Practice focused topic-wise MCQs with answers and explanations for quick revision and exam preparation.
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Choose the range of (f(x)=\sqrt{2x-x-2}).
A. ([0,1])
(2x-x-2=1-(x-1)2), जिसका अधिकतम मान (1) है। वर्गमूल से परिसर ([0,1]) मिलता है। / (2x-x-2=1-(x-1)2), whose maximum value is (1). The square root gives the range ([0,1]).
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If (f(x)=\frac{1}{x-1}+\frac{1}{x+1}), what is the domain?
A. \(\mathbb{R}-{-1,1}\)
दोनों हर शून्य नहीं होने चाहिए, इसलिए \(x\ne 1\) और \(x\ne -1\)। योग वाले रैशनल फलन में सभी हर जांचें। / Both denominators must be non-zero, so \(x\ne 1\) and \(x\ne -1\). In sums of rational functions, check every denominator.
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Which is the range of (f(x)=\frac{1}{\sqrt{x-2+6x+10}})?
A. ((0,1])
क्योंकि (x-2+6x+10=(x+3)2+1\ge 1), हर का न्यूनतम मान (1) है। इसलिए (0<f(x)\le 1)। / Since (x-2+6x+10=(x+3)2+1\ge 1), the denominator has minimum value (1). Therefore (0<f(x)\le 1).
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What is the range of (f(x)=\frac{x}{\sqrt{x-2+1}})?
A. ((-1,1))
हर हमेशा (|x|) से बड़ा होता है, इसलिए (-1<f(x)<1)। (-1) और (1) केवल सीमा मान हैं। / The denominator is always greater than (|x|), so (-1<f(x)<1). The values (-1) and (1) are only limiting values.
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If (f(x)=\sqrt{x}+\sqrt{1-x}), what is the domain?
A. ([0,1])
शर्तें \(x\ge 0\) और \(1-x\ge 0\) हैं। दोनों मिलाकर \(x\in[0,1]\) मिलता है। / The conditions are \(x\ge 0\) and \(1-x\ge 0\). Together they give \(x\in[0,1]\).
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Choose the range of (f(x)=\sqrt{x}+\sqrt{1-x}).
A. \([1,\sqrt{2}]\)
प्रांत ([0,1]) में न्यूनतम मान किनारों पर (1) और अधिकतम \(x=\frac{1}{2}\) पर \(\sqrt{2}\) है। इसलिए परिसर \([1,\sqrt{2}]\) है। / On ([0,1]), the minimum at endpoints is (1) and the maximum at \(x=\frac{1}{2}\) is \(\sqrt{2}\). Hence the range is \([1,\sqrt{2}]\).
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What is the maximum value of (f(x)=\frac{1}{x-2+2x+2})?
A. (1)
हर (x-2+2x+2=(x+1)2+1) का न्यूनतम मान (1) है। इसलिए फलन का अधिकतम मान (1) है। / The denominator (x-2+2x+2=(x+1)2+1) has minimum value (1). Hence the maximum value of the function is (1).
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If the real function (f(x)=\frac{\sqrt{x-2}}{x-2-9}) is given, what is its domain?
A. \([2,\infty\)-{3})
वर्गमूल के लिए \(x\ge 2\) और हर के लिए \(x\ne -3,3\) चाहिए। लेकिन (-3) पहले ही \(x\ge 2\) में नहीं है, इसलिए केवल (3) हटेगा। / For the square root \(x\ge 2\), and for the denominator \(x\ne -3,3\) are required. But (-3) is already outside \(x\ge 2\), so only (3) is removed.
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What is the domain of (f(x)=\sqrt{\frac{x-2-4}{9-x-2}})?
A. (\(-3,-2]\cup[2,3\))
शर्त \(\frac{x^2-4}{9-x^2}\ge 0\) और \(x\ne -3,3\) है। साइन चार्ट से (\(-3,-2]\cup[2,3\)) मिलता है। / The condition is \(\frac{x^2-4}{9-x^2}\ge 0\) and \(x\ne -3,3\). A sign chart gives (\(-3,-2]\cup[2,3\)).
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If (f(x)=\frac{2x-2+3}{x-2+1}), what is the range of (f)?
A. ((2,3])
मान लें \(t=x^2\ge 0\), तब \(f=\frac{2t+3}{t+1}\) घटता है। (t=0) पर (3) मिलता है और (2) प्राप्त नहीं होता। / Let \(t=x^2\ge 0\), then \(f=\frac{2t+3}{t+1}\) is decreasing. At (t=0), (3) is obtained and (2) is not attained.
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