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 301-310 of 600 questions.
What is the domain of the function (f(x)=\frac{1}{\sqrt{x-2}})?
A. (x>2)
हर में \(\sqrt{x-2}\) है इसलिए (x-2>0) होना चाहिए। परीक्षा में हर वाले मूल में शून्य भी हटाएं। / The denominator has \(\sqrt{x-2}\), so (x-2>0) is required. In exams, also exclude zero in a denominator.
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What is the domain of (f(x)=\frac{\sqrt{x-3}}{\sqrt{x-2-25}})?
A. (\(5,\infty\))
अंश के लिए \(x-3\ge 0\) और हर के लिए \(x^2-25>0\) चाहिए। दोनों शर्तों का प्रतिच्छेद लेने पर (\(5,\infty\)) मिलता है। / The numerator needs \(x-3\ge 0\), and the denominator needs \(x^2-25>0\). Intersecting both conditions gives (\(5,\infty\)).
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What is the domain of the function (f(x)=\sqrt{5-2x})?
A. \(x\le \frac{5}{2}\)
वर्गमूल के अंदर \(5-2x\ge 0\) होना चाहिए। चिन्ह बदलते समय असमता की दिशा ध्यान रखें। / The radicand must satisfy \(5-2x\ge 0\). Remember to reverse the inequality when dividing by a negative number.
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If (f(x)=\frac{x-2-9}{x-2+9}) and \(x\in\mathbb{R}\), what is the range of (f)?
A. ([-1,1))
(f(x)=1-\frac{18}{x-2+9}) है और \(x^2+9\ge 9\) है। इसलिए न्यूनतम मान (-1) मिलता है और (1) प्राप्त नहीं होता। / Here (f(x)=1-\frac{18}{x-2+9}), and \(x^2+9\ge 9\). So the minimum value is (-1), and (1) is not attained.
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What is the domain of (f(x)=\frac{x+1}{x-2-x-6})?
A. \(\mathbb{R}\setminus{-2,3}\)
हर (x-2-x-6=(x-3)(x+2)) शून्य नहीं हो सकता। हर के शून्य देने वाले मान हमेशा प्रांत से हटते हैं। / The denominator (x-2-x-6=(x-3)(x+2)) cannot be zero. Values that make the denominator zero are excluded from the domain.
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What is the domain of (f(x)=\frac{1}{\sqrt{|x-2|-3}})?
A. (\(-\infty,-1\)\cup\(5,\infty\))
हर में वर्गमूल है इसलिए (|x-2|-3>0) चाहिए। इससे (|x-2|>3) और प्रांत (\(-\infty,-1\)\cup\(5,\infty\)) मिलता है। / The square root is in the denominator, so (|x-2|-3>0) is needed. This gives (|x-2|>3), so the domain is (\(-\infty,-1\)\cup\(5,\infty\)).
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What is the domain of (f(x)=\frac{1}{x-2+4x+4})?
A. \(\mathbb{R}\setminus{-2}\)
हर (x-2+4x+4=(x+2)2) है इसलिए (x=-2) वर्जित है। पूर्ण वर्ग को पहचानना समय बचाता है। / The denominator is (x-2+4x+4=(x+2)2), so (x=-2) is excluded. Recognizing perfect squares saves time.
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If (f(x)=\frac{2x+7}{3x-4}), which value is not in the range of (f)?
A. \(\frac{2}{3}\)
\(y=\frac{2x+7}{3x-4}\) से \(x=\frac{4y+7}{3y-2}\) मिलता है। इसलिए \(3y-2\ne 0\), अतः \(y\ne \frac{2}{3}\) होगा। / From \(y=\frac{2x+7}{3x-4}\), we get \(x=\frac{4y+7}{3y-2}\). Hence \(3y-2\ne 0\), so \(y\ne \frac{2}{3}\).
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What is the domain of (f(x)=\frac{1}{\sqrt{x-2-4}})?
A. (\(-\infty,-2\)\cup\(2,\infty\))
हर में वर्गमूल है इसलिए \(x^2-4>0\) चाहिए। हर के कारण बराबरी वाले मान शामिल नहीं होंगे। / The square root is in the denominator, so \(x^2-4>0\) is required. Equality values are not included because of the denominator.
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What is the domain of (f(x)=\sqrt{(x-1)(x+4)})?
A. (\(-\infty,-4]\cup[1,\infty\))
वर्गमूल के लिए ((x-1)(x+4)\ge 0) चाहिए। दो गुणनखंडों की असमता में अंतराल जांचें। / The radicand must satisfy ((x-1)(x+4)\ge 0). Check intervals when solving a product inequality.
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