This identity follows from \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\). Its domain is \(x\in[-1,1]\).
Step 2
Why this answer is correct
The correct answer is A. जब \(x\in[-1,1]\) / When \(x\in[-1,1]\). This identity follows from \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\). Its domain is \(x\in[-1,1]\).
Step 3
Exam Tip
यह identity \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\) से मिलती है। इसका domain \(x\in[-1,1]\) है।
A. real में परिभाषित नहीं/Not defined in real numbers
Step 1
Concept
\(\cos^{-1}x\) needs \(x\in[-1,1]\). Since \(\frac{3}{2}>1\), there is no real value.
Step 2
Why this answer is correct
The correct answer is A. real में परिभाषित नहीं / Not defined in real numbers. \(\cos^{-1}x\) needs \(x\in[-1,1]\). Since \(\frac{3}{2}>1\), there is no real value.
Step 3
Exam Tip
\(\cos^{-1}x\) के लिए \(x\in[-1,1]\) चाहिए। \(\frac{3}{2}>1\), इसलिए real value नहीं है।
C. यह real numbers में परिभाषित नहीं है/It is not defined in real numbers
Step 1
Concept
\(\sin^{-1}x\) is real-defined only when \(x\in[-1,1]\). The number (2) is not in this interval.
Step 2
Why this answer is correct
The correct answer is C. यह real numbers में परिभाषित नहीं है / It is not defined in real numbers. \(\sin^{-1}x\) is real-defined only when \(x\in[-1,1]\). The number (2) is not in this interval.
Step 3
Exam Tip
\(\sin^{-1}x\) real में तभी defined है जब \(x\in[-1,1]\)। (2) इस interval में नहीं है।
\(cosec y=\frac{1}{sin y}\), so values of \(cosec y\) satisfy \(|x|\ge1\). Use the reciprocal nature to remember the domain.
Step 2
Why this answer is correct
The correct answer is B. (\(-\infty,-1]\cup[1,\infty\)). \(cosec y=\frac{1}{sin y}\), so values of \(cosec y\) satisfy \(|x|\ge1\). Use the reciprocal nature to remember the domain.
Step 3
Exam Tip
\(cosec y=\frac{1}{sin y}\), इसलिए \(cosec y\) के values \(|x|\ge1\) होते हैं। domain याद करते समय reciprocal nature देखें।
From \(-1\le x^2-1\le1\), we get \(0\le x^2\le2\). Thus \(x\in[-\sqrt{2},\sqrt{2}]\), so none of the first three is exact.
Step 2
Why this answer is correct
The correct answer is D. ([-2,2]). From \(-1\le x^2-1\le1\), we get \(0\le x^2\le2\). Thus \(x\in[-\sqrt{2},\sqrt{2}]\), so none of the first three is exact.
Step 3
Exam Tip
\(-1\le x^2-1\le1\) से \(0\le x^2\le2\) मिलता है। इसलिए \(x\in[-\sqrt{2},\sqrt{2}]\), जो दिए विकल्पों में नहीं है।
For \(\sin^{-1}(2x)\), we need \(-1\le 2x\le1\). This gives \(x\in[-\frac{1}{2},\frac{1}{2}]\).
Step 2
Why this answer is correct
The correct answer is A. \([-\frac{1}{2},\frac{1}{2}]\). For \(\sin^{-1}(2x)\), we need \(-1\le 2x\le1\). This gives \(x\in[-\frac{1}{2},\frac{1}{2}]\).
Step 3
Exam Tip
\(\sin^{-1}(2x)\) के लिए \(-1\le 2x\le1\) होना चाहिए। इससे \(x\in[-\frac{1}{2},\frac{1}{2}]\) मिलता है।
B. केवल \(x\in[-1,1]\) के लिए/Only for \(x\in[-1,1]\)
Step 1
Concept
The domain of \(\sin^{-1}x\) is \([-1,1]\), so \(\sin(\sin^{-1}x)=x\) there. In composition, check the inner function domain.
Step 2
Why this answer is correct
The correct answer is B. केवल \(x\in[-1,1]\) के लिए / Only for \(x\in[-1,1]\). The domain of \(\sin^{-1}x\) is \([-1,1]\), so \(\sin(\sin^{-1}x)=x\) there. In composition, check the inner function domain.
Step 3
Exam Tip
\(\sin^{-1}x\) का डोमेन \([-1,1]\) है, इसलिए इसी पर \(\sin(\sin^{-1}x)=x\) होगा। composition में inner function का domain देखें।
For \(\sin^{-1}(2x)\), we need \(-1\le 2x\le 1\). Hence \(-\frac{1}{2}\le x\le \frac{1}{2}\).
Step 2
Why this answer is correct
The correct answer is B. \(-\frac{1}{2}\le x\le \frac{1}{2}\). For \(\sin^{-1}(2x)\), we need \(-1\le 2x\le 1\). Hence \(-\frac{1}{2}\le x\le \frac{1}{2}\).
Step 3
Exam Tip
\(\sin^{-1}(2x)\) के लिए \(-1\le 2x\le 1\) चाहिए। अतः \(-\frac{1}{2}\le x\le \frac{1}{2}\) मिलेगा।
Values of \(\sec y\) do not lie in (\left\(-1,1\right\)), so the domain of \(\sec^{-1}x\) is (\mathbb{R}\setminus\left\(-1,1\right\)). Zero is also not in the domain.
Step 2
Why this answer is correct
The correct answer is B. (\mathbb{R}\setminus\left\(-1,1\right\)). Values of \(\sec y\) do not lie in (\left\(-1,1\right\)), so the domain of \(\sec^{-1}x\) is (\mathbb{R}\setminus\left\(-1,1\right\)). Zero is also not in the domain.
Step 3
Exam Tip
\(\sec y\) के मान (\left\(-1,1\right\)) में नहीं आते, इसलिए \(\sec^{-1}x\) का प्रांत (\mathbb{R}\setminus\left\(-1,1\right\)) है। शून्य भी प्रांत में नहीं है।
For (\sin^{-1}(2x)), we need \(-1\le 2x\le1\). This gives \(x\in\left[-\frac{1}{2},\frac{1}{2}\right]\).
Step 2
Why this answer is correct
The correct answer is A. \(\left[-\frac{1}{2},\frac{1}{2}\right]\). For (\sin^{-1}(2x)), we need \(-1\le 2x\le1\). This gives \(x\in\left[-\frac{1}{2},\frac{1}{2}\right]\).
Step 3
Exam Tip
(\sin^{-1}(2x)) के लिए \(-1\le 2x\le1\) होना चाहिए। इससे \(x\in\left[-\frac{1}{2},\frac{1}{2}\right]\) मिलता है।
A. \(\sin^{-1}x\) का डोमेन ([-1,1]) है/The domain of \(\sin^{-1}x\) is ([-1,1])
Step 1
Concept
For \(\sin^{-1}x\), the value of (x) must lie in ([-1,1]). The other statements are wrong in domain or range.
Step 2
Why this answer is correct
The correct answer is A. \(\sin^{-1}x\) का डोमेन ([-1,1]) है / The domain of \(\sin^{-1}x\) is ([-1,1]). For \(\sin^{-1}x\), the value of (x) must lie in ([-1,1]). The other statements are wrong in domain or range.
Step 3
Exam Tip
\(\sin^{-1}x\) के लिए (x) का मान ([-1,1]) में होना चाहिए। बाकी कथन डोमेन या सीमा में गलत हैं।
A. समस्त वास्तविक संख्याएं \(\mathbb{R}\)/All real numbers \(\mathbb{R}\)
Step 1
Concept
Since \(\cot x\) can take all real values the domain of \(\cot^{-1}x\) is \(\mathbb{R}\). Do not mix it with the domain of \(\sin^{-1}x\).
Step 2
Why this answer is correct
The correct answer is A. समस्त वास्तविक संख्याएं \(\mathbb{R}\) / All real numbers \(\mathbb{R}\). Since \(\cot x\) can take all real values the domain of \(\cot^{-1}x\) is \(\mathbb{R}\). Do not mix it with the domain of \(\sin^{-1}x\).
Step 3
Exam Tip
\(\cot x\) सभी वास्तविक मान ले सकता है इसलिए \(\cot^{-1}x\) का डोमेन \(\mathbb{R}\) है। इसे \(\sin^{-1}\) के डोमेन से न मिलाएं।
The value of \(\cos x\) also lies in ([-1,1]) so the domain of \(\cos^{-1}x\) is ([-1,1]). Check the range of (x) before substitution.
Step 2
Why this answer is correct
The correct answer is A. अंतराल ([-1,1]) / Interval ([-1,1]). The value of \(\cos x\) also lies in ([-1,1]) so the domain of \(\cos^{-1}x\) is ([-1,1]). Check the range of (x) before substitution.
Step 3
Exam Tip
\(\cos x\) का मान भी ([-1,1]) में रहता है इसलिए \(\cos^{-1}x\) का डोमेन ([-1,1]) है। मान रखने से पहले (x) की सीमा देखें।
The value of \(\sin x\) lies only in ([-1,1]) so the domain of \(\sin^{-1}x\) is ([-1,1]). In exams check the domain first.
Step 2
Why this answer is correct
The correct answer is A. अंतराल ([-1,1]) / Interval ([-1,1]). The value of \(\sin x\) lies only in ([-1,1]) so the domain of \(\sin^{-1}x\) is ([-1,1]). In exams check the domain first.
Step 3
Exam Tip
\(\sin x\) का मान केवल ([-1,1]) में आता है इसलिए \(\sin^{-1}x\) का डोमेन ([-1,1]) है। परीक्षा में पहले डोमेन जरूर जांचें।
The domain of \(\sec^{-1}x\) is (\left\(-\infty,-1\right]\cup\left[1,\infty\right\)). The number (0) 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 domain of \(\sec^{-1}x\) is (\left\(-\infty,-1\right]\cup\left[1,\infty\right\)). The number (0) is not in this domain.
Step 3
Exam Tip
\(\sec^{-1}x\) का प्रांत (\left\(-\infty,-1\right]\cup\left[1,\infty\right\)) है। (0) इस प्रांत में नहीं है।
The function \(cosec^{-1}x\) is defined when \(\left|x\right|\ge1\). Here \(\left|-\frac{5}{4}\right|\ge1\), so it is defined.
Step 2
Why this answer is correct
The correct answer is A. यह परिभाषित है / It is defined. The function \(cosec^{-1}x\) is defined when \(\left|x\right|\ge1\). Here \(\left|-\frac{5}{4}\right|\ge1\), so it is defined.
Step 3
Exam Tip
\(cosec^{-1}x\) तभी परिभाषित है जब \(\left|x\right|\ge1\)। यहाँ \(\left|-\frac{5}{4}\right|\ge1\), इसलिए यह परिभाषित है।
For \(\sec^{-1}x\), we need \(\left|x\right|\ge1\). Here \(\left|\frac{3}{2}\right|\ge1\), so it is defined.
Step 2
Why this answer is correct
The correct answer is A. यह परिभाषित है / It is defined. For \(\sec^{-1}x\), we need \(\left|x\right|\ge1\). Here \(\left|\frac{3}{2}\right|\ge1\), so it is defined.
Step 3
Exam Tip
\(\sec^{-1}x\) के लिए \(\left|x\right|\ge1\) होना चाहिए। यहाँ \(\left|\frac{3}{2}\right|\ge1\), इसलिए यह परिभाषित है।
The domain of \(\cos^{-1}x\) is \(\left[-1,1\right]\). Since \(\frac{11}{10}>1\), it is not defined.
Step 2
Why this answer is correct
The correct answer is A. परिभाषित नहीं / Not defined. The domain of \(\cos^{-1}x\) is \(\left[-1,1\right]\). Since \(\frac{11}{10}>1\), it is not defined.
Step 3
Exam Tip
\(\cos^{-1}x\) का प्रांत \(\left[-1,1\right]\) है। क्योंकि \(\frac{11}{10}>1\), यह परिभाषित नहीं है।
Since \(\frac{7}{6}\notin\left[-1,1\right]\), \(\sin^{-1}\left(\frac{7}{6}\right)\) is not defined. Check the domain first.
Step 2
Why this answer is correct
The correct answer is B. यह परिभाषित नहीं है / It is not defined. Since \(\frac{7}{6}\notin\left[-1,1\right]\), \(\sin^{-1}\left(\frac{7}{6}\right)\) is not defined. Check the domain first.
Step 3
Exam Tip
\(\frac{7}{6}\notin\left[-1,1\right]\), इसलिए \(\sin^{-1}\left(\frac{7}{6}\right)\) परिभाषित नहीं है। प्रांत पहले जांचें।
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।
