Mathematics Test

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Mathematics Test - 65

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Weekly Quiz Competition

Question 1
1 / -0

The value of following is, cos24° + cos55° + cos125° + cos204° + cos300° ?

A
-1/2

B
1/2

C
2

D
1

Solution

The value of \(, \cos 24^{\circ}+\cos 55^{\circ}+\cos 125^{\circ}+\cos 204^{\circ}+\cos 300^{\circ}\)
We know that, cos \(\left(180^{\circ} \pm \theta\right)\)
\(=-\cos \theta\)
\(\Rightarrow \cos 24^{\circ}\)
\(+\cos 55^{\circ}\)
\(+\cos \left(180^{\circ}-55^{\circ}\right)\)
\(+\cos \left(180^{\circ}-24^{\circ}\right)\)
\(+\cos\)
\(\left(360^{\circ}-60^{\circ}\right)\)
\(\Rightarrow \cos 24^{\circ}\)
\(+\cos 55^{\circ}\)
\(-\cos 55^{\circ}\)
\(-\cos 24^{\circ}\)
\(+\cos 60^{\circ}\)
\(\Rightarrow \cos 60^{\circ}\)
\(\Rightarrow \frac{1}{2}\)
Question 2
1 / -0

\(\frac{d}{d x}\left(x^{2} \sin \frac{1}{x}\right)=\)

A
cos (1/x) +2x sin(1/x)

B
cos (1/x) -cos (1/x)

C
cos (1/x) -2x sin (1/x)

D
None of these

Solution

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

\(+2 x \sin \left(\frac{1}{x}\right)\)

\(=-\frac{1}{x^{2}} \cdot x^{2} \cos \left(\frac{1}{x}\right)+2 x \sin \left(\frac{1}{x}\right)=2 x \sin \left(\frac{1}{x}\right)-\cos \left(\frac{1}{x}\right)\)
Question 3
1 / -0

The differential coefficient of a^x+ logx. sinx is

A
a^xlog_ea + sinx/x + logx.cosx

B
a^x+ sinx/x + cosx. logx

C
a^xloga + cosx/x + sinx.logx.

D
None of these

Solution

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

we get \(\frac{d y}{d x}=a^{x} \log _{e}(a)+\frac{1}{x} \sin x+\log x \cdot \cos x\)
Question 4
1 / -0

A train covers the first 16 km at a speed of 20 km per hour another 20 km at 40 km per hour and the last 10 km at 15 km per hour. Find the average speed for the entire journey.

A
24 km

B
26 km

C
21 km

D

\(=23 \frac{23}{59} \mathrm{km} / \mathrm{hr}\)

Solution

Average speed \(=\frac{\text { total distance covered }}{\text { total time }}\) Total Distance
\(=(16+20+10)=46 \mathrm{km}\)
Time taken
\(=\left(\frac{16}{20}\right)+\left(\frac{20}{40}\right)+\left(\frac{10}{15}\right)\)
\(=\frac{59}{30}\)
Average speed
\(=\frac{46 \times 30}{59}\)
\(=23 \frac{23}{59} \mathrm{km} / \mathrm{hr}\)
Question 5
1 / -0

A card is drawn from a pack of 52 cards. The probability of getting a queen of club or a king of heart is:

A
1/13

B
2/13

C
1/26

D
1/52

Solution

Here, \(n(S)=52\) Let \(E=\) event of getting a queen of club or a king of heart Then \(n(E)=2\)

\(\therefore P(E)=\frac{n(E)}{n(S)}\)
\(=\frac{2}{52}\)
\(=\frac{1}{26}\)
Question 6
1 / -0

If x = 3t, y = 1/2(t + 1), then the value of t for which x = 2y is?

A
1

B
1/2

C
-1

D
2/3

Solution

\(x=3 t \ldots \ldots(i)\)
\(y=\frac{1}{2}(t+1)\)
\(x=2 y\)
\(\quad \Rightarrow x=2\)
\(\quad \times \frac{1}{2}(t+1)\)
\(\quad \Rightarrow x=t+1 \ldots\)
\(\quad \ldots .(i i)\)
\(\therefore 3 t=t+1\)
(From equation
(i) and (ii))
\(\Rightarrow 2 t=1\)
\(\Rightarrow t=\frac{1}{2}\)
Question 7
1 / -0

The first term of an Arithmetic Progression is 22 and the last term is -11. If the sum is 66, the number of terms in the sequence are:

A
10

B
12

C
9

D
8

Solution

Number of terms \(=n\) (let) First term (a) \(=22\) Last term \((\mathrm{l})=-11\)
Sum \(=66\)
Sum of an AP is given by:
\(=\) Number of terms \(\times\left\{\frac{(\text { First term }+\text { Last term })}{2}\right\}\) \(66=\mathbf{n} \times\left\{\frac{(\mathbf{a}+1)}{2}\right\}\)
\(66=n \times \frac{(22-11)}{2}\)
\(66=\mathrm{n} \times\left(\frac{11}{2}\right)\)
\(\mathrm{n}=\frac{(66 \times 2)}{11}\)
\(\mathbf{n}=12\)
No. of terms \(=12\)
Question 8
1 / -0

If A = tan11°. tan29°, B = 2cot61°. cot79° then -

A
A = 2B

B
A = -2B

C
2A = B

D
2A = -B

Solution

\(\Leftrightarrow \frac{A}{B}=\frac{\tan 11^{\circ} \cdot \tan 29^{\circ}}{2 \cot 61^{\circ} \cdot \cot 79^{\circ}}\)
\(\Leftrightarrow \frac{A}{B}=\frac{\tan 11^{\circ} \tan 29^{\circ}}{2\left[\cot \left(90^{\circ}-29^{\circ}\right) \cdot \cot \left(90^{\circ}-11^{\circ}\right)\right]}\)
\(\Leftrightarrow \frac{A}{B}=\frac{\tan 11^{\circ} \cdot \tan 29^{\circ}}{2 \tan 11^{\circ} \cdot \tan 29^{\circ}}\)
\(\Leftrightarrow \frac{\mathrm{A}}{\mathrm{B}}=\frac{1}{2}\)
\(\Leftrightarrow 2 \mathrm{A}=\mathrm{B}\)
Question 9
1 / -0

After striking the floor, a rubber ball rebounds to 4/5th of the height from which it has fallen. Find the total distance that it travels before coming to rest if it has been gently dropped from a height of 120 metres.

A
540 m

B
960 m

C
1080 m

D
1020 m

Solution

\(\frac{a}{(1-r)}\)
Now, \(\left[\left\{\frac{120}{\left(\frac{1}{5}\right)}\right\}+\left\{\frac{96}{\left(\frac{1}{5}\right)}\right\}\right]\)
\(=1080 \mathrm{m}\)
Question 10
1 / -0

There are 10 points in a plane out of which 4 are collinear. Find the number of triangles formed by the points as vertices.

A
120

B
116

C
140

D
20

Solution

The number of triangle can be formed by 10 points \(={ }^{10} \mathrm{C}_{3}\)
Similarly, the number of triangle can be formed by 4 points when no one is collinear \(={ }^{4} \mathrm{C}_{3}\)
In the question, given 4 points are collinear, Thus, required number of triangle can be formed,
\(={ }^{10} \mathrm{C}_{3}-{ }^{4} \mathrm{C}_{3}\)
\(=120-4\)
\(=116\)

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Mathematics Test - 65

Result

Mathematics Test - 65

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SHARING IS CARING

Question 1

-1/2

1/2

2

1

Solution

Question 2

cos (1/x) +2x sin(1/x)

cos (1/x) -cos (1/x)

cos (1/x) -2x sin (1/x)

None of these

Solution

Question 3

axloge a + sinx/x + logx.cosx

ax + sinx/x + cosx. logx

axloga + cosx/x + sinx.logx.

None of these

Solution

Question 4

24 km

26 km

21 km

\(=23 \frac{23}{59} \mathrm{km} / \mathrm{hr}\)

Solution

Question 5

1/13

2/13

1/26

1/52

Solution

Question 6

1

1/2

-1

2/3

Solution

Question 7

10

12

9

8

Solution

Question 8

A = 2B

A = -2B

2A = B

2A = -B

Solution

Question 9

540 m

960 m

1080 m

1020 m

Solution

Question 10

120

116

140

20

Solution

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a^xlog_ea + sinx/x + logx.cosx

a^x+ sinx/x + cosx. logx

a^xloga + cosx/x + sinx.logx.