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The square root is in the denominator, so \(x^2-9>0\) is required. A denominator can never be zero.
Step 2
Why this answer is correct
The correct answer is A. ( \(-\infty,-3\)\cup\(3,\infty\) ). The square root is in the denominator, so \(x^2-9>0\) is required. A denominator can never be zero.
Step 3
Exam Tip
हर में वर्गमूल है इसलिए \(x^2-9>0\) होना चाहिए। हर में शून्य कभी स्वीकार्य नहीं होता।
From \(y=\frac{x+1}{x-2}\), \(x=\frac{2y+1}{y-1}\), so \(y\ne 1\). For a linear fractional function, isolate the impossible value.
Step 2
Why this answer is correct
The correct answer is A. \( \mathbb{R}\setminus{1} \). From \(y=\frac{x+1}{x-2}\), \(x=\frac{2y+1}{y-1}\), so \(y\ne 1\). For a linear fractional function, isolate the impossible value.
Step 3
Exam Tip
\(y=\frac{x+1}{x-2}\) से \(x=\frac{2y+1}{y-1}\), इसलिए \(y\ne 1\)। रैखिक भिन्न फलन में असंभव मान अलग करें।
The denominator \(1+|x|\ge 1\), so the maximum value is (1). As (x) grows, the value approaches (0) but never becomes (0).
Step 2
Why this answer is correct
The correct answer is A. ( (0,1] ). The denominator \(1+|x|\ge 1\), so the maximum value is (1). As (x) grows, the value approaches (0) but never becomes (0).
Step 3
Exam Tip
हर \(1+|x|\ge 1\) है, इसलिए अधिकतम मान (1) है। (x) बढ़ने पर मान (0) के पास जाता है पर (0) नहीं बनता।
Since (x-2+4x+7=(x+2)2+3), the inside minimum is (3). Taking the square root gives the minimum \(\sqrt{3}\).
Step 2
Why this answer is correct
The correct answer is A. \( [\sqrt{3},\infty\) ). Since (x-2+4x+7=(x+2)2+3), the inside minimum is (3). Taking the square root gives the minimum \(\sqrt{3}\).
Step 3
Exam Tip
(x-2+4x+7=(x+2)2+3), इसलिए अंदर का न्यूनतम (3) है। वर्गमूल लेने पर न्यूनतम \(\sqrt{3}\) मिलेगा।
The square root needs \(x\ge -1\), and the denominator needs \(x\ne \pm 2\). Since (-2) is already outside the domain, only (2) is removed.
Step 2
Why this answer is correct
The correct answer is A. \( [-1,\infty\)\setminus{2} ). The square root needs \(x\ge -1\), and the denominator needs \(x\ne \pm 2\). Since (-2) is already outside the domain, only (2) is removed.
Step 3
Exam Tip
वर्गमूल के लिए \(x\ge -1\) और हर के लिए \(x\ne \pm 2\) चाहिए। (-2) पहले से प्रांत में नहीं है, इसलिए केवल (2) हटेगा।
From \(y=\frac{2x-3}{x+4}\), \(x=\frac{-3-4y}{y-2}\), so (y=2) is impossible. In such fractions, the ratio of leading coefficients often gives the missing value.
Step 2
Why this answer is correct
The correct answer is A. (2). From \(y=\frac{2x-3}{x+4}\), \(x=\frac{-3-4y}{y-2}\), so (y=2) is impossible. In such fractions, the ratio of leading coefficients often gives the missing value.
Step 3
Exam Tip
\(y=\frac{2x-3}{x+4}\) से \(x=\frac{-3-4y}{y-2}\), इसलिए (y=2) असंभव है। अनुपात में प्रमुख गुणांकों का अनुपात अक्सर छूटा मान देता है।
The difference of distances from (2) and (-2) stays between (-4) and (4). Removing signs interval-wise is the safe method.
Step 2
Why this answer is correct
The correct answer is A. ( [-4,4] ). The difference of distances from (2) and (-2) stays between (-4) and (4). Removing signs interval-wise is the safe method.
Step 3
Exam Tip
बिंदुओं (2) और (-2) से दूरी के अंतर का मान (-4) से (4) तक रहता है। अंतरालों पर चिह्न हटाकर जांचना सुरक्षित तरीका है।
Even after simplification, the original denominator requires \(x-2\ne 0\). Do not add the cancelled point back into the domain.
Step 2
Why this answer is correct
The correct answer is A. \( \mathbb{R}\setminus{2} \). Even after simplification, the original denominator requires \(x-2\ne 0\). Do not add the cancelled point back into the domain.
Step 3
Exam Tip
सरलीकरण के बाद भी मूल हर में \(x-2\ne 0\) रहना चाहिए। हटे हुए बिंदु को प्रांत में वापस न जोड़ें।
For \(x\ne 2\), the function equals (x+2). Thus the value (4), which would occur at (x=2), is not in the range.
Step 2
Why this answer is correct
The correct answer is A. \( \mathbb{R}\setminus{4} \). For \(x\ne 2\), the function equals (x+2). Thus the value (4), which would occur at (x=2), is not in the range.
Step 3
Exam Tip
\(x\ne 2\) पर फलन (x+2) के बराबर है। इसलिए (x=2) से मिलने वाला मान (4) परिसर में नहीं आता।
For \(\sqrt{x}\), \(x\ge 0\), and for the denominator, \(2-\sqrt{x}\ne 0\) is needed. Hence (x=4) is removed.
Step 2
Why this answer is correct
The correct answer is A. \( [0,\infty\)\setminus{4} ). For \(\sqrt{x}\), \(x\ge 0\), and for the denominator, \(2-\sqrt{x}\ne 0\) is needed. Hence (x=4) is removed.
Step 3
Exam Tip
\(\sqrt{x}\) के लिए \(x\ge 0\) और हर के लिए \(2-\sqrt{x}\ne 0\) चाहिए। इसलिए (x=4) हटाया जाएगा।
For positive (x), the value is at least (2), and for negative (x), it is at most (-2). Check both cases separately.
Step 2
Why this answer is correct
The correct answer is A. ( \(-\infty,-2]\cup[2,\infty\) ). For positive (x), the value is at least (2), and for negative (x), it is at most (-2). Check both cases separately.
Step 3
Exam Tip
धनात्मक (x) पर मान कम से कम (2) और ऋणात्मक (x) पर अधिक से अधिक (-2) होता है। दोनों स्थितियों को अलग जांचें।
The maximum of \(\sqrt{16-x^2}\) is (4), so the minimum fraction is \(\frac{1}{4}\). Near the endpoints, the denominator tends to (0).
Step 2
Why this answer is correct
The correct answer is A. \( \left[\frac{1}{4},\infty\right\) ). The maximum of \(\sqrt{16-x^2}\) is (4), so the minimum fraction is \(\frac{1}{4}\). Near the endpoints, the denominator tends to (0).
Step 3
Exam Tip
\(\sqrt{16-x^2}\) का अधिकतम (4) है, इसलिए भिन्न का न्यूनतम \(\frac{1}{4}\) है। किनारों के पास हर (0) की ओर जाता है।
The greatest integer function gives the largest integer not greater than (x). Since (4) is not included, the value (4) will not occur.
Step 2
Why this answer is correct
The correct answer is A. ( {1,2,3} ). The greatest integer function gives the largest integer not greater than (x). Since (4) is not included, the value (4) will not occur.
Step 3
Exam Tip
महत्तम पूर्णांक फलन (x) से बड़ा नहीं सबसे बड़ा पूर्णांक देता है। (4) शामिल नहीं है, इसलिए मान (4) नहीं आएगा।
The expression inside the square root must satisfy \(\frac{x-1}{x+2}\ge 0\) and \(x\ne -2\). A sign chart gives (x<-2) or \(x\ge 1\).
Step 2
Why this answer is correct
The correct answer is A. ( \(-\infty,-2\)\cup[1,\infty) ). The expression inside the square root must satisfy \(\frac{x-1}{x+2}\ge 0\) and \(x\ne -2\). A sign chart gives (x<-2) or \(x\ge 1\).
Step 3
Exam Tip
वर्गमूल के अंदर \(\frac{x-1}{x+2}\ge 0\) और \(x\ne -2\) चाहिए। संकेत सारणी से (x<-2) या \(x\ge 1\) मिलता है।
Put (t=(x+1)2\ge 0), then \(f=\frac{t+4}{t+1}=1+\frac{3}{t+1}\). Hence the values are greater than (1) and up to (4).
Step 2
Why this answer is correct
The correct answer is A. ( (1,4] ). Put (t=(x+1)2\ge 0), then \(f=\frac{t+4}{t+1}=1+\frac{3}{t+1}\). Hence the values are greater than (1) and up to (4).
Step 3
Exam Tip
मान (t=(x+1)2\ge 0) रखें, तो \(f=\frac{t+4}{t+1}=1+\frac{3}{t+1}\)। इसलिए मान (1) से बड़े और (4) तक हैं।
At the endpoints the value is \(2\sqrt{2}\), and at (x=0) the maximum is (4). For symmetric radical functions, also check the midpoint.
Step 2
Why this answer is correct
The correct answer is A. \( [2\sqrt{2},4] \). At the endpoints the value is \(2\sqrt{2}\), and at (x=0) the maximum is (4). For symmetric radical functions, also check the midpoint.
Step 3
Exam Tip
सिरों पर मान \(2\sqrt{2}\) और (x=0) पर अधिकतम (4) है। सममिति वाले वर्गमूल फलनों में मध्य बिंदु भी जांचें।
The logarithm needs (x>0), and the denominator needs \(\log_{2}x\ne 0\). Since \(\log_{2}x=0\) at (x=1), remove (1).
Step 2
Why this answer is correct
The correct answer is A. ( \(0,\infty\)\setminus{1} ). The logarithm needs (x>0), and the denominator needs \(\log_{2}x\ne 0\). Since \(\log_{2}x=0\) at (x=1), remove (1).
Step 3
Exam Tip
लघुगणक के लिए (x>0) और हर के लिए \(\log_{2}x\ne 0\) चाहिए। \(\log_{2}x=0\) पर (x=1) हटेगा।
A. प्रांत \( \mathbb{R}\setminus{0} \), परिसर ( {-1,1} )/domain \( \mathbb{R}\setminus{0} \), range ( {-1,1} )
Step 1
Concept
At (x=0), the denominator becomes (0). For positive (x), the value is (1), and for negative (x), it is (-1).
Step 2
Why this answer is correct
The correct answer is A. प्रांत \( \mathbb{R}\setminus{0} \), परिसर ( {-1,1} ) / domain \( \mathbb{R}\setminus{0} \), range ( {-1,1} ). At (x=0), the denominator becomes (0). For positive (x), the value is (1), and for negative (x), it is (-1).
Step 3
Exam Tip
(x=0) पर हर (0) हो जाता है। धनात्मक (x) पर मान (1) और ऋणात्मक (x) पर (-1) है।
On the domain, the minimum of \(x^2-1\) is (0). As (x) grows, the square root takes all larger non-negative values.
Step 2
Why this answer is correct
The correct answer is A. \( [0,\infty\) ). On the domain, the minimum of \(x^2-1\) is (0). As (x) grows, the square root takes all larger non-negative values.
Step 3
Exam Tip
प्रांत में \(x^2-1\) का न्यूनतम (0) है। (x) बड़ा होने पर वर्गमूल सभी बड़े अनऋणात्मक मान लेता है।
\(|4x|\le x^2+4\) because ((x-2)2\ge 0) and ((x+2)2\ge 0). Hence values lie from (-1) to (1), and both endpoints occur.
Step 2
Why this answer is correct
The correct answer is A. ( [-1,1] ). \(|4x|\le x^2+4\) because ((x-2)2\ge 0) and ((x+2)2\ge 0). Hence values lie from (-1) to (1), and both endpoints occur.
Step 3
Exam Tip
\(|4x|\le x^2+4\) क्योंकि ((x-2)2\ge 0) और ((x+2)2\ge 0)। इसलिए मान (-1) से (1) तक हैं और दोनों मिलते हैं।
Put \(t=x^2\ge 0\) and \(t\ne 4\). Then \(f=\frac{t}{t-4}\), giving (\(-\infty,0]\) for \(t\in[0,4\)) and (\(1,\infty\)) for (t>4).
Step 2
Why this answer is correct
The correct answer is A. ( (-\infty,0]\cup\(1,\infty\) ). Put \(t=x^2\ge 0\) and \(t\ne 4\). Then \(f=\frac{t}{t-4}\), giving (\(-\infty,0]\) for \(t\in[0,4\)) and (\(1,\infty\)) for (t>4).
Step 3
Exam Tip
मान \(t=x^2\ge 0\) रखें और \(t\ne 4\)। तब \(f=\frac{t}{t-4}\), जिससे \(t\in[0,4\)) पर (\(-\infty,0]\) और (t>4) पर (\(1,\infty\)) मिलता है।
Put \(t=\sqrt{x+1}\ge 0\), then \(f=\frac{t}{t+2}\). At (t=0), the value is (0), and as \(t\to\infty\), it approaches (1) but never equals (1).
Step 2
Why this answer is correct
The correct answer is A. ( [0,1) ). Put \(t=\sqrt{x+1}\ge 0\), then \(f=\frac{t}{t+2}\). At (t=0), the value is (0), and as \(t\to\infty\), it approaches (1) but never equals (1).
Step 3
Exam Tip
\(t=\sqrt{x+1}\ge 0\) रखने पर \(f=\frac{t}{t+2}\)। (t=0) पर मान (0) और \(t\to\infty\) पर मान (1) के पास जाता है पर (1) नहीं होता।
The expression inside the square root must satisfy \(\frac{x-3}{x+1}\ge 0\) and \(x\ne -1\). A sign chart gives (x<-1) or \(x\ge 3\).
Step 2
Why this answer is correct
The correct answer is B. ( \(-\infty,-1\)\cup[3,\infty) ). The expression inside the square root must satisfy \(\frac{x-3}{x+1}\ge 0\) and \(x\ne -1\). A sign chart gives (x<-1) or \(x\ge 3\).
Step 3
Exam Tip
वर्गमूल के अंदर \(\frac{x-3}{x+1}\ge 0\) और \(x\ne -1\) चाहिए। संकेत सारणी से (x<-1) या \(x\ge 3\) मिलता है।
From \(y=\frac{5x-1}{x+2}\), \(x=\frac{-1-2y}{y-5}\), so \(y\ne 5\). In a linear fractional function, the (y) making the denominator zero is excluded.
Step 2
Why this answer is correct
The correct answer is C. \( \mathbb{R}\setminus{5} \). From \(y=\frac{5x-1}{x+2}\), \(x=\frac{-1-2y}{y-5}\), so \(y\ne 5\). In a linear fractional function, the (y) making the denominator zero is excluded.
Step 3
Exam Tip
\(y=\frac{5x-1}{x+2}\) से \(x=\frac{-1-2y}{y-5}\), इसलिए \(y\ne 5\)। रैखिक भिन्न फलन में हर शून्य करने वाला (y) परिसर से हटता है।
The logarithm input must satisfy \(\frac{x+4}{x-1}>0\). A sign check gives (x<-4) or (x>1).
Step 2
Why this answer is correct
The correct answer is C. ( \(-\infty,-4\)\cup\(1,\infty\) ). The logarithm input must satisfy \(\frac{x+4}{x-1}>0\). A sign check gives (x<-4) or (x>1).
Step 3
Exam Tip
लघुगणक के अंदर \(\frac{x+4}{x-1}>0\) होना चाहिए। संकेत जांच से (x<-4) या (x>1) मिलता है।
The denominator ((x+3)2+4\ge 4), so the maximum is \(\frac{1}{2}\). As the denominator grows, the value approaches (0) but never becomes (0).
Step 2
Why this answer is correct
The correct answer is A. ( \(0,\frac{1}{2}] \). The denominator ((x+3)2+4\ge 4), so the maximum is \(\frac{1}{2}\). As the denominator grows, the value approaches (0) but never becomes (0).
Step 3
Exam Tip
हर ((x+3)2+4\ge 4) है, इसलिए अधिकतम \(\frac{1}{2}\) है। हर असीम होने पर मान (0) के पास जाता है पर (0) नहीं होता।
((x+2)(5-x)=\frac{49}{4}-\left\(x-\frac{3}{2}\right\)2) has maximum \(\frac{49}{4}\). The square root gives maximum \(\frac{7}{2}\) and minimum (0).
Step 2
Why this answer is correct
The correct answer is A. \( [0,\frac{7}{2}] \). ((x+2)(5-x)=\frac{49}{4}-\left\(x-\frac{3}{2}\right\)2) has maximum \(\frac{49}{4}\). The square root gives maximum \(\frac{7}{2}\) and minimum (0).
Step 3
Exam Tip
((x+2)(5-x)=\frac{49}{4}-\left\(x-\frac{3}{2}\right\)2) का अधिकतम \(\frac{49}{4}\) है। वर्गमूल से अधिकतम \(\frac{7}{2}\) और न्यूनतम (0) मिलता है।
Put \(t=|x-1|\ge 0\), then \(f=\frac{t}{t+3}\). At (t=0), the value is (0), and as \(t\to\infty\), it approaches (1).
Step 2
Why this answer is correct
The correct answer is B. ( [0,1) ). Put \(t=|x-1|\ge 0\), then \(f=\frac{t}{t+3}\). At (t=0), the value is (0), and as \(t\to\infty\), it approaches (1).
Step 3
Exam Tip
\(t=|x-1|\ge 0\) रखने पर \(f=\frac{t}{t+3}\)। (t=0) पर (0) मिलता है और \(t\to\infty\) पर मान (1) के पास जाता है।
The square root is in the denominator, so \(x^2-10x+24>0\) is required. From ((x-4)(x-6)>0), the outer intervals are obtained.
Step 2
Why this answer is correct
The correct answer is A. ( \(-\infty,4\)\cup\(6,\infty\) ). The square root is in the denominator, so \(x^2-10x+24>0\) is required. From ((x-4)(x-6)>0), the outer intervals are obtained.
Step 3
Exam Tip
हर में वर्गमूल है, इसलिए \(x^2-10x+24>0\) चाहिए। ((x-4)(x-6)>0) से बाहरी अंतराल मिलते हैं।
For \(\sqrt{x-2}\), \(x\ge 2\), and for the denominator square root, (x-5>0) is needed. The combined condition is (x>5).
Step 2
Why this answer is correct
The correct answer is B. ( \(5,\infty\) ). For \(\sqrt{x-2}\), \(x\ge 2\), and for the denominator square root, (x-5>0) is needed. The combined condition is (x>5).
Step 3
Exam Tip
\(\sqrt{x-2}\) के लिए \(x\ge 2\) और हर वाले वर्गमूल के लिए (x-5>0) चाहिए। संयुक्त शर्त (x>5) है।
Put \(t=x^2\ge 0\), then \(f=\frac{3t+1}{t+1}=3-\frac{2}{t+1}\). At (t=0), the value is (1), and (3) is only a limiting value.
Step 2
Why this answer is correct
The correct answer is A. ( [1,3) ). Put \(t=x^2\ge 0\), then \(f=\frac{3t+1}{t+1}=3-\frac{2}{t+1}\). At (t=0), the value is (1), and (3) is only a limiting value.
Step 3
Exam Tip
\(t=x^2\ge 0\) रखने पर \(f=\frac{3t+1}{t+1}=3-\frac{2}{t+1}\)। (t=0) पर (1) मिलता है और (3) केवल सीमा मान है।
Taking \(t=x^2-1\), we get \(t\in[-1,\infty\)) and \(t\ne 0\). Hence \(\frac{1}{t}\) has values ((-\infty,-1]\cup\(0,\infty\)).
Step 2
Why this answer is correct
The correct answer is A. ( (-\infty,-1]\cup\(0,\infty\) ). Taking \(t=x^2-1\), we get \(t\in[-1,\infty\)) and \(t\ne 0\). Hence \(\frac{1}{t}\) has values ((-\infty,-1]\cup\(0,\infty\)).
Step 3
Exam Tip
\(t=x^2-1\) लेने पर \(t\in[-1,\infty\)) और \(t\ne 0\)। इसलिए \(\frac{1}{t}\) के मान ((-\infty,-1]\cup\(0,\infty\)) हैं।
The square root is in the denominator, so \(\log_{5}x>0\) is required. Since the base (5>1), this gives (x>1).
Step 2
Why this answer is correct
The correct answer is C. ( \(1,\infty\) ). The square root is in the denominator, so \(\log_{5}x>0\) is required. Since the base (5>1), this gives (x>1).
Step 3
Exam Tip
हर में वर्गमूल है, इसलिए \(\log_{5}x>0\) चाहिए। आधार (5>1) होने से (x>1) मिलता है।
The sum of distances from (-2) and (4) has minimum (6). Any (x) between these points gives the same minimum value.
Step 2
Why this answer is correct
The correct answer is A. \( [6,\infty\) ). The sum of distances from (-2) and (4) has minimum (6). Any (x) between these points gives the same minimum value.
Step 3
Exam Tip
दो बिंदुओं (-2) और (4) से दूरी का योग न्यूनतम (6) होता है। इनके बीच किसी भी (x) पर वही न्यूनतम मान मिलता है।
The difference of distances from fixed points (1) and (5) lies from (-4) to (4). Remove signs interval-wise and check endpoint values.
Step 2
Why this answer is correct
The correct answer is A. ( [-4,4] ). The difference of distances from fixed points (1) and (5) lies from (-4) to (4). Remove signs interval-wise and check endpoint values.
Step 3
Exam Tip
दो निश्चित बिंदुओं (1) और (5) से दूरियों का अंतर (-4) से (4) तक रहता है। अंतरालों पर चिह्न हटाकर सीमा मान जांचें।
The denominator \(|x+3|+2\ge 2\), so the maximum value is \(\frac{1}{2}\). As the denominator grows, the value approaches (0).
Step 2
Why this answer is correct
The correct answer is A. ( \(0,\frac{1}{2}] \). The denominator \(|x+3|+2\ge 2\), so the maximum value is \(\frac{1}{2}\). As the denominator grows, the value approaches (0).
Step 3
Exam Tip
हर \(|x+3|+2\ge 2\), इसलिए अधिकतम मान \(\frac{1}{2}\) है। हर असीम होने पर मान (0) के पास जाता है।
The original domain has \(x\ne 3\), and simplification gives (f(x)=x+3). Hence the value (6), which would occur at (x=3), is excluded from the range.
Step 2
Why this answer is correct
The correct answer is A. \( \mathbb{R}\setminus{6} \). The original domain has \(x\ne 3\), and simplification gives (f(x)=x+3). Hence the value (6), which would occur at (x=3), is excluded from the range.
Step 3
Exam Tip
मूल प्रांत में \(x\ne 3\) है और सरलीकरण से (f(x)=x+3)। इसलिए (x=3) पर आने वाला मान (6) परिसर से हटेगा।
The original denominator is (x-3), so (x=3) is not allowed. Even after simplification, do not add the removed point to the domain.
Step 2
Why this answer is correct
The correct answer is B. \( \mathbb{R}\setminus{3} \). The original denominator is (x-3), so (x=3) is not allowed. Even after simplification, do not add the removed point to the domain.
Step 3
Exam Tip
मूल हर (x-3) है, इसलिए (x=3) स्वीकार्य नहीं है। सरलीकरण के बाद भी हटे हुए बिंदु को प्रांत में न जोड़ें।
On the closed interval ([-2,8]), the minimum occurs at the endpoints. At both endpoints, the value is \(\sqrt{10}\).
Step 2
Why this answer is correct
The correct answer is A. \( \sqrt{10} \). On the closed interval ([-2,8]), the minimum occurs at the endpoints. At both endpoints, the value is \(\sqrt{10}\).
Step 3
Exam Tip
बंद अंतराल ([-2,8]) में न्यूनतम सिरों पर आता है। दोनों सिरों पर मान \(\sqrt{10}\) है।
Since \(|3x|\le \frac{x^2+9}{2}\), \(\left|\frac{3x}{x^2+9}\right|\le \frac{1}{2}\). Equality occurs at \(x=\pm 3\).
Step 2
Why this answer is correct
The correct answer is A. \( [-\frac{1}{2},\frac{1}{2}] \). Since \(|3x|\le \frac{x^2+9}{2}\), \(\left|\frac{3x}{x^2+9}\right|\le \frac{1}{2}\). Equality occurs at \(x=\pm 3\).
Step 3
Exam Tip
\(|3x|\le \frac{x^2+9}{2}\), इसलिए \(\left|\frac{3x}{x^2+9}\right|\le \frac{1}{2}\)। समानता \(x=\pm 3\) पर मिलती है।
The square root is in the denominator, so \(x^2-4>0\) is required. This gives (x<-2) or (x>2).
Step 2
Why this answer is correct
The correct answer is A. ( \(-\infty,-2\)\cup\(2,\infty\) ). The square root is in the denominator, so \(x^2-4>0\) is required. This gives (x<-2) or (x>2).
Step 3
Exam Tip
हर में वर्गमूल है, इसलिए \(x^2-4>0\) चाहिए। इससे (x<-2) या (x>2) मिलता है।
Since \(\sqrt{x}\ge 0\), the denominator is at least (1). At (x=0), the value is (1), and as \(x\to\infty\), it approaches (0).
Step 2
Why this answer is correct
The correct answer is A. ( (0,1] ). Since \(\sqrt{x}\ge 0\), the denominator is at least (1). At (x=0), the value is (1), and as \(x\to\infty\), it approaches (0).
Step 3
Exam Tip
\(\sqrt{x}\ge 0\), इसलिए हर कम से कम (1) है। (x=0) पर (1) मिलता है और \(x\to\infty\) पर मान (0) के पास जाता है।
The ceiling function gives the smallest integer not less than (x). Since (x>-2), the value (-2) will not occur.
Step 2
Why this answer is correct
The correct answer is A. ( {-1,0,1,2} ). The ceiling function gives the smallest integer not less than (x). Since (x>-2), the value (-2) will not occur.
Step 3
Exam Tip
छत फलन (x) से छोटा नहीं सबसे छोटा पूर्णांक देता है। (x>-2) है, इसलिए (-2) मान नहीं आएगा।
Put (t=(x-1)2\ge 0), then \(f=\frac{t+4}{t+1}=1+\frac{3}{t+1}\). The maximum is (4), and (1) is only a limiting value.
Step 2
Why this answer is correct
The correct answer is A. ( (1,4] ). Put (t=(x-1)2\ge 0), then \(f=\frac{t+4}{t+1}=1+\frac{3}{t+1}\). The maximum is (4), and (1) is only a limiting value.
Step 3
Exam Tip
(t=(x-1)2\ge 0) रखने पर \(f=\frac{t+4}{t+1}=1+\frac{3}{t+1}\)। अधिकतम (4) है और (1) केवल सीमा मान है।
At the endpoints the value is \(2\sqrt{2}\), and at the midpoint (x=5), the maximum is (4). For symmetric radicals, check both endpoints and the midpoint.
Step 2
Why this answer is correct
The correct answer is A. \( [2\sqrt{2},4] \). At the endpoints the value is \(2\sqrt{2}\), and at the midpoint (x=5), the maximum is (4). For symmetric radicals, check both endpoints and the midpoint.
Step 3
Exam Tip
सिरों पर मान \(2\sqrt{2}\) और मध्य (x=5) पर अधिकतम (4) है। सममित वर्गमूलों में सिरों और मध्य दोनों जांचें।
The expression inside the square root must satisfy \(\frac{x+5}{2-x}\ge 0\) and \(x\ne 2\). A sign chart gives ([-5,2)).
Step 2
Why this answer is correct
The correct answer is A. ( [-5,2) ). The expression inside the square root must satisfy \(\frac{x+5}{2-x}\ge 0\) and \(x\ne 2\). A sign chart gives ([-5,2)).
Step 3
Exam Tip
वर्गमूल के अंदर \(\frac{x+5}{2-x}\ge 0\) और \(x\ne 2\) चाहिए। संकेत सारणी से ([-5,2)) मिलता है।
The square root needs \(x\ge 2\), and the denominator needs \(x\ne 7\). Hence the domain is \([2,\infty\)\setminus{7}).
Step 2
Why this answer is correct
The correct answer is A. \( [2,\infty\)\setminus{7} ). The square root needs \(x\ge 2\), and the denominator needs \(x\ne 7\). Hence the domain is \([2,\infty\)\setminus{7}).
Step 3
Exam Tip
वर्गमूल के लिए \(x\ge 2\) और हर के लिए \(x\ne 7\) चाहिए। इसलिए प्रांत \([2,\infty\)\setminus{7}) है।
The square root needs \(x^2-6x+5\ge 0\). From ((x-1)(x-5)\ge 0), the outer intervals are obtained.
Step 2
Why this answer is correct
The correct answer is A. ( \(-\infty,1]\cup[5,\infty\) ). The square root needs \(x^2-6x+5\ge 0\). From ((x-1)(x-5)\ge 0), the outer intervals are obtained.
Step 3
Exam Tip
वर्गमूल के लिए \(x^2-6x+5\ge 0\) चाहिए। ((x-1)(x-5)\ge 0) से बाहरी अंतराल मिलते हैं।
On the domain, the inside expression starts at (0) and goes to infinity. Hence the square root range is \([0,\infty\)).
Step 2
Why this answer is correct
The correct answer is A. \( [0,\infty\) ). On the domain, the inside expression starts at (0) and goes to infinity. Hence the square root range is \([0,\infty\)).
Step 3
Exam Tip
प्रांत में अंदर का व्यंजक (0) से शुरू होकर असीम तक जाता है। इसलिए वर्गमूल का परिसर \([0,\infty\)) है।
Here (f(x)=x+\frac{4}{x}). For (x>0), values are \([4,\infty\)), and for (x<0), values are (\(-\infty,-4]\).
Step 2
Why this answer is correct
The correct answer is A. ( \(-\infty,-4]\cup[4,\infty\) ). Here (f(x)=x+\frac{4}{x}). For (x>0), values are \([4,\infty\)), and for (x<0), values are (\(-\infty,-4]\).
Step 3
Exam Tip
(f(x)=x+\frac{4}{x}) है। (x>0) पर मान \([4,\infty\)) और (x<0) पर (\(-\infty,-4]\) मिलता है।
The square root needs (\log_{\frac{1}{2}}(x-4)\ge 0) and (x-4>0). Since the base \(\frac{1}{2}<1\), \(0<x-4\le 1\), so the domain is ((4,5]).
Step 2
Why this answer is correct
The correct answer is A. ( (4,5] ). The square root needs (\log_{\frac{1}{2}}(x-4)\ge 0) and (x-4>0). Since the base \(\frac{1}{2}<1\), \(0<x-4\le 1\), so the domain is ((4,5]).
Step 3
Exam Tip
वर्गमूल के लिए (\log_{\frac{1}{2}}(x-4)\ge 0) और (x-4>0) चाहिए। आधार \(\frac{1}{2}<1\) होने से \(0<x-4\le 1\), इसलिए प्रांत ((4,5]) है।
The expression inside the square root must satisfy \(\frac{x-5}{x-2}\ge 0\) and \(x\ne 2\). A sign chart gives (x<2) or \(x\ge 5\).
Step 2
Why this answer is correct
The correct answer is A. ( \(-\infty,2\)\cup[5,\infty) ). The expression inside the square root must satisfy \(\frac{x-5}{x-2}\ge 0\) and \(x\ne 2\). A sign chart gives (x<2) or \(x\ge 5\).
Step 3
Exam Tip
वर्गमूल के अंदर \(\frac{x-5}{x-2}\ge 0\) और \(x\ne 2\) चाहिए। संकेत सारणी से (x<2) या \(x\ge 5\) मिलता है।
From \(y=\frac{7x+4}{2x-3}\), \(x=\frac{3y+4}{2y-7}\), so \(y\ne \frac{7}{2}\). In a linear fractional function, remove the impossible (y).
Step 2
Why this answer is correct
The correct answer is A. \( \mathbb{R}\setminus{\frac{7}{2}} \). From \(y=\frac{7x+4}{2x-3}\), \(x=\frac{3y+4}{2y-7}\), so \(y\ne \frac{7}{2}\). In a linear fractional function, remove the impossible (y).
Step 3
Exam Tip
\(y=\frac{7x+4}{2x-3}\) से \(x=\frac{3y+4}{2y-7}\), इसलिए \(y\ne \frac{7}{2}\)। रैखिक भिन्न फलन में असंभव (y) को हटाएं।
Since (12-x-x-2=\frac{49}{4}-\left\(x+\frac{1}{2}\right\)2), the inside maximum is \(\frac{49}{4}\). Taking square root gives maximum \(\frac{7}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \( [0,\frac{7}{2}] \). Since (12-x-x-2=\frac{49}{4}-\left\(x+\frac{1}{2}\right\)2), the inside maximum is \(\frac{49}{4}\). Taking square root gives maximum \(\frac{7}{2}\).
Step 3
Exam Tip
(12-x-x-2=\frac{49}{4}-\left\(x+\frac{1}{2}\right\)2), इसलिए अंदर का अधिकतम \(\frac{49}{4}\) है। वर्गमूल से अधिकतम \(\frac{7}{2}\) मिलता है।
The square root is in the denominator, so \(x^2-2x-8>0\) is required. From ((x-4)(x+2)>0), the outer open intervals are obtained.
Step 2
Why this answer is correct
The correct answer is A. ( \(-\infty,-2\)\cup\(4,\infty\) ). The square root is in the denominator, so \(x^2-2x-8>0\) is required. From ((x-4)(x+2)>0), the outer open intervals are obtained.
Step 3
Exam Tip
हर में वर्गमूल है, इसलिए \(x^2-2x-8>0\) चाहिए। ((x-4)(x+2)>0) से बाहरी खुले अंतराल मिलते हैं।
Put \(t=x^2\ge 0\), then \(f=\frac{t+5}{t+2}=1+\frac{3}{t+2}\). The maximum is \(\frac{5}{2}\), and (1) is only a limiting value.
Step 2
Why this answer is correct
The correct answer is A. ( \(1,\frac{5}{2}] \). Put \(t=x^2\ge 0\), then \(f=\frac{t+5}{t+2}=1+\frac{3}{t+2}\). The maximum is \(\frac{5}{2}\), and (1) is only a limiting value.
Step 3
Exam Tip
\(t=x^2\ge 0\) रखने पर \(f=\frac{t+5}{t+2}=1+\frac{3}{t+2}\)। अधिकतम \(\frac{5}{2}\) है और (1) केवल सीमा मान है।
A logarithmic function has all real numbers as range when its argument can take all positive values. The base \(\frac{1}{3}\) only changes direction, not the range.
Step 2
Why this answer is correct
The correct answer is A. \( \mathbb{R} \). A logarithmic function has all real numbers as range when its argument can take all positive values. The base \(\frac{1}{3}\) only changes direction, not the range.
Step 3
Exam Tip
लघुगणकीय फलन का परिसर सभी वास्तविक संख्याएं होता है जब उसका आर्गुमेंट सभी धनात्मक मान ले सकता है। आधार \(\frac{1}{3}\) केवल दिशा बदलता है, परिसर नहीं।
Put \(t=\sqrt{x-1}\ge 0\), then \(f=\frac{t}{t+4}\). At (t=0), the value is (0), and as \(t\to\infty\), it approaches (1).
Step 2
Why this answer is correct
The correct answer is A. ( [0,1) ). Put \(t=\sqrt{x-1}\ge 0\), then \(f=\frac{t}{t+4}\). At (t=0), the value is (0), and as \(t\to\infty\), it approaches (1).
Step 3
Exam Tip
\(t=\sqrt{x-1}\ge 0\) रखने पर \(f=\frac{t}{t+4}\)। (t=0) पर (0) मिलता है और \(t\to\infty\) पर मान (1) के पास जाता है।
The square root needs \(|x+2|-5\ge 0\). Hence \(|x+2|\ge 5\), giving \(x\le -7\) or \(x\ge 3\).
Step 2
Why this answer is correct
The correct answer is A. ( \(-\infty,-7]\cup[3,\infty\) ). The square root needs \(|x+2|-5\ge 0\). Hence \(|x+2|\ge 5\), giving \(x\le -7\) or \(x\ge 3\).
Step 3
Exam Tip
वर्गमूल के लिए \(|x+2|-5\ge 0\) चाहिए। इसलिए \(|x+2|\ge 5\), जिससे \(x\le -7\) या \(x\ge 3\) मिलता है।
The denominator must not be zero, so \(|x-4|-1\ne 0\). This gives \(|x-4|\ne 1\), hence \(x\ne 3,5\).
Step 2
Why this answer is correct
The correct answer is A. \( \mathbb{R}\setminus{3,5} \). The denominator must not be zero, so \(|x-4|-1\ne 0\). This gives \(|x-4|\ne 1\), hence \(x\ne 3,5\).
Step 3
Exam Tip
हर शून्य नहीं होना चाहिए, इसलिए \(|x-4|-1\ne 0\)। इससे \(|x-4|\ne 1\), अतः \(x\ne 3,5\)।
The original denominator is (x-4), so (x=4) is not allowed. Even after simplification, do not add the removed point to the domain.
Step 2
Why this answer is correct
The correct answer is A. \( \mathbb{R}\setminus{4} \). The original denominator is (x-4), so (x=4) is not allowed. Even after simplification, do not add the removed point to the domain.
Step 3
Exam Tip
मूल हर (x-4) है, इसलिए (x=4) स्वीकार्य नहीं है। सरलीकरण के बाद भी हटे बिंदु को प्रांत में न जोड़ें।
For \(x\ne 4\), the function equals (x+4). The value (8), which would come from (x=4), will not be in the range.
Step 2
Why this answer is correct
The correct answer is A. \( \mathbb{R}\setminus{8} \). For \(x\ne 4\), the function equals (x+4). The value (8), which would come from (x=4), will not be in the range.
Step 3
Exam Tip
\(x\ne 4\) पर फलन (x+4) के बराबर है। (x=4) से आने वाला मान (8) परिसर में नहीं आएगा।
\(Here (f(x)=-(x-3)^2+4), so the maximum is (4). A downward-opening parabola has range ((-\infty\),maximum]).
Step 2
Why this answer is correct
\(The correct answer is A. ( (-\infty,4] ). Here (f(x)=-(x-3)^2+4), so the maximum is (4). A downward-opening parabola has range ((-\infty\),maximum]).
Step 3
Exam Tip
(f(x)=-(x-3)2+4), इसलिए अधिकतम (4) है। \(नीचे खुलने वाले परवलय का परिसर ((-\infty\),अधिकतम]) होता है।
The denominator ((x+2)2+4\ge 4), so the maximum value is \(\frac{1}{4}\). As the denominator grows, the value approaches (0).
Step 2
Why this answer is correct
The correct answer is A. ( \(0,\frac{1}{4}] \). The denominator ((x+2)2+4\ge 4), so the maximum value is \(\frac{1}{4}\). As the denominator grows, the value approaches (0).
Step 3
Exam Tip
हर ((x+2)2+4\ge 4) है, इसलिए अधिकतम मान \(\frac{1}{4}\) है। हर असीम होने पर मान (0) के पास जाता है।
From \(y=\frac{x^2}{x^2+4x+8}\), real (x) is needed in ((y-1)x-2+4yx+8y=0). The discriminant gives \(0\le y\le 2\), and (y=1) is also possible, so the range is ([0,2]).
Step 2
Why this answer is correct
The correct answer is A. \( [0,\infty\) ). From \(y=\frac{x^2}{x^2+4x+8}\), real (x) is needed in ((y-1)x-2+4yx+8y=0). The discriminant gives \(0\le y\le 2\), and (y=1) is also possible, so the range is ([0,2]).
Step 3
Exam Tip
\(y=\frac{x^2}{x^2+4x+8}\) से ((y-1)x-2+4yx+8y=0) में वास्तविक (x) चाहिए। विविक्तकर (16y-2-32y(y-1)=16y(2-y)) से \(0\le y\le 2\), पर (y=1) भी संभव है, इसलिए विस्तृत जांच से ([0,2]) मिलता है।
From \(y=\frac{x^2+4x+8}{x^2+4}\), real (x) is required in ((y-1)x-2-4x+4y-8=0). The discriminant condition (16-4(y-1)(4y-8)\ge 0) gives \(0\le y\le 2\).
Step 2
Why this answer is correct
The correct answer is A. ( [0,2] ). From \(y=\frac{x^2+4x+8}{x^2+4}\), real (x) is required in ((y-1)x-2-4x+4y-8=0). The discriminant condition (16-4(y-1)(4y-8)\ge 0) gives \(0\le y\le 2\).
Step 3
Exam Tip
\(y=\frac{x^2+4x+8}{x^2+4}\) से ((y-1)x-2-4x+4y-8=0) में वास्तविक (x) चाहिए। विविक्तकर (16-4(y-1)(4y-8)\ge 0) से \(0\le y\le 2\) मिलता है।
At the endpoints the value is \(\sqrt{5}\), and at the midpoint \(x=\frac{9}{2}\), the maximum is \(\sqrt{10}\). In symmetric questions, always check the midpoint.
Step 2
Why this answer is correct
The correct answer is A. \( [\sqrt{5},\sqrt{10}] \). At the endpoints the value is \(\sqrt{5}\), and at the midpoint \(x=\frac{9}{2}\), the maximum is \(\sqrt{10}\). In symmetric questions, always check the midpoint.
Step 3
Exam Tip
सिरों पर मान \(\sqrt{5}\) और मध्य \(x=\frac{9}{2}\) पर अधिकतम \(\sqrt{10}\) है। सममिति वाले प्रश्नों में मध्य बिंदु जरूर जांचें।
The denominator square root needs (x+2>0), and the second square root needs \(6-x\ge 0\). Therefore \(-2<x\le 6\).
Step 2
Why this answer is correct
The correct answer is A. ( (-2,6] ). The denominator square root needs (x+2>0), and the second square root needs \(6-x\ge 0\). Therefore \(-2<x\le 6\).
Step 3
Exam Tip
हर वाले वर्गमूल के लिए (x+2>0) और दूसरे वर्गमूल के लिए \(6-x\ge 0\) चाहिए। इसलिए \(-2<x\le 6\)।
In the given interval, \(\sin x\) attains both maximum (1) and minimum (-1). For trigonometric ranges, check key angles.
Step 2
Why this answer is correct
The correct answer is A. ( [-1,1] ). In the given interval, \(\sin x\) attains both maximum (1) and minimum (-1). For trigonometric ranges, check key angles.
Step 3
Exam Tip
दिए गए अंतराल में \(\sin x\) का अधिकतम (1) और न्यूनतम (-1) दोनों आते हैं। त्रिकोणमितीय परिसर के लिए प्रमुख कोणों की जांच करें।
A. \( \mathbb{R}\setminus\left{\frac{3\pi}{2}+2n\pi:n\in\mathbb{Z}\right} \)
Step 1
Concept
The denominator must not be zero, so \(1+\sin x\ne 0\). When \(\sin x=-1\), \(x=\frac{3\pi}{2}+2n\pi\) is excluded.
Step 2
Why this answer is correct
The correct answer is A. \( \mathbb{R}\setminus\left{\frac{3\pi}{2}+2n\pi:n\in\mathbb{Z}\right} \). The denominator must not be zero, so \(1+\sin x\ne 0\). When \(\sin x=-1\), \(x=\frac{3\pi}{2}+2n\pi\) is excluded.
Step 3
Exam Tip
हर शून्य नहीं होना चाहिए, इसलिए \(1+\sin x\ne 0\)। \(\sin x=-1\) पर \(x=\frac{3\pi}{2}+2n\pi\) हटेगा।
A. \( \left[0,\frac{\pi}{2}\right]\cup\left[\frac{3\pi}{2},2\pi\right] \)
Step 1
Concept
The square root needs \(\cos x\ge 0\). In \([0,2\pi]\), this is true in the first and fourth quadrants.
Step 2
Why this answer is correct
The correct answer is A. \( \left[0,\frac{\pi}{2}\right]\cup\left[\frac{3\pi}{2},2\pi\right] \). The square root needs \(\cos x\ge 0\). In \([0,2\pi]\), this is true in the first and fourth quadrants.
Step 3
Exam Tip
वर्गमूल के लिए \(\cos x\ge 0\) चाहिए। \([0,2\pi]\) में यह पहले और चौथे चतुर्थांश में सत्य है।
Here \(x+2\in[1,5\)), so the greatest integer values are (1,2,3,4). The value (5) does not occur because the upper endpoint is open.
Step 2
Why this answer is correct
The correct answer is A. ( {1,2,3,4} ). Here \(x+2\in[1,5\)), so the greatest integer values are (1,2,3,4). The value (5) does not occur because the upper endpoint is open.
Step 3
Exam Tip
\(x+2\in[1,5\)), इसलिए महत्तम पूर्णांक मान (1,2,3,4) होंगे। (5) नहीं आता क्योंकि ऊपरी सिरा खुला है।
The fractional part of any real number is at least (0) and less than (1). When (2x) is an integer, the value is (0).
Step 2
Why this answer is correct
The correct answer is A. ( [0,1) ). The fractional part of any real number is at least (0) and less than (1). When (2x) is an integer, the value is (0).
Step 3
Exam Tip
किसी भी वास्तविक संख्या का भिन्नांश भाग (0) से बड़ा या बराबर और (1) से छोटा होता है। (2x) पूर्णांक होने पर मान (0) मिलता है।
Since \(\sqrt{x^2+1}\ge 1\), the denominator is at least (3). The maximum is \(\frac{1}{3}\), and the value approaches (0).
Step 2
Why this answer is correct
The correct answer is A. ( \(0,\frac{1}{3}] \). Since \(\sqrt{x^2+1}\ge 1\), the denominator is at least (3). The maximum is \(\frac{1}{3}\), and the value approaches (0).
Step 3
Exam Tip
\(\sqrt{x^2+1}\ge 1\), इसलिए हर कम से कम (3) है। अधिकतम \(\frac{1}{3}\) है और मान (0) के पास जाता है।
The denominator is always greater than (|x|), so \(\left|\frac{x}{\sqrt{x^2+9}}\right|<1\). For large (|x|), the value approaches \(\pm 1\) but never equals them.
Step 2
Why this answer is correct
The correct answer is A. ( (-1,1) ). The denominator is always greater than (|x|), so \(\left|\frac{x}{\sqrt{x^2+9}}\right|<1\). For large (|x|), the value approaches \(\pm 1\) but never equals them.
Step 3
Exam Tip
हर हमेशा (|x|) से बड़ा है, इसलिए \(\left|\frac{x}{\sqrt{x^2+9}}\right|<1\)। बड़े (|x|) पर मान \(\pm 1\) के पास जाता है पर बराबर नहीं होता।
The denominator is (\sqrt{(x-2)2+4}), which is greater than (|x-2|). Hence the ratio stays strictly between (-1) and (1).
Step 2
Why this answer is correct
The correct answer is A. ( (-1,1) ). The denominator is (\sqrt{(x-2)2+4}), which is greater than (|x-2|). Hence the ratio stays strictly between (-1) and (1).
Step 3
Exam Tip
हर (\sqrt{(x-2)2+4}) है, जो (|x-2|) से बड़ा है। इसलिए अनुपात का मान (-1) और (1) के बीच ही रहता है।
Here (f(x)=x+\frac{1}{x}). For (x>0), values are \([2,\infty\)), and for (x<0), values are (\(-\infty,-2]\).
Step 2
Why this answer is correct
The correct answer is A. ( \(-\infty,-2]\cup[2,\infty\) ). Here (f(x)=x+\frac{1}{x}). For (x>0), values are \([2,\infty\)), and for (x<0), values are (\(-\infty,-2]\).
Step 3
Exam Tip
(f(x)=x+\frac{1}{x}) है। (x>0) पर मान \([2,\infty\)) और (x<0) पर (\(-\infty,-2]\) मिलते हैं।
Here (f(x)=\frac{x}{3}+\frac{3}{x}). For positive (x), the minimum is (2), and for negative (x), the maximum is (-2).
Step 2
Why this answer is correct
The correct answer is A. ( \(-\infty,-2]\cup[2,\infty\) ). Here (f(x)=\frac{x}{3}+\frac{3}{x}). For positive (x), the minimum is (2), and for negative (x), the maximum is (-2).
Step 3
Exam Tip
(f(x)=\frac{x}{3}+\frac{3}{x}) है। धनात्मक (x) पर न्यूनतम (2) और ऋणात्मक (x) पर अधिकतम (-2) मिलता है।
The square root needs \(\frac{9-x^2}{x^2-4}\ge 0\) and \(x\ne\pm2\). A sign check gives ([-3,-2)\cup(2,3]).
Step 2
Why this answer is correct
The correct answer is B. ( [-3,-2)\cup(2,3] ). The square root needs \(\frac{9-x^2}{x^2-4}\ge 0\) and \(x\ne\pm2\). A sign check gives ([-3,-2)\cup(2,3]).
Step 3
Exam Tip
वर्गमूल के अंदर \(\frac{9-x^2}{x^2-4}\ge 0\) और \(x\ne\pm2\) चाहिए। संकेत जांच से ([-3,-2)\cup(2,3]) मिलता है।
The square root needs \(\frac{x^2-1}{4-x^2}\ge 0\) and \(x\ne\pm2\). A sign check gives (\(-2,-1]\cup[1,2\)).
Step 2
Why this answer is correct
The correct answer is B. ( \(-2,-1]\cup[1,2\) ). The square root needs \(\frac{x^2-1}{4-x^2}\ge 0\) and \(x\ne\pm2\). A sign check gives (\(-2,-1]\cup[1,2\)).
Step 3
Exam Tip
वर्गमूल के अंदर \(\frac{x^2-1}{4-x^2}\ge 0\) और \(x\ne\pm2\) चाहिए। संकेत जांच से (\(-2,-1]\cup[1,2\)) मिलता है।
The function is \(\sqrt{\frac{x+3}{2-x}}\), where the fraction takes all values in \([0,\infty\)). Hence the square root range is also \([0,\infty\)).
Step 2
Why this answer is correct
The correct answer is A. \( [0,\infty\) ). The function is \(\sqrt{\frac{x+3}{2-x}}\), where the fraction takes all values in \([0,\infty\)). Hence the square root range is also \([0,\infty\)).
Step 3
Exam Tip
फलन \(\sqrt{\frac{x+3}{2-x}}\) है, जहां भिन्न \([0,\infty\)) के सभी मान लेता है। इसलिए वर्गमूल का परिसर भी \([0,\infty\)) है।
The square root needs \(1-\frac{1}{x^2}\ge 0\) and \(x\ne 0\). This gives \(x^2\ge 1\), so \(|x|\ge 1\).
Step 2
Why this answer is correct
The correct answer is A. ( \(-\infty,-1]\cup[1,\infty\) ). The square root needs \(1-\frac{1}{x^2}\ge 0\) and \(x\ne 0\). This gives \(x^2\ge 1\), so \(|x|\ge 1\).
Step 3
Exam Tip
वर्गमूल के लिए \(1-\frac{1}{x^2}\ge 0\) और \(x\ne 0\) चाहिए। इससे \(x^2\ge 1\), यानी \(|x|\ge 1\) मिलता है।
On the domain, \(\frac{1}{x^2}\in(0,1]), so the inside value lies in ([0,1)\). After square root, the range remains ([0,1)).
Step 2
Why this answer is correct
The correct answer is A. ( [0,1) ). On the domain, \(\frac{1}{x^2}\in(0,1]), so the inside value lies in ([0,1)\). After square root, the range remains ([0,1)).
Step 3
Exam Tip
प्रांत में \(\frac{1}{x^2}\in(0,1]), इसलिए अंदर का मान ([0,1)\) में है। वर्गमूल के बाद भी परिसर ([0,1)) रहता है।
The square root needs (\log_{\frac{1}{4}}(9-x)\ge 0) and (9-x>0). Since the base \(\frac{1}{4}<1\), \(0<9-x\le 1\), so the domain is ([8,9)).
Step 2
Why this answer is correct
The correct answer is A. ( [8,9) ). The square root needs (\log_{\frac{1}{4}}(9-x)\ge 0) and (9-x>0). Since the base \(\frac{1}{4}<1\), \(0<9-x\le 1\), so the domain is ([8,9)).
Step 3
Exam Tip
वर्गमूल के लिए (\log_{\frac{1}{4}}(9-x)\ge 0) और (9-x>0) चाहिए। आधार \(\frac{1}{4}<1\) होने से \(0<9-x\le 1\), इसलिए प्रांत ([8,9)) है।
Since \(\sqrt{x^2+4}\ge 2\), the denominator is at least (5), so the maximum value is \(\frac{2}{5}\). As \(|x|\to\infty\), the value approaches (0) but never becomes (0).
Step 2
Why this answer is correct
The correct answer is A. ( \left\(0,\frac{2}{5}\right] \). Since \(\sqrt{x^2+4}\ge 2\), the denominator is at least (5), so the maximum value is \(\frac{2}{5}\). As \(|x|\to\infty\), the value approaches (0) but never becomes (0).
Step 3
Exam Tip
\(\sqrt{x^2+4}\ge 2\), इसलिए हर कम से कम (5) है और अधिकतम मान \(\frac{2}{5}\) है। \(|x|\to\infty\) पर मान (0) के पास जाता है पर (0) नहीं होता।
