Trigonometric Functions Test 28

Result

Trigonometric Functions Test 28

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The trigonometric equation is-
$$\sin x+3 \sin 2x+\sin 3x=\cos x+3 \cos 2x+\cos 3x$$
when $$x$$ lies in first four quadrants. It means $$x\epsilon [0, 2\pi]$$, then-

How many solutions are there-

A
$$2$$

B
$$3$$

C
$$4$$

D
$$5$$

Solution

$$\sin x+3 \sin 2x+\sin 3x=\cos x+3 \cos 2x+\cos 3x$$

$$2\sin { 2x } \cos { x } +3\sin { 2x } =2\cos { 2x } \cos { x } +3\cos { 2x } $$

$$\Rightarrow (\sin { 2x-\cos { 2x } ) } (2\cos { x } +3)=0$$

$$\Rightarrow \tan 2x=1, \cos x=-\displaystyle \frac{3}{2}$$ (not possible)

Since, $$0 \le x\le 2\pi$$

$$0 \le 2x \le 4\pi$$

Hence, there are a number of solutions $$ = 4$$
Question 2
1 / -0

If $$\sec$$ $$\beta=\alpha+\dfrac{1}{4a}$$,then the value of $$\sec\beta+\tan\beta$$ is

A
$$a$$ or $$\dfrac{1}{a}$$

B
$$2a$$ or$$\dfrac{1}{2a}$$

C
$$4a$$ or $$\dfrac{1}{4a}$$

D
$$1$$

Solution

We know that

$$\sec^2\beta-\tan^2\beta=1$$

Therefore

$$\tan\beta=\pm\sqrt{\sec^2\beta-1}$$

$$\tan\beta=\pm\sqrt{\alpha^2+(\dfrac{1}{4\alpha})\:^2+\dfrac{1}{2}}-1$$

$$\tan\beta=\pm\sqrt{\alpha^2+(\dfrac{1}{4\alpha})\:^2-\dfrac{1}{2}}$$

$$\tan\beta=\pm\sqrt{(\alpha-\dfrac{1}{4\alpha})\:^2}$$

$$\tan\beta=\pm(\alpha-\dfrac{1}{4\alpha})$$ ...(i)

Hence $$sec\alpha+\tan\alpha=2\alpha\:or\:\dfrac{1}{2\alpha}$$

The answer is B
Question 3
1 / -0

$$|\tan\theta+\sec\theta|=|\tan\theta|+|\sec\theta|, 0\leq \theta \leq 2\pi$$ is possible only if-

A
$$\theta \epsilon [0, \pi]-\left \{\dfrac {\pi}{2}\right \}$$

B
$$\theta \epsilon [0, \pi]$$

C
$$\theta \epsilon [0, \dfrac {\pi}{2})$$

D
$$(0, \dfrac {\pi}{2}]$$

Solution

$$\mid \tan\theta+\sec\theta \mid=\mid \tan\theta \mid +\sec\theta$$
Case 1:-
If $$\theta \in [0, \cfrac{\pi}{2})$$ then $$\tan\theta=+ve , \,\, \sec\theta=+ve$$
$$\tan\theta+\sec\theta=+ve$$
$$\therefore \mid \sec\theta+\tan\theta \mid =\mid \tan\theta \mid +\mid \sec\theta \mid$$

Case 2:-
$$\theta \in (\cfrac{\pi}{2}, \pi]$$
$$\tan\theta=-ve , \,\, \sec\theta=-ve$$
$$\mid \sec\theta+\tan\theta \mid=+ve$$
$$\mid \sec\theta \mid=+ve \,\,\, \mid \tan\theta \mid=+ve$$
$$\mid \sec\theta+\tan\theta \mid=\mid \sec\theta \mid +\mid \tan\theta \mid$$

Case 3:-
$$\theta \in [ \pi,\cfrac{3\pi}{2}]$$
$$\tan\theta=+ve , \,\, \sec\theta=-ve$$
$$\mid \sec\theta+\tan\theta \mid=\tan\theta-\sec\theta$$
$$\mid \tan\theta \mid +\mid \sec\theta \mid=\sec\theta+\tan\theta$$
$$\mid \sec\theta+\tan\theta \mid \ne \mid \sec\theta \mid +\mid \tan\theta \mid$$

Case 4:-
$$\theta \in (\cfrac{3\pi}{2}, 2\pi]$$
$$\tan\theta=-ve , \,\, \sec\theta=+ve$$
$$\mid \sec\theta+\tan\theta \mid=\sec\theta -\tan\theta$$
$$\mid \sec\theta \mid=\sec\theta \,\,\, \mid \tan\theta \mid=\tan\theta$$
$$\mid \sec\theta+\tan\theta \mid \ne \mid \sec\theta \mid +\mid \tan\theta \mid$$

$$\theta \ne \cfrac{\pi}{2}$$ because in that case $$\tan\theta=\infty$$
Question 4
1 / -0

Directions For Questions

The trigonometric equation is-
$$\sin x+3 \sin 2x+\sin 3x=\cos x+3 \cos 2x+\cos 3x$$
when $$x$$ lies in first four quadrants. It means $$x\epsilon [0, 2\pi]$$, then-

...view full instructions

The sum of the solution of $$x$$ is-

A
$$\frac {14\pi}{23}$$

B
$$\frac {3\pi}{4}$$

C
$$\frac {9\pi}{8}$$

D
$$6\pi$$

Solution

Given equation is
$$\sin\,x+3\,\sin\,2x+\sin\,3x=\cos\,x+3\,\cos\,2x+\cos\,3x$$
$$(\sin x+\sin\,3x)+3\sin\,2x=(\cos x+\cos\,3x)+3\cos\,2x$$
$$(2\sin\,2x\,\cos x+3\sin\,2x)-(2\cos\,2x\,\cos x+3\cos\,2x)=0$$
$$\sin\,2x(2\cos\,x+3)-\cos\,2x(2\cos\,x+3)=0$$
$$\Rightarrow (\sin 2x-\cos 2x)(2\cos x+3)=0$$
$$cos\,x=-\displaystyle\frac{3}{2}$$
which is not possible
$$\sin\,2x=\cos\,2x$$
$$\tan\,2x=1$$
$$\tan 2x=tan\displaystyle\frac{\pi}{4}, \tan\frac{5\pi}{4}$$
$$x=\displaystyle\frac{\pi}{8},\dfrac{5\pi}{8}$$
Required sum $$=\dfrac{5\pi}{8}+\dfrac{\pi}{8}=\frac{3\pi}{4}$$
Question 5
1 / -0

Consider the system of equations $$\displaystyle \sin x \cos 2y= (a^{2}-1)^{2}+1,\ \cos x\sin 2y= a+1$$, then the number of values of $$\displaystyle y\in [0,2\pi]$$ when the system has solution for permissible values of $$a$$ are,

A
$$2$$

B
$$3$$

C
$$4$$

D
$$5$$

Solution
Question 6
1 / -0

The value of $$\displaystyle \sin ^{2}1^{\circ}+\sin ^{2}2^{\circ}+\sin ^{2}2^{\circ}+...+\sin ^{2}89^{\circ}+\sin ^{2}90^{\circ}$$

A
1

B
0

C
45.5

D
44

Solution

$$\displaystyle \sin ^{2}1^{\circ}+\sin ^{2}2^{\circ}+\sin ^{2}2^{\circ}+...+\sin ^{2}89^{\circ}+\sin ^{2}90^{\circ}$$
= $$(\sin^2 1 + \sin^2 89) + (\sin^2 + \sin^288) + ....+ (\sin^2 44 + \sin^2 46) + (\sin^245 + \sin^ 90)$$
= $$(\sin^2 1 + \cos^2 (90 - 1)) + (\sin^2 + \cos^2(90 - 2)) + ....+ (\sin^2 44 + \sin^2 (90 - 44)) + (\sin^245 + \sin^ 90)$$
= $$(\sin^2 1 + \sin^2 1) + (\sin^2 + \sin^2 2) + ....+ (\sin^2 44 + \cos^2 44) + (\sin^245 + \sin^ 90)$$
= $$1 + 1 ...44 times + (\dfrac{1}{\sqrt{2}})^2 +1$$
= $$45 + \dfrac{1}{2}$$
= $$45.5$$
Question 7
1 / -0

Consider the system of equations $$\displaystyle \sin x. \cos 2y= (a^{2}-1)^{2}+1,\ \cos x.\sin 2y= a+1$$, then the number of values of $$\displaystyle x\epsilon [0,2\pi]$$ when the system has a solution for permissible values of a is/are,

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution
Question 8
1 / -0

The total number of solutions of $$\displaystyle \sin \left \{ x \right \}=\cos \left \{ x \right \}$$ (where $$\displaystyle \left \{ . \right \}$$ denotes the fractional part) in $$\displaystyle \left [ 0,2\pi \right ]$$ is equal to

A
$$5$$

B
$$6$$

C
$$8$$

D
None of these

Solution

$$ \sin\{x\} = \cos\{x\} $$

$$\Rightarrow \tan\{x\} = 1 $$

thus general solution is,

$$ \{x\} = x - [x] = n\pi + \dfrac{\pi}{4} $$, where n is any integer

hence solution in the given interval is,

$$ x = \dfrac{\pi}{4}, 1 + \dfrac{\pi}{4}, 2 + \dfrac{\pi}{4}, 3 + \dfrac{\pi}{4}, 4 + \dfrac{\pi}{4}, 5 + \dfrac{\pi}{4}$$

Hence, option 'B' is correct.
Question 9
1 / -0

In the given figure, $$\displaystyle \angle B =90^{\circ}$$ and $$\displaystyle \angle ADB=x^{\circ}$$, then find $$\displaystyle \cos^{2} C^{\circ}+\sin^{2} C^{\circ} $$.

A
1

B
2

C
0

D
-1

Solution

In $$\triangle CAB$$,

$$CA = 20$$ and $$AB = 12$$

Using Pythagoras theorem,
$$CA^2 = AB^2 + BC^2$$
$$20^2 = 12^2 + BC^2$$
$$BC = 16$$

$$\cos^2 C + \sin^2 C = \left (\dfrac{BC}{AC}\right )^2 + \left (\dfrac{AB}{AC}\right )^2$$

= $$\left (\dfrac{16}{20}\right )^2 + \left (\dfrac{12}{20}\right )^2$$

= $$\dfrac{256 + 144}{400}$$

= $$\dfrac{400}{400}$$

= $$1$$
Question 10
1 / -0

The number of real solutions of $$\sin e^{x}.\cos e^{x}=2^{x-2}+2^{-x-2}$$ is

A
zero

B
one

C
two

D
infinite

Solution

$$\sin e^{x}.\cos e^{x}=2^{x-2}+2^{-x-2}$$
$$\Rightarrow \sin { 2e^{ x } } =2^{ x-1 }+2^{ -x-1 }$$
$$\Rightarrow 2\sin { 2e^{ x } } =2^{ x }+2^{ -x }$$
$$\Rightarrow \displaystyle \frac { 1 }{ 2 } \sin { 2e^{ x } } =\frac { 1 }{ 4 } (2^{ x }+2^{ -x })$$
For any real value of $$x$$, $$2^x+2^{-x} \ge 2$$
$$\Rightarrow \displaystyle \frac { 1 }{ 2 } \sin { 2e^{ x } } \ge \frac { 1 }{ 2 } $$
$$\Rightarrow \sin { 2e^{ x } } \ge 1$$
But $$\sin 2e^x \le 1$$
So, $$\sin 2e^x=1$$
But equality can hold only if RHS also equals $$1$$ , which is possible only when $$x=0$$
Hence, $$\sin 2e^x=1$$ is not true
Hence, there is no solution.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Trigonometric Functions Test 28

Result

Trigonometric Functions Test 28

Score

-

Rank

-

SHARING IS CARING

Question 1

$$2$$

$$3$$

$$4$$

$$5$$

Solution

Question 2

$$a$$ or $$\dfrac{1}{a}$$

$$2a$$ or$$\dfrac{1}{2a}$$

$$4a$$ or $$\dfrac{1}{4a}$$

$$1$$

Solution

Question 3

$$\theta \epsilon [0, \pi]-\left \{\dfrac {\pi}{2}\right \}$$

$$\theta \epsilon [0, \pi]$$

$$\theta \epsilon [0, \dfrac {\pi}{2})$$

$$(0, \dfrac {\pi}{2}]$$

Solution

Question 4

$$\frac {14\pi}{23}$$

$$\frac {3\pi}{4}$$

$$\frac {9\pi}{8}$$

$$6\pi$$

Solution

Question 5

$$2$$

$$3$$

$$4$$

$$5$$

Solution

Question 6

1

0

45.5

44

Solution

Question 7

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 8

$$5$$

$$6$$

$$8$$

None of these

Solution

Question 9

1

2

0

-1

Solution

Question 10

zero

one

two

infinite

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.