Mathematics Test

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Mathematics Test - 10

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Weekly Quiz Competition

Question 1
4 / -1

If f(x) = | x |, then f′(2) =

A
2

B
1

C
Does not exist

D
None of these

Solution

\(R f^{\prime}(2)=\frac{10}{1-0} \frac{f(2+h)-f(2)}{h}\)
\(=\lim _{h \rightarrow 0} \frac{|2+h|-|2|}{h}=\lim _{h \rightarrow 0} \frac{2+h-2}{h}=1\)
\(L(2)=\lim _{h \rightarrow 0} \frac{f(2-h)-\pi(2)}{-h}\)
\(=\lim _{h \rightarrow 0} \frac{|2-h|-|2|}{-h}=\lim _{h \rightarrow 0} \frac{2-h-2}{-h}=1\)
\(R f^{\prime}(2)=L f^{\prime}(2)=1\)
So. \(f^{\prime}(2)=1\)
Question 2
4 / -1

\(\int_{0}^{\pi / 2} \sin ^{2} x d x=\)

A
π/4

B
π/3

C
π/2

D
none of these

Solution

\(n=\frac{1}{2} \int_{0}^{\pi / 2}(1-\cos 2 x) d x\)
\(=\frac{1}{2}\left[x-\frac{\sin 2 x}{2}\right]_{0}^{\pi / 2}=\frac{\pi}{4}\)
Question 3
4 / -1

The area of the region bounded by y = | x - 1 | and y = 1 is

A
1

B
2

C
1/2

D
None of these

Solution

Required Area \(=\int_{0}^{1}\left(x_{2}-x_{1}\right) d y=\int_{0}^{1}\{(x+1)-(-x+1)\} d x\)
\(=\int_{0}^{1} 2 x d x=\left[x^{2}\right]=1\)
Question 4
4 / -1

If \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) be the angles of a triangle, then \(\Sigma \frac{\cot A+\cot B}{\tan A+\tan B}=\)

A
1

B
2

C
-1

D
-2

Solution

To find, \(\Sigma \frac{\cot A+\cot B}{\tan A+\tan B}\)
\(\cot A+\cot B=\frac{1}{\tan A}+\frac{1}{\tan B}=\frac{\tan A+\tan B}{\tan A \tan B}\)
Putting the value in (1). we get \(\Rightarrow \quad \Sigma \frac{\frac{\tan A+\tan B}{\tan A \tan B}}{\tan A+\tan B}=\Sigma \frac{1}{\tan A \tan B}\)
\(=\frac{1}{\tan A \tan B}+\frac{1}{\tan B \tan C}+\frac{1}{\tan A \tan C}\)
Let \(A=B=C=60^{\circ}\)
Putting the values in the above equation, we get \(\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1\)
Question 5
4 / -1

In how many ways can 8 students be arranged in a row?

A
8!

B
7!

C
8

D
7

Solution

Required number of ways = 8!

Hence option A is correct.
Question 6
4 / -1

India plays two matches each with West Indies and Australia. In any match, the probabilities of India getting points 0, 1 and 2 are 0.45, 0.05 and 0.50 respectively. Assuming that the outcomes are independent, the probability of India getting atleast 7 points is

A
0.8750

B
0.0875

C
0.0625

D
0.0250

Solution

Probability of getting atleast seven points

= Probability of getting 7 points or 8 points

= Prob. of getting 7 points + Prob. of getting 8 points.

Seven points in four matches can be obtained in the following four different ways :

2, 2, 2, 1;

2, 2, 1, 2;

2, 1, 2, 2;

1, 2, 2, 2

The probability of each of these ways = (0.50)^{3 x} (0.05) (by multiplication theorem for independent events)

= 0.00625

Hence, probability of getting 7 points = ^{4 x} (0.00625) (by add. Theorem)

= 0.0250.

Eight points in four matches can be obtained only in one way i.e. 2, 2, 2, 2.

Hence, Prob. of getting 8 points = (0.50)⁴ = 0.0625

Thus, the required prob. = 0.250 + 0.0625 = 0.0875.
Question 7
4 / -1

Find shortest distance between lines \(\overrightarrow{\mathrm{r}}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}+\lambda(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+4 \hat{\mathbf{k}})\) and \(\overrightarrow{\mathrm{r}}=2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}+\mu(3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+5 \hat{\mathrm{k}})\)

A
2

B
\(\frac{1}{6}\)

C
\(\frac{1}{\sqrt{6}}\)

D
\(\sqrt{6}\)

Solution

Given equation of lines:

\(\overrightarrow{\mathbf{r}}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}+\lambda(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})\)

\(\overrightarrow{\mathbf{r}}=2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+\mathbf{5} \hat{\mathbf{k}}+\boldsymbol{\mu}(3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})^{-\cdots}-(2)\)

position vector and normal vector of line (1)

\(\overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\)

\(\overrightarrow{\mathbf{n}_{1}}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}\)

position vector and normal vector of line (2)

\(\overrightarrow{\mathbf{c}}=2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}\)

\(\overrightarrow{\mathbf{n}_{2}}=3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+\mathbf{5} \hat{\mathbf{k}}\)

shortest between two skews line \(\mathrm{SD}=\frac{\overrightarrow{\mathbf{A C}} \cdot\left(\overrightarrow{\mathbf{n}_{1}} \times \overrightarrow{\mathbf{n}_{2}}\right)}{\left|\overrightarrow{\mathbf{n}_{1}} \times \overrightarrow{\mathbf{n}_{2}}\right|}\)

\(\mathbf{A} \mathbf{C}=\overrightarrow{\mathbf{c}}-\overrightarrow{\mathbf{a}}\)

\(\mathbf{A} \mathbf{C}=2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+\mathbf{5} \hat{\mathbf{k}}-(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})\)

\(\mathbf{A} \mathbf{C}=2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}-\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}\)

\(\mathbf{A} \mathbf{C}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}\)

\(\overrightarrow{\mathbf{n}_{1}} \times \overrightarrow{\mathbf{n}_{2}}=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & 3 & 4 \\ 3 & 4 & 5\end{array}\right|\)

\(\overrightarrow{\mathbf{n}_{1}} \times \overrightarrow{\mathbf{n}_{2}}=\hat{\mathbf{i}}(15-16)-\hat{\mathbf{j}}(\mathbf{1 0}-\mathbf{1 2})+\hat{\mathbf{k}}(\mathbf{8}-\mathbf{9})\)

\(\overrightarrow{\mathbf{n}}_{1} \times \overrightarrow{\mathbf{n}}_{2}=-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}\)

\(\left|\overrightarrow{\mathrm{n}}_{1} \times \overrightarrow{\mathrm{n}}_{2}\right|=\sqrt{(-1)^{2}+2^{2}+(-1)^{2}}\)

\(\left|\overrightarrow{\mathbf{n}_{1}} \times \overrightarrow{\mathbf{n}_{2}}\right|=\sqrt{6}\)

putting \(\mathrm{AC}.\overrightarrow{\mathrm{n}}_{1} \times \overrightarrow{\mathrm{n}}_{2},\left|\overrightarrow{\mathrm{n}}_{1} \times \overrightarrow{\mathrm{n}}_{2}\right|\) in formula

\(\mathrm{SD}=\frac{\overrightarrow{\mathbf{A C}} \cdot\left(\overrightarrow{\mathrm{n}}_{1} \times \overrightarrow{\mathbf{n}}_{2}\right)}{\left|\overrightarrow{\mathbf{n}_{1}} \times \overrightarrow{\mathbf{n}_{2}}\right|}\)

\(\mathbf{S D}=\frac{(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \cdot(-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}})}{\sqrt{\mathbf{6}}}\)

\(\mathrm{SD}=\frac{-1+4-2}{\sqrt{6}}\)

\(\mathrm{SD}=\frac{1}{\sqrt{6}}\)
Question 8
4 / -1

A man of 2 m height walks at a uniform speed of 6 km/h away from a lamp post of 6 m height. The rate at which the length of his shadow increases is

A
2 km/h

B
1 km/h

C
3 km/h

D
6 km/h

Solution

In a ADC

\(\quad \tan \theta=\frac{6}{x+y}\)
and in \(\Delta B C E,\)
\(\begin{aligned} \tan \theta=& \frac{2}{x} \\ \frac{2}{x}=\frac{6}{x+y} & \Rightarrow x+y=3 x \\ \Rightarrow & y=2 x \end{aligned}\)
On differentiating w.r.t. \(t\), we get
\(\frac{d y}{d t}=2 \frac{d x}{d t}\)
\(\Rightarrow \quad 6=2 \frac{d x}{d t} \quad\left(\because \frac{d y}{d t}=6 \operatorname{given}\right)\)
\(\Rightarrow \quad \frac{d x}{d t}=3 \mathrm{km} / \mathrm{h}\)
Question 9
4 / -1

If \(A+B+C=180^{\circ}\), then \(\frac{\tan A+\tan \theta+\tan C}{\tan A . \tan B . \tan C}=\)

A
0

B
2

C
1

D
-1

Solution

\(A+B+C=180^{\circ}\)
\(A+B=180-C\)
\(\tan (A+B)=\tan (180-C)\)
\(\tan A+\tan B\)
\(1-\tan A \tan B^{\prime}=-\tan C\)
\(\tan A+\tan B+\tan C=\tan A \tan B \tan C\)
\(\operatorname{So.} \frac{\tan A+\tan B+\tan C}{\tan A \tan B \tan C}=1\)
Question 10
4 / -1

Three of six vertices of a regular hexagon are chosen at random. The probability that the triangle with these three vertices will be equilateral is

A
1/10

B
1/5

C
2/15

D
None of these

Solution

Let ABCDEF be a regular hexagon.

Three vertices out of its 6 vertices can be chosen in ⁶C₃ways. Therefore, 20 triangles can be made by joining three vertices at a time.

Out of these 20 triangles, only ACE and BDF are equilateral. Hence, the required probability = 1/10.

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Mathematics Test - 10

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Mathematics Test - 10

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SHARING IS CARING

Question 1

2

1

Does not exist

None of these

Solution

Question 2

π/4

π/3

π/2

none of these

Solution

Question 3

1

2

1/2

None of these

Solution

Question 4

1

2

-1

-2

Solution

Question 5

8!

7!

8

7

Solution

Question 6

0.8750

0.0875

0.0625

0.0250

Solution

Question 7

2

\(\frac{1}{6}\)

\(\frac{1}{\sqrt{6}}\)

\(\sqrt{6}\)

Solution

Question 8

2 km/h

1 km/h

3 km/h

6 km/h

Solution

Question 9

0

2

1

-1

Solution

Question 10

1/10

1/5

2/15

None of these

Solution

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