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।
