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 521-530 of 600 questions.
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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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 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)=\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 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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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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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 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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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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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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