Trigonometric Functions Test 0

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Trigonometric Functions Test 0

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Weekly Quiz Competition

Question 1
1 / -0

Number of values of $$\theta \epsilon [0, 2\pi]$$ satisfying the equation $$\cot x-\cos x=1-\cot x.\cos x$$

A
1

B
2

C
3

D
4

Solution

$$\dfrac{cosx}{sinx}-cosx = 1 - \dfrac{cos^2x}{sinx}$$

$$\Rightarrow cosx-cosx*sinx=sinx-cos^2x$$

$$\Rightarrow cos^2x+cosx-cosxsinx-sinx=0$$

$$\Rightarrow cosx(1+cosx)-sinx(1+cosx)=0$$

$$\Rightarrow (cosx-sinx)(1+cosx)=0$$

$$\Rightarrow x = \dfrac{\pi}{4} , \dfrac{5\pi}{4} , {\pi} $$

Hence, 3 solutions exist.
Question 2
1 / -0

The smallest positive integral value of p for which the equation $$\cos \left ( p\sin x \right )=\sin \left ( p\cos x \right )$$ in x has a solution in $$\left [ 0, 2\pi \right ]$$ is

A
2

B
1

C
3

D
none of these

Solution

$$\cos(p \sin x) = \sin(p \cos x)$$

$$\Rightarrow \cos( p \sin x) = \cos\left(\dfrac{\pi}{2} - p \cos x \right)$$

$$\Rightarrow p \sin x = \dfrac{\pi}{2} - p \cos x \Rightarrow \cos x + \sin x = \dfrac{\pi}{2p}$$

condition for this equation to have solution is,
$$ \dfrac{\pi}{2p} \leq \sqrt{2} \Rightarrow p \geq\dfrac{\pi}{2\sqrt{2}} $$

Hence smallest positive integer is $$2$$.
Question 3
1 / -0

The number of solutions of equation $$5\sec \theta-13=12\tan \theta $$ in $$\left [ 0, 2\pi \right ]$$ is

A
2

B
1

C
4

D
0

Solution

$$ 5 sec\theta - 13 = 12 tan\theta$$

$$\Rightarrow 12 sin\theta + 13 cos\theta = 5$$

$$\Rightarrow \dfrac{12}{\sqrt{313}} sin\theta + \dfrac{13}{\sqrt{313}} cos\theta = \dfrac{5}{\sqrt{313}}$$

Let $$ cos\alpha = \dfrac{13}{\sqrt{313}} \Rightarrow sin\alpha = \dfrac{12}{\sqrt{313}}$$

$$ \Rightarrow cos\theta.cos\alpha + sin\theta.sin\alpha = \dfrac{5}{\sqrt{313}}$$

$$ \Rightarrow cos(\theta - \alpha) = \dfrac{5}{\sqrt{313}}$$

Hence solution in the given interval is,

$$ \theta = \alpha - cos^{-1}(\dfrac{5}{\sqrt{313}}), \alpha + cos^{-1}(\dfrac{5}{\sqrt{313}}) $$

Hence, option 'A' is correct.
Question 4
1 / -0

The number of solutions of the equation $$x^3+2x^2+5x+2 cos x=0$$ in $$[0, 2\pi]$$ is:

A
0

B
1

C
2

D
3

Solution

Let $$f(x)=x^3+2x^2+5x+2 \cos x$$

$$\Rightarrow f'(x)=3x^2+4x+5-2 sinx=2(x+\frac {2}{3})^2+\frac {11}{3}-2 sin x$$

Now $$\frac {11}{3}-2 sinx > 0$$ $$x(as -1 \leq sin x \leq 1)$$

$$\Rightarrow f'(x) > 0$$ $$x\Rightarrow f(x)$$ is an increasing function.

Now $$f(0)=2$$

$$\Rightarrow f(x)=0$$ has no solution in $$[0, 2\pi]$$.

Hence (A) is the correct answer.
Question 5
1 / -0

The number of solutions of the equation $$\tan x+\sec x=2\cos x$$ lying in the interval $$\left [ 0, 2\pi \right ]$$ is

A
0

B
1

C
2

D
3

Solution

$$ tanx + secx = 2 cosx$$

$$ sinx + 1 = 2 cos^2x = 2 - 2 sin^2x$$

$$ 2 sin^2x + sinx -1 = 0$$

$$ (2 sinx - 1) (sinx + 1) = 0$$

but $$ sinx = -1 \Rightarrow x = \dfrac{3\pi}{2}$$, which does not satisfy the given equation.

$$ sinx = \dfrac{1}{2} = sin\dfrac{\pi}{6}$$

therefore general solution is,
$$ x = n\pi + (-1)^n.\dfrac{\pi}{6}$$

$$ x = ......, \dfrac{\pi}{6}, \dfrac{5\pi}{6}, ......$$

hence number of solution in the given interval is 2.
Question 6
1 / -0

For each integer $$n>1$$, let $$S(n)$$ denote the number of solution of the equation $$\sin x=\sin nx$$ on the interval $$[0,\pi]$$ Find the value of $$S(2)+S(3)+S(4)$$.

A
12

B
11

C
15

D
10

Solution

$$sin(x)=sin2x$$
$$sin(x)=2sin(x).cos(x)$$
$$sin(x)[2cos(x)-1]=0$$
$$sin(x)=0$$ and $$cos(x)=\dfrac{1}{2}$$
$$x=0,\pi$$ and $$x=\dfrac{\pi}{3}$$ ...(i)
Hence there are 3 solutions.

$$S(3)$$
$$sin(x)=sin3x$$
$$sin3x-sinx=0$$
$$2cos(2x).sinx=0$$
$$sin(x)=0$$ and $$cos(2x)=0$$
$$x=0,\pi$$ and $$x=\dfrac{\pi}{4},\dfrac{3\pi}{4}$$
Hence number of solutions are 4.

$$S(4)$$
$$sin(x)=sin4x$$
$$sin4x-sinx=0$$
$$2cos(\dfrac{5x}{2}).sin(\dfrac{3x}{2})=0$$
$$x=0,\dfrac{2\pi}{3}$$ and $$x=\dfrac{\pi}{5},\dfrac{3\pi}{5},\pi$$
Hence number of solutions are 5.
Thus $$S(2)+S(3)+S(4)$$
$$=3+4+5$$
$$=12$$.
Question 7
1 / -0

If $$0\leq x\leq 2\pi $$, $$0\leq y\leq 2\pi $$ and $$\sin x+\sin y=2$$ then the value of $$x+y$$ is

A
$$\pi $$

B
$$\dfrac{\pi }{2}$$

C
$$3\pi $$

D
none of these

Solution

We have, $$ \sin x + \sin y = 2$$

but we know that maximum value of \sin x is $$1$$

hence only solution is $$ \sin x = 1 $$ and $$\sin y = 1$$

so in $$0 \leq x \leq 2\pi, 0 \leq y \leq 2\pi $$

solution is $$ x = \dfrac{\pi}{2}, y = \dfrac{\pi}{2} $$

therefore,
$$ x + y = \pi$$

Hence, option 'A' is correct.
Question 8
1 / -0

If $$3\sin ^{2}\theta +2\sin ^{2}\phi=0 $$ and $$3\sin 2\theta +2\sin 2\phi=0$$, $$0< \theta < \frac{\pi }{2}$$ and $$0< \phi < \frac{\pi }{2}$$, then the value of $$\theta +2\phi $$

A
$$\dfrac{\pi }{2}$$

B
$$\dfrac{\pi }{4}$$

C
$$0$$

D
none of these

Solution
Question 9
1 / -0

if $$\displaystyle x,y\epsilon \left [ 0,2\pi \right ]$$ and $$\displaystyle \sin x+\sin y=2$$ then the value of $$ x+y$$ is

A
$$\displaystyle \pi $$

B
$$\displaystyle \frac{\pi}{2} $$

C
$$\displaystyle3 \pi $$

D
None of these

Solution

$$ sinx + siny = 2$$

but we know that maximum value of $$sinx$$ is 1

hence only solution is $$ sinx = 1 $$ and $$siny = 1$$

so in $$0 \leq x \leq 2\pi, 0 \leq y \leq 2\pi $$

solution is $$ x = \dfrac{\pi}{2}, y = \dfrac{\pi}{2} $$

therefore,

$$ x + y = \pi$$
Question 10
1 / -0

The number of solutions of the equation $$\displaystyle \cos 6x+\tan ^{2}x+\cos 6x\cdot \tan ^{2}x= 1$$ in the interval $$\displaystyle \left [ 0 ,2\pi \right ]$$ is

A
4

B
5

C
6

D
7

Solution

$$ cos6x + tan^2x +cos6x.tan^2x = 1$$

$$\Rightarrow cos6x(1 + tan^2x) = (1 - tan^2x) \Rightarrow cos6x = \dfrac{1 - tan^2x}{1 + tan^2x} $$

$$\Rightarrow cos6x = cos2x $$ $$[\because \sin 2A=\dfrac {2 \tan A}{1+ \tan^2A}$$, $$\cos 2A=\dfrac {1- \tan^2A}{1+ \tan^2A}]$$

Threrefore general solution is,

$$ 6x = 2n\pi \pm 2x $$, where $$n$$ is any integer

$$ x = \dfrac{n\pi}{2} $$ and $$ x = \dfrac{n\pi}{4}$$

Hence solutions in the given interval are,

$$ x = 0, \dfrac{\pi}{4}, \dfrac{\pi}{2}, \dfrac{3\pi}{4}, \pi, \dfrac{3\pi}{4}, 2\pi $$ , total $$7$$ solutions.

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Trigonometric Functions Test 0

Result

Trigonometric Functions Test 0

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Rank

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SHARING IS CARING

Question 1

1

2

3

4

Solution

Question 2

2

1

3

none of these

Solution

Question 3

2

1

4

0

Solution

Question 4

0

1

2

3

Solution

Question 5

0

1

2

3

Solution

Question 6

12

11

15

10

Solution

Question 7

$$\pi $$

$$\dfrac{\pi }{2}$$

$$3\pi $$

none of these

Solution

Question 8

$$\dfrac{\pi }{2}$$

$$\dfrac{\pi }{4}$$

$$0$$

none of these

Solution

Question 9

$$\displaystyle \pi $$

$$\displaystyle \frac{\pi}{2} $$

$$\displaystyle3 \pi $$

None of these

Solution

Question 10

4

5

6

7

Solution

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Rank

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