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CBSE Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 13 - MCQExams.com
CBSE
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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$$\displaystyle \theta = n\pi \pm \alpha, n \in l$$
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$$\displaystyle \theta = 2n\pi \pm \alpha, n \in l$$
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
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$$\displaystyle \theta = n\pi \pm \frac{2\pi}{3}, n \in I$$
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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
$$\displaystyle cos \theta = \frac{-1}{2} = -cos \frac{\pi}{3} = cos \left(\pi - \frac{pi}{3}\right) = cos \frac{2\pi}{3}$$$$\displaystyle \therefore \theta = \left(2n\pi \pm \frac{2\pi}{3}\right), n\in l$$.
If $$ x + 1 / x = 2 $$ , 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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$$ 3\pi / 2 $$
If $$ x_{1} $$ 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, $$\tan \left(\dfrac{x_{1}+x_{2}}{2}\right)=\dfrac { \tan \dfrac{x_{1}}{2}+\tan \dfrac{x_{2}}{2}}{1-\tan \dfrac{x_{1}}{2} \tan \dfrac{x_{2}}{2}}$$
$$ \Rightarrow \tan \left(\dfrac{x_{1}+x_{2}}{2}\right)=\dfrac{\dfrac{2 b}{c+a}}{1-\dfrac{c-a}{c+a}}=\dfrac{2 b}{2 a}=\dfrac{b}{a} $$
If $$ \operatorname{cosec} \theta-\cot \theta=q $$, 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
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$$4$$
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$$1$$
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$$2$$
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$$6$$
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
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$$(1, 0)$$
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$$(1, \dfrac{7\pi}2)$$
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$$(-1,\dfrac{7\pi}2)$$
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$$(-1,0)$$
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
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$$2$$
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$$4$$
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$$0$$
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$$1$$
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. $$x=120 ^{o}$$
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