Mathematics Test

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Mathematics Test - 21

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Weekly Quiz Competition

Question 1
1 / -0

$If \frac{a^{n + 1} + b^{n + 1}}{a^{n} + b^{n}}$ is the A.M between a and b. Then, find the value of $n$

A
1

B
2

C
0

D
3

Solution

We have,
A. M between a and b = $\frac{a+b}{2}$
It is given that $\frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}$ is the A.M between a & b
$\therefore \frac{a^{n+2}+b^{n}+1}{a^{n}+b^{n}}=\frac{a+b}{2}$
$\Rightarrow 2\left(a^{n+1}+b^{n+1}\right)=\left(a^{n}+b^{n}\right)(a+b)$
$\Rightarrow 2 a^{n+1}+2 b^{n+1}=a^{n+1}+a b^{n}+a^{n} b+b^{n}+1$
$\Rightarrow a^{n+1}+b^{n+1}=a b^{n}+a^{n} b$
$\Rightarrow a^{n+1}-a^{n} b=a b^{n}-b^{n+1}$
$\Rightarrow a^{n}(a-b)=b^{n}(a-b)$
$\Rightarrow a^{n}=b^{n}$
$\Rightarrow \frac{a^{n}}{b^{n}}=1$
$\Rightarrow\left(\frac{a}{b}\right)^{n}=\left(\frac{a}{b}\right)^{0}$
$\Rightarrow n=0$
Question 2
1 / -0

Find the condition that the point (x, y) may lie on the line joining (3, 4) and (–5, 6).

A
5x–4y+1=0

B
2x+3y+6=0

C
4x+5y–3=0

D
3x–2y–5=0

Solution
Question 3
1 / -0

The equivalent discount with respect to two successive discounts 30% and 40% will be?

A
74% discount

B
70% discount

C
58% discount

D
66% discount

Solution

$\begin{array}{l} = 100 - \frac{(100 \times d_{1}) \times (100 - d_{2})}{100} \\ = 100 - \frac{(100 - 30) (100 - 40)}{100} \\ = 100 - \frac{70 \times 60}{100} \\ = 100 - 42 \\ = 58 % \end{array}$
Question 4
1 / -0

The maximum value of $\frac{(\log x)}{x}$ is?

A
1

B
2/e

C
e

D
1/e

Solution
Question 5
1 / -0

Find the values of $k$ for which $2 x^{2}-10 x+k=0$
has real and equal roots:

A
25

B
50

C
25/2

D
20

Solution

The given equation is 2x2−10x+k=0
Here a=2,b=−10 and c=k
$\begin{array}{l} ∴ D = b^{2} - 4 a c \\ = (- 10)^{2} - 4 \times 2 \times k \\ = 100 - 8 k \end{array}$
The given equation will have real and equal roots, if D=0
$\begin{array}{l} \Rightarrow 100 - 8 k = 0 \\ \Rightarrow k = \frac{100}{8} = \frac{25}{2} \end{array}$
Question 6
1 / -0

If the 8th term of an A.P is 31 and the 15th term is 16 more than the 11th term, find the A.P.

A
2, 5, 8, 11, 14, 17, -----

B
3, 6, 9, 12, 15, --------

C
2, 6, 10, 14, 18, ---------

D
3, 7, 11, 15, 19, --------

Solution

Let a be the first term and $d$ be the common difference of the A.P. We have,
$a_{a}=31$ and $a_{15}=16+a_{12}$
$\Rightarrow a+7 d=31$ and $a+14 d=16+a+10 d$
$\Rightarrow a+7 d=31$ and $4 d=16$
$\Rightarrow a+7 d=31$ and $d=4$
$\Rightarrow a+7 \times 4=31 \Rightarrow a+28=31 \Rightarrow a=3$
Hence the A.P. Is $a, a+d, a+2 d, a+3 d$ L.e. $3,7,11,15,19, \ldots$
Question 7
1 / -0

$If (1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots . + C_{n} x^{n}, C_{0} + \frac{c_{2}}{2} + \frac{c_{2}}{3} + \dots \dots . + \frac{C_{n}}{n + 1} is equal to$ ?

A
$2^{n + 1}$

B
$\frac{2^{n + 1} - 1}{n + 1}$

C
$\frac{2^{n + 1}}{n + 1}$

D
$2^{n + 1} - 1$

Solution
Question 8
1 / -0

Direction : If f (–x) = –f (x), then f (x) is an odd function & if f (–x) = f (x) then f (x) is an even function.
$Also \int_{- a}^{a} f (x) d x = \{\begin{array}{cl} 2 \int_{0}^{a} f (x), & if f (x) is even \\ 0 & if f (x) is odd \end{array}$
The value of the integral $\int_{-\pi / 2}^{\pi / 2} \sin ^{7} x d x$ is

A
0

B
$π / 2$

C
$- π / 2$

D
1

Solution

$I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \sin^{7} x d x$
$Here f^{} (x) = \sin^{7} x$
$\begin{array}{l} f (- x) = \sin^{7} (- x) = [- \sin x]^{7} = - \sin^{7} x \\ ∴ f (- x) = - f (x) \end{array}$
So is an odd function

$∴ \int_{- π / 2}^{π / 2} \sin^{7} x d x = 0$
Question 9
1 / -0

In the given figure, there are three semicircles A, B and C having diameter 3 cm each and another semi circle E having a circle D with diameter 4.5 cm are shown. Calculate the area of the shaded region.

A
$23.645 {cm}^{2}$

B
$10.01 {cm}^{2}$

C
$51.066 {cm}^{2}$

D
$12.375 {cm}^{2}$

Solution
Question 10
1 / -0

What is the value of $\int (\frac{1}{\cos^{2} x} - \frac{1}{\sin^{2} x}) d x ?$

A
$2 cosec 2 x + c$

B
$- 2 \cot 2 x + c$

C
$2 \sec 2 x + c$

D
$- 2 \tan 2 x + c$

Solution

$\begin{array}{l} \int (\frac{1}{\cos^{2} x} - \frac{1}{\sin^{2} x}) d x \\ = \int (\sec^{2} x - {cosec}^{2} x) d x \\ = \tan x + \cot x + C \\ = (\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}) + C \\ = (\frac{1}{\sin x \cdot \cos x}) + C \\ = \frac{2}{\sin 2 x} + C = 2 cosec 2 x + C \end{array}$

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Mathematics Test - 21

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Mathematics Test - 21

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SHARING IS CARING

Question 1

1

2

0

3

Solution

Question 2

5x–4y+1=0

2x+3y+6=0

4x+5y–3=0

3x–2y–5=0

Solution

Question 3

74% discount

70% discount

58% discount

66% discount

Solution

Question 4

1

2/e

e

1/e

Solution

Question 5

25

50

25/2

20

Solution

Question 6

2, 5, 8, 11, 14, 17, -----

3, 6, 9, 12, 15, --------

2, 6, 10, 14, 18, ---------

3, 7, 11, 15, 19, --------

Solution

Question 7

2n+1

2n+1-1n+1

2n+1n+1

2n+1-1

Solution

Question 8

0

π/2

-π/2

1

Solution

Question 9

23.645cm2

10.01cm2

51.066cm2

12.375cm2

Solution

Question 10

2cosec2x+c

-2cot2x+c

2sec2x+c

-2tan2x+c

Solution

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$2^{n + 1}$

$\frac{2^{n + 1} - 1}{n + 1}$

$\frac{2^{n + 1}}{n + 1}$

$2^{n + 1} - 1$

$π / 2$

$- π / 2$

$23.645 {cm}^{2}$

$10.01 {cm}^{2}$

$51.066 {cm}^{2}$

$12.375 {cm}^{2}$

$2 cosec 2 x + c$

$- 2 \cot 2 x + c$

$2 \sec 2 x + c$

$- 2 \tan 2 x + c$