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 71-80 of 600 questions.
What is the real domain of (f(x)=\sqrt{x+10})?
A. \([-10,\infty\))
वर्गमूल के अंदर \(x+10 \ge 0\) चाहिए। इससे \(x \ge -10\) मिलता है। / The expression inside the square root must satisfy \(x+10 \ge 0\). This gives \(x \ge -10\).
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What is the real domain of (f(x)=\sqrt{12-3x})?
B. (\(-\infty,4]\)
\(12-3x \ge 0\) से \(x \le 4\) मिलता है। इसलिए प्रांत (\(-\infty,4]\) है। / From \(12-3x \ge 0\), we get \(x \le 4\). Therefore the domain is (\(-\infty,4]\).
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Which values are excluded from the domain of (f(x)=\frac{5}{x-2-9})?
C. ({-3,3})
हर \(x^2-9\) है और \(x^2-9=0\) से \(x=\pm3\) मिलता है। इसलिए (-3) और (3) हटेंगे। / The denominator is \(x^2-9\), and \(x^2-9=0\) gives \(x=\pm3\). So (-3) and (3) are excluded.
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What is the domain of (f(x)=\frac{5}{x-2-9})?
A. \(\mathbb{R}-{-3,3}\)
हर (0) नहीं होना चाहिए और \(x^2-9=0\) पर (x=-3,3) मिलते हैं। इसलिए प्रांत \(\mathbb{R}-{-3,3}\) है। / The denominator must not be (0), and \(x^2-9=0\) gives (x=-3,3). Hence the domain is \(\mathbb{R}-{-3,3}\).
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What is the domain of (f(x)=\frac{1}{x-2+6})?
D. \(\mathbb{R}\)
\(x^2+6\) हर वास्तविक (x) के लिए धनात्मक है। इसलिए हर कभी (0) नहीं बनता। / \(x^2+6\) is positive for every real (x). Therefore the denominator never becomes (0).
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For (f(x)=\frac{1}{x-2+6}), when is the greatest value obtained?
A. जब (x=0)When (x=0)
हर \(x^2+6\) सबसे छोटा (6) तब होता है जब (x=0)। इसलिए भिन्न का मान सबसे बड़ा तब मिलता है। / The denominator \(x^2+6\) is smallest at (6) when (x=0). Therefore the fraction is greatest then.
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What is the range of (f(x)=\frac{1}{x-2+6})?
A. (\(0,\frac{1}{6}]\)
हर का न्यूनतम मान (6) है, इसलिए अधिकतम मान \(\frac{1}{6}\) है। (0) कभी नहीं मिलता पर मान (0) के पास जा सकता है। / The minimum denominator is (6), so the maximum value is \(\frac{1}{6}\). The value (0) is never reached but can be approached.
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Which is the correct domain for (f(x)=\sqrt{16-x-2})?
B. ([-4,4])
वर्गमूल के लिए \(16-x^2 \ge 0\) चाहिए। इससे \(-4 \le x \le 4\) मिलता है। / For the square root, \(16-x^2 \ge 0\) is needed. This gives \(-4 \le x \le 4\).
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What is the greatest value of (f(x)=\sqrt{16-x-2})?
C. (4)
\(16-x^2\) सबसे बड़ा (16) तब होता है जब (x=0)। इसलिए वर्गमूल का सबसे बड़ा मान (4) है। / \(16-x^2\) is greatest at (16) when (x=0). Therefore the greatest square-root value is (4).
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What is the range of (f(x)=\sqrt{16-x-2})?
A. ([0,4])
वर्गमूल का मान ऋणात्मक नहीं होता और अधिकतम (4) है। इसलिए परिसर ([0,4]) है। / A square-root value is not negative, and the maximum is (4). Hence the range is ([0,4]).
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