Practice focused topic-wise MCQs with answers and explanations for quick revision and exam preparation.
Difficulty select karke sirf is topic ke focused MCQs practice karein. Score, timer aur explanations student-friendly flow me milenge.
Showing 531-540 of 600 questions.
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 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 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 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)=\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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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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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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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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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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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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