Class 12 Mathematics Trigonometric Functions Practice
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Introduction to inverse trigonometric functions
116 MCQs
One-one function
21 MCQs
Onto function
6 MCQs
Functions
5 MCQs
Equivalence relation
2 MCQs
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\(\sin 0=0\) and \(\sin \pi=0\), while \(0\neq\pi\).
Step 3
Exam Tip
Periodic functions are usually not one-one on the whole real domain. चरण 1: \(\sin x\) आवर्ती फलन है। चरण 2: \(\sin 0=0\) और \(\sin \pi=0\) जबकि \(0\neq\pi\)। चरण 3: पूरे वास्तविक प्रांत पर आवर्ती फलन अक्सर एकैकी नहीं होते।
In this interval, two different (x) values do not give the same sine value.
Step 3
Exam Tip
Trigonometric functions can be one-one on suitable restricted intervals. चरण 1: दिए गए अंतराल पर \(\sin x\) लगातार बढ़ता है। चरण 2: इस अंतराल में दो अलग (x) समान ज्या मान नहीं देते। चरण 3: त्रिकोणमितीय फलन उचित अंतराल पर एकैकी हो सकते हैं।
\(\cos 0=1\) and \(\cos 2\pi=1\), while \(0\neq2\pi\).
Step 3
Exam Tip
If different angles give the same value, the function is not one-one. चरण 1: \(\cos x\) आवर्ती फलन है। चरण 2: \(\cos 0=1\) और \(\cos 2\pi=1\) जबकि \(0\neq2\pi\)। चरण 3: समान मान वाले अलग कोण मिल जाएँ तो फलन एकैकी नहीं होता।
A decreasing function gives different outputs for different inputs.
Step 3
Exam Tip
Being strictly decreasing is also a strong sign of one-one property. चरण 1: \([0,\pi]\) पर \(\cos x\) घटने वाला फलन है। चरण 2: घटने वाला फलन अलग आगतों पर अलग मान देता है। चरण 3: घटता हुआ होना भी एकैकीपन की मजबूत पहचान है।
In periodic functions, different angles can give the same value. चरण 1: \(\sin 0=0\) है। चरण 2: \(\sin \pi=0\) है जबकि \(0\neq\pi\)। चरण 3: आवर्ती फलनों में अलग कोण समान मान दे सकते हैं।
On the given interval, \(\tan x\) is strictly increasing.
Step 2
Why this answer is correct
Therefore two different (x) values cannot give the same tangent value.
Step 3
Exam Tip
Restricting a trigonometric function to a suitable interval can make it one-one. चरण 1: दिए गए अंतराल पर \(\tan x\) लगातार बढ़ता है। चरण 2: इसलिए दो अलग (x) समान स्पर्शज्या मान नहीं दे सकते। चरण 3: त्रिकोणमितीय फलनों को सही अंतराल पर सीमित करने से एकैकीपन मिल सकता है।
Every value in the codomain ([-1,1]) is obtained for some (x).
Step 3
Exam Tip
For onto nature, every codomain value must be hit; however, the function must also be well-defined with outputs in the codomain. चरण 1: \(2\sin x\) का परास ([-2,2]) है। चरण 2: सहप्रांत ([-1,1]) का हर मान इस परास में आता है और किसी न किसी (x) से मिल जाता है। चरण 3: आच्छादी के लिए परास का सहप्रांत के बराबर होना जरूरी नहीं, सहप्रांत का हर मान परास में होना चाहिए; फलन की परिभाषा में वास्तविक मान सहप्रांत में होने चाहिए, इसलिए ऐसे प्रश्न में सहप्रांत अनुकूल होना चाहिए।
On \([0,2\pi]\), \(\sin x\) takes all values from (-1) to (1).
Step 2
Why this answer is correct
The codomain is ([-1,1]), so every codomain value is obtained.
Step 3
Exam Tip
On a full period, identifying trigonometric range is easy. चरण 1: \([0,2\pi]\) पर \(\sin x\) (-1) से (1) तक सभी मान लेता है। चरण 2: सहप्रांत ([-1,1]) है, इसलिए हर सहप्रांत मान प्राप्त होता है। चरण 3: पूर्ण आवर्त पर त्रिकोणमितीय परास पहचानना आसान होता है।
The principal value of \(\sin^{-1}x\) lies in \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\). Remember the range in exams.
Step 2
Why this answer is correct
The correct answer is A. \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\). The principal value of \(\sin^{-1}x\) lies in \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\). Remember the range in exams.
Step 3
Exam Tip
\(\sin^{-1}x\) का मुख्य मान \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]) में लिया जाता है। परीक्षा में परिसर याद रखें।
The principal value of \(\tan^{-1}x\) lies in (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)). The endpoints are not included.
Step 2
Why this answer is correct
The correct answer is B. \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\). The principal value of \(\tan^{-1}x\) lies in (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)). The endpoints are not included.
Step 3
Exam Tip
\(\tan^{-1}x\) का मुख्य मान (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)) में होता है। सिरों को शामिल नहीं किया जाता।
The principal value of \(\cot^{-1}x\) is taken in (\left\(0,\pi\right\)). Remember that both endpoints are excluded.
Step 2
Why this answer is correct
The correct answer is A. \(\left(0,\pi\right)\). The principal value of \(\cot^{-1}x\) is taken in (\left\(0,\pi\right\)). Remember that both endpoints are excluded.
Step 3
Exam Tip
\(\cot^{-1}x\) का मुख्य मान (\left\(0,\pi\right\)) में लिया जाता है। याद रखें कि \(\cot^{-1}x\) में दोनों सिरे नहीं आते।
C. \(\left(-\infty,-1\right]\cup\left[1,\infty\right)\)
Step 1
Concept
Values of \(\sec y\) lie in (\left\(-\infty,-1\right]\cup\left[1,\infty\right\)). Hence this is the domain of \(\sec^{-1}x\).
Step 2
Why this answer is correct
The correct answer is C. \(\left(-\infty,-1\right]\cup\left[1,\infty\right)\). Values of \(\sec y\) lie in (\left\(-\infty,-1\right]\cup\left[1,\infty\right\)). Hence this is the domain of \(\sec^{-1}x\).
Step 3
Exam Tip
\(\sec y\) के मान (\left\(-\infty,-1\right]\cup\left[1,\infty\right\)) में होते हैं। इसलिए \(\sec^{-1}x\) का प्रांत यही है।
B. \(\left(-\infty,-1\right]\cup\left[1,\infty\right)\)
Step 1
Concept
The value of \(\cosec y\) never lies in (\left\(-1,1\right\)). So the domain of \(\cosec^{-1}x\) is (\left\(-\infty,-1\right]\cup\left[1,\infty\right\)).
Step 2
Why this answer is correct
The correct answer is B. \(\left(-\infty,-1\right]\cup\left[1,\infty\right)\). The value of \(\cosec y\) never lies in (\left\(-1,1\right\)). So the domain of \(\cosec^{-1}x\) is (\left\(-\infty,-1\right]\cup\left[1,\infty\right\)).
Step 3
Exam Tip
\(\cosec y\) कभी (\left\(-1,1\right\)) में नहीं आता। इसलिए \(\cosec^{-1}x\) का प्रांत (\left\(-\infty,-1\right]\cup\left[1,\infty\right\)) है।
Values of \(\sin y\) lie only in \(\left[-1,1\right]\). So \(\sin^{-1}x\) is defined when \(x\in\left[-1,1\right]\).
Step 2
Why this answer is correct
The correct answer is B. \(\left[-1,1\right]\). Values of \(\sin y\) lie only in \(\left[-1,1\right]\). So \(\sin^{-1}x\) is defined when \(x\in\left[-1,1\right]\).
Step 3
Exam Tip
\(\sin y\) के मान केवल \(\left[-1,1\right]\) में होते हैं। इसलिए \(\sin^{-1}x\) तभी परिभाषित है जब \(x\in\left[-1,1\right]\)।
The value of \(\cos y\) does not go outside \(\left[-1,1\right]\). Hence the domain of \(\cos^{-1}x\) is \(\left[-1,1\right]\).
Step 2
Why this answer is correct
The correct answer is A. \(\left[-1,1\right]\). The value of \(\cos y\) does not go outside \(\left[-1,1\right]\). Hence the domain of \(\cos^{-1}x\) is \(\left[-1,1\right]\).
Step 3
Exam Tip
\(\cos y\) का मान \(\left[-1,1\right]\) से बाहर नहीं जाता। इसलिए \(\cos^{-1}x\) का प्रांत \(\left[-1,1\right]\) है।
Since \(\sin\frac{\pi}{2}=1\) and \(\frac{\pi}{2}\) is in the principal range, the answer is \(\frac{\pi}{2}\).
Step 2
Why this answer is correct
The correct answer is B. \(\frac{\pi}{2}\). Since \(\sin\frac{\pi}{2}=1\) and \(\frac{\pi}{2}\) is in the principal range, the answer is \(\frac{\pi}{2}\).
Step 3
Exam Tip
\(\sin\frac{\pi}{2}=1\) और \(\frac{\pi}{2}\) मुख्य परिसर में है। इसलिए उत्तर \(\frac{\pi}{2}\) है।
Since (\tan\left\(-\frac{\pi}{4}\right\)=-1) and \(-\frac{\pi}{4}\) lies in the principal range, the answer is \(-\frac{\pi}{4}\).
Step 2
Why this answer is correct
The correct answer is B. -\(\frac{\pi}{4}\). Since (\tan\left\(-\frac{\pi}{4}\right\)=-1) and \(-\frac{\pi}{4}\) lies in the principal range, the answer is \(-\frac{\pi}{4}\).
Step 3
Exam Tip
(\tan\left\(-\frac{\pi}{4}\right\)=-1) और \(-\frac{\pi}{4}\) मुख्य परिसर में है। इसलिए उत्तर \(-\frac{\pi}{4}\) है।
Since \(\cot\frac{\pi}{2}=0\) and \(\frac{\pi}{2}\) lies in (\left\(0,\pi\right\)), the value is \(\frac{\pi}{2}\).
Step 2
Why this answer is correct
The correct answer is C. \(\frac{\pi}{2}\). Since \(\cot\frac{\pi}{2}=0\) and \(\frac{\pi}{2}\) lies in (\left\(0,\pi\right\)), the value is \(\frac{\pi}{2}\).
Step 3
Exam Tip
\(\cot\frac{\pi}{2}=0\) और \(\frac{\pi}{2}\) परिसर (\left\(0,\pi\right\)) में है। इसलिए मान \(\frac{\pi}{2}\) है।
The expression \(\sin^{-1}x\) gives the angle whose \(\sin\) value is (x). So \(y=\sin^{-1}x\) means \(\sin y=x\).
Step 2
Why this answer is correct
The correct answer is A. \(\sin y=x\). The expression \(\sin^{-1}x\) gives the angle whose \(\sin\) value is (x). So \(y=\sin^{-1}x\) means \(\sin y=x\).
Step 3
Exam Tip
\(\sin^{-1}x\) वह कोण देता है जिसका \(\sin\) मान (x) हो। इसलिए \(y=\sin^{-1}x\) का अर्थ \(\sin y=x\) है।
The angle \(\frac{\pi}{6}\) lies in the principal range \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\). Hence the value is \(\frac{\pi}{6}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{\pi}{6}\). The angle \(\frac{\pi}{6}\) lies in the principal range \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\). Hence the value is \(\frac{\pi}{6}\).
Step 3
Exam Tip
\(\frac{\pi}{6}\) मुख्य परिसर \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\) में है। इसलिए मान \(\frac{\pi}{6}\) है।
The angle \(\frac{2\pi}{3}\) lies in the principal range \(\left[0,\pi\right]\) of \(\cos^{-1}x\). Hence the value is \(\frac{2\pi}{3}\).
Step 2
Why this answer is correct
The correct answer is B. \(\frac{2\pi}{3}\). The angle \(\frac{2\pi}{3}\) lies in the principal range \(\left[0,\pi\right]\) of \(\cos^{-1}x\). Hence the value is \(\frac{2\pi}{3}\).
Step 3
Exam Tip
\(\frac{2\pi}{3}\) \(\cos^{-1}x\) के मुख्य परिसर \(\left[0,\pi\right]\) में है। इसलिए मान \(\frac{2\pi}{3}\) है।
The angle \(\frac{\pi}{6}\) lies in the principal range of \(\tan^{-1}x\). So the value remains \(\frac{\pi}{6}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{\pi}{6}\). The angle \(\frac{\pi}{6}\) lies in the principal range of \(\tan^{-1}x\). So the value remains \(\frac{\pi}{6}\).
Step 3
Exam Tip
\(\frac{\pi}{6}\) \(\tan^{-1}x\) के मुख्य परिसर में है। इसलिए मान \(\frac{\pi}{6}\) रहेगा।
Since \(\cos\frac{2\pi}{3}=-\frac{1}{2}\) and \(\frac{2\pi}{3}\) is in the principal range, the answer is \(\frac{2\pi}{3}\).
Step 2
Why this answer is correct
The correct answer is B. \(\frac{2\pi}{3}\). Since \(\cos\frac{2\pi}{3}=-\frac{1}{2}\) and \(\frac{2\pi}{3}\) is in the principal range, the answer is \(\frac{2\pi}{3}\).
Step 3
Exam Tip
\(\cos\frac{2\pi}{3}=-\frac{1}{2}\) और \(\frac{2\pi}{3}\) मुख्य परिसर में है। इसलिए उत्तर \(\frac{2\pi}{3}\) है।
The domain of \(\sin^{-1}x\) is \(\left[-1,1\right]\) and \(2\notin\left[-1,1\right]\). Hence it is not defined.
Step 2
Why this answer is correct
The correct answer is B. परिभाषित नहीं होगा / It is not defined. The domain of \(\sin^{-1}x\) is \(\left[-1,1\right]\) and \(2\notin\left[-1,1\right]\). Hence it is not defined.
Step 3
Exam Tip
\(\sin^{-1}x\) का प्रांत \(\left[-1,1\right]\) है और \(2\notin\left[-1,1\right]\)। इसलिए यह परिभाषित नहीं है।
The function \(\cos^{-1}x\) is defined only for \(x\in\left[-1,1\right]\). The number (-2) is not in this domain.
Step 2
Why this answer is correct
The correct answer is B. यह परिभाषित नहीं है / It is not defined. The function \(\cos^{-1}x\) is defined only for \(x\in\left[-1,1\right]\). The number (-2) is not in this domain.
Step 3
Exam Tip
\(\cos^{-1}x\) केवल \(x\in\left[-1,1\right]\) के लिए परिभाषित है। (-2) इस प्रांत में नहीं है।
Related questions grouped automatically for chapter-wise practice. Topics include Introduction to inverse trigonometric functions, One-one function, Onto function, Functions, Equivalence relation.
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