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।
