Mathematics Test

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Mathematics Test - 64

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Weekly Quiz Competition

Question 1
1 / -0

If tan15° = 2 - √3,then the value of tan15° cot75° + tan75° cot15° is?

A
14

B
12

C
10

D
8

Solution

$\tan 15^{\circ} \cot 75^{\circ} + \tan 75^{\circ} \cot 15^{\circ}$
$= \tan 15^{\circ} \cot (90^{\circ} - 15^{\circ}) + \tan {(90^{\circ} - 15)}^{\circ} \cot 15^{\circ}$
$= \tan^{2} 15^{\circ} + \cot^{2} 15^{\circ}$
$= \tan^{2} 15^{\circ} + \cot^{2} 15^{\circ} \dots \dots (i)$
[Formula] $\cot (90^{\circ} - θ) = \tan θ$
$\tan (90^{\circ} - θ) = \cot θ$
Put value of $\tan 15^{\circ}$
$\cot 15^{\circ} = \frac{1}{\tan 15^{\circ}}$
$\cot 15^{\circ} = \frac{1}{(2 - \sqrt{3})}$
$\cot 15^{\circ} = \frac{1}{(2 - \sqrt{3})} \times \frac{(2 + \sqrt{3})}{(2 + \sqrt{3})}$
$\cot 15^{\circ} = 2 + \sqrt{3}$
Now put value in equation (i) $\tan 15^{\circ} + \cot 15^{\circ}$
$= (2 - \sqrt{3})^{2} + (2 + \sqrt{3})^{2}$
$= 4 + 3 - 4 \sqrt{3} + 4 + 3 + 4 \sqrt{3}$
$= 14$
Question 2
1 / -0

A square is drawn by joining the mid points of the sides of a given square in the same way and this process continues indefinitely. If a side of the first square is 4 cm, determine the sum of the areas all the square.

A
32 Cm²

B
16 Cm²

C
20 Cm²

D
64 Cm²

Solution

Side of the first square is $4 cm$ . side of second square $= 2 \sqrt{2} cm$ Side of third square $= 2 cm$ and so on, i.e. $4, 2, \sqrt{2}, \sqrt{2}, 1 \dots \dots$
Thus, area of these square will be $= 16, 8, 4, 2, 1, \frac{1}{2}$ Hence, Sum of the area of first, second, third square $= 16 + 8 + 4 + 2 + 1 + \dots \dots$
$= [\frac{16}{{1 - (\frac{1}{2})}}]$
$= 32 {cm}^{2}$
Question 3
1 / -0

The 7^thand 21^stterms of an AP are 6 and -22 respectively. Find the 26th term

A
-34

B
-32

C
-12

D
-10

Solution

7^th term = 6
21^st term = -22
That means, 14 times common difference or -28 is added to 6 to get -22
Thus, d = -2
7^st term = 6 = a + 6d
Or, a + (6 × -2) = 6
Or, a = 18
26^st term = a + 25d = 18 -25 × 2 = -32
Question 4
1 / -0

If $\frac{a}{b} = \frac{4}{5}$ and $\frac{b}{c} = \frac{15}{16},$ then $\frac{18 c^{2} - 7 a^{2}}{45 c^{2} + 20 a^{2}}$ is equal to?

A
1/3

B
2/5

C
3/4

D
1/4

Solution

$\frac{a}{b} = \frac{4}{5}$ and $\frac{b}{c} = \frac{15}{16}$
$\Rightarrow \frac{a}{b} \times \frac{b}{c} = \frac{4}{5} \times \frac{15}{16}$
$\Rightarrow \frac{a}{c} = \frac{3}{4}$
$∴ \frac{18 c^{2} - 7 a^{2}}{45 c^{2} + 20 a^{2}}$
$= \frac{c^{2} (18 - 7 \frac{a^{2}}{c^{2}})}{c^{2} (45 + 20 \frac{a^{2}}{c^{2}})}$
$= \frac{18 - 7 {(\frac{a}{c})}^{2}}{45 + 20 {(\frac{a}{c})}^{2}}$
$= \frac{18 - 7 \times \frac{9}{16}}{45 + 20 \times \frac{9}{16}}$
$= \frac{18 - \frac{63}{16}}{45 + \frac{45}{4}}$
$= \frac{225 \times 4}{16 \times 225}$
$= \frac{1}{4}$
Question 5
1 / -0

(d/dx)log|x| =......,(x≠0)

A
1/x

B
-1/x

C
x

D
-x

Solution

$\log | x | = \log x,$ if $x > 0 = \log (- x)$ , if $x < 0$

Hence $\frac{d}{d x} {\log | x |} = \frac{1}{x}$ if $x > 0$

$= (\frac{1}{- x}) (- 1) = \frac{1}{x}$ if $x < 0$ Thus

$\frac{d}{d x} {\log | x |} = \frac{1}{x},$ if $x \neq 0$
Question 6
1 / -0

If y = 3x⁵+ 4x⁴+ 2x + 3, then

A
y₄=0

B
y₅=0

C
y₆=0

D
None of these

Solution

Since highest power of x is 5, therefore y₆= 0.
Question 7
1 / -0

Students of three different classes appeared in common examination. Pass average of 10 students of first class was 70%, pass average of 15 students of second class was 60% and pass average of 25 students of third class was 80% then what will be the pass average of all students of three classes?

A
74%

B
75%

C
69%

D
72%

Solution

Sum of pass student first, second and third class,
=(70% of 10)+(60% of 15)+(80% of 25)
=7+9+20=36
Total students appeared,
=10+15+25=50
Pass average,
=36×100/50=72%
Question 8
1 / -0

The value of (1/log₃60+1/log₄60+1/log₅60)is:

A
0

B
1

C
5

D
60

Solution

Given expression

$= \log_{60} 3 + \log_{60} 4 + \log_{60} 5$
$= \log_{60} (3 \times 4 \times 5)$
$= \log_{60} 60$
$= 1$
Question 9
1 / -0

If y=sin⁻¹(x√1−x+√x√1−x²),then dy/dx =

A

$\frac{- 2 x}{\sqrt{1 - x^{2}}} + \frac{1}{2 \sqrt{x - x^{2}}}$

B

$\frac{- 1}{\sqrt{1 - x^{2}}} - \frac{1}{2 \sqrt{x - x^{2}}}$

C

$\frac{1}{\sqrt{1 - x^{2}}} + \frac{1}{2 \sqrt{x - x^{2}}}$

D
None of these

Solution

Putting $x = \sin A$ and $\sqrt{x} = \sin B$
$y = \sin^{- 1} (\sin A \sqrt{1 - \sin^{2} B} + \sin B \sqrt{1 - \sin^{2} A})$
$= \sin^{- 1} [\sin (A + B)] = A + B = \sin^{- 1} x + \sin^{- 1} \sqrt{x}$
$\frac{d y}{d x} = \frac{1}{\sqrt{1 - x^{2}}} + \frac{1}{2 \sqrt{x - x^{2}}}$
Question 10
1 / -0

If f(x) = mx + c, f(0) = f′(0) = 1 then f(2) =

A
1

B
2

C
3

D
? 3

Solution

Let $y = a^{x} + \log x \cdot \sin x$ Differentiating w.r.t. $x$

we get $\frac{d y}{d x} = a^{x} \log_{e} (a) + \frac{1}{x} \sin x + \log x \cdot \cos x$

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Re-Attempt Weekly Quiz Competition

Mathematics Test - 64

Result

Mathematics Test - 64

Score

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Rank

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SHARING IS CARING

Question 1

14

12

10

8

Solution

Question 2

32 Cm2

16 Cm2

20 Cm2

64 Cm2

Solution

Question 3

-34

-32

-12

-10

Solution

Question 4

1/3

2/5

3/4

1/4

Solution

Question 5

1/x

-1/x

x

-x

Solution

Question 6

y4=0

y5=0

y6=0

None of these

Solution

Question 7

74%

75%

69%

72%

Solution

Question 8

0

1

5

60

Solution

Question 9

−2x1−x2+12x−x2

−11−x2−12x−x2

11−x2+12x−x2

None of these

Solution

Question 10

1

2

3

? 3

Solution

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32 Cm²

16 Cm²

20 Cm²

64 Cm²

y₄=0

y₅=0

y₆=0

$\frac{- 2 x}{\sqrt{1 - x^{2}}} + \frac{1}{2 \sqrt{x - x^{2}}}$

$\frac{- 1}{\sqrt{1 - x^{2}}} - \frac{1}{2 \sqrt{x - x^{2}}}$

$\frac{1}{\sqrt{1 - x^{2}}} + \frac{1}{2 \sqrt{x - x^{2}}}$