Mathematics Test

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Mathematics Test - 52

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Question 1
1 / -0

The value of the expression: sin²1° + sin²11° + sin²21° + sin²31° + sin²41° + sin²49° + sin²59° + sin²69° + sin²79° + sin²89° is?

A
0

B
5(1/2)

C
4(1/2)

D
5

Solution

$\sin ^{2} 1^{\circ}+\sin ^{2} 11^{\circ}+\sin ^{2} 21^{\circ}+\sin ^{2} 31^{\circ}+\sin ^{2} 41^{\circ}+\sin ^{2} 45^{\circ}+\sin ^{2} 49^{\circ}+\sin ^{2} 59^{\circ}+$
$=\sin ^{2} 1^{\circ}+\sin ^{2} 11^{\circ}+\sin ^{2} 21^{\circ}+\sin ^{2} 31^{\circ}+\sin ^{2} 41^{\circ}+\sin ^{2} 45^{\circ}+\sin ^{2} 49^{\circ}+\sin ^{2} 59^{\circ}$
$=\left(\sin ^{2} 1^{\circ}+\sin ^{2} 89^{\circ}\right)+\left(\sin ^{2} 11^{\circ}+\sin ^{2} 79^{\circ}\right)+\left(\sin ^{2} 21^{\circ}+\sin ^{2} 69^{\circ}\right)+\left(\sin ^{2} 31^{\circ}+\right.$
$=1+1+1+1+1+\frac{1}{2}$
$\left[\sin ^{2} \mathrm{A}+\sin ^{2} \mathrm{B}=1 . \text { If }, \mathrm{A}+\mathrm{B}=90^{\circ}\right]$
$=5 \frac{1}{2}$
Question 2
1 / -0

How many terms are there in 20, 25, 30......... 140

A
22

B
25

C
23

D
24

Solution

Number of terms, $=\left\{\frac{\left(1^{\text {st}} \text {term }-\text { last term }\right)}{\text { common difference }}\right\}+1$ $=\left(140-\frac{20}{5}\right)+1$
$=\left(\frac{120}{5}\right)+1$
$=24+1$
$=25$
Question 3
1 / -0

How many terms are there in the GP 5, 20, 80, 320........... 20480?

A
5

B
6

C
8

D
7

Solution

Common ratio, $r=\frac{20}{5}=4$
Last term or $n^{\text {th }}$ term of $G P=a r^{n-1}$
$20480=5 \times\left(4^{n-1}\right)$
Or, $4^{n-1}=\frac{20480}{5}=4^{8}$
So, comparing the power, Thus, $n-1=8$
Or, $n=7$
Number of terms $=7$
Question 4
1 / -0

If a + b + c = 9 (where a, b, c are real numbers) then the minimum value of a² + b² + c² is?

A
3

B
$3 \frac{1}{4}$

C
2

D
$2 \frac{1}{4}$

Solution

$a+b+c=9$
For minimum value
$a=b=c$
$\Rightarrow 3 a=9$
$\Rightarrow a=\frac{9}{3}$
$\Rightarrow a=3$
For minimum value
$a=b=c=3$
$\therefore a^{2}+b^{2}+c^{2}$
$\Rightarrow 3^{2}+3^{2}+3^{2}$
$\Rightarrow 9+9+9$
$\Rightarrow 27$
Question 5
1 / -0

If $y=\frac{1}{a-z},$ then $\frac{d z}{d y}=$

A
(z−a)²

B
−(z−a)²

C
(z+a)²

D
−(z+a)²

Solution

Here $z=a-\frac{1}{y}$ p $\frac{d z}{d y}=\frac{1}{y^{2}}=(a-z)^{2}$
Question 6
1 / -0

$\frac{d}{d x}\left[\cos \left(1-x^{2}\right)^{2}\right]=$

A

$-2 x\left(1-x^{2}\right) \sin \left(1-x^{2}\right)^{2}$

B

$-4 x\left(1-x^{2}\right) \sin \left(1-x^{2}\right)^{2}$

C

$4 x\left(1-x^{2}\right) \sin \left(1-x^{2}\right)^{2}$

D

$-2\left(1-x^{2}\right) \sin \left(1-x^{2}\right)^{2}$

Solution

$\frac{d}{d x}\left[\cos \left(1-x^{2}\right)^{2}\right]=-\sin \left(1-x^{2}\right)^{2} \frac{d}{d x}\left(1-x^{2}\right)^{2}$

$=4 x\left(1-x^{2}\right) \sin \left(1-x^{2}\right)^{2}$
Question 7
1 / -0

The average age of three boys is 15 years. If their ages are in ratio 3 : 5 : 7, the age of the youngest boy is

A
21 years

B
18 years

C
15 years

D
9 years

Solution

Sum of ages of three boys = 45 years
Now, (3x + 5x + 7x) = 45
⇒ 15x = 45
⇒ x = 3
So, age of youngest boy = 3x
= 3 × 3
= 9 years
Question 8
1 / -0

If log a/b+log b/a = log(a+b),then:

A
a + b = 1

B
a - b = 1

C
a = b

D
a² - b²= 1

Solution

$\log \frac{a}{b}+\log \frac{b}{a}=\log (a+b)$
$\Rightarrow \log (a+b)=\log \left(\frac{a}{b} \times \frac{b}{a}\right)=\log 1$
So$, a+b=1$
Question 9
1 / -0

If $y=1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\ldots \ldots \infty,$ then $\frac{d y}{d x}=$

A
y

B
y+xnn!

C
y−xnn!

D
y−1−xnn!

Solution

$y=1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\ldots \ldots \infty^{\mathrm{}} y=e^{x}$

respect to $x,$ we get $\frac{d y}{d x}=e^{x}=y$
Question 10
1 / -0

$\frac{d}{d x}\left(\tan ^{-1} \frac{\cos x}{1+\sin x}\right)=$

A
-1/2

B
1/2

C
-1

D
1

Solution

$\frac{d}{d x}\left[\tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)\right]$

$=\frac{d}{d x}\left[\tan ^{-1}\left(\frac{\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}}{\cos ^{2} \frac{x}{2}+\sin ^{2} \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}}\right)\right]$
$=\frac{d}{d x}\left[\tan ^{-1}\left(\frac{1-\tan \left(\frac{x}{2}\right)}{1+\tan \left(\frac{x}{2}\right)}\right)\right]=\frac{d}{d x}\left[\tan ^{-1} \tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\right]=-\frac{1}{2}$

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Re-Attempt Weekly Quiz Competition

Mathematics Test - 52

Result

Mathematics Test - 52

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SHARING IS CARING

Question 1

0

5(1/2)

4(1/2)

5

Solution

Question 2

22

25

23

24

Solution

Question 3

5

6

8

7

Solution

Question 4

3

314

2

214

Solution

Question 5

(z−a)2

−(z−a)2

(z+a)2

−(z+a)2

Solution

Question 6

\(-2 x\left(1-x^{2}\right) \sin \left(1-x^{2}\right)^{2}\)

\(-4 x\left(1-x^{2}\right) \sin \left(1-x^{2}\right)^{2}\)

\(4 x\left(1-x^{2}\right) \sin \left(1-x^{2}\right)^{2}\)

\(-2\left(1-x^{2}\right) \sin \left(1-x^{2}\right)^{2}\)

Solution

Question 7

21 years

18 years

15 years

9 years

Solution

Question 8

a + b = 1

a - b = 1

a = b

a2 - b2 = 1

Solution

Question 9

y

y+xnn!

y−xnn!

y−1−xnn!

Solution

Question 10

-1/2

1/2

-1

1

Solution

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$3 \frac{1}{4}$

$2 \frac{1}{4}$

(z−a)²

−(z−a)²

(z+a)²

−(z+a)²

a² - b²= 1