Mathematics Test

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Mathematics Test - 26

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Weekly Quiz Competition

Question 1
4 / -1

If no husband and wife play in the same game, then the number of ways in which a mixed double game can be arranged from among nine married couples is

A
756

B
1512

C
3024

D
None of these

Solution

Out of 9 men, 2 men can be chosen in \({ }^{9} \mathrm{C}_{2}\) ways.
since no husband and wife are to play in the same game, we have to select 2 women from the remaining 7 women.
This can be done in \({ }^{7} \mathrm{C}_{2}\) ways.
If \(\mathrm{M}_{1}, \mathrm{M}_{2}\) and \(\mathrm{W}_{1}, \mathrm{W}_{2}\) are chosen, then a team can be constituted in 2 ways, viz. \(\mathrm{M}_{1}\)
\(\mathrm{W}_{1}\) or \(\mathrm{M}_{1} \mathrm{W}_{2}\)
Thus, number of ways of arranging the game \(={ }^{9} \mathrm{C}_{2} \times{ }^{7} \mathrm{C}_{2} \times 2=1512\)
Question 2
4 / -1

Which of the following pieces of data does not uniquely determine the acute-angled ΔABC (R = circumradius)?

A
a, sin A, sin B

B
a, b, c

C
a, sin B, R

D
a, sin A, R

Solution

\(\because \ln a \Delta A B C, \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin (\pi-(A+B)\}}=2 R\)
\(\Rightarrow \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin (A+B)}=2 R\)
(a) If we know a, sin \(A\), sin \(B\), then we can find \(b, c, A, B\) and \(C\).
(b) We can find \(A, B, C\) by using cosine rule.
(c) \(\because a, \sin B, R\) are given, we can find \(\sin A, b\) and hence \(\sin C=\sin (\pi-(A+B)\}=\sin C\)
(d) \(a, \sin A, R\) are given, we know only the ratio \(\frac{b}{\sin B}\) or \(\frac{c}{\sin (A+B)} ;\) we cannot determine the values of \(\mathrm{b}, \mathrm{c}, \sin \mathrm{B}, \sin \mathrm{C}\) separately.
: Triangle ABC cannot be determined in this case.
Question 3
4 / -1

A chord of the circle x² + y² – 4x – 6y = 0 passing through the origin subtends an angle tan^–1(7/4) at the point where the circle meets the positive y-axis. The equation of the chord is

A
2x + 3y = 0

B
x + 2y = 0

C
x – 2y = 0

D
2x – 3y = 0

Solution

The given circle passes through the origin \(\mathrm{O}\) and meets the positive \(\mathrm{y}\) -axis at \(\mathrm{B}(0,6)\). Let OP be the chord of the circle passing through the origin, subtending an angle \(\theta\) at
B, where tan \(\theta=7 / 4\)
\[
\Rightarrow \quad \angle \mathrm{OBP}=\theta
\]
Equation of the tangent OT at 0 to the given circle is \(2 x+3 y=0\)
\(\Rightarrow \quad\) Slope of the tangent \(=-2 / 3\) So that, if \(\angle \mathrm{XOT}=\alpha,\) then \(\tan \alpha=2 / 3\)
From geometry, \(\angle \mathrm{POT}=\angle \mathrm{OBP}=\theta\)
\(\Rightarrow \quad \angle \mathrm{POX}=\theta-\alpha\)
and \(\quad \tan (\theta-\alpha)=\frac{\tan \theta-\tan \alpha}{1+\tan \theta \tan \alpha}\)
\[
=\frac{\frac{7}{4}-\frac{2}{3}}{1+\frac{7}{4} \times \frac{2}{3}}=\frac{13}{26}=\frac{1}{2}
\]

Hence, the equation of \(\mathrm{OP}\) is \(\mathrm{y}=\mathrm{x}\) tan \((\theta-\alpha)\)
\(\Rightarrow \quad x-2 y=0\)
Question 4
4 / -1

Find the equation of the line passing through the intersection of the lines x + 2y - 3 = 0 and 4x – y + 7 = 0 and which is parallel to y – x + 10 = 0.

A
3x + 3y + 10 = 0

B
- 3x + 3y + 10 = 0

C
- 3x + 3y - 10 = 0

D
None of these

Solution

Let the equation be x + 2y - 3 + k (4x - y + 7) = 0 ............ (i)
i.e., (1 + 4k) x + (2 - k) y + 7k - 3 = 0
i.e., y = (-1 - 4k)/(2-k) + 3 - 7k
Since it is parallel to y - x + 10 = 0 (i.e., y = x - 10), slopes are equal.
So, (- 1 - 4k)/(2 - k) = 1
- 1 - 4k = 2 - k
k = - 1
Substituting this in eq (i), we get:
The required line is x + 2y - 3 + (-1)(4x - y + 7) = 0
i.e. - 3x + 3y - 10 = 0
Question 5
4 / -1

It is given that n arithmetic means are inserted between two sets of numbers a, 2b and 2a, b where a, b ∈ R. If the m^th mean between these two sets of numbers is the same, then a : b equals

A
n - m + 1 : m

B
n + m + 1 : m

C
m : n - m + 1

D
n : n - m + 1
Question 6
4 / -1

If \(a, b\) and \(c\) are the sides of \(\Delta A B C\) and equations \(a x^{2}+b x+c=0\) and \(5 x^{2}+12 x+13=0\) have both the roots common, then the measure of \(\angle \mathrm{C}\) is

A
60º

B
90º

C
120º

D
45º

Solution

Given: Both roots of \(a x^{2}+b x+c=0\) and \(5 x^{2}+12 x+13=0\) are common.
\(\therefore \frac{a}{5}=\frac{b}{12}=\frac{c}{13}\)
\(\Rightarrow a^{2}+b^{2}=c^{2}\)
\(\Rightarrow \angle C=90^{\circ}\)
Question 7
4 / -1

A line which is parallel to the x-axis and crosses the curve y =√x at an angle of 45^ois

A
x = 1/4

B
y = 1/4

C
y = 1/2

D
y = 1

Solution

Equation of a line parallel to the \(\mathrm{x}\) -axis is \(\mathrm{y}=\mathrm{k}\). It meets the curve \(\mathrm{y}=\sqrt{\mathrm{x}}\) at the point \(\left(k^{2}, k\right) .\) since the line \(y=k\) crosses the curve at an angle of \(45^{\circ},\) the angle between the tangent to the curve \(y=\sqrt{x}\) at \(\left(k^{2}, k\right)\) and the line \(y=k\) is \(45^{\circ},\) the line \(y=k\) being parallel to the \(\mathrm{x}\) -axis. We have
\[
\begin{aligned}
&\left[\frac{d y}{d x}\right]_{\left(x^{2}, k\right)}=\pm 1 \text { on the curve } y=\sqrt{x} \\
\Rightarrow \quad\left[\frac{1}{2 \sqrt{x}}\right]_{\left(x^{2}, k\right)} &=\pm 1 \Rightarrow k=\frac{1}{2}
\end{aligned}
\]
\(\mathrm{k}>0,\) as \(\mathrm{y}=\sqrt{\mathrm{x}}\) lies in the first quadrant.
Question 8
4 / -1

The function \(F(x)=\int_{0}^{x} \log \frac{1-t}{1+t} d t\) is

A
an even function

B
an odd function

C
a periodic function

D
None of these

Solution

\(\begin{aligned} F(-x) &=\int_{0}^{-x} \log \frac{1-t}{1+t} d t=-\int_{0}^{x} \log \frac{1+u}{1-u} d u(u=-t) \\ &=\int_{0}^{x} \log \frac{1-u}{1+u} d u=F(x) \end{aligned}\)
Question 9
4 / -1

For \(0<\alpha<\pi,\) the value of the definite integral \(\int_{0}^{1} d x /\left(x^{2}+2 x \cos \alpha+1\right)\) is equal to

A
sinα

B
tan^-1(sinα)

C
α sinα

D
α/2 sinα

Solution

\(\begin{aligned} & \int_{0}^{1} \frac{d x}{x^{2}+2 x \cos \alpha+1} \\=& \int_{0}^{1} \frac{d x}{(x+\cos \alpha)^{2}+1-\cos ^{2} \alpha} \\=& \int_{0}^{1} \frac{d x}{(x+\cos \alpha)^{2}+\sin ^{2} \alpha} \\=&\left.\frac{1}{\sin \alpha} \tan ^{-1} \frac{x+\cos \alpha}{\sin \alpha}\right|_{0} ^{1} \\=& \frac{1}{\sin \alpha}\left(\tan ^{-1}\left(\frac{1+\cos \alpha}{\sin \alpha}\right)-\tan ^{-1} \frac{\cos \alpha}{\sin \alpha}\right) \\=& \frac{1}{\sin \alpha}\left(\tan ^{-1}\left(\cot \frac{\alpha}{2}\right)-\tan ^{-1} \cot \alpha\right) \\=& \frac{1}{\sin \alpha}\left(\frac{\pi}{2}-\frac{\alpha}{2}-\frac{\pi}{2}+\alpha\right)=\frac{\alpha}{2 \sin \alpha} \end{aligned}\)
Question 10
4 / -1

The coefficient of x^k in the expansion of E = 1 + (1 + x) + (1 + x)² + … + (1 + x)ⁿ is

A
ⁿC_k

B
^{n +1}C_k

C
^{n +1}C_{k +1}

D
None of these

Solution

We have
\[
\begin{aligned}
E &=1+(1+x)+(1+x)^{2}+\ldots+(1+x)^{n} \\
&=\frac{1-(1+x)^{n+1}}{1-(1+x)}=\frac{1}{x}\left[(1+x)^{n+1}-1\right]
\end{aligned}
\]
Thus, coefficient of \(x^{k}\) in \(E\)
\[
\begin{array}{l}
=\text { Coefficient of } x^{k+1} \text { in }\left[(1+x)^{n+1}-1\right] \\
={ }^{n+1} C_{k+1}
\end{array}
\]

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Re-Attempt Weekly Quiz Competition

Mathematics Test - 26

Result

Mathematics Test - 26

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SHARING IS CARING

Question 1

756

1512

3024

None of these

Solution

Question 2

a, sin A, sin B

a, b, c

a, sin B, R

a, sin A, R

Solution

Question 3

2x + 3y = 0

x + 2y = 0

x – 2y = 0

2x – 3y = 0

Solution

Question 4

3x + 3y + 10 = 0

- 3x + 3y + 10 = 0

- 3x + 3y - 10 = 0

None of these

Solution

Question 5

n - m + 1 : m

n + m + 1 : m

m : n - m + 1

n : n - m + 1

Question 6

60º

90º

120º

45º

Solution

Question 7

x = 1/4

y = 1/4

y = 1/2

y = 1

Solution

Question 8

an even function

an odd function

a periodic function

None of these

Solution

Question 9

sinα

tan-1 (sinα)

α sinα

α/2 sinα

Solution

Question 10

nCk

n +1Ck

n +1Ck +1

None of these

Solution

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tan^-1(sinα)

ⁿC_k

^{n +1}C_k

^{n +1}C_{k +1}