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Jee Main Advanced Math Chapter Wise Mock Test
Trigonometry
Quiz 1
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Q.1
The number of integral values of k for which equation 7cos(x) + 5 sin(x) =2k +1 has a solution
0%
4
0%
10
0%
12
0%
8
Q.2
if $\alpha + \beta = \frac {\pi}{2}$ and $ \beta + \gamma = \alpha$,then $tan \alpha$ equals to
0%
$ tan \beta + tan \gamma$
0%
$2 tan \beta + tan \gamma$
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$2( tan \beta + tan \gamma)$
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$ tan \beta + 2 tan \gamma$
Q.3
Minimum value of the expression $ \frac {1}{cos^2 ( \pi/4 +x) + sin^2( \pi/4 -x)}$ equals is
0%
$\frac {1}{\sqrt 2}$
0%
$\sqrt 2$
0%
$\frac {1}{2}$
0%
2
Q.4
The solutions of the trigonometric equation tan x + tan 2x + tan x tan 2x=1
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$\frac {n \pi}{2} + \frac {\pi}{12}, n \in Z$
0%
$\frac {n \pi}{3} + \frac {\pi}{12}, n \in Z$
0%
$\frac {n \pi}{3} + \frac {\pi}{4}, n \in Z$
0%
$\frac {n \pi}{3} + \frac {\pi}{8}, n \in Z$
Q.5
if $2 \sin^2 (x) – 5 \sin (x) + 2 > 0$, x \in ( 0 ,2 \pi) then $x \in $
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$(0, \frac {\pi}{6} ) \cup (\frac {5\pi}{6} , 2 \pi)$
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$(\frac {\pi}{80}, \frac {\pi}{6} )$
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$(0, \frac {\pi}{6} )$
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$(\frac {5\pi}{6} , 2 \pi)$
Q.6
for a positive integer n , let $f_n \theta = ( tan \frac {\theta}{2}) ( 1+ sec \theta) (1 + sec 2 \theta) .....(1+ sec 2^n \theta)$. Now four values are given as (i) $f_2 ( \frac {\pi}{16})=1$ (ii) $f_3 ( \frac {\pi}{32})=1$ (iii) $f_4 ( \frac {\pi}{64})=1$ (iv) $f_5 ( \frac {\pi}{128})=1$ Which of these are correct
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(i) and (ii)
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(i) and (iv)
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All are correct
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(i) ,(ii) and (iii)
Q.7
The minimum value of 3 cosx + 4 sinx + 5 is
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0
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-5
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5
0%
4
Q.8
which of these is not the solution for the equation $7 cos^2 x + 3 sin^2 x =4$
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$\frac {4 \pi}{5}$
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$\frac {4 \pi}{3}$
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$\frac {2 \pi}{3}$
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$\frac {11 \pi}{3}$
Q.9
which of the following number is rational?
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$ cos 15^0$
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$ sin 15^0$
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$ sin 15^0 cos 75^0$
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$ sin 15^0 cos 15^0$
Q.10
if y = |sin (x) | + |cos (x)| and $x \in R$ then
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$y \in [0,2]$
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$y \in [0,1]$
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$y \in [1, \sqrt 2]$
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$y \in [0,\sqrt 2]$
Q.11
The maximum value of sin (cos x) is equal to
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0
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sin 1
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$sin ( \frac {1}{\sqrt 2}) $
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1
Q.12
If sin A and cos A are the roots of the equation $ax^2 – bx + c = 0$, then a, b and c satisfy the relation
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$a^2 – b^2 – 2ac = 0$
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$a^2 + b^2 + 2ac = 0$
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$a^2 + c^2 + 2ab = 0$
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$a^2 – b^2 + 2ac = 0$
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