Trigonometric Functions Test 29

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Trigonometric Functions Test 29

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Weekly Quiz Competition

Question 1
1 / -0

The number of solutions of $$\displaystyle \sum_{r=1}^{5} \cos r x=5 $$ in the interval of $$\displaystyle \left [ 0,2\pi \right ]$$ is

A
$$0$$

B
$$2$$

C
$$5$$

D
$$10$$

Solution

$$ \cos x + \cos 2x + \cos 3x + \cos 4x + \cos 5x = 5$$
we know, $$ \cos x, \cos 2x, \cos 3x, \cos 4x, \cos 5x \leq 1 $$
considering both,
$$ \cos x, \cos 2x, \cos 3x, \cos 4x, \cos 5x = 1$$
hence general solutions in the given interval are,
$$ \cos x = 1 \Rightarrow x = 0, 2\pi$$
$$ \cos 2x = 1 \Rightarrow 2x = 0, 2\pi, 4\pi \Rightarrow x = 0, \pi, 2\pi $$
$$ \cos 3x = 1 \Rightarrow 3x = 0, 2\pi, 4\pi, 6\pi \Rightarrow x = 0, \dfrac{2\pi}{3}, 2\pi $$
$$ \cos 4x = 1 \Rightarrow 4x = 0, 2\pi, 4\pi, 6\pi, 8\pi \Rightarrow x = 0, \dfrac{\pi}{2},\pi, \dfrac{3\pi}{2}, 2\pi $$
$$ \cos 5x = 1 \Rightarrow 5x = 0, 2\pi, 4\pi, 6\pi, 8\pi, 10\pi \Rightarrow x = 0, \dfrac{2\pi}{5}, \dfrac{4\pi}{5}, \dfrac{6\pi}{5}, \dfrac{8\pi}{5}, 2\pi $$
thus common solutions are,
$$ x = 0, 2\pi$$
Hence, option 'B' is correct.
Question 2
1 / -0

If $$3x = \text{cosec } \theta$$ and $$\dfrac {3}{x} = \cot \theta$$, then $$\left (x^{2} - \dfrac {1}{x^{2}}\right ) =$$

A
$$\dfrac {1}{27}$$

B
$$\dfrac {1}{81}$$

C
$$\dfrac {1}{3}$$

D
$$\dfrac {1}{9}$$

Solution

Since $$cosec^{2} \theta - \cot^{2}\theta = 1$$
$$\therefore (3x)^{2} - \left (\dfrac {3}{x}\right )^{2} = 1 $$, substitute the given values
$$\Rightarrow 9x^{2} - \dfrac {9}{x^{2}} = 1\Rightarrow 9 \left (x^{2} - \dfrac {1}{x^{2}}\right ) = 1$$
$$\Rightarrow x^{2} - \dfrac {1}{x^{2}} = \dfrac {1}{9}$$.
Question 3
1 / -0

If $$\displaystyle \sin x+\mathrm{cosec}\: x=2, $$ then $$\displaystyle \sin ^{n} x + \mathrm{cosec} ^{n} \: x$$ is equal to

A
$$2$$

B
$$\displaystyle 2^{n}$$

C
$$\displaystyle 2^{n-1}$$

D
$$\displaystyle 2^{n-2}$$

Solution

We have $$ sin x + cosec x = 2 $$
$$ => sin x + \dfrac {1}{sin x } = 2 $$

This means $$ x = 90^0 $$ as maximum value of $$ sin x = 1 $$ such that $$ 1 + 1 = 2 $$

Now, for any $$ n $$, we have $$ sin^n 90^0 + cosec ^n 90^0 = 1^2 + 1^2 = 2 $$
Question 4
1 / -0

The value of $$\displaystyle \frac { \sin { \theta } \cos { \theta } .\sin { \left( { 90 }^{ o }-\theta \right) } }{ \cos { \left( { 90 }^{ o }-\theta \right) } } +\frac { \cos { \theta } .\sin { \theta } .\cos { \left( { 90 }^{ o }-\theta \right) } }{ \sin { \left( { 90 }^{ o }-\theta \right) } } +\frac { { \sin }^{ 2 }{ 27 }^{ o }+{ \sin }^{ 2 }{ 63 }^{ o } }{ { \cos }^{ 2 }{ 40 }^{ o }+{ \cos }^{ 2 }{ 50 }^{ o } } $$ is :

A
$$1$$

B
$$2$$

C
$$3$$

D
$$0$$

Solution

$$\displaystyle \sin { \left( { 90 }^{ o }-\theta \right) } =\cos { \theta } ;\cos { \left( { 90 }^{ o }-\theta \right) } =\sin { \theta } $$
$$\displaystyle \sin { { 27 }^{ o } } =\sin { \left( { 90 }^{ o }-{ 63 }^{ o } \right) } ={ \cos63 }^{ o }$$
$$\displaystyle { \cos40 }^{ o }=\cos { \left( { 90 }^{ o }-{ 50 }^{ o } \right) } =\sin { { 50 }^{ o } } $$
$$\displaystyle \therefore \quad Given\quad exp.$$
$$\displaystyle =\frac { \sin { \theta } .\cos { \theta } .\cos { \theta } }{ \sin { \theta } } +\frac { \cos { \theta } .\sin { \theta } \sin { \theta } }{ \cos { \theta } } +\frac { \left( { \cos }^{ 2 }{ 63 }^{ o }+{ \sin }^{ 2 }{ 63 }^{ o } \right) }{ \left( { \sin }^{ 2 }{ 50 }^{ o }+{ \cos }^{ 2 }{ 50 }^{ o } \right) } $$
$$\displaystyle ={ \cos }^{ 2 }\theta +{ \sin }^{ 2 }\theta +\frac { 1 }{ 1 } =1+1=2$$
Question 5
1 / -0

If $$\displaystyle p=\sqrt{\frac{1-\sin x}{1+\sin x}},q=\frac{1-\sin x}{\cos x},r=\frac{\cos x}{1+\sin x}$$
Which one of the following statement is correct ?

A
$$\displaystyle p=q\neq r$$

B
$$\displaystyle q=r\neq p$$

C
$$\displaystyle r=p\neq q$$

D
$$\displaystyle p=q=r$$

Solution

Consider, $$\displaystyle p=\sqrt{\frac{1-\sin x}{1+\sin x}}$$

$$=\displaystyle \sqrt{\frac{1-\sin x}{1+\sin x}\times\frac{1-\sin x}{1+\sin x}}$$

$$=\displaystyle \sqrt{\frac{(1-\sin x)^2}{1-\sin^2 x}}$$

$$=\displaystyle \sqrt{\frac{(1-\sin x)^2}{\cos^2 x}}$$

$$q=\displaystyle \frac{1-\sin x}{\cos x}$$

Again consider, $$ r=\displaystyle \frac{\cos x}{1+\sin x}=\displaystyle \frac{\cos x}{1+\sin x} \times \frac{1-\sin x}{1-\sin x}$$

$$=\displaystyle \frac{\cos x(1-\sin x)}{1-\sin^2 x} $$

$$=\displaystyle \frac{\cos x(1-sinx)}{]cos^2 x} $$

$$=\displaystyle \frac{1-\sin x}{\cos x} $$

$$\text{Clearly, p=q=r}$$

Option D is correct.
Question 6
1 / -0

The value of $$\alpha \varepsilon (- \pi, 0)$$ satisfying $$sin \alpha + \int_{\alpha}^{2 \alpha} . cos 2x dx = 0$$ is

A
$$0$$

B
$$- \dfrac{\pi}{3}$$

C
$$-\pi$$

D
All of these

Solution

Consider,
$$sin\alpha + \int^{2\alpha}_\alpha cos2x\ dx = sin\alpha +\left [\dfrac{sin2\alpha}{2} \right]_{\alpha}^{2\alpha}$$
$$\Rightarrow sin \alpha+\dfrac{1}{2}[sin 4\alpha-sin 2\alpha]\\= sin 2\alpha(2cos 2\alpha-1)+sin\alpha\\= (4cos^3\alpha-3cos\alpha +1)sin\alpha $$

And it is given that $$(\alpha \ne 0)$$

$$cos \alpha=1/2 $$ is satisfying the above equation

$$\alpha =\dfrac{-\pi}{3}$$
Question 7
1 / -0

$$\dfrac12\sin{(2x)}(1+\cot ^{ 2 }{ (x) } )$$ is equal to

A
$$\tan(x)$$

B
$$\sin (x)$$

C
$$\cos(x)$$

D
$$\cot(x)$$

E
$$\sec(x)$$

Solution

The given identity can be solved as:
$$\left( \dfrac { 1 }{ 2 } \right) (\sin { 2x) } (1+\cot ^{ 2 }{ x } )$$

$$=\left( \dfrac { 1 }{ 2 } \right) (2\sin { x } \cos { x } )(cosec ^{ 2 }{ x } )$$ .... (because $$\sin { 2x } =2\sin { x } \cos { x }$$ and $$1+\cot ^{ 2 }{ x } =cosec ^{ 2 }{ x }$$)

$$=(\sin { x } \cos { x } )\left( \dfrac { 1 }{ \sin ^{ 2 }{ x } } \right)$$ ...... (because $$cosec ^{ 2 }{ x } =\dfrac { 1 }{ \sin ^{ 2 }{ x } }$$)

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

$$=\cot { x }$$

Hence, $$\left( \dfrac { 1 }{ 2 } \right) (\sin { 2x) } (1+\cot ^{ 2 }{ x } )=\cot { x }$$.
Question 8
1 / -0

The number of $$x\epsilon [0, 2\pi]$$ for which $$|\sqrt {2\sin^{4} x + 18\cos^{2}x} - \sqrt {2\cos^{4} x + 18\sin^{2} x}| = 1$$ is:

A
$$4$$

B
$$2$$

C
$$6$$

D
$$8$$

Solution

$$sin^2x+cos^2x=1$$
$$sin^4x+cos^4x+2sin^2xcos^2x=1$$
$$sin^4x+cos^4x=1-2sin^2xcos^2x$$ ...[1]

$$sin^2x+cos^2x=1$$
$$sin^6x+cos^6x+3sin^2xcos^2x(sin^2x+cos^2x)=1$$
$$sin^6x+cos^6x=1-3sin^2xcos^2x$$ ...[2]

$$|\sqrt{2sin^4x+18cos^2x}-\sqrt{2cos^4x+18sin^2x}|=1$$

Squaring above equation
$$\therefore 2sin^4x+18cos^2x+2cos^4x+18sin^2x-2(\sqrt{2sin^4x+18cos^2x})(\sqrt{2cos^4x+18sin^2x})=1$$

$$\therefore 2(sin^4x+cos^4x)+18(cos^2x+sin^2x)-1=2(\sqrt{2sin^4x+18cos^2x})(\sqrt{2cos^4x+18sin^2x})$$

$$\therefore 2(1-2sin^2xcos^2x)+18-1=2(\sqrt{2sin^4x+18cos^2x})(\sqrt{2cos^4x+18sin^2x})$$ ..( [Using[1] )

$$\therefore 19-4sin^2xcos^2x=2(\sqrt{2sin^4x+18cos^2x})(\sqrt{2cos^4x+18sin^2x})$$

Squaring above equation
$$\therefore 361+16sin^4xcos^4x-152sin^2xcos^2x=4(4sin^4xcos^4x+36cos^6x+36sin^6x+324sin^2xcos^2x)$$

$$\therefore 361+16sin^4xcos^4x-152sin^2xcos^2x=16sin^4xcos^4x+144cos^6x+144sin^6x+1296sin^2xcos^2x$$

$$\therefore 361=144(cos^6x+sin^6x)+1448sin^2xcos^2x$$

$$\therefore 361=144(1-3sin^2xcos^2x)+1448sin^2xcos^2x$$ ....( Using [2] )

$$\therefore 217=1016sin^2xcos^2x$$

$$\therefore 217=254sin^22x$$

$$\therefore sin2x=\pm \sqrt{\dfrac{217}{254}}$$

Let $$sin^{-1} \sqrt{\dfrac{217}{254}}=\theta$$

$$ sin2x=\pm \sqrt{\dfrac{217}{254}}$$

$$\therefore 2x=sin^{-1} \sqrt{\dfrac{217}{254}}$$

$$\therefore 2x=\theta,\pi-\theta,2\pi+\theta,3\pi-theta$$

$$\therefore x=\dfrac{\theta}{2},\dfrac{\pi-\theta}{2},\dfrac{2\pi+\theta}{2},\dfrac{3\pi-\theta}{2}$$

$$ 2x=sin^{-1} (-\sqrt{\dfrac{217}{254}})$$

$$\therefore 2x=\pi+\theta,2\pi-\theta,3\pi+\theta,4\pi-\theta$$

$$\therefore x=\dfrac{\pi+\theta}{2},\dfrac{2\pi-\theta}{2},\dfrac{3\pi+\theta}{2},\dfrac{4\pi-\theta}{2}$$

So, total number of solutions =8
So, the answer is option (D)
Question 9
1 / -0

The range of $$f(x)=\cfrac { 1 }{ |\sin\ x| } +\cfrac { 1 }{ |\cos\ x| } $$ is

A
$$(2\sqrt { 2 } ,\quad \infty )$$

B
$$(\sqrt { 2 } ,\quad 2\sqrt { 2 } )$$

C
$$(0,\quad 2\sqrt { 2 } )$$

D
$$[2\sqrt { 2 } ,\quad 4]$$

Solution

Given, $$f(x)=\cfrac{1}{\mid \sin x \mid}$$ + $$\cfrac{1}{\mid \cos x \mid}$$
If either of $$\sin x$$ or $$\cos x=0$$
Then $$f(x)=\infty$$
$$\therefore$$ maximum of $$f(x)=\infty$$
$$f(x)$$ will be minimum if $$\sin x=\cos x \Rightarrow x=\cfrac{\pi}{4}$$
$$\therefore \mid \sin x \mid=\mid \cos x \mid=\cfrac{!}{\sqrt 2}$$
$$\therefore$$ minimum of $$f(x)=\sqrt 2+\sqrt 2=s\sqrt 2$$
$$\therefore$$ Range of $$f(x)=[2\sqrt 2,\infty)$$
Question 10
1 / -0

$$\cos (2001) \pi + \cot (2001)\dfrac {\pi}{2} + \sec (2001) \dfrac {\pi}{3} + \tan (2001) \dfrac {\pi}{4} + cosec (2001) \dfrac {\pi}{6}$$ equal to

A
$$0$$

B
$$1$$

C
$$-2$$

D
Not defined

Solution

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Trigonometric Functions Test 29

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Trigonometric Functions Test 29

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SHARING IS CARING

Question 1

$$0$$

$$2$$

$$5$$

$$10$$

Solution

Question 2

$$\dfrac {1}{27}$$

$$\dfrac {1}{81}$$

$$\dfrac {1}{3}$$

$$\dfrac {1}{9}$$

Solution

Question 3

$$2$$

$$\displaystyle 2^{n}$$

$$\displaystyle 2^{n-1}$$

$$\displaystyle 2^{n-2}$$

Solution

Question 4

$$1$$

$$2$$

$$3$$

$$0$$

Solution

Question 5

$$\displaystyle p=q\neq r$$

$$\displaystyle q=r\neq p$$

$$\displaystyle r=p\neq q$$

$$\displaystyle p=q=r$$

Solution

Question 6

$$0$$

$$- \dfrac{\pi}{3}$$

$$-\pi$$

All of these

Solution

Question 7

$$\tan(x)$$

$$\sin (x)$$

$$\cos(x)$$

$$\cot(x)$$

$$\sec(x)$$

Solution

Question 8

$$4$$

$$2$$

$$6$$

$$8$$

Solution

Question 9

$$(2\sqrt { 2 } ,\quad \infty )$$

$$(\sqrt { 2 } ,\quad 2\sqrt { 2 } )$$

$$(0,\quad 2\sqrt { 2 } )$$

$$[2\sqrt { 2 } ,\quad 4]$$

Solution

Question 10

$$0$$

$$1$$

$$-2$$

Not defined

Solution

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