 Median –2 Mean

Mode = 3 median + 2 Mean

Solution

Question 5

1

-1

0

E

Solution

Question 6

6 : 7

16 : 81

36 : 49

4 : 9

Solution

Question 7

4cm, 2cm, 3cm

7cm, 5cm, 3cm

3cm, 2cm, 1cm.

7cm, 6cm, 3cm

Solution

Question 8

2 : 3

3 : 2

–2 : 3

4 : –3

Solution

Question 9

3.5

4

4.5

2.8

Solution

Question 10

5

4

6

7

Solution

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$(2, \frac{1}{3})$

$(3, \frac{11}{2})$

$(\frac{9}{7}, \frac{12}{7})$

$(\frac{3}{2}, \frac{11}{2})$

$\frac{\cos \sqrt{x}}{2} + c$

$2 \cos \sqrt{x} + c$

$\frac{- \cos \sqrt{x}}{2} + c$

$- 2 \cos \sqrt{x} + c$