The denominator has a square root, so \(x^2-9>0\) is needed. Remember zero is not allowed in a denominator.
Step 2
Why this answer is correct
The correct answer is A. ( \(-\infty,-3\)\cup\(3,\infty\) ). The denominator has a square root, so \(x^2-9>0\) is needed. Remember zero is not allowed in a denominator.
Step 3
Exam Tip
हर में वर्गमूल है इसलिए \(x^2-9>0\) चाहिए। ध्यान रखें हर में शून्य स्वीकार नहीं होता।
The denominator is (x-2-5x+6=(x-2)(x-3)), so (x=2,3) are excluded. In exams set the denominator equal to zero first.
Step 2
Why this answer is correct
The correct answer is A. \( \mathbb{R}\setminus{2,3} \). The denominator is (x-2-5x+6=(x-2)(x-3)), so (x=2,3) are excluded. In exams set the denominator equal to zero first.
Step 3
Exam Tip
हर (x-2-5x+6=(x-2)(x-3)) है इसलिए (x=2,3) हटेंगे। परीक्षा में पहले हर को शून्य के बराबर करें।
Because \(x^2+4\ge4\), the maximum value is \(\frac{1}{4}\), and (0) is only approached. In exams do not close an unattained endpoint.
Step 2
Why this answer is correct
The correct answer is A. ( \(0,\frac{1}{4}] \). Because \(x^2+4\ge4\), the maximum value is \(\frac{1}{4}\), and (0) is only approached. In exams do not close an unattained endpoint.
Step 3
Exam Tip
क्योंकि \(x^2+4\ge4\), अधिकतम मान \(\frac{1}{4}\) है और (0) केवल निकट आता है। परीक्षा में असम्प्राप्त सीमा को बंद न करें।
The sum of distances from (-1) and (3) is at least their distance (4). The distance idea is useful for absolute value questions.
Step 2
Why this answer is correct
The correct answer is A. (4). The sum of distances from (-1) and (3) is at least their distance (4). The distance idea is useful for absolute value questions.
Step 3
Exam Tip
दो बिंदुओं ( -1 ) और (3) से दूरी का योग कम से कम उनके बीच की दूरी (4) होता है। निरपेक्ष मान में दूरी वाला विचार उपयोगी है।
From \(y=\frac{x-1}{x+2}\), we get \(x=\frac{1+2y}{1-y}\), so \(y\ne1\). Writing (x) in terms of (y) works well for range.
Step 2
Why this answer is correct
The correct answer is A. \( \mathbb{R}\setminus{1} \). From \(y=\frac{x-1}{x+2}\), we get \(x=\frac{1+2y}{1-y}\), so \(y\ne1\). Writing (x) in terms of (y) works well for range.
Step 3
Exam Tip
\(y=\frac{x-1}{x+2}\) से \(x=\frac{1+2y}{1-y}\) मिलता है इसलिए \(y\ne1\)। परिसर के लिए (x) को (y) के रूप में लिखना कारगर है।
The square root gives \(x\ge-4\), and the denominator gives \(x\ne1\). Apply all restrictions together in mixed functions.
Step 2
Why this answer is correct
The correct answer is A. \( [-4,\infty\)\setminus{1} ). The square root gives \(x\ge-4\), and the denominator gives \(x\ne1\). Apply all restrictions together in mixed functions.
Step 3
Exam Tip
वर्गमूल से \(x\ge-4\) और हर से \(x\ne1\) चाहिए। मिश्रित फलन में सभी प्रतिबंध साथ लागू करें।
The logarithm argument must satisfy (7-2x>0). A logarithm does not accept zero or negative input.
Step 2
Why this answer is correct
The correct answer is A. ( \(-\infty,\frac{7}{2}\) ). The logarithm argument must satisfy (7-2x>0). A logarithm does not accept zero or negative input.
Step 3
Exam Tip
लघुगणक के अंदर (7-2x>0) होना चाहिए। \(\log\) में शून्य या ऋणात्मक मान स्वीकार नहीं होता।
We need \(\frac{x-1}{x+3}\ge0\) and \(x\ne-3\). A sign chart on the number line is the safest method.
Step 2
Why this answer is correct
The correct answer is A. ( \(-\infty,-3\)\cup[1,\infty) ). We need \(\frac{x-1}{x+3}\ge0\) and \(x\ne-3\). A sign chart on the number line is the safest method.
Step 3
Exam Tip
हमें \(\frac{x-1}{x+3}\ge0\) और \(x\ne-3\) चाहिए। संख्या रेखा पर चिन्ह जाँच सबसे सुरक्षित है।
Since \(\sqrt{2x-1}\ge0\), \(3-\sqrt{2x-1}\le3\) and is unbounded below. The negative sign reverses the direction of the range.
Step 2
Why this answer is correct
The correct answer is A. ( \(-\infty,3] \). Since \(\sqrt{2x-1}\ge0\), \(3-\sqrt{2x-1}\le3\) and is unbounded below. The negative sign reverses the direction of the range.
Step 3
Exam Tip
\(\sqrt{2x-1}\ge0\) इसलिए \(3-\sqrt{2x-1}\le3\) और नीचे अनबाउंड है। ऋण चिह्न सीमा की दिशा बदल देता है।
The denominator is zero when (|x|-2=0), so \(x=\pm2\). Remember both solutions of an absolute value equation.
Step 2
Why this answer is correct
The correct answer is A. \( \mathbb{R}\setminus{-2,2} \). The denominator is zero when (|x|-2=0), so \(x=\pm2\). Remember both solutions of an absolute value equation.
Step 3
Exam Tip
हर शून्य तब होगा जब (|x|-2=0), यानी \(x=\pm2\)। निरपेक्ष मान समीकरण के दोनों हल याद रखें।
Because \(|x-2|\ge0\), the minimum value is (3). In such questions keep the minimum of absolute value as (0).
Step 2
Why this answer is correct
The correct answer is A. \( [3,\infty\) ). Because \(|x-2|\ge0\), the minimum value is (3). In such questions keep the minimum of absolute value as (0).
Step 3
Exam Tip
क्योंकि \(|x-2|\ge0\), न्यूनतम मान (3) है। ऐसे प्रश्न में निरपेक्ष मान का न्यूनतम (0) रखें।
The denominator becomes zero at \(x^2-1=0\), so \(x=\pm1\) are excluded. The numerator does not restrict the domain.
Step 2
Why this answer is correct
The correct answer is A. \( \mathbb{R}\setminus{-1,1} \). The denominator becomes zero at \(x^2-1=0\), so \(x=\pm1\) are excluded. The numerator does not restrict the domain.
Step 3
Exam Tip
हर \(x^2-1=0\) पर शून्य होता है इसलिए \(x=\pm1\) हटेंगे। अंश से प्रांत पर प्रतिबंध नहीं आता।
For negative (x), the values are (-2) or less, with equality at (x=-1). The sign of the domain changes the range.
Step 2
Why this answer is correct
The correct answer is A. ( \(-\infty,-2] \). For negative (x), the values are (-2) or less, with equality at (x=-1). The sign of the domain changes the range.
Step 3
Exam Tip
ऋणात्मक (x) के लिए मान (-2) या उससे कम होते हैं और समानता (x=-1) पर आती है। प्रांत का चिन्ह परिसर बदल देता है।
Since (x-2-4x+3=(x-1)(x-3)), it must be \(\ge0\). For an upward quadratic, the outside intervals are selected.
Step 2
Why this answer is correct
The correct answer is A. ( \(-\infty,1]\cup[3,\infty\) ). Since (x-2-4x+3=(x-1)(x-3)), it must be \(\ge0\). For an upward quadratic, the outside intervals are selected.
Step 3
Exam Tip
(x-2-4x+3=(x-1)(x-3)) और इसे \(\ge0\) चाहिए। ऊपर खुलने वाले द्विघात में बाहरी अंतराल चुने जाते हैं।
The square root is in the denominator, so \(16-x^2>0\) is required. The endpoints \(x=\pm4\) are not included here.
Step 2
Why this answer is correct
The correct answer is A. ( (-4,4) ). The square root is in the denominator, so \(16-x^2>0\) is required. The endpoints \(x=\pm4\) are not included here.
Step 3
Exam Tip
हर में वर्गमूल है इसलिए \(16-x^2>0\) चाहिए। यहाँ सिरों \(x=\pm4\) को शामिल नहीं किया जाएगा।
Since \(\cos x\in[-1,1]\), \(2+\cos x\in[1,3]\), so \(f(x)\in[1,3]\). The order may reverse when taking reciprocals.
Step 2
Why this answer is correct
The correct answer is A. ( [1,3] ). Since \(\cos x\in[-1,1]\), \(2+\cos x\in[1,3]\), so \(f(x)\in[1,3]\). The order may reverse when taking reciprocals.
Step 3
Exam Tip
\(\cos x\in[-1,1]\) इसलिए \(2+\cos x\in[1,3]\), अतः \(f(x)\in[1,3]\)। reciprocal लेते समय क्रम उलट सकता है।
Since \(\sin x\in[-1,1]\), \(-2\sin x\in[-2,2]\), giving total range ([3,7]). Apply multiplication and addition step by step.
Step 2
Why this answer is correct
The correct answer is A. ( [3,7] ). Since \(\sin x\in[-1,1]\), \(-2\sin x\in[-2,2]\), giving total range ([3,7]). Apply multiplication and addition step by step.
Step 3
Exam Tip
\(\sin x\in[-1,1]\) होने से \(-2\sin x\in[-2,2]\) और कुल परिसर ( [3,7] ) है। गुणा और जोड़ को क्रम से लागू करें।
A. \( \mathbb{R}\setminus{\frac{\pi}{2}+n\pi:n\in\mathbb{Z}} \)
Step 1
Concept
Since \(\tan x=\frac{\sin x}{\cos x}\), we need \(\cos x\ne0\). For \(\tan x\), odd multiples around \(\frac{\pi}{2}\) are excluded.
Step 2
Why this answer is correct
The correct answer is A. \( \mathbb{R}\setminus{\frac{\pi}{2}+n\pi:n\in\mathbb{Z}} \). Since \(\tan x=\frac{\sin x}{\cos x}\), we need \(\cos x\ne0\). For \(\tan x\), odd multiples around \(\frac{\pi}{2}\) are excluded.
Step 3
Exam Tip
\(\tan x=\frac{\sin x}{\cos x}\) है इसलिए \(\cos x\ne0\) होना चाहिए। \(\tan x\) में विषम \(\frac{\pi}{2}\) वाले कोण हटते हैं।
For \(\log x\), (x>0), and for the denominator \(\log x\ne0\). Therefore (x=1) must be excluded separately.
Step 2
Why this answer is correct
The correct answer is A. ( \(0,\infty\)\setminus{1} ). For \(\log x\), (x>0), and for the denominator \(\log x\ne0\). Therefore (x=1) must be excluded separately.
Step 3
Exam Tip
\(\log x\) के लिए (x>0) और हर के लिए \(\log x\ne0\) चाहिए। इसलिए (x=1) अलग से हटेगा।
For \(\sqrt{x-1}\), \(x\ge1\), and for the denominator \(x\ne5\). The endpoint (x=1) is valid because the denominator is not zero.
Step 2
Why this answer is correct
The correct answer is A. \( [1,\infty\)\setminus{5} ). For \(\sqrt{x-1}\), \(x\ge1\), and for the denominator \(x\ne5\). The endpoint (x=1) is valid because the denominator is not zero.
Step 3
Exam Tip
\(\sqrt{x-1}\) के लिए \(x\ge1\) और हर के लिए \(x\ne5\) चाहिए। सिरा (x=1) मान्य है क्योंकि हर शून्य नहीं है।
The denominator ((x-3)2+1\ge1), so the maximum is \(\frac{1}{1}=1\), and (0) is not attained. Completing the square gives the bounds clearly.
Step 2
Why this answer is correct
The correct answer is A. ( (0,1] ). The denominator ((x-3)2+1\ge1), so the maximum is \(\frac{1}{1}=1\), and (0) is not attained. Completing the square gives the bounds clearly.
Step 3
Exam Tip
हर ((x-3)2+1\ge1) है इसलिए अधिकतम \(\frac{1}{1}=1\) है और (0) प्राप्त नहीं होता। पूर्ण वर्ग से सीमा साफ मिलती है।
By symmetry the maximum occurs at (x=3), and the value is \(2\sqrt{3}\). In such forms try making the two terms equal.
Step 2
Why this answer is correct
The correct answer is A. \(2\sqrt{3}\). By symmetry the maximum occurs at (x=3), and the value is \(2\sqrt{3}\). In such forms try making the two terms equal.
Step 3
Exam Tip
सममिति के कारण अधिकतम (x=3) पर है और मान \(2\sqrt{3}\) है। ऐसे रूप में दोनों पद बराबर करने की कोशिश करें।
On this interval, the minimum of (|x|) is (2) and the maximum is (5). On a negative domain \(x^2\) may decrease, but values remain positive.
Step 2
Why this answer is correct
The correct answer is A. ( [4,25] ). On this interval, the minimum of (|x|) is (2) and the maximum is (5). On a negative domain \(x^2\) may decrease, but values remain positive.
Step 3
Exam Tip
इस अंतराल में (|x|) का न्यूनतम (2) और अधिकतम (5) है। ऋणात्मक प्रांत में \(x^2\) घटता दिख सकता है पर मान धनात्मक रहते हैं।
From \(y=\frac{x}{x^2+1}\), we get \(yx^2-x+y=0\), and discriminant \(\ge0\) gives \(|y|\le\frac{1}{2}\). The discriminant method is useful for range.
Step 2
Why this answer is correct
The correct answer is A. \( [-\frac{1}{2},\frac{1}{2}] \). From \(y=\frac{x}{x^2+1}\), we get \(yx^2-x+y=0\), and discriminant \(\ge0\) gives \(|y|\le\frac{1}{2}\). The discriminant method is useful for range.
Step 3
Exam Tip
\(y=\frac{x}{x^2+1}\) से \(yx^2-x+y=0\) मिलता है और विविक्तकर \(\ge0\) देने पर \(|y|\le\frac{1}{2}\)। परिसर के लिए विविक्तकर विधि उपयोगी है।
The denominator is ((x+1)2+1), and (f(x)=1+\frac{3}{(x+1)2+1}). Hence values are greater than (1) and up to \(\frac{5}{2}\).
Step 2
Why this answer is correct
The correct answer is A. ( \(1,\frac{5}{2}] \). The denominator is ((x+1)2+1), and (f(x)=1+\frac{3}{(x+1)2+1}). Hence values are greater than (1) and up to \(\frac{5}{2}\).
Step 3
Exam Tip
हर ((x+1)2+1) और (f(x)=1+\frac{3}{(x+1)2+1}) है। इसलिए मान (1) से बड़े और \(\frac{5}{2}\) तक हैं।
Since \(|x|+1\ge1\), the maximum is (1), and (0) is not attained. As the denominator grows without bound, the fraction approaches (0).
Step 2
Why this answer is correct
The correct answer is A. ( (0,1] ). Since \(|x|+1\ge1\), the maximum is (1), and (0) is not attained. As the denominator grows without bound, the fraction approaches (0).
Step 3
Exam Tip
\(|x|+1\ge1\) इसलिए अधिकतम (1) है और (0) प्राप्त नहीं होता। हर के अनंत होने पर भिन्न (0) के पास जाती है।
The square root requires \(|x|-3\ge0\), so \(|x|\ge3\). For this absolute inequality, the outside intervals are selected.
Step 2
Why this answer is correct
The correct answer is A. ( \(-\infty,-3]\cup[3,\infty\) ). The square root requires \(|x|-3\ge0\), so \(|x|\ge3\). For this absolute inequality, the outside intervals are selected.
Step 3
Exam Tip
वर्गमूल के लिए \(|x|-3\ge0\) यानी \(|x|\ge3\) चाहिए। निरपेक्ष असमानता में बाहरी भाग चुने जाते हैं।
The difference of distances from two fixed points cannot exceed their distance (3). Therefore the range is ([-3,3]).
Step 2
Why this answer is correct
The correct answer is A. ( [-3,3] ). The difference of distances from two fixed points cannot exceed their distance (3). Therefore the range is ([-3,3]).
Step 3
Exam Tip
दो स्थिर बिंदुओं से दूरी के अंतर का मान उनकी दूरी (3) से अधिक नहीं होता। इसलिए परिसर ( [-3,3] ) है।
Since \(\sqrt{x-2}\) is in the denominator, (x>2), and \(\sqrt{8-x}\) gives \(x\le8\). Because of the denominator (2) is excluded but (8) remains.
Step 2
Why this answer is correct
The correct answer is A. ( (2,8] ). Since \(\sqrt{x-2}\) is in the denominator, (x>2), and \(\sqrt{8-x}\) gives \(x\le8\). Because of the denominator (2) is excluded but (8) remains.
Step 3
Exam Tip
हर में \(\sqrt{x-2}\) है इसलिए (x>2), और \(\sqrt{8-x}\) से \(x\le8\) है। हर के कारण (2) हटेगा पर (8) रहेगा।
Here (f(x)=1-\frac{2}{x-2+1}), so the minimum is (-1), and the upper bound (1) is not attained. Rewriting the expression quickly gives the range.
Step 2
Why this answer is correct
The correct answer is A. ( [-1,1) ). Here (f(x)=1-\frac{2}{x-2+1}), so the minimum is (-1), and the upper bound (1) is not attained. Rewriting the expression quickly gives the range.
Step 3
Exam Tip
(f(x)=1-\frac{2}{x-2+1}) है इसलिए न्यूनतम (-1) और ऊपरी सीमा (1) अप्राप्त है। रूप बदलकर परिसर जल्दी मिलता है।
The domain is \(x\ge4\), and at (x=4) the value is \(\sqrt{3}\). Both terms increase, so this is the minimum.
Step 2
Why this answer is correct
The correct answer is A. \( [\sqrt{3},\infty\) ). The domain is \(x\ge4\), and at (x=4) the value is \(\sqrt{3}\). Both terms increase, so this is the minimum.
Step 3
Exam Tip
प्रांत \(x\ge4\) है और (x=4) पर मान \(\sqrt{3}\) मिलता है। दोनों पद बढ़ते हैं इसलिए यही न्यूनतम है।
The denominator ((x+1)2+4) has minimum (4). The fraction is maximum when the positive denominator is minimum.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{1}{4}\). The denominator ((x+1)2+4) has minimum (4). The fraction is maximum when the positive denominator is minimum.
Step 3
Exam Tip
हर ((x+1)2+4) का न्यूनतम (4) है। भिन्न का अधिकतम तब होता है जब धनात्मक हर न्यूनतम हो।
The inside expression (2-|x-1|) has maximum (2) and minimum (0). Taking square root gives the range \([0,\sqrt{2}]\).
Step 2
Why this answer is correct
The correct answer is A. \( [0,\sqrt{2}] \). The inside expression (2-|x-1|) has maximum (2) and minimum (0). Taking square root gives the range \([0,\sqrt{2}]\).
Step 3
Exam Tip
अंदर (2-|x-1|) का अधिकतम (2) और न्यूनतम (0) है। वर्गमूल से परिसर \( [0,\sqrt{2}] \) मिलता है।
The square root is in the denominator, so (|x+2|-5>0) is required. Thus (|x+2|>5) gives the outside open intervals.
Step 2
Why this answer is correct
The correct answer is A. ( \(-\infty,-7\)\cup\(3,\infty\) ). The square root is in the denominator, so (|x+2|-5>0) is required. Thus (|x+2|>5) gives the outside open intervals.
Step 3
Exam Tip
हर में वर्गमूल है इसलिए (|x+2|-5>0) चाहिए। अतः (|x+2|>5) से बाहरी खुले अंतराल मिलते हैं।
From \(y=\frac{4x-1}{2x+5}\), we get \(x=\frac{-1-5y}{2y-4}\), so \(y\ne2\). For a linear fractional function, the missing value is often the ratio of leading coefficients.
Step 2
Why this answer is correct
The correct answer is A. \( \mathbb{R}\setminus{2} \). From \(y=\frac{4x-1}{2x+5}\), we get \(x=\frac{-1-5y}{2y-4}\), so \(y\ne2\). For a linear fractional function, the missing value is often the ratio of leading coefficients.
Step 3
Exam Tip
\(y=\frac{4x-1}{2x+5}\) से \(x=\frac{-1-5y}{2y-4}\) मिलता है इसलिए \(y\ne2\)। रैखिक भिन्नात्मक फलन में अनुपलब्ध मान अक्सर अग्र गुणांकों का अनुपात होता है।
For the square root, ((x-2)(7-x)\ge0) gives \(x\in[2,7]\), and the denominator removes (x=3). Taking the intersection of all restrictions is essential.
Step 2
Why this answer is correct
The correct answer is A. ( [2,3)\cup(3,7] ). For the square root, ((x-2)(7-x)\ge0) gives \(x\in[2,7]\), and the denominator removes (x=3). Taking the intersection of all restrictions is essential.
Step 3
Exam Tip
वर्गमूल के लिए ((x-2)(7-x)\ge0) से \(x\in[2,7]\) मिलता है और हर से \(x\ne3\) हटेगा। संयुक्त प्रतिबंधों का प्रतिच्छेद लेना जरूरी है।
The denominator is (x-2-4x+8=(x-2)2+4), so the fraction has maximum \(\frac{1}{2}\) and never attains (0). Hence the total range is (\(1,\frac{3}{2}]\).
Step 2
Why this answer is correct
The correct answer is A. ( \(1,\frac{3}{2}] \). The denominator is (x-2-4x+8=(x-2)2+4), so the fraction has maximum \(\frac{1}{2}\) and never attains (0). Hence the total range is (\(1,\frac{3}{2}]\).
Step 3
Exam Tip
हर (x-2-4x+8=(x-2)2+4) है इसलिए भिन्न का अधिकतम \(\frac{1}{2}\) है और न्यूनतम (0) प्राप्त नहीं होता। इसलिए कुल परिसर ( \(1,\frac{3}{2}] \) है।
