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

\(\tan 15^{\circ} \cot 75^{\circ}+\tan 75^{\circ} \cot 15^{\circ}\)
\(=\tan 15^{\circ} \cot \left(90^{\circ}-15^{\circ}\right)+\tan \left(90^{\circ}-15\right)^{\circ} \cot 15^{\circ}\)
\(=\tan ^{2} 15^{\circ}+\cot ^{2} 15^{\circ}\)
\(=\tan ^{2} 15^{\circ}+\cot ^{2} 15^{\circ} \ldots \ldots(i)\)
[Formula] \(\cot \left(90^{\circ}-\theta\right)=\tan \theta\)
\(\tan \left(90^{\circ}-\theta\right)=\cot \theta\)
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 \(4 \mathrm{cm}\). side of second square \(=2 \sqrt{2} \mathrm{cm}\) Side of third square \(=2 \mathrm{cm}\) and so on, i.e. \(4,2, \sqrt{2}, \sqrt{2}, 1 \ldots \ldots\)
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 \(=16+8+4+2+1+\ldots \ldots\)
\(=\left[\frac{16}{\left\{1-\left(\frac{1}{2}\right)\right\}}\right]\)
\(=32 \mathrm{cm}^{2}\)
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 \(\frac{b}{c}=\frac{15}{16}\)
\(\Rightarrow \frac{a}{b} \times \frac{b}{c}=\frac{4}{5} \times \frac{15}{16}\)
\(\Rightarrow \frac{a}{c}=\frac{3}{4}\)
\(\therefore \frac{18 c^{2}-7 a^{2}}{45 c^{2}+20 a^{2}}\)
\(=\frac{c^{2}\left(18-7 \frac{a^{2}}{c^{2}}\right)}{c^{2}\left(45+20 \frac{a^{2}}{c^{2}}\right)}\)
\(=\frac{18-7\left(\frac{a}{c}\right)^{2}}{45+20\left(\frac{a}{c}\right)^{2}}\)
\(=\frac{18-7 \times \frac{9}{16}}{45+20 \times \frac{9}{16}}\)
\(=\frac{18-\frac{63}{16}}{45+\frac{45}{4}}\)
\(=\frac{225 \times 4}{16 \times 225}\)
\(=\frac{1}{4}\)
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 \(x>0\)

\(=\left(\frac{1}{-x}\right)(-1)=\frac{1}{x}\) if \(x<0 \quad\) 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 \(y=a^{x}+\log x \cdot \sin x \quad\) 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

-

Rank

-

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

\(\frac{-2 x}{\sqrt{1-x^{2}}}+\frac{1}{2 \sqrt{x-x^{2}}}\)

\(\frac{-1}{\sqrt{1-x^{2}}}-\frac{1}{2 \sqrt{x-x^{2}}}\)

\(\frac{1}{\sqrt{1-x^{2}}}+\frac{1}{2 \sqrt{x-x^{2}}}\)

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