Mathematics Test

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Mathematics Test - 11

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Weekly Quiz Competition

Question 1
1 / -0

What is the value of cos 36°?

A
(√5 - 1)/4

B
(√5 + 1)/4

C
$\frac{\sqrt{10+2 \sqrt{5}}}{4}$

D
$\frac{\sqrt{10-2 \sqrt{5}}}{4}$

Solution

Let $A=18^{\circ}$
Therefore, $5 A=90^{\circ}$
$\Rightarrow 2 A+3 A=90^{\circ}$
$\Rightarrow 2 \theta=90^{\circ}-3 A$
Taking sine on both sides, we get
$\sin 2 A=\sin \left(90^{\circ}-3 A\right)=\cos 3 A$
$\Rightarrow 2 \sin A \cos A=4 \cos ^{3} A-3 \cos A$
$\Rightarrow 2 \sin A \cos A-4 \cos ^{3} A+3 \cos A=0$
$\Rightarrow \cos A\left(2 \sin A-4 \cos ^{2} A+3\right)=0$
Dividing both sides by $\cos A=\cos 18^{\circ} \neq 0,$ we get
$\Rightarrow 2 \sin \theta-4\left(1-\sin ^{2} A\right)+3=0$
$\Rightarrow 4 \sin ^{2} A+2 \sin A-1=0,$ which is a quadratic in $\sin A$
Therefore, $\sin \theta=\frac{-21 \sqrt{4(4)(-1)}}{2(4)}$
$\Rightarrow \sin \theta=\frac{-2 \pm \sqrt{4+10}}{8}$
$\Rightarrow \sin \theta=\frac{-212 \sqrt{5}}{8}$
$\Rightarrow \sin \theta=\frac{11 \sqrt{5}}{4}$
Now $\sin 18^{\circ}$ is positive, as $18^{\circ}$ lies in first quadrant.
Therefore, $\sin 18^{\circ}=\sin A=\frac{-11 \sqrt{5}}{4}$
Now, $\cos 36^{\circ}=\cos 2 \cdot 18^{\circ}$
$\Rightarrow \cos 36^{\circ}=1-2 \sin ^{2} 18^{\circ}$
$\Rightarrow \cos 36^{\circ}=1-2\left(\frac{\sqrt{5}-1}{4}\right)^{2}$
$\Rightarrow \cos 36^{\circ}=\frac{10-2(5+1-2 \sqrt{5})}{10}$
$\Rightarrow \cos 36^{\circ}=\frac{1+4 \sqrt{5}}{16}$
$\Rightarrow \cos 36^{\circ}=\frac{\sqrt{5}+1}{4}$
Therefore, $\cos 36^{\circ}=\frac{\sqrt{5}+1}{4}$
Question 2
1 / -0

Consider the following statements:
1. Value of sin θ oscillates between –1 and 1.
2. Value of cos θ oscillates between 0 and 1.
Which of the above statements is/are correct?

A
1 only

B
2 only

C
Both 1 and 2

D
Neither 1 nor 2

Solution

$-1 . \sin \theta \in[-1,1] ; Q \in R,$ the value of $\sin \theta$ lies between -1 to 1
2. $\cos \theta \in[-1,1] ; Q \in R,$ the value of $\sin \theta$ lies between -1 to 1.
Question 3
1 / -0

If x and y are positive and xy > 1, then what is tan^-1x + tan^-1y equal to?

A
$\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$

B
$\pi+\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$

C
$\pi-\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$

D
$\tan ^{-1}\left(\frac{x-y}{1+x y}\right)$

Solution

$\tan ^{-1} x+\tan ^{1} y=\pi+\tan ^{-1}\left[\frac{x+y}{1+x y}\right]$
if $x<0, y<0$ and $x y>1$, then $\tan ^{-1} x+\tan ^{-1} y=-\pi+\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$
Question 4
1 / -0

Consider the following statements:
1. $n\left(\sin ^{2} 67 \frac{1^{\circ}}{2}-\sin ^{2} 22 \frac{1^{\circ}}{2}\right)>1$ for all positive integers $n \geq 2$
2. If $x$ is any positive real number, then $n x>1$ for all positive integers $n \geq 2$ Which of the above statements is/are correct?

A
1 only

B
2 only

C
Both 1 and 2

D
Neither 1 nor 2

Solution

Give $n\left(\sin ^{2} 67 \frac{1^{\circ}}{2}-\sin ^{2} 22 \frac{1^{\circ}}{2}\right)$
$\operatorname{or} n\left(\sin ^{2} \frac{135^{\circ}}{2}-\sin ^{2} \frac{45^{\circ}}{2}\right)$
$=n\left(\sin \frac{135^{\circ}}{2}+\sin \frac{45^{\circ}}{2}\right)\left(\sin \frac{135^{\circ}}{2}-\sin \frac{45^{\circ}}{2}\right)$
$=n\left[2 \sin \frac{\left(\frac{135^{\circ}}{2}+\frac{45^{\circ}}{2}\right)}{2} \cdot \cos \frac{\left(\frac{135}{2}-\frac{45^{\circ}}{2}\right)}{2}\right]$
$\left[2 \cdot \cos \left(\frac{\frac{135^{\circ}+45^{\circ}}{2}}{2}\right) \cdot \sin \left(\frac{\frac{135^{\circ}-45^{\circ}}{2}}{2}\right)\right]$
$=n\left[2 . \sin \left(\frac{90^{\circ}}{2}\right) \cdot \cos \left(\frac{45^{\circ}}{2}\right)\right]\left[2 . \cos \left(\frac{90^{\circ}}{2}\right) \cdot \sin \left(\frac{45^{\circ}}{2}\right)\right]$
$=2 n\left(2 \sin \frac{45^{\circ}}{2} \cdot \cos \frac{45^{\circ}}{2}\right)\left(\sin 45^{\circ} \cdot \cos 45^{\circ}\right)$
$=2 n \cdot \sin \left(2 \times \frac{45^{\circ}}{2}\right)\left(\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}\right)$
$=2 n . \sin 45 \times \frac{1}{2}=n \cdot \frac{1}{\sqrt{2}}=\frac{n}{\sqrt{2}}$
$\therefore \frac{n}{\sqrt{2}}>1$ for all positive integers $n \geq 2$
$\therefore$ Statement 1 is true
Statement 2 $n x>1, \forall n \geq 2$
$\Rightarrow n>\frac{1}{x}, \forall n \geq 2$
$\mathrm{x} \epsilon(0, \infty),$ then we take $\mathrm{x}=1$
$\mathrm{n}>1,$ but $\mathrm{n}$ is always greater or equal to 2 for all $\mathrm{x}$ positive real number.
$\therefore$ Statement 2 is false.
Question 5
1 / -0

Consider the following statements:
1. If 3θ is an acute angle such that sin 3θ = cos 2θ, then the measurement of θ in radian equals toπ/10
2. One radian is the angle subtended at the centre of a circle by an arc of the same circle whose length is equal to the diameter of that circle.
Which of the above statements is/are correct?

A
1 only

B
2 only

C
Both 1 and 2

D
Neither 1 nor 2

Solution

$\sin 3 θ = \cos 2 θ$ $\sin 3 θ = \sin (\frac{π}{2} - 2 θ)$ $3 θ = \frac{π}{2} - 2 θ$ $5 θ = \frac{π}{2} \Rightarrow θ = \frac{π}{10}$
One radian is the angle subtended at the center of a circle by an arc of a circle by an arc of the same circle whose length is equal to radius of that circle.
hence statement 1 is correct.
Question 6
1 / -0

From an aeroplane above a straight road the angle of depression of two positions at a distance 20 m apart on the road are observed to be 30° and 45°. The height of the aeroplane above the ground is:

A
10√3 m

B
10(√3-1)m

C
10(√3+1)m

D
20m

Solution

$\operatorname{In} \Delta \mathrm{ABC}, \tan 45^{\circ}=\frac{A B}{B C}=\frac{h}{x}$

$1=\frac{h}{x}$

$h=x$

$\tan 30^{\circ}=\frac{A B}{B D}$

$\frac{1}{\sqrt{3}}=\frac{h}{x+20}$

$x+20=\sqrt{3} h$

$h+20=\sqrt{3} h$

$20=(\sqrt{3}-1) h$

$\mathrm{h}=\frac{20}{\sqrt{3}-1}$

$=\frac{20}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}$

$=\frac{20(\sqrt{3}+1)}{2}=10(\sqrt{3}+1) m$

Hence the height is $10(\sqrt{3}+1) m$
Question 7
1 / -0

Consider the following statements:
1. There exists no triangle ABC for which sin A + sin B = sin C.
2. If the angle of a triangle are in the ratio 1 : 2 : 3, then its sides will be in the ratio 1 : √3 : 2.
Which of the above statements is/are correct?

A
1 only

B
2 only

C
Both 1 and 2

D
Neither 1 nor 2

Solution

1. Given, $\sin A+\sin B=\sin C$
$\mathrm{a}+\mathrm{b}+\mathrm{c}\left(\because B y \sin e \operatorname{law}, \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=K\right.$
Here, the sum of two sides of $\Delta A B C$ is equal to the third side, but it is not possible (Because by triangle inequality, the sum of the length of two sides of a triangle is always greater than the length of the third side)
$a+b>c$
2. Ratio of angles of a triangle $A: B: C=1: 2: 3$
$A+B+C=180^{\circ}$
$\therefore A=30^{\circ}$
$\mathrm{B}=60^{\circ}$
$\mathrm{C}=90^{\circ}$
The ratio in sides according to sine rule $a: b: c=\sin A: \sin B: \sin C$
$=\sin 30^{\circ}: \sin 60^{\circ}: \sin 90^{\circ}$
$=\frac{1}{2}, \frac{\sqrt{3}}{2}, 1=1: \sqrt{3}: 2$
Question 8
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Four points (0, 4, 1) (2, 3, -1) (4, 5, 0) (2, 6, 2) Which of the following are the vertices of the figure.

A
Rhombus

B
Rectangle

C
Square

D
Parallelogram

Solution

The constant of four points A, B, C and D is (0, 4, 1) (2, 3, -1) (4, 5, 0) (2, 6, 2), AB = BC = CD = DA = 3 unit
And AC = BD = $\sqrt{18}$ units
Therefore, the four sides and diagonals of the quadrilateral are equal,
Hence, quadrilateral ABCD will be square.
Question 9
1 / -0

What is $\frac{1+\sin A}{1-\sin A}-\frac{1-\sin A}{1+\sin A}$ equal to?

A
sec A – tan A

B
2 sec A. tan A

C
4 sec A. tan A

D
4 cosec A. cot A

Solution

$\frac{1 + \sin (A)}{1 - \sin (A)} - \frac{1 - \sin (A)}{1 + \sin (A)}$

$= \frac{{(1 + \sin (A))}^{2} - {(1 - \sin (A))}^{2}}{(1 - \sin (A)) (1 + \sin (A))}$

$= \frac{4 \times \sin (A) \times 1}{(1^{2}) - \sin {(A)}^{2}}$

$= \frac{4 \sin (A)}{\cos A^{2}}$

$= 4 \tan A s e c A$
Question 10
1 / -0

What is $\frac{\cot 224^{\circ}-\cot 134^{\circ}}{\cot 226^{\circ}+\cot 316^{\circ}}$ equal to?

A
– cosec 88°

B
– cosec 2°

C
– cosec 44°

D
–cosec 46°

Solution

$\frac{\cot 224^{\circ}-\cot 134^{\circ}}{\cot 226^{\circ}+\cot 316^{\circ}}$
$=\frac{\cot \left(180^{\circ}+44^{\circ}\right)-\cot \left(180^{\circ}-46^{\circ}\right)}{\cot \left(180+46^{\circ}\right)+\cot \left(270^{\circ}+46\right)^{\circ}}$
$=\frac{\cot 44^{\circ}+\cot 46^{\circ}}{\cot 46^{\circ}-\tan 46^{\circ}}=\frac{\tan 46^{\circ}+\tan 44^{\circ}}{\tan 44^{\circ}-\tan 46^{\circ}}$
$=\frac{\sin \left(46^{\circ}+44^{\circ}\right)}{\sin \left(44^{\circ}-46^{\circ}\right)}=-\operatorname{cosec} 2^{\circ}$

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Mathematics Test - 11

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SHARING IS CARING

Question 1

(√5 - 1)/4

(√5 + 1)/4

\(\frac{\sqrt{10+2 \sqrt{5}}}{4}\)

\(\frac{\sqrt{10-2 \sqrt{5}}}{4}\)

Solution

Question 2

1 only

2 only

Both 1 and 2

Neither 1 nor 2

Solution

Question 3

\(\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\)

\(\pi+\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\)

\(\pi-\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\)

\(\tan ^{-1}\left(\frac{x-y}{1+x y}\right)\)

Solution

Question 4

1 only

2 only

Both 1 and 2

Neither 1 nor 2

Solution

Question 5

1 only

2 only

Both 1 and 2

Neither 1 nor 2

Solution

Question 6

10√3 m

10(√3-1)m

10(√3+1)m

20m

Solution

Question 7

1 only

2 only

Both 1 and 2

Neither 1 nor 2

Solution

Question 8

Rhombus

Rectangle

Square

Parallelogram

Solution

Question 9

sec A – tan A

2 sec A. tan A

4 sec A. tan A

4 cosec A. cot A

Solution

Question 10

– cosec 88°

– cosec 2°

– cosec 44°

–cosec 46°

Solution

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