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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।
The function \(\sec^{-1}x\) is defined when \(\left|x\right|\ge1\). Here \(\left|\frac{1}{2}\right|<1\), so it is not defined.
Step 2
Why this answer is correct
The correct answer is B. यह परिभाषित नहीं है / It is not defined. The function \(\sec^{-1}x\) is defined when \(\left|x\right|\ge1\). Here \(\left|\frac{1}{2}\right|<1\), so it is not defined.
Step 3
Exam Tip
\(\sec^{-1}x\) तभी परिभाषित है जब \(\left|x\right|\ge1\)। यहाँ \(\left|\frac{1}{2}\right|<1\), इसलिए यह परिभाषित नहीं है।
Since \(sin\frac{2\pi}{3}=\frac{\sqrt{3}}{2}\) and \(sin^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{3}\). Always choose the principal range value.
Step 2
Why this answer is correct
The correct answer is B. \(\frac{\pi}{3}\). Since \(sin\frac{2\pi}{3}=\frac{\sqrt{3}}{2}\) and \(sin^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{3}\). Always choose the principal range value.
Step 3
Exam Tip
\(sin\frac{2\pi}{3}=\frac{\sqrt{3}}{2}\) और \(sin^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{3}\)। मुख्य मान हमेशा निर्धारित परिसर में लेना चाहिए।
We have \(tan\frac{3\pi}{4}=-1\) and \(tan^{-1}\left(-1\right)=-\frac{\pi}{4}\). Remember the principal range \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\).
Step 2
Why this answer is correct
The correct answer is B. -\(\frac{\pi}{4}\). We have \(tan\frac{3\pi}{4}=-1\) and \(tan^{-1}\left(-1\right)=-\frac{\pi}{4}\). Remember the principal range \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\).
Step 3
Exam Tip
\(tan\frac{3\pi}{4}=-1\) है और \(tan^{-1}\left(-1\right)=-\frac{\pi}{4}\)। मुख्य परिसर \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\) याद रखें।
A. व्युत्क्रम त्रिकोणमितीय फलन/Inverse trigonometric function
Step 1
Concept
The function \(\sin^{-1}x\) is the inverse of \(\sin x\) on a suitable branch. Hence it is an inverse trigonometric function.
Step 2
Why this answer is correct
The correct answer is A. व्युत्क्रम त्रिकोणमितीय फलन / Inverse trigonometric function. The function \(\sin^{-1}x\) is the inverse of \(\sin x\) on a suitable branch. Hence it is an inverse trigonometric function.
Step 3
Exam Tip
\(\sin^{-1}x\) \(\sin x\) का उपयुक्त शाखा पर व्युत्क्रम है। इसलिए यह व्युत्क्रम त्रिकोणमितीय फलन है।
B. यह \(\sin x\) का व्युत्क्रम फलन है/It is the inverse function of \(\sin x\)
Step 1
Concept
The notation \(\sin^{-1}x\) means the inverse function of \(\sin x\), not the reciprocal \(\frac{1}{\sin x}\). This is an important mistake to avoid.
Step 2
Why this answer is correct
The correct answer is B. यह \(\sin x\) का व्युत्क्रम फलन है / It is the inverse function of \(\sin x\). The notation \(\sin^{-1}x\) means the inverse function of \(\sin x\), not the reciprocal \(\frac{1}{\sin x}\). This is an important mistake to avoid.
Step 3
Exam Tip
\(\sin^{-1}x\) का अर्थ \(\sin x\) का व्युत्क्रम फलन है, प्रतिलोम \(\frac{1}{\sin x}\) नहीं। यह एक महत्वपूर्ण गलती है।
