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CBSE Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 16 - MCQExams.com
CBSE
Class 11 Engineering Maths
Trigonometric Functions
Quiz 16
If $$sin \alpha + sin \beta = \dfrac{1}{2} $$ and $$cos \alpha + cos \beta = \dfrac{\sqrt{3}}{2}$$ then $$3 \beta + \alpha = $$
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$$0^o$$
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$$60^o$$
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$$120^o$$
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$$90^o$$
The value of $$\csc \dfrac{\pi}{13}-\sqrt{3} \sec \dfrac{\pi}{18}$$ is a
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Surd
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Rational which is not integral
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Negative integer
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Natural number
The number of solution of the equation $${x^3} + 2{x^2} + 5x + 2\cos x = 0\,in\,[0,2\pi ]$$ is
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$$0$$
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$$1$$
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$$2$$
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$$3$$
$$\displaystyle \frac { \tan ^{ 2 }{ \theta } }{ \tan ^{ 2 }{ \theta } -1 } +\frac { \csc ^{ 2 }{ \theta } }{ \sec ^{ 2 }{ \theta } -\csc ^{ 2 }{ \theta } } =\frac { 1 }{ \sin ^{ 2 }{ \theta } -\cos ^{ 2 }{ \theta } } $$.
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0%
True
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False
$$sec^{2}(Tan^{-1} 2) +Cosec^{2}(Cot^{-1}3)=$$
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5
0%
10
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15
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20
$$\dfrac { 1 }{ \text{cosec}\theta +\cot\theta } -\dfrac { 1 }{ \sin\theta } =\dfrac { 1 }{ \sin\theta } -\dfrac { 1 }{ \text{cosec}\theta -\cot\theta } $$
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0%
True
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False
If y = acosx + bsinx then find the maximum value of y
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$$\sqrt{a^2+b^2}$$
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$${a^2}+{b^2}$$
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$$\frac{a+b}{2}$$
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$$\frac{\sqrt{a^2+b^2}}{2}$$
The value of $$\sum\limits_{r = 0}^9 {{{\sin }^2}\dfrac{{\pi r}}{{18}}} \;is\;equal\;to\;$$
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$$\dfrac{9}{2}$$
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$$\dfrac{7}{2}$$
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5
0%
None of these
If $$\csc\theta=\dfrac{x^{2}-y^{2}}{x^{2}+y^{2}}$$ where $$x$$ and $$y$$ are two unequal non-zero real numbers, then the number of real values of $$\theta$$ is
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$$0$$
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$$1$$
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$$2$$
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$$infinite$$
$$\displaystyle\sum_{r=1}^{n-1} \cos^2\left ( \dfrac{r\pi}{n} \right )=? $$
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$$\dfrac{n}{2}$$
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$$\dfrac{n-2}{2}$$
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$$\dfrac{n-1}{2}$$
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$$\dfrac{n-2}{4}$$
Explanation
As we know that
$$\cos 2 A=2 \cos ^{2} A-1$$
$$\cos ^{2} A=\frac{1+\cos 2 A}{2}$$
say , $$A=\frac{\pi \pi}{n}$$
$$\Rightarrow \sum_{r=1}^{n-1} 1+\cos 2\left(\frac{r \pi}{n}\right)=\frac{1}{2}\left[\sum_{r=1}^{n-1} \cos \frac{2 \pi \pi}{n}+\sum_{n=1}^{n-1} 1\right]$$
$$\Rightarrow \frac{1}{2}\left[\cos \frac{2 \pi}{n}+\cos \frac{4 \pi}{n}+\ldots+\cos \frac{2(n-1)}{n} \pi+(n-1)\right]$$
As we know that
$$\cos \alpha+\cos (\alpha+\beta)+\cos (\alpha+2 \beta)+\cdots+\cos (\alpha+(n-1) \beta)=\frac{\sin \left(\frac{n \beta}{2}\right)}{\sin \frac{\beta}{2}} \cdot \cos \left(\frac{2 \alpha+(n-1) \beta}{2}\right) \\$$
$$= \frac{1}{2}\left[\frac{\sin \left(\frac{(n-1) \pi}{n}\right)}{\sin \left(\frac{\pi}{n}\right)} \cdot \cos \left(\frac{\frac{4\pi}{n}+(n-2) \frac{2 \pi}{n}}{2}\right)+n-1\right]$$
$$= \frac{1}{2}\left[\frac{-\sin \left(\frac{n-1}{n} \pi\right)}{\sin \frac{\pi}{n}}+n-1\right]$$
$$\left.= \frac{1}{2}\left[-\left\{\sin \frac{\sin \left(\pi-\frac{(n-1)}{n}\right) \pi}{\sin \frac{\pi}{n}}\right\}\right\}+n-1\right]$$
$$= \frac{1}{2}[-\frac{sin\frac{\pi}{n}}{sin\frac{\pi}{n}}+n-1]$$
$$= \frac{1}{2}(-1+n-1)=\frac{n-2}{2}$$
Hence, $$(B)$$ is the correct option.
If $$sinx+{ sin }^{ 2 }x=1,$$ then value of $${ cos }^{ 2 }x+{ cos }^{ 4 }x$$ is
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1
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2
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1.5
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none of these
If $$\frac { \cos { \theta } }{ p } =\frac { \sin { \theta } }{ q } $$, then $$\frac { p }{ \sec { 2\theta } } +\frac { q }{ cosec2\theta } $$ is equal to
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0%
p
0%
q
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qp
0%
none of these
The value of $$\sum _{ r=1 }^{ 10 }{ \cos ^{ 3 }{ \frac { r\pi }{ 3 } } } =$$?
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$$-\frac { 1 }{ 8 } $$
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$$-\frac { 7 }{ 8 } $$
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$$-\frac { 9 }{ 8 } $$
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None of the above
If $$x+y=\dfrac { 2\pi }{ 3 } and\quad cosx+cosy=\dfrac { \sqrt { 3 } }{ 2 } ,\quad then\quad x,y\quad is\quad equal\quad to$$
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$$\dfrac { \pi }{ 3 } ,\dfrac { \pi }{ 6 } $$
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$$\dfrac { \pi }{ 2 } ,\dfrac { \pi }{ 6 } $$
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$$\dfrac { \pi }{ 4 } ,\dfrac { \pi }{ 3 } $$
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no solution
$$\sqrt{\dfrac{1+ cosx }{1-cos x}}$$
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$$\dfrac{1+ cosx }{1-cos x}$$
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$$cosec x+ cot x$$
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$$cosec x- cot x$$
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$$cosec x$$
If $$f(x) = \frac{{{{\cos }^2}x + {{\sin }^4}x}}{{{{\sin }^2}x + {{\cos }^4}x}}$$ for $$x \in R$$ then $$f(2002)=$$
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1
0%
2
0%
3
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4
If $${T_n} = \left( {{{\sin }^n}\theta + {{\cos }^n}\theta } \right),\;then\;for\;permissible\;values\;of\;\theta $$, $$\dfrac{{{T_5} - {T_3}}}{{{T_7} - {T_5}}}$$ is always equal to
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$$\dfrac{{{T_1}}}{{{T_3}}}$$
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$$\dfrac{{{T_2}}}{{{T_4}}}$$
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$$\dfrac{{{T_5}}}{{{T_7}}}$$
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$$\dfrac{{{T_3}}}{{{T_7}}}$$
The smallest value of 0 satisfying the equation $$\sqrt { 3(cot0+\quad tan0)\quad =\quad 4 } $$ is
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$$2x/3$$
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$$\pi /3$$
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$$\pi /6$$
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$$\pi /12$$
The number of real values of x such that $$\left( {{2^{ - x}} + {2^x} - 2\cos x} \right)\;\left( {{3^{x + \pi }} + {3^{ - x - \pi }} + 2\cos x} \right)\left( {{5^{\pi - x}} - 2\cos x + {5^{x - \pi }}} \right) = 0,is$$
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0%
1
0%
2
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3
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infinite
The number of solutions of the equation sin $$\left( \dfrac { \pi x }{ 2\sqrt { 3 } } \right) ={ x }^{ 2 }-2\sqrt { 3 } x+4$$ is
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0
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2
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1
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none of these
Find the fix point through which the line $$(2cos\Theta +3sin\Theta )x+(3 cos\Theta-5sin\Theta )y-(5cos\Theta -2sin\Theta )=0$$ passes for all values of $$\Theta $$-
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(0,0)
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(1,1)
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(2,1)
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None of these
The principal solutions of $$secx= \frac{2}{\sqrt{3}}$$ are _________
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$$\frac{\pi }{3},\frac{||\pi }{6}$$
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$$\frac{\pi }{6},\frac{11\pi }{6}$$
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$$\frac{\pi }{4},\frac{||\pi }{4}$$
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$$\frac{\pi }{6},\frac{||\pi }{4}$$
$$\dfrac { \tan\theta }{ 1-\cot\theta } +\dfrac { \cot\theta }{ 1-\tan\theta } $$ is equal to
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$$1+\tan\theta +\cot\theta $$
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$$1-\tan\theta -\cot\theta $$
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$$1+\tan\theta -\cot\theta $$
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none of these
$$cos^{4}\Theta - sin^{4}\Theta $$ is equal
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$$\frac{1-tan^{2}\Theta }{1+ tan^{2}\Theta }$$
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$$2 cos^{2}\Theta -1$$
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$$1-2 sin^{2}\Theta $$
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$$sin^{2}\Theta - cos^{2}\Theta $$
32 $$sin^{6}15^{0}-48sin^{4}15^{0}+18sin^{2}15^{0}=$$
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1
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2
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3
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-1
Choose the correct answer.
$$(1+{ \tan }^{ 2 }\theta ){ \sin }^{ 2 }\theta =$$
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$${ \tan }^{ 2 }\theta $$
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$${ \cot }^{ 2 }\theta $$
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$${ \sin }^{ 2 }\theta $$
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$${ \cos }^{ 2 }\theta $$
The number of solutions of the equation $$\sin (9x) + \sin (3x) = 0$$ in the closed interval $$[0, 2\pi]$$ is
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$$7$$
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$$13$$
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$$19$$
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$$25$$
Explanation
$$\sin{9x}+\sin{3x}=0,\,\,x\in\left[0,2\pi\right]$$
$$\Rightarrow\,3\sin{3x}-4{\sin}^{3}{3x}+\sin{3x}=0$$
$$\Rightarrow\,4\sin{3x}-4{\sin}^{3}{3x}=0$$
$$\Rightarrow\,4\sin{3x}\left(1-{\sin}^{2}{3x}\right)=0$$
$$\Rightarrow\,\sin{3x}=0,\,\,\left(1-{\sin}^{2}{3x}\right)=0$$
$$\Rightarrow\,\sin{3x}=0,\,\,{\sin}^{2}{3x}=1$$
$$\Rightarrow\,3x=n\pi,\,n\in I$$ or $$3x=k\pi+\dfrac{\pi}{2},\,k\in I$$
$$\Rightarrow\,x=\dfrac{n\pi}{3},\,x=\dfrac{k\pi}{3}+\dfrac{\pi}{6},\,k\in I$$
For $$n=0,1,2,3,4,5,6$$ we have
$$x=0,\dfrac{\pi}{3},\dfrac{2\pi}{3},\pi,\dfrac{4\pi}{3},\dfrac{5\pi}{3},2\pi$$
For $$k=0,1,2,3,4,5$$ we have
$$x=\dfrac{\pi}{6},\,\dfrac{\pi}{3}+\dfrac{\pi}{6},\,\dfrac{2\pi}{3}+\dfrac{\pi}{6},\,\pi+\dfrac{\pi}{6},\,\dfrac{4\pi}{3}+\dfrac{\pi}{6},\,\dfrac{5\pi}{3}+\dfrac{\pi}{6}$$
or $$x=\dfrac{\pi}{6},\,\dfrac{2\pi}{6}+\dfrac{\pi}{6},\,\dfrac{4\pi}{6}+\dfrac{\pi}{6},\,\dfrac{6\pi+\pi}{6},\,\dfrac{8\pi}{6}+\dfrac{\pi}{6},\,\dfrac{10\pi}{6}+\dfrac{\pi}{6}$$
or $$x=\dfrac{\pi}{6},\,\dfrac{3\pi}{6},\,\dfrac{5\pi}{6},\,\dfrac{7\pi}{6},\,\dfrac{9\pi}{6},\,\dfrac{11\pi}{6}$$
or $$x=\dfrac{\pi}{6},\,\dfrac{\pi}{2},\,\dfrac{5\pi}{6},\,\dfrac{7\pi}{6},\,\dfrac{3\pi}{2},\,\dfrac{11\pi}{6}$$
Thus, the solutions are $$\left\{0,\dfrac{\pi}{3},\dfrac{2\pi}{3},\pi,\dfrac{4\pi}{3},\dfrac{5\pi}{3},2\pi,\dfrac{\pi}{6},\,\dfrac{\pi}{2},\,\dfrac{5\pi}{6},\,\dfrac{7\pi}{6},\,\dfrac{3\pi}{2},\,\dfrac{11\pi}{6}\right\}$$
or $$\left\{0,\,\dfrac{\pi}{2},\dfrac{\pi}{6},\dfrac{\pi}{3},\dfrac{2\pi}{3},\dfrac{3\pi}{2},\pi,\dfrac{4\pi}{3},\dfrac{5\pi}{3},\dfrac{5\pi}{6},\,\dfrac{7\pi}{6},\dfrac{11\pi}{6},2\pi\right\}$$
$$=13$$ solutions
If $$sec A=a+(\frac{1}{4a})$$ , then sec A + tan A is equal to
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$$2a \;or\; \frac{1}{2a}$$
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$$a or \frac{1}{a}$$
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$$2a or\frac{1}{a}$$
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$$a or \frac{1}{2a}$$
If $$\displaystyle sin^{-1} x + sin ^{-1} y + sin^{-1} z = 3\pi / 2 $$ then the value of
$$ x^{100} + y ^{100} + z^{100} - \dfrac{3}{x^{101} + y^{101} + z^{101}} $$ is
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$$ 0 $$
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$$ 1 $$
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$$ 2 $$
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$$ 3 $$
Indicate the relation which is true
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$$ tan \left | tan^{-1} x \right | = \left | x \right | $$
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$$ cot \left | cot^{-1}x \right | = x $$
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$$ tan^{-1} \left | tan\,x \right | = \left | x \right | $$
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$$ sin \left | sin^{-1} \right | = \left | x \right |$$
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