MCQ Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 13

The general solution of the equation

$\displaystyle sin\theta = \frac{1}{\sqrt2}$ is

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$\displaystyle \theta = n\pi + \frac{\pi}{4}, n \in l$
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$\displaystyle \theta = 2 n\pi + \frac{\pi}{4}, n \in l$
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$\displaystyle \theta = n\pi + (-1)^n \frac{\pi}{4}, n \in l$
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none of these.

Explanation

$\displaystyle sin \theta = \frac{1}{\sqrt2} = sin \frac{\pi}{4} \Rightarrow \theta = n\pi +(-1)^n \frac{\pi}{4}, n \in l$ .

The general solution of the equation

$\displaystyle sin^2\theta = sin^2\alpha$ is

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$\displaystyle \theta = n\pi + \alpha \in l$
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$\displaystyle \theta = n\pi \pm \alpha, n \in l$
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$\displaystyle \theta = 2n\pi + \alpha, n \in l$
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$\displaystyle \theta = 2n\pi \pm \alpha, n \in l$

Explanation

$\displaystyle sin^2 \theta = sin^2 \alpha \Rightarrow 2 sin^2 \theta = 2 sin^2 \alpha$

$\displaystyle \Rightarrow (1 - cos 2\theta) = (1 - cos 2\alpha) \Rightarrow cos 2\theta = cos 2\alpha$ .

$\displaystyle 2 \theta = 2n\pi \pm 2 \alpha \Rightarrow \theta = n\pi \pm \alpha, n \in l$ .

The general solution of the equation

$\displaystyle tan^2\theta = tan^2\alpha$ is

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$\displaystyle \theta = n\pi + \alpha \in l$
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$\displaystyle \theta = 2n\pi + \alpha, n \in l$
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$\displaystyle \theta = n\pi \pm \alpha, n \in l$
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$\displaystyle \theta = 2n\pi \pm \alpha, n \in l$

Explanation

$\displaystyle tan^2\theta = tan^2\alpha \Rightarrow \frac{1 - tan^2\theta}{1 + tan^2 \theta} = \frac{1 - tan^2\alpha}{1 + tan^2 \alpha}$

$\displaystyle \Rightarrow cos2 \theta = cos 2\alpha \Rightarrow 2\theta = 2n\pi \pm 2\alpha \Rightarrow \theta = n\pi \pm \alpha, n\in l$ .

The general solution of the equation

$\displaystyle tan\theta = \frac{1}{\sqrt3}$ is

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$\displaystyle \theta = n\pi + \frac{\pi}{6}, n \in I$
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$\displaystyle \theta = 2 n\pi + \frac{\pi}{6}, n \in I$
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$\displaystyle \theta = 2 n\pi \pm \frac{\pi}{6}, n \in I$
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none of these.

Explanation

$\displaystyle tan\theta = \frac{1}{\sqrt3} = tan \frac{\pi}{6} \Rightarrow \theta = n\pi + \frac{\pi}{6}, n \in l$ .

The general solution of the equation

$\displaystyle tan \theta = tan \alpha$ is

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$\displaystyle \theta = n\pi + \alpha \in l$
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$\displaystyle \theta = 2n\pi + \alpha, n \in l$
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$\displaystyle \theta = 2n\pi \pm \alpha, n \in l$
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$\displaystyle \theta = 2n\pi \pm \alpha, n \in l$

The general solution of the equation

$\displaystyle cos^2\theta = cos^2\alpha$ is

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$\displaystyle \theta = n\pi + \alpha \in l$
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$\displaystyle \theta = 2n\pi + \alpha, n \in l$
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$\displaystyle \theta = n\pi \pm \alpha, n \in l$
0%
$\displaystyle \theta = 2n\pi \pm \alpha, n \in l$

Explanation

$\displaystyle cos^2 \theta = cos^2 \alpha \Rightarrow 2 cos^2\theta = 2 cos^2\alpha$

$\displaystyle (1 + cos 2\theta) = (1 + cos 2\alpha) \Rightarrow cos 2\theta = cos 2\alpha$

$\displaystyle \Rightarrow 2\theta = 2 n\pi \pm 2 \alpha \Rightarrow \theta = n\pi \pm \alpha, n \in l$ .

The general solution of the equation

$\displaystyle sin\theta = \frac{-\sqrt3}{2}$ is

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$\displaystyle \theta = n\pi + \frac{4\pi}{3}, n \in I$
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$\displaystyle \theta = 2 n\pi + \frac{4\pi}{3}, n \in I$
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$\displaystyle \theta = n\pi + (-1)^n \frac{4\pi}{3}, n \in I$
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none of these.

Explanation

$\displaystyle sin\theta = \frac{\sqrt3}{2} = -sin\frac{\pi}{3} = sin\left(\pi + \frac{\pi}{3}\right) = sin \frac{4\pi}{3}$

$\displaystyle \therefore \theta = n\pi + (-1)^n \frac{4\pi}{3},n \in l$ .

The angles of a triangle are in

$\displaystyle AP$ and the ratio of the number of degrees in the least to the number of radius in the greatest is

$\displaystyle 60 : \pi$ . The smallest angle is

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$\displaystyle 15^0$
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$\displaystyle 30^0$
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$\displaystyle 45^0$
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$\displaystyle 60^0$

Explanation

Let the angles be

$\displaystyle (a - d)^0 , a^0$ and

$\displaystyle (a + d)^0$ . Then,

$\displaystyle a - d + a + a + d = 180 \Rightarrow 3a = 180 \Rightarrow a = 60$ .

So, the angles are

$\displaystyle (60 - d)^0, 60^0$ and

$\displaystyle(60 + d)^0$ .

$\displaystyle 180^0 = \pi^c \Rightarrow (60 + d)^0 =\left \{\frac{\pi}{180} \times (60 + d) \right\}^c = \left\{\frac{(60 + d) \pi}{180}\right\}^c$

$\therefore$

$\displaystyle \frac{(60 - d)}{(60 + d) \frac{\pi}{180}} = \frac{60}{\pi}\Rightarrow \frac{3(60 - d)}{(60 + d)} = 1$

$\displaystyle \Rightarrow 180 - 3d = 60 + d \Rightarrow 4d = 120 \Rightarrow d = 30$ .

$\therefore$ the smallest angle =

$\displaystyle (60 - 30)^0 = 30^0$ .

The general solution of the equation

$\displaystyle cos\theta = \frac{1}{2}$ is

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$\displaystyle \theta = n\pi + \frac{\pi}{3}, n \in I$
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$\displaystyle \theta = 2 n\pi + \frac{\pi}{3}, n \in I$
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$\displaystyle \theta = 2 n\pi \pm \frac{\pi}{3}, n \in I$
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none of these.

Explanation

$\displaystyle cos \theta = \frac{1}{2} = cos \frac{\pi}{3}\Rightarrow \theta = 2n\pi \pm \frac{\pi}{3}, n \in l$ .

The general solution of the equation

$\displaystyle cos \theta = cos \alpha$ is

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$\displaystyle \theta = \alpha$
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$\displaystyle \theta = n\pi \pm \alpha, n \in l$
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$\displaystyle \theta = 2n\pi \pm \alpha, n \in l$
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none of these

The general solution of the equation

$\displaystyle cosec \theta + \sqrt2 = 0$ is

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$\displaystyle \theta = n\pi + \frac{5\pi}{4}, n \in I$
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$\displaystyle \theta = n\pi - \frac{5\pi}{4}, n \in I$
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$\displaystyle \theta = n\pi + (-1)^n \frac{5\pi}{4}, n \in I$
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none of these.

Explanation

$\displaystyle cosec \theta = -\sqrt2 \Rightarrow sin \theta = \frac{-1}{\sqrt2} = - sin \frac{\pi}{4} = sin \left(\pi + \frac{\pi}{4}\right) = sin \frac{5\pi}{4}$

$\displaystyle \Rightarrow \theta = n\pi + (-1)^n . \frac{5\pi}{4}, n \in I.$

The general solution of the equation

$\displaystyle 4 sin^2\theta = 1$ is

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$\displaystyle \theta = n\pi \pm \frac{\pi}{6}, n \in I$
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$\displaystyle \theta = 2n\pi \pm \frac{\pi}{6}, n \in I$
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$\displaystyle \theta = \frac{n\pi}{4} + \frac{\pi}{24}, n \in I$
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none of these.

Explanation

$\displaystyle 4sin^2 \theta = 1 \Rightarrow sin^2 \theta = \frac{1}{4} = \left(\frac{1}{2}\right)^2 = sin^2 \frac{\pi}{6}$

$\displaystyle \Rightarrow x = n\pi \pm \frac{\pi}{6}, n \in I$

The general solution of the equation

$\displaystyle cot \theta = -\sqrt3$ is

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$\displaystyle \theta = n\pi +\frac{5\pi}{6}, n \in l$
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$\displaystyle \theta = 2 n\pi + \frac{5\pi}{6}, n \in l$
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$\displaystyle \theta = n\pi + \frac{2\pi}{3}, n \in l$
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none of these.

Explanation

$cot\theta = -\sqrt3 \Rightarrow tan \theta = \frac{-1}{\sqrt3} = -tan\frac{\pi}{6} = tan \left(\pi - \frac{\pi}{6}\right) = tan \frac{5\pi}{6}$

$\displaystyle \Rightarrow \theta = \left(n\pi + \frac{5\pi}{6}\right), n\in l$ .

The general solution of the equation

$\displaystyle sin 2\theta = \frac{-1}{2}$ is

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$\displaystyle \theta = \frac{n\pi}{4} + \frac{\pi}{24}, n \in I$
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$\displaystyle \frac{n\pi}{2} + (-1)^n \frac{7\pi}{12}, n \in N$
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$\displaystyle \theta = \frac{n\pi}{4} \pm \frac{\pi}{24}, n \in I$
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none of these.

Explanation

$\displaystyle sin 2\theta = \frac {-1}{2} = -sin \frac{\pi}{6} = sin \left(\pi + \frac{\pi}{6}\right) = sin \frac{7\pi}{6}$

$\displaystyle \Rightarrow 2\theta = nx + (-1)^n . \frac{7\pi}{6}, n \in I \Rightarrow \theta = \frac {n\pi}{2} +(-1)^n \frac{7\pi}{12}, n \in I.$

The general solution of the equation

$\displaystyle 2 cos^2\theta = 1$ is

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$\displaystyle \theta = 2n\pi \pm \frac{\pi}{4}, n \in I$
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$\displaystyle \theta = \frac{n\pi}{2} + \frac{\pi}{8}, n \in I$
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$\displaystyle \theta = n\pi \pm \frac{\pi}{4}, n \in I$
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none of these.

Explanation

$\displaystyle 2cos^2 \theta = 1 \Rightarrow cos^2 \theta = \frac{1}{2} = \left(\frac{1}{\sqrt2}\right)^2 = cos^2 \frac{\pi}{4}$

$\displaystyle \therefore \theta = n\pi \pm \frac{\pi}{4}, n \in N.$

The general solution of the equation

$\displaystyle cot^2\theta = 3$ is

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$\displaystyle \theta = n\pi + \frac{\pi}{6},n \in I$
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$\displaystyle \theta = n\pi \pm \frac{\pi}{6},n \in I$
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$\displaystyle \theta = 2n\pi + \frac{\pi}{6},n \in I$
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none of these.

Explanation

$\displaystyle cot^2 \theta = 3 \Rightarrow tan^2 \theta = \frac{1}{3}$

$\displaystyle = \left( \frac{1}{\sqrt3}\right)^2 = tan^2 \frac{\pi}{6} \Rightarrow \theta = nx \pm \frac{\pi}{6}, n \in I.$

The general solution of the equation

$\displaystyle cos\theta = \frac{-1}{2}$ is

0%
$\displaystyle \theta = n\pi \pm \frac{2\pi}{3}, n \in I$
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$\displaystyle \theta = 2 n\pi + \frac{\pi}{3}, n \in I$
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$\displaystyle \theta = 2 n\pi \pm \frac{2\pi}{3}, n \in I$
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none of these.

Explanation

$\displaystyle cos \theta = \frac{-1}{2} = -cos \frac{\pi}{3} = cos \left(\pi - \frac{pi}{3}\right) = cos \frac{2\pi}{3}$

$\displaystyle \therefore \theta = \left(2n\pi \pm \frac{2\pi}{3}\right), n\in l$ .

$x + 1 / x = 2$ , the principal value of

$sin^{-1}x$ is

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$\pi / 4$
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$\pi / 2$
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$\pi$
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$3\pi / 2$

$x_{1}$ and

$x_{2}$ are two distinct roots of the equation

$a \cos x+b \sin x=c,$ then

$\tan \dfrac{x_{1}+x_{2}}{2}$ is equal to

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$\dfrac{a}{b}$
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$\dfrac{b}{a}$
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$\dfrac{c}{a}$
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$\dfrac{a}{c}$

Explanation

$a \cos x+b \sin x=c$

$\Rightarrow \dfrac{a\left(1-\tan ^{2} \dfrac{x}{2}\right)}{1+\tan ^{2} \dfrac{x}{2}}+\dfrac{2 b \tan \dfrac{x}{2}}{1+\tan ^{2} \dfrac{x}{2}}=c$

$\Rightarrow \quad(c+a) \tan ^{2} \dfrac{x}{2}-2 b \tan \dfrac{x}{2}+c-a=0$

$\Rightarrow \quad \tan \dfrac{x_{1}}{2}+\tan \dfrac{x_{2}}{2}=\dfrac{2 b}{c+a}$ and

$\tan \dfrac{x_{1}}{2} \tan \dfrac{x_{2}}{2}=\dfrac{c-a}{c+a}$

Using,

$\tan \left(\dfrac{x_{1}+x_{2}}{2}\right)=\dfrac { \tan \dfrac{x_{1}}{2}+\tan \dfrac{x_{2}}{2}}{1-\tan \dfrac{x_{1}}{2} \tan \dfrac{x_{2}}{2}}$

$\Rightarrow \tan \left(\dfrac{x_{1}+x_{2}}{2}\right)=\dfrac{\dfrac{2 b}{c+a}}{1-\dfrac{c-a}{c+a}}=\dfrac{2 b}{2 a}=\dfrac{b}{a}$

$\operatorname{cosec} \theta-\cot \theta=q$ , then the value of

$\operatorname{cosec} \theta$ is

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$q+\dfrac{1}{q}$
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$q-\dfrac{1}{q}$
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$\dfrac{1}{2}\left(q+\dfrac{1}{q}\right)$
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none of these

Explanation

$\operatorname{cosec} \theta-\cot \theta=q$

$\operatorname{cosec}^2 \theta-\cot^2 \theta=1\Rightarrow (\operatorname{cosec} \theta-\cot \theta ) (\operatorname{cosec} \theta+\cot \theta)=1 \Rightarrow q \cdot (\operatorname{cosec} \theta+\cot \theta)=1$

$\therefore \quad \operatorname{cosec} \theta+\cot \theta=\dfrac{1}{q}$

$\therefore \quad \operatorname{cosec} \theta=\dfrac{1}{2}[q+(1 / q)](\text { on addition })$

The principal value of

$cos^{-1} \left (cos\dfrac{2\pi}{3} \right ) + sin^{-1} \left (sin\dfrac{2\pi}{3} \right )$ is

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$\pi$
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$\pi/2$
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$\pi/3$
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$4\pi/3$

Evaluate :

$tan \left [ 2\, tan^{-1}\dfrac{1}{5} - \dfrac{\pi}{4} \right ]$

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$\dfrac{5}{4}$
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$\dfrac{5}{16}$
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$- \dfrac{7}{17}$
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$\dfrac{7}{17}$

The value of

$\dfrac{5}{16}$ right angles in sexagesimal system is equal to

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$28^{\circ} 30' 7''$
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$27^{\circ} 5' 26''$
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$28^{\circ} 7' 30''$
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$29^{\circ} 3' 27''$

The value of

$\dfrac{3 \pi}{4}$ in sexagesimal system is:

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$75^{\circ}$
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$135^{\circ}$
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$120^{\circ}$
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$220^{\circ}$

1 radian is equal to:

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$180^{\circ}$
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$200^{\circ}$
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$100^{\circ}$
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None of these

Let

$f(X)=\sin (\pi\cos x)$ and

$g(x) =\cos (2\pi\sin x)$ be two function defined for

$x>0$ . Define the following sets whose elements are written in increasing order

$X=\{x:f(x)=0\},Y=\{x:f'(x)=0\}$

$Z=\{x:g(x)=0\},W=\{x:g'(x)=0$
List I contains sets

$X,Y,Z$ and

$W$ List II contains some information regarding these set.
Which of the following is the only correct combination ?

Sr.No	List I	Sr.No	List II
(I)	X	(P)	$\supseteq\left\{\dfrac{\pi}{2},\dfrac{3\pi}{2},4\pi,7\pi\right\}$
(II)	Y	(Q)	an arithmetic progression
(III)	Z	(R)	NOT an arithmetic progression
(IV)	W	(S)	$\supseteq \left\{\dfrac{\pi}{6},\dfrac{7\pi}{6},\dfrac{13\pi}{6}\right\}$
		(T)	$\supseteq \left\{\dfrac{\pi}{3},\dfrac{2\pi}{3},\pi\right\}$
		(U)	$\supseteq \left\{\dfrac{\pi}{6},\dfrac{3\pi}{4}\right\}$

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(I) (P) (R)
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(II) (Q) (T)
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(I) (Q) (U)
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(II) (R) (S)

Explanation

$f(x)=0\Rightarrow \sin (\pi \cos x )=0$

$=\pi \cos x =n\pi \Rightarrow \cos x =n$

$= \cos x =-1,0,1 \Rightarrow X=\{n\pi,(2n+1)\dfrac{\pi}{2}\}=\{n\dfrac{\pi}{2},n\in I\}$

$f'(x)=0\Rightarrow \cos (\pi\cos x )(-\pi \sin x )=0$

$\pi \cos x =(2n+1)\dfrac{\pi}{2}$ or

$x=n\pi$

$\Rightarrow \cos x =n+\dfrac{1}{2}$ or

$x=n\pi$

$\Rightarrow \cos x =\pm \dfrac{1}{2}$ or

$x=n\pi$

$\Rightarrow Y=\left\{.....-\dfrac{2\pi}{3},\dfrac{\pi}{3},0,\dfrac{\pi}{3},\dfrac{2\pi}{3},\pi,\dfrac{4\pi}{3},.....\right\}$ which is an arithmetic progression

$g(x) =0$

$\Rightarrow (2\pi \sin x)=0$

$\Rightarrow 2\pi \sin x =(2n+1)\dfrac{\pi}{2}$

$\Rightarrow \sin x=\dfrac{2n+1}{4}=\pm \dfrac{1}{4},\pm \dfrac{3}{4}$

$\Rightarrow \sin x =\dfrac{2n+1}{4}=\pm\dfrac{1}{4},\pm\dfrac{3}{4}$

$\Rightarrow z=\left\{n\pi\pm\sin^{-1}\dfrac{1}{4},n\pi \pm \sin^{-1}\dfrac{3}{4},n\in I\right\}$

$g'(x)=0$

$\Rightarrow -\sin (2\pi \sin x )(2\pi\cos x )=0$

$\Rightarrow 2\pi \sin x= n\pi$ or

$x=(2n+)\dfrac{\pi}{2}$

$\Rightarrow \sin x =\dfrac{n}{2}=0,\pm \dfrac{1}{2},\pm 1$ or

$x=(2n+1)\dfrac{\pi}{2}$

$W=\left\{n\pi ,(2n+1)\dfrac{\pi}{2},n\pi \pm \dfrac{\pi}{6},n\in I\right\}$

so, correct combination is

$(II)$

$\rightarrow$

$(Q), (T)$

How many right angles is equal to

$56^{\circ} 15'$ ?

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$\dfrac{8}{5}$ right angles
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$\dfrac{5}{8}$ right angles
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$\dfrac{3}{5}$ right angles
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$\dfrac{5}{4}$ right angles

The number of solution of

$|\tan x |= \ tanx + \displaystyle \dfrac{1}{\cos x}$ in

$[0,2\pi]$ is

0%
$4$
0%
$1$
0%
$2$
0%
$6$

Explanation

Let split domain in two parts

$\displaystyle [0,\frac{\pi }{2}] \cup [\pi , \frac{3\pi }{2}]$

$\ and \ [\frac{\pi }{2} , \pi ] \cup [\frac{3\pi }{2} , 2\pi ]$
Case - 1

$\displaystyle x\epsilon [0,\frac{\pi }{2}] \cup [\pi , \frac{3\pi }{2}] , |tan x|=tan x=tan x+\frac{1}{cos x}$

$\displaystyle \frac{1}{cos x}=0$

$\Rightarrow$ no solution.
Case - 2

$\displaystyle x\epsilon [\frac{\pi }{2} , \pi ] \cup [\frac{3\pi }{2} , 2\pi ]$

$|tan x|=-tan x=tan x+\frac{1}{cos x}$

$\displaystyle 0=\frac{2sin x+1}{cos x}$

$\displaystyle \Rightarrow sin x=-\frac{1}{2}$

only one solution i.e.,

$x=330^{o}$

A solution

$(\mathrm{x},\mathrm{y})$ of

$\mathrm{x}^{2}+2\mathrm{x}$

$\sin(xy)+1=0$ is

0%
$(1, 0)$
0%
$(1, \dfrac{7\pi}2)$
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$(-1,\dfrac{7\pi}2)$
0%
$(-1,0)$

Explanation

we get

$(x + \sin(xy))^{2} + \cos^{2}(xy) = 0$
Now we verify with the options

The number of solution of the equation

$|\cot x|= cotx +\displaystyle \frac{1}{sin x}$ in

$[0,2\pi]$ is

0%
$2$
0%
$4$
0%
$0$
0%
$1$

Explanation

Let split the domain in two parts

$\displaystyle [0, \frac{\pi }{2}] \cup [\pi , \frac{3\pi }{2}] \ and \ [\frac{\pi }{2} , \pi ]\cup [\frac{3\pi }{2} , 2\pi ]$

Case -1

$\displaystyle x\epsilon [0,\frac{\pi }{2}] \cup [\pi , \frac{3\pi }{2}]$

$\displaystyle |cot x|=cot x=cot x+\frac{1}{sin x}$

$\displaystyle \frac{1}{sin x}=0$

$\Rightarrow$ no solution
Case -2

$x\epsilon [\frac{\pi }{2} , \pi] \cup [\frac{3\pi }{2} , 2\pi]$

$\displaystyle |cot x|=-cot x=cot x+\frac{1}{sin x}$

$\displaystyle \frac{2cos x+1}{sin x}=0$

$cos x=-\frac{1}{2}$
only 1 solution i.e.

$x=120 ^{o}$

CBSE Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 13 - 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 13 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

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