real-valued-functions se related questions ko ek jagah revise karein. Har question me bilingual content, answer feedback aur explanation available hai.
308 questions tagged with real-valued-functions.
If (f(x)=\sqrt{x-1}+\sqrt{9-x}), what is the domain of (f)?
A. ( [1,9] )
दोनों वर्गमूलों के लिए \(x-1\ge 0\) और \(9-x\ge 0\) चाहिए। इसलिए \(1\le x\le 9\)। / Both square roots need \(x-1\ge 0\) and \(9-x\ge 0\). Therefore \(1\le x\le 9\).
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What is the range of (f(x)=\frac{x-2-2x+5}{x-2-2x+2})?
A. ( (1,4] )
(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। / 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.
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What is the range of (f(x)=\frac{x-2+4x+8}{x-2+4x+5})?
A. ( (1,4] )
(t=(x+2)2\ge 0) रखने पर \(f=\frac{t+4}{t+1}=1+\frac{3}{t+1}\)। इसलिए परिसर ((1,4]) है। / Put (t=(x+2)2\ge 0), then \(f=\frac{t+4}{t+1}=1+\frac{3}{t+1}\). Hence the range is ((1,4]).
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If (f(x)=\lceil x\rceil) and (x\in(-2,2]), what is the range of (f)?
A. ( {-1,0,1,2} )
छत फलन (x) से छोटा नहीं सबसे छोटा पूर्णांक देता है। (x>-2) है, इसलिए (-2) मान नहीं आएगा। / The ceiling function gives the smallest integer not less than (x). Since (x>-2), the value (-2) will not occur.
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If (f(x)=\lfloor 2x\rfloor) and \(x\in[0,2\)), what is the range of (f)?
A. ( {0,1,2,3} )
\(x\in[0,2\)) होने से \(2x\in[0,4\))। महत्तम पूर्णांक मान (0,1,2,3) होंगे। / Since \(x\in[0,2\)), \(2x\in[0,4\)). The greatest integer values are (0,1,2,3).
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What is the range of (f(x)=\log_{2}\(x^2+2x+2\))?
A. \( [0,\infty\) )
(x-2+2x+2=(x+1)2+1\ge 1)। इसलिए (\log_{2}\(x^2+2x+2\)\ge 0) और मान ऊपर असीम हैं। / Since (x-2+2x+2=(x+1)2+1\ge 1). Hence (\log_{2}\(x^2+2x+2\)\ge 0), and values are unbounded above.
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What is the range of (f(x)=\sqrt{x-2+2x+2})?
A. \( [1,\infty\) )
(x-2+2x+2=(x+1)2+1), इसलिए अंदर का न्यूनतम (1) है। वर्गमूल के बाद परिसर \([1,\infty\)) होगा। / Since (x-2+2x+2=(x+1)2+1), the inside minimum is (1). After square root, the range is \([1,\infty\)).
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What is the range of (f(x)=\frac{1}{1+\sqrt{x}})?
A. ( (0,1] )
\(\sqrt{x}\ge 0\), इसलिए हर कम से कम (1) है। (x=0) पर (1) मिलता है और \(x\to\infty\) पर मान (0) के पास जाता है। / 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).
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What is the domain of (f(x)=\frac{\sqrt{9-x-2}}{x+1})?
A. \( [-3,3]\setminus{-1} \)
वर्गमूल के लिए \(9-x^2\ge 0\), यानी \(-3\le x\le 3\)। हर के कारण (x=-1) हटेगा। / The square root needs \(9-x^2\ge 0\), so \(-3\le x\le 3\). The denominator removes (x=-1).
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What is the domain of (f(x)=\frac{x+1}{\sqrt{x-2-4}})?
A. ( \(-\infty,-2\)\cup\(2,\infty\) )
हर में वर्गमूल है, इसलिए \(x^2-4>0\) चाहिए। इससे (x<-2) या (x>2) मिलता है। / The square root is in the denominator, so \(x^2-4>0\) is required. This gives (x<-2) or (x>2).
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What is the range of (f(x)=\frac{3x}{x-2+9})?
A. \( [-\frac{1}{2},\frac{1}{2}] \)
\(|3x|\le \frac{x^2+9}{2}\), इसलिए \(\left|\frac{3x}{x^2+9}\right|\le \frac{1}{2}\)। समानता \(x=\pm 3\) पर मिलती है। / Since \(|3x|\le \frac{x^2+9}{2}\), \(\left|\frac{3x}{x^2+9}\right|\le \frac{1}{2}\). Equality occurs at \(x=\pm 3\).
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What is the range of (f(x)=\frac{2x}{x-2+1})?
B. ( [-1,1] )
\(|2x|\le x^2+1\) क्योंकि ((x-1)2\ge 0) और ((x+1)2\ge 0)। इसलिए मान (-1) से (1) तक हैं और दोनों मिलते हैं। / \(|2x|\le x^2+1\) because ((x-1)2\ge 0) and ((x+1)2\ge 0). Hence the values lie from (-1) to (1), and both occur.
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If (f(x)=\sqrt{x+2}+\sqrt{8-x}), what is the minimum value of (f)?
A. \( \sqrt{10} \)
बंद अंतराल ([-2,8]) में न्यूनतम सिरों पर आता है। दोनों सिरों पर मान \(\sqrt{10}\) है। / On the closed interval ([-2,8]), the minimum occurs at the endpoints. At both endpoints, the value is \(\sqrt{10}\).
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If (f(x)=\sqrt{x+2}+\sqrt{8-x}), what is the maximum value of (f)?
A. \(2\sqrt{5}\)
सममिति के कारण अधिकतम (x=3) पर आता है। तब (f(3)=\sqrt{5}+\sqrt{5}=2\sqrt{5})। / By symmetry, the maximum occurs at (x=3). Then (f(3)=\sqrt{5}+\sqrt{5}=2\sqrt{5}).
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If (f(x)=\sqrt{x+2}+\sqrt{8-x}), what is the domain of (f)?
A. ( [-2,8] )
दोनों वर्गमूलों के लिए \(x+2\ge 0\) और \(8-x\ge 0\) चाहिए। इनका छेदन ([-2,8]) है। / Both square roots need \(x+2\ge 0\) and \(8-x\ge 0\). Their intersection is ([-2,8]).
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What is the domain of (f(x)=\frac{x-2-9}{x-3})?
B. \( \mathbb{R}\setminus{3} \)
मूल हर (x-3) है, इसलिए (x=3) स्वीकार्य नहीं है। सरलीकरण के बाद भी हटे हुए बिंदु को प्रांत में न जोड़ें। / The original denominator is (x-3), so (x=3) is not allowed. Even after simplification, do not add the removed point to the domain.
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What is the range of (f(x)=\frac{x-2-9}{x-3})?
A. \( \mathbb{R}\setminus{6} \)
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 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.
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What is the range of (f(x)=\frac{1}{|x+3|+2})?
A. ( \(0,\frac{1}{2}] \)
हर \(|x+3|+2\ge 2\), इसलिए अधिकतम मान \(\frac{1}{2}\) है। हर असीम होने पर मान (0) के पास जाता है। / The denominator \(|x+3|+2\ge 2\), so the maximum value is \(\frac{1}{2}\). As the denominator grows, the value approaches (0).
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What is the range of (f(x)=\sqrt{6x-x-2-5})?
A. ( [0,2] )
(6x-x-2-5=4-(x-3)2), जिसका अधिकतम (4) है। वर्गमूल से परिसर ([0,2]) मिलता है। / Since (6x-x-2-5=4-(x-3)2), its maximum is (4). Taking the square root gives the range ([0,2]).
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What is the domain of (f(x)=\sqrt{6x-x-2-5})?
A. ( [1,5] )
वर्गमूल के लिए \(6x-x^2-5\ge 0\) चाहिए। इसे (-(x-1)(x-5)\ge 0) लिखने पर \(1\le x\le 5\) मिलता है। / The square root needs \(6x-x^2-5\ge 0\). Writing it as (-(x-1)(x-5)\ge 0) gives \(1\le x\le 5\).
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If the domain of (f(x)=\frac{1}{x-2-a}) is \(\mathbb{R}\setminus{-3,3}\), what is (a)?
C. (9)
हर शून्य तब होगा जब \(x^2=a\)। हटे हुए मान \(\pm 3\) हैं, इसलिए (a=9)। / The denominator is zero when \(x^2=a\). The excluded values are \(\pm 3\), so (a=9).
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If the domain of (f(x)=\sqrt{a-x}) is (\(-\infty,9]\), what is the value of (a)?
A. (9)
\(\sqrt{a-x}\) के लिए \(a-x\ge 0\), यानी \(x\le a\)। दिए गए प्रांत से (a=9) है। / For \(\sqrt{a-x}\), \(a-x\ge 0\), so \(x\le a\). Comparing with the given domain gives (a=9).
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What is the range of (f(x)=|x-5|-|x-1|)?
A. ( [-4,4] )
दो निश्चित बिंदुओं (1) और (5) से दूरियों का अंतर (-4) से (4) तक रहता है। अंतरालों पर चिह्न हटाकर सीमा मान जांचें। / The difference of distances from fixed points (1) and (5) lies from (-4) to (4). Remove signs interval-wise and check endpoint values.
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What is the range of (f(x)=|x-4|+|x+2|)?
A. \( [6,\infty\) )
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 sum of distances from (-2) and (4) has minimum (6). Any (x) between these points gives the same minimum value.
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What is the domain of (f(x)=\frac{1}{\sqrt{\log_{5}x}})?
C. ( \(1,\infty\) )
हर में वर्गमूल है, इसलिए \(\log_{5}x>0\) चाहिए। आधार (5>1) होने से (x>1) मिलता है। / The square root is in the denominator, so \(\log_{5}x>0\) is required. Since the base (5>1), this gives (x>1).
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If (f(x)=\frac{1}{x-2-1}), what is the range of (f)?
A. ( (-\infty,-1]\cup\(0,\infty\) )
\(t=x^2-1\) लेने पर \(t\in[-1,\infty\)) और \(t\ne 0\)। इसलिए \(\frac{1}{t}\) के मान ((-\infty,-1]\cup\(0,\infty\)) हैं। / Taking \(t=x^2-1\), we get \(t\in[-1,\infty\)) and \(t\ne 0\). Hence \(\frac{1}{t}\) has values ((-\infty,-1]\cup\(0,\infty\)).
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What is the range of (f(x)=\frac{3x-2+1}{x-2+1})?
A. ( [1,3) )
\(t=x^2\ge 0\) 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। / 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.
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What is the domain of (f(x)=\sqrt{x-2}+\frac{1}{\sqrt{x-5}})?
B. ( \(5,\infty\) )
\(\sqrt{x-2}\) के लिए \(x\ge 2\) और हर वाले वर्गमूल के लिए (x-5>0) चाहिए। संयुक्त शर्त (x>5) है। / For \(\sqrt{x-2}\), \(x\ge 2\), and for the denominator square root, (x-5>0) is needed. The combined condition is (x>5).
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What is the range of (f(x)=\frac{x-2+3}{x-2+1})?
A. ( (1,3] )
\(t=x^2\ge 0\) रखने पर \(f=\frac{t+3}{t+1}=1+\frac{2}{t+1}\)। इसलिए मान (1) से बड़े और (3) तक हैं। / Put \(t=x^2\ge 0\), then \(f=\frac{t+3}{t+1}=1+\frac{2}{t+1}\). Hence values are greater than (1) and up to (3).
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If (f(x)=x-2-4x+8) and \(x\in[0,5]\), what is the range of (f)?
B. ( [4,13] )
(f(x)=(x-2)2+4), इसलिए न्यूनतम (4) है। सिरों पर (f(0)=8) और (f(5)=13), इसलिए अधिकतम (13) है। / Since (f(x)=(x-2)2+4), the minimum is (4). At the endpoints, (f(0)=8) and (f(5)=13), so the maximum is (13).
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