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Class 12 Maths
Inverse Trigonometric Functions
Quiz 2
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tan\(\frac{1}{2}\) + tan\(\frac{1}{3}\) =
0%
\(\frac{π}{4}\)
0%
\(\frac{π}{2}\)
0%
\(\frac{π}{3}\)
0%
π
Explanation
\(\frac{π}{4}\)
If sin x + sin y = \(\frac{2π}{3}\) then cos x + cos y =
0%
\(\frac{2π}{3}\)
0%
\(\frac{π}{4}\)
0%
\(\frac{π}{3}\)
0%
\(\frac{π}{2}\)
Explanation
\(\frac{π}{3}\)
The principal value of cosec (-2) is
0%
–\(\frac{2π}{3}\)
0%
\(\frac{π}{6}\)
0%
\(\frac{2π}{3}\)
0%
–\(\frac{π}{6}\)
Explanation
–\(\frac{π}{6}\)
Which of the following is the principal value branch of cos x?
0%
[\(\frac{-π}{2}\), \(\frac{π}{2}\)]
0%
(0, π)
0%
(π, 0)
0%
(0, π) – {\(\frac{π}{2}\)}
Explanation
(0, π)
Which of the following is the principal value branch of cosec x?
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(\(\frac{-π}{2}\), \(\frac{π}{2}\))
0%
(0, π) – {\(\frac{π}{2}\)}
0%
[\(\frac{-π}{2}\), \(\frac{π}{2}\)]
0%
[\(\frac{-π}{2}\), \(\frac{π}{2}\)] – [0]
Explanation
[\(\frac{-π}{2}\), \(\frac{π}{2}\)] – [0]
The value of cos[cos(\(\frac{33π}{5}\))] is
0%
\(\frac{3π}{5}\)
0%
\(\frac{-3π}{5}\)
0%
\(\frac{π}{10}\)
0%
–\(\frac{-π}{10}\)
Explanation
\(\frac{3π}{5}\)
The domain of the function cos(2x – 1) is
0%
[0, 1]
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[-1, 1]
0%
[-1, -1]
0%
[0, π]
Explanation
[0, 1]
The domain of the function defined by f (x) = sin \(\sqrt{x-1}\) is
0%
[1, 2]
0%
[-1, 1]
0%
[0, 1]
0%
None of these
Explanation
[1, 2]
If cos(sin\(\frac{2}{5}\) + cosx) = 0 then x is equal to
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\(\frac{1}{5}\)
0%
\(\frac{2}{5}\)
0%
0
0%
1
Explanation
\(\frac{2}{5}\)
The value of sin (2 tan(.75)) is equal to
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.75
0%
1.5
0%
.96
0%
sin 1.5
Explanation
.96
The value of cos (cos\(\frac{3π}{2}\)) is equal to
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\(\frac{π}{2}\)
0%
\(\frac{3π}{2}\)
0%
\(\frac{5π}{2}\)
0%
–\(\frac{7π}{2}\)
Explanation
\(\frac{π}{2}\)
The value of expression 2 sec 2 + sin(\(\frac{1}{2}\)) is
0%
\(\frac{π}{6}\)
0%
\(\frac{5π}{6}\)
0%
\(\frac{7π}{6}\)
0%
1
Explanation
\(\frac{5π}{6}\)
The value of sin [cos
0%
\(\frac{25}{24}\)
0%
\(\frac{25}{7}\)
0%
\(\frac{24}{25}\)
0%
\(\frac{7}{24}\)
Explanation
\(\frac{24}{25}\)
The number of real solution of the equation is \(\sqrt{1+cos 2x}\) = √2 cos (cos x) in [\(\frac{π}{2}\), π] is
0%
0
0%
1
0%
2
0%
None of these
Explanation
2
If cos x > sin x, then
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\(\frac{1}{√2}\) < x ≤ 1
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0 ≤ x < \(\frac{1}{√2}\)
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-1 ≤ x < \(\frac{1}{√2}\) (d) x > 0
Explanation
0 ≤ x < \(\frac{1}{√2}\)
sin(\(\frac{-1}{2}\))
0%
\(\frac{π}{3}\)
0%
–\(\frac{π}{3}\)
0%
\(\frac{π}{6}\)
0%
–\(\frac{π}{6}\)
Explanation
–\(\frac{π}{6}\)
sec(\(\frac{-2}{√3}\))
0%
\(\frac{π}{6}\)
0%
\(\frac{π}{3}\)
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\(\frac{5π}{6}\)
0%
–\(\frac{2π}{3}\)
Explanation
\(\frac{5π}{6}\)
cos(\(\frac{1}{2}\))
0%
–\(\frac{π}{3}\)
0%
\(\frac{π}{3}\)
0%
\(\frac{π}{2}\)
0%
\(\frac{2π}{3}\)
Explanation
\(\frac{π}{3}\)
cosec(\(\frac{-2}{√3}\))
0%
–\(\frac{π}{3}\)
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\(\frac{π}{3}\)
0%
\(\frac{π}{2}\)
0%
–\(\frac{π}{2}\)
Explanation
–\(\frac{π}{3}\)
cot(1)
0%
\(\frac{π}{3}\)
0%
\(\frac{π}{4}\)
0%
\(\frac{π}{2}\)
0%
0
Explanation
\(\frac{π}{4}\)
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