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 411-420 of 600 questions.
What is the domain of (f(x)=\sqrt{9-(x-1)2})?
A. ([-2,4])
शर्त (9-(x-1)2\ge 0) से ((x-1)2\le 9) मिलता है। इसलिए \(x\in[-2,4]\)। / The condition (9-(x-1)2\ge 0) gives ((x-1)2\le 9). Hence \(x\in[-2,4]\).
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Choose the range of (f(x)=\sqrt{9-(x-1)2}).
A. ([0,3])
वर्गमूल का मान ऋणात्मक नहीं होता और अंदर का अधिकतम मान (9) है। इसलिए परिसर ([0,3]) है। / A square root is never negative and the maximum inside value is (9). Hence the range is ([0,3]).
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If (f(x)=|x-4|+2), what is its range?
A. \([2,\infty\))
क्योंकि \(|x-4|\ge 0\), न्यूनतम मान (2) है। अतः परिसर \([2,\infty\)) है। / Since \(|x-4|\ge 0\), the minimum value is (2). Hence the range is \([2,\infty\)).
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Which is the range of (f(x)=\frac{1}{|x|+1})?
A. ((0,1])
क्योंकि \(|x|+1\ge 1\), इसलिए (0<f(x)\le 1)। (0) केवल सीमा मान है, प्राप्त नहीं होता। / Since \(|x|+1\ge 1\), (0<f(x)\le 1). The value (0) is only a limiting value and is not attained.
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What is the range of (f(x)=\frac{x-2-1}{x-2+1})?
A. ([-1,1))
मान लें \(t=x^2\ge 0\), तब \(f=\frac{t-1}{t+1}\)। (t=0) पर (-1) मिलता है और (1) प्राप्त नहीं होता। / Let \(t=x^2\ge 0\), then \(f=\frac{t-1}{t+1}\). At (t=0), (-1) occurs and (1) is not attained.
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Find the domain of (f(x)=\sqrt{x-2-16}).
A. (\(-\infty,-4]\cup[4,\infty\))
वर्गमूल के लिए \(x^2-16\ge 0\), इसलिए \(x^2\ge 16\)। अतः \(x\le -4\) या \(x\ge 4\)। / For the square root \(x^2-16\ge 0\), so \(x^2\ge 16\). Hence \(x\le -4\) or \(x\ge 4\).
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Choose the range of (f(x)=\sqrt{x-2-16}).
A. \([0,\infty\))
प्रांत में \(x^2-16\ge 0\) और वर्गमूल का न्यूनतम मान (0) है। बड़े (|x|) पर मान असीमित बढ़ता है। / On the domain, \(x^2-16\ge 0\) and the minimum square-root value is (0). For large (|x|), the value grows without bound.
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What is the domain of (f(x)=\log_2\(x^2-5x+6\))?
A. (\(-\infty,2\)\cup\(3,\infty\))
लघुगणक के लिए \(x^2-5x+6>0\) चाहिए। ((x-2)(x-3)>0), इसलिए (x<2) या (x>3)। / For a logarithm, \(x^2-5x+6>0\) is needed. Since ((x-2)(x-3)>0), (x<2) or (x>3).
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Which is the domain of (f(x)=\log_5(4-x))?
A. (\(-\infty,4\))
लघुगणक का आर्गुमेंट (4-x>0) होना चाहिए। इसलिए (x<4)। / The logarithm argument must satisfy (4-x>0). Therefore (x<4).
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What is the domain of (f(x)=\frac{1}{\log_3 x})?
A. (\(0,\infty\)-{1})
लघुगणक के लिए (x>0) और हर के लिए \(\log_3 x\ne 0\), यानी \(x\ne 1\)। दोनों शर्तें साथ रखें। / For the logarithm (x>0), and for the denominator \(\log_3 x\ne 0\), so \(x\ne 1\). Keep both conditions together.
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