If $$\sin ^ { - 1 } x = y ,$$ then

$$- \frac { \pi } { 2 } \leq y \leq \frac { \pi } { 2 }$$

$$- \frac { \pi } { 2 } < y < \frac { \pi } { 2 }$$

If $$x = \tan \theta + \cot \theta, y = \cos \theta - \sin \theta$$, then

$$\dfrac {y^2-1}{2} = \dfrac{1}{x}$$

MCQ Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 11

$\cos\ \theta+\sin\ \theta=a$ , then

$\sin\ 2\theta=$

0%
$a^{2}+1$
0%
$a^{2}-1$
0%
$a^{2}$
0%
$1$

Explanation

$\cos{\theta}+\sin{\theta}=a$

Squaring both sides, we get

${\cos}^{2}{\theta}+{\sin}^{2}{\theta}+2\cos{\theta}\sin{\theta}={a}^{2}$

$\Rightarrow\,1+2\cos{\theta}\sin{\theta}={a}^{2}$ since

${\cos}^{2}{\theta}+{\sin}^{2}{\theta}=1$

$\Rightarrow\,\sin{2\theta}={a}^{2}-1$

$\sin^{2} x- \cos x=1/4$ , then the value of

$x$ between

$0$ and

$2\pi$ are :

0%
$\pi /3, 5\pi /3$
0%
$\pi /3, -\pi/3$
0%
$2 \pi/3, \pi/3$
0%
$None\ of\ these$

Explanation

Given

$\sin ^2x-\cos x=\dfrac 14$

$\implies 1-\cos ^2x-\cos x=\dfrac 14$

$\implies 4-4\cos ^2x-4\cos x=1$

$\implies 4\cos ^2x+4\cos x-3=0$

$\implies 4\cos ^2x+6\cos x-2\cos x-3=0$

$\implies 2\cos x(2\cos x+3)-1(2\cos x+3)=0$

$\implies (2\cos x-1)(2\cos x+3)=0$

$\implies \cos x=\dfrac 12$ and

$\cos x\neq -\dfrac 32$

$\implies x=\dfrac \pi 3,\dfrac {5\pi}3$

$\dfrac{1+cot\alpha+cosec\alpha}{1-cot\alpha+cosec \alpha}=$

0%
$\dfrac{sin\alpha}{1+cos\alpha}$
0%
$\dfrac{sin\alpha}{1-cos\alpha}$
0%
$\dfrac{1+cos\alpha}{sin\alpha}$
0%
$\dfrac{1-sin\alpha}{cos\alpha}$

$\sin ^ { - 1 } x = y ,$ then

0%
$0 \leq y \leq \pi$
0%
$- \frac { \pi } { 2 } \leq y \leq \frac { \pi } { 2 }$
0%
$0 < y < \pi$
0%
$- \frac { \pi } { 2 } < y < \frac { \pi } { 2 }$

Explanation

$\begin{array}{l} { \sin ^{ -1 } }x=y \\ or\, \, y={ \sin ^{ -1 } }x \end{array}$

we know that

Range of principle Value of

$\sin$ is

$\left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]$

So,

$- \frac{\pi }{2} \le y \le \frac{\pi }{2}$

The equation

$\sin x + \sin y + \sin z = -3$ for

$0 \le x \le 2\pi, 0 \le y \le 2\pi, 0 \le z \le 2\pi$ , has

0%
One solution
0%
Two sets of solutions
0%
Four sets of solutions
0%
No solution

Explanation

$sinx+siny+sinz=-3$

only possible then

$x=y=z={ sin }^{ -1 }\left( -1 \right)$

So,

$x=y=z=3\Pi /2$

for

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

So, Only one solution is possible.

The general solution to the equation

$\sin 3\alpha = 4\sin \alpha \sin(x + \alpha) \sin(x - \alpha)$ is.

0%
$n\pi \pm \dfrac{\pi}{4}, \forall n \in I$
0%
$n\pi \pm \dfrac{\pi}{3}, \forall n \in I$
0%
$n\pi \pm \dfrac{\pi}{9}, \forall n \in I$
0%
$n\pi \pm \dfrac{\pi}{12}, \forall n \in I$

Number of solutions of the equation

$4\sin^{2}x+\tan^{2}x+\cot^{2}x+\csc^{2}=6$ in

$[0, \pi]$ is

0%
$2$
0%
$8$
0%
$0$
0%
$4$

Solution of

$7 \sin ^ { 2 } x + 3 \cos ^ { 2 } x = 4$ is

0%
$n \pi \pm \frac { \pi } { 2 }$
0%
$n \pi \pm \frac { \pi } { 4 }$
0%
$n \pi \pm \frac { \pi } {3 }$
0%
$n \pi \pm \frac { \pi } { 6 }$

Explanation

$\\7sin^2x+3cos^2x=4\\4sin^2x+3sin^2x+3cos^2x=4\\4sin^2x+3=4\\4sin^2x=1\\sin^2x=(\frac{1}{4})\\sinx=\pm (\frac{1}{2})\\\therefore\>x=n\pi\pm (\frac{\pi}{6})$

Genral solution of the equation

$\theta -cosec\theta =\dfrac { 4 }{ 3 }$ is :

0%
$\frac { 1 }{ 2 } \left( \left[ n\pi +{ \left( -1 \right) }^{ n }{ sin }^{ -1 }\left( 3/4 \right) \right] \right) n\epsilon Z$
0%
$n\pi +{ \left( -1 \right) }^{ n }{ sin }^{ -1 }\left( 3/4 \right) n\epsilon Z$
0%
$\frac { n\pi }{ 2 } +{ \left( -1 \right) }^{ n }{ sin }^{ -1 }\left( 3/4 \right) n\epsilon Z$
0%
none of these

$\sin \alpha = a \sin \beta$ and

$\cos \alpha = b \cos \beta$ then

$\tan \alpha =$

0%
$\pm \sqrt { \left( \frac { a ^ { 2 } b ^ { 2 } - a ^ { 2 } } { b ^ { 2 } - a ^ { 2 } b ^ { 2 } } \right) }$
0%
$\pm \sqrt { \left( \frac { a ^ { 2 } - a ^ { 2 } b ^ { 2 } } { b ^ { 2 } +a ^ { 2 } b ^ { 2 } } \right) }$
0%
$\pm \sqrt { \left( \frac { 1 - a ^ { 2 } } { 1 - b ^ { 2 } } \right) }$
0%
$\pm \sqrt { \left( \frac { 1 + a ^ { 2 } } { 1 + b ^ { 2 } } \right. ) }$

$x = \tan \theta + \cot \theta, y = \cos \theta - \sin \theta$ , then

0%
$x = y$
0%
$\dfrac {1-y^2}{2} = \dfrac{1}{x}$
0%
$\dfrac {y^2-1}{2} = \dfrac{1}{x}$
0%
$\dfrac {1+y^2}{2} = \dfrac{1}{x}$

Explanation

${\textbf{Step - 1: Deduce and simplify value of x.}}$

${\text{Given x}} = {\text{tan}}\theta + {\text{cot}}\theta .$

$\Rightarrow {\text{x}} = \dfrac{{{\text{sin}}\theta }}{{{\text{cos}}\theta }} + \dfrac{{{\text{cos}}\theta }}{{{\text{sin}}\theta }}.$

$\Rightarrow {\text{x}} = \dfrac{{{\text{si}}{{\text{n}}^2}\theta + {\text{co}}{{\text{s}}^2}\theta }}{{{\text{sin}}\theta {\text{cos}}\theta }}.$

$\Rightarrow {\text{x}} = \dfrac{1}{{{\text{sin}}\theta {\text{cos}}\theta }}{\text{ - - - - - - - (i)}}$

${\textbf{Step - 2: Simplify value of y}}{\text{.}}$

${\text{Given y}} = {\text{cos}}\theta - {\text{sin}}\theta .$

${\text{so , }}\dfrac{{1 - {y^2}}}{2} = \dfrac{{1 - {{\left( {{\text{cos}}\theta - {\text{sin}}\theta } \right)}^2}}}{2}.$

$\Rightarrow \dfrac{{1 - {y^2}}}{2} = \dfrac{{1 - {\text{co}}{{\text{s}}^2}\theta - {\text{si}}{{\text{n}}^2}\theta + 2{\text{sin}}\theta {\text{cos}}\theta }}{2}$

$\Rightarrow \dfrac{{1 - {y^2}}}{2} = \dfrac{{1 - \left( {{\text{si}}{{\text{n}}^2}\theta + {\text{co}}{{\text{s}}^2}\theta } \right) + 2{\text{sin}}\theta {\text{cos}}\theta }}{2}$

$\Rightarrow \dfrac{{1 - {y^2}}}{2} = \dfrac{{1 - 1 + 2{\text{sin}}\theta {\text{cos}}\theta }}{2}$

$\Rightarrow \dfrac{{1 - {y^2}}}{2} = \dfrac{{{\text{2sin}}\theta {\text{cos}}\theta }}{2}$

$\Rightarrow \dfrac{{1 - {y^2}}}{2} = {\text{sin}}\theta {\text{cos}}\theta$

$\Rightarrow \dfrac{{1 - {y^2}}}{2} = \dfrac{1}{x}{\text{ }}\left[ {{\text{Put value of sin}}\theta c{\text{os}}\theta {\text{ from (i)}}} \right]$

${\textbf{So , as per given value of x and y they have a relation of }}\mathbf{\dfrac{{1 - {y^2}}}{2} = \dfrac{1}{x}(B)}.$

$\cosh x=\sec \theta$ , then

$\coth^{2}(x/2)=$

0%
$\tan^{2}(\theta/2)$
0%
$\tan^{2}\theta$
0%
$\cot^{2}(\theta/2)$
0%
$\cot^{2}\theta$

Explanation

Let

$t=\tanh^2 (x/2)$

Given

$\cosh x=\text{sec}\theta$

$\implies \dfrac{1-\tanh^2 x}{1+\tanh^2 x}=\dfrac{1}{\cos \theta}$

$\implies \dfrac{1-t^2}{1+t^2}=\dfrac{1}{\cos \theta}$

$\implies (1-t^2)\cos \theta=1+t^2$

$\implies t^2(1+\cos \theta)=1-\cos \theta$

$\implies t^2=\dfrac{1-\cos \theta}{1+\cos \theta}=\dfrac{2\sin^2 \theta/2}{2\cos^2 \theta/2}=\tan^2 \theta/2$

$\coth^2(x/2)=\dfrac{1}{t^2}=\cot^2 \dfrac{\theta}{2}$

Choose the correct answer the following:
2) The principle solution of

$\sec{x}+2=0$ is____

0%
$\dfrac{4\pi}{3}$
0%
$-\dfrac{4\pi}{3}$
0%
$\dfrac{2\pi}{3}$
0%
$-\dfrac{2\pi}{3}$

One of principal solution of

$\sqrt 3 \sec x = - 2$ is equal to

0%
$5\dfrac{\pi }{6}$
0%
$5\dfrac{\pi }{3}$
0%
$5\dfrac{\pi }{4}$
0%
none

$\sqrt{2} cos \theta$ =

$\sqrt { \dfrac { 2+\sqrt { 2+\sqrt { 2 } } }{ 2 } }$ then the value of

$\theta$ is

0%
$\dfrac{\pi}{16}$
0%
$\dfrac{\pi}{32}$
0%
$\dfrac{\pi}{64}$
0%
$\dfrac{\pi}{128}$

Explanation

Given

$\sqrt{2}\cos \theta=\sqrt{\dfrac{2+\sqrt{2+\sqrt{2}}}{2}}$

Squaring on both sides

$2\cos ^2 \theta=1+\dfrac{\sqrt{2+\sqrt{2}}}{2}$

$\implies (2\cos^2 \theta-1)=\dfrac{\sqrt{2+\sqrt{2}}}{2}$

$\implies \cos 2\theta=\dfrac{\sqrt{2+\sqrt{2}}}{2}$

Squaring on both sides

$\implies \cos^2 2\theta=\dfrac{2+\sqrt{2}}{4}$

$\implies 2\cos^2 2\theta=1+\dfrac{1}{\sqrt{2}}$

$\implies \cos 4\theta=\dfrac{1}{\sqrt{2}}$

$\implies 4\theta=\dfrac{\pi}{4}$

$\implies \theta=\dfrac{\pi}{16}$

$sin A=\dfrac{3}{5},\tan B=\dfrac{1}{2}$ and

$\dfrac{\pi}{2}<A<\pi<B<\dfrac{3\pi}{2}$ , then the value of

$8\tan A-\sqrt{5}sec B=$

0%
$5/2$
0%
$-5/2$
0%
$-7/2$
0%
$7/2$

Explanation

Given

$\dfrac{\pi}{2}<A<\pi,\implies \tan A<0$

$\sin A=\dfrac{3}{5}\implies \tan A=-\dfrac{3}{\sqrt{5^2-3^2}}=-\dfrac{3}{4}$

$\pi<B<\dfrac{3\pi}{2}\implies \text{sec} B<0$

$\tan B=\dfrac{1}{2}\implies \text{sec} B=-\sqrt{1+\dfrac{1}{4}}=-\dfrac{\sqrt{5}}{2}$

$8\tan A-\sqrt{5}\text{sec}B=8\times \bigg(-\dfrac{3}{4}\bigg)-\sqrt{5}\times \bigg(-\dfrac{\sqrt{5}}{2}\bigg)=-6+\dfrac{5}{2}=-\dfrac{7}{2}$

$cos (4 \pi + \theta)$ = ...

0%
$sin \theta$
0%
$cos \theta$
0%
$- sin \theta$
0%
$-cos \theta$

Explanation

As we know that

$\cos(2 n\pi+\theta)=\cos \theta$

$\cos(4\pi+\theta)=\cos \theta$

$4\cos{\theta}-3\sec{\theta}=2\tan{\theta}$ , then

${\theta}$ is equal to

0%
$n\pi+(-1)^{n}\dfrac{\pi}{10}$
0%
$n\pi+(-1)^{n}\dfrac{\pi}{6}$
0%
$n\pi+(-1)^{n}\dfrac{3\pi}{10}$
0%
$n\pi$

Explanation

Given,

$4\cos \theta -3\sec \theta =2\tan \theta$

$4\cos \theta -3\times \dfrac{1}{\cos \theta} =2\times \dfrac{\sin \theta}{\cos \theta}$

multiply by

$\cos \theta$

$4\cos ^2\theta -3=2\sin \theta$ ...........

$(i)$

$\because \left (\cos ^2\theta =1-\sin ^2\theta\right )$

$4-4\sin ^2\theta -3-2\sin \theta =0$ ............

$from (i)$

$4\sin ^2\theta +2\sin \theta -1=0$

$\therefore \sin \theta =\dfrac{-1\pm \sqrt{5}}{4}$

$\Rightarrow \theta =n\pi +(-1)^n\dfrac{\pi}{10}$

The value(s) of

$\theta$ , which satisfy

$3-2cos\theta-4sin\theta -cos2\theta +sin2\theta =0$ is/are

0%
$\theta =2n\pi ;n\in 1$
0%
$2n\pi +\dfrac { \pi }{ 2 } ;n\in 1$
0%
$2n\pi -\dfrac { \pi }{ 2 } ;n\in 1$
0%
$n\pi ;n\in 1$

Consider the operator a = X +

$\dfrac{d}{dx}$ acting on smooth function of x. The commutator [a, cos x] is

0%
(A) - sin x
0%
(B) cos x
0%
(C) -cos x
0%
0

The value of the expression

$\dfrac { 1 - 4 \sin 10 ^ { \circ } \sin 70 ^ { \circ } } { 2 \sin 10 ^ { \circ } }$ is

0%
$\dfrac { 1 } { 2 }$
0%
$1$
0%
$2$
0%
None if these

Explanation

$\dfrac {1-4 sin 10^o sin 70^o}{2 son 10^o}=\dfrac {1-4 sin 10^o sin(90-20)}{2 sin 10^o}$

$\dfrac {1-4 sin 10^o cos 20^o}{2 sin 10^o}$

$\left \{ since \,\, sin (90-\theta)=cos \theta \right.$

$\dfrac {1-2 (sin 10^o cos 20^o)}{2 sin 10^o}$

$\dfrac {1-2 sin 30 +sin(-10^o)}{2 sin 10^o}$

$\left \{ since\,\, sin(A+B)+Sin (A-B)= 2 Sin A \, Cos B \right.$

$\dfrac {1-2(1/2 -sin 10^o)}{2 sin 10^o}$

$\dfrac {1-2\times \frac{1}{2} + 2sin 10^o)}{2 sin 10^o}=1$

Number of solutions to the equation

$2 ^ { \sin ^ { 2 } x } + 5.2 ^ { \cos ^ { 2 } x } = 7 ,$ in the interval

$[ - \pi , \pi ] ,$ is

0%
$2$
0%
$4$
0%
$6$
0%
none of these

Explanation

$2^{\sin^2x}+5.2^{\cos^2x}=7$

$2^{\sin^2x}+5.2^{1-\sin^2x}=7$

Let

$2^{\sin^2x}=k$

$k+5\cdot\dfrac{2}{k}=7$

$k+\dfrac{10}{k}=7$

$k^2+10=7h$

$k^2-7k+10=0$

$(k-2)(k-5)=0$

But

$2^{\sin^2x}\leq 2'=2$

Hence

$2^{\sin^2x}\neq 5$

$\sin^2x=1$

$\Rightarrow x=\dfrac{\pi}{2}, \dfrac{-\pi}{2}$

Hence there are only two solution in

$[-\pi, \pi]$ i.e.,

$\dfrac{-\pi}{2}$ and

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

$5\left({\tan}^{2}x-{\cos}^{2}x\right)=2\cos2x+9$ , then the value of

$\cos4x$ is:

0%
$-\dfrac{3}{5}$
0%
$\dfrac{1}{3}$
0%
$\dfrac{2}{9}$
0%
$-\dfrac{7}{9}$

$\tan^4 \theta + \tan^2 \theta = \sec^4 \theta -\sec^2 \theta$

0%
True
0%
False

Explanation

L.H.S. :

$\tan ^4 \theta +\tan^2 \theta$

$=\tan ^2 \theta (\tan^2 \theta +1)$

$=(\sec^2 \theta -1)(\tan ^2 \theta +1)$

$=(\sec^2 \theta -1)\sec^2 \theta$

$=\sec^4 \theta -\sec^2 \theta$ = RHS

$\left(\dfrac {1+\tan^{2}A}{1+\cot^{2}A}\right)=\left(\dfrac {1-\tan^{2}A}{1-\cot^{2}A}\right)^{2}=\tan^{2}A$

0%
True
0%
False

Explanation

Given

$\left(\dfrac {1+\tan ^2A}{1+\cot ^2A}\right)=\left(\dfrac {1-\tan ^2A}{1-\cot ^2A}\right)^2=\tan ^2A$

Considering the term

$\left(\dfrac {1+\tan ^2A}{1+\cot ^2A}\right)$

$\implies \left(\dfrac {1+\tan ^2A}{1+\dfrac 1{\tan ^2A}}\right)$

$\bigg( \cot \theta =\dfrac 1 {\tan \theta }\bigg)$

$\implies \dfrac {\tan ^2A+1}{\left(\dfrac {1+\tan ^2A}{\tan ^2A}\right)}$

$\implies \dfrac 1{\dfrac 1{\tan ^2A}}$

$\implies \tan ^2A \cdots(1)$

Considering the term

$\left(\dfrac {1-\tan ^2A}{1-\cot ^2A}\right)^2$

$\implies \left(\dfrac {1-\tan ^2A}{1-\dfrac 1{\tan ^2A}}\right)^2$

$\bigg( \cot \theta =\dfrac 1 {\tan \theta }\bigg)$

$\implies \left(\dfrac {1-\tan ^2A}{\dfrac {\tan ^2A-1}{\tan ^2A}}\right)^2$

$\implies \left(\dfrac {-1}{\dfrac 1{\tan ^2A}}\right)^2$

$\implies \tan ^4A \cdots(2)$

$\therefore$ From

$(1),(2)$

The equations are not true.

State whether the following statement is true or false

$\sin^{2}\theta .\cos \theta +\tan \theta \sin \theta +\cos^{3}\theta =\sec \theta$

0%
True
0%
False

Explanation

Given

$\sin ^2\theta \cos \theta +\tan \theta\sin \theta +\cos ^3\theta$

Multiplying and Dividing by

$\cos \theta$

$\implies \dfrac {\sin ^2\theta \cos^2 \theta + \tan \theta \sin \theta \cos \theta +\cos ^4\theta}{\cos \theta}$

$\implies \dfrac {\sin ^2\theta \cos^2 \theta + \left(\dfrac {\sin \theta }{\cos \theta }\right)\sin \theta \cos \theta +\cos ^4\theta}{\cos \theta}$

$\left( \tan \theta =\dfrac {\sin \theta }{\cos \theta } \right)$

$\implies \dfrac {\sin ^2\theta \cos^2 \theta +\sin^2 \theta +\cos ^4\theta}{\cos \theta}$

$\implies \dfrac {\left(\sin ^2\theta + \cos^2 \theta\right)\cos ^2\theta +\sin^2 \theta}{\cos \theta}$

$\bigg (\sin ^2\theta +\cos ^2 \theta =1 \bigg)$

$\implies \dfrac {\cos ^2\theta +\sin^2 \theta}{\cos \theta}$

$\bigg (\sin ^2\theta +\cos ^2 \theta =1 \bigg)$

$\implies \dfrac 1 {\cos \theta }$

$\implies \sec \theta$

$(\sin \theta +\cos \theta)^{2}+(\sin \theta -\cos \theta)^{2}=2$

0%
True
0%
False

Explanation

As we know that

$(a+b)^2+(a-b)^2=2(a^2+b^2)$

$\implies (\sin \theta+\cos \theta)^2+(\sin \theta-\cos \theta)^2=2(\sin^2 \theta+\cos^2 \theta)=2(1)=2$

So the relation is

$\text{True}$

$\sin A(1+\tan A)+\cos A(1+\cot A)=\sec A+\csc A$ .

0%
True
0%
False

Explanation

$\sin A(1+\tan A)+\cos A(1+\cot A)=\sin A(1+\dfrac{\sin A}{\cos A})+\cos A(1+\dfrac{\cos A}{\sin A})$

$=\dfrac{\sin A}{\cos A}(\sin A+\cos A)+\dfrac{\cos A}{\sin A}(\sin A+\cos A)$

$=(\sin A+\cos A)(\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A})$

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

$=\dfrac{\sin A+\cos A}{\sin A\cos A}$

$=\text{sec}A+\text{csc}A$

So the relation is

$\text{True}$

$\dfrac {\sec A-\tan A}{\sec A+\tan A}=1-2\sec A\tan A+2\tan^{2}A$ .

0%
True
0%
False

Explanation

$\dfrac{\text{sec}A-\tan A}{\text{sec}A+\tan A}=\dfrac{(\text{sec}A-\tan A)(\text{sec}A-\tan A)}{(\text{sec}A+\tan A)(\text{sec}A-\tan A)}$

$=\dfrac{(\text{sec}A-\tan A)^2}{\text{sec}^2 A-\tan^2 A}$

$=(\text{sec}A-\tan A)^2$

$=\text{sec}^2 A+\tan^2 A-2\text{sec}A\tan A$

$=1+\tan^2 A+\tan^2 A-2\text{sec}A\tan A$

$=1-2\text{sec}A\tan A+2\tan^2 A$

So the relation is

$\text{True}$

State true or false.

$\sec^{4}\theta-\tan^{4}\theta=1+2\tan^{2}\theta$ .

0%
True
0%
False

Explanation

$\text{sec}^4 \theta-\tan^4 \theta=(\text{sec}^2 \theta)^2-(\tan^2 \theta)^2$

$=(\text{sec}^2 \theta-\tan^2 \theta)(\text{sec}^2 \theta+\tan^2 \theta)$

$(\because a^2-b^2=(a+b)(a-b))$

$=\text{sec}^2 \theta+\tan^2 \theta$

$=1+\tan^2 \theta+\tan ^2 \theta$

$=1+2\tan^2 \theta$

So the relation is

$\text{True}$

CBSE Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 11 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 11 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

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