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 21-30 of 600 questions.
What is the domain of (f(x)=\frac{2}{x+1})?
A. \(\mathbb{R}-{-1}\)
हर (x+1) है, इसलिए (x=-1) पर हर (0) हो जाता है। इस मान को प्रांत से हटाते हैं। / The denominator is (x+1), so it becomes (0) at (x=-1). This value is removed from the domain.
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Which condition is correct for the real domain of (f(x)=\sqrt{9-x-2})?
A. \(-3 \le x \le 3\)
वर्गमूल के लिए \(9-x^2 \ge 0\) चाहिए। इससे \(x^2 \le 9\), यानी \(-3 \le x \le 3\) मिलता है। / For the square root, \(9-x^2 \ge 0\) is needed. This gives \(x^2 \le 9\), that is \(-3 \le x \le 3\).
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What is the real domain of (f(x)=\sqrt{9-x-2})?
A. ([-3,3])
शर्त \(9-x^2 \ge 0\) से ([-3,3]) मिलता है। सिरों को शामिल करें क्योंकि वर्गमूल (0) हो सकता है। / The condition \(9-x^2 \ge 0\) gives ([-3,3]). Include the endpoints because the square root can be (0).
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What is the minimum value of (f(x)=x-2-4) when \(x \in \mathbb{R}\)?
A. (-4)
क्योंकि \(x^2 \ge 0\), इसलिए \(x^2-4 \ge -4\) है। न्यूनतम मान (x=0) पर (-4) है। / Since \(x^2 \ge 0\), \(x^2-4 \ge -4\). The minimum value is (-4) at (x=0).
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What is the range of (f(x)=x-2-4)?
A. \([-4,\infty\))
\(x^2\) का न्यूनतम मान (0) है, इसलिए \(x^2-4\) का न्यूनतम मान (-4) है। अतः परिसर \([-4,\infty\)) है। / The minimum value of \(x^2\) is (0), so the minimum value of \(x^2-4\) is (-4). Hence the range is \([-4,\infty\)).
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If \(A=\{-1,0,1\}\) and (f(x)=x-2), what is the range?
A. ({0,1})
((-1)2=1), \(0^2=0\), और \(1^2=1\) हैं। अलग-अलग प्राप्त मान ({0,1}) हैं। / ((-1)2=1), \(0^2=0\), and \(1^2=1\). The distinct output values are ({0,1}).
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What is the domain of (f(x)=\frac{1}{x-2})?
A. \(\mathbb{R}-{0}\)
हर \(x^2\) है और (x=0) पर यह (0) बनता है। इसलिए (0) प्रांत में नहीं होगा। / The denominator is \(x^2\), and it becomes (0) at (x=0). Therefore (0) is not in the domain.
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What is the range of (f(x)=\frac{1}{x-2})?
A. (\(0,\infty\))
\(x \ne 0\) होने पर \(x^2>0\), इसलिए \(\frac{1}{x^2}>0\) है। मान (0) कभी नहीं मिलता। / When \(x \ne 0\), \(x^2>0\), so \(\frac{1}{x^2}>0\). The value (0) is never obtained.
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What is the least value of (f(x)=x-2+4)?
A. (4)
\(x^2\) का सबसे छोटा मान (0) होता है। इसलिए \(x^2+4\) का सबसे छोटा मान (4) है। / The least value of \(x^2\) is (0). Therefore the least value of \(x^2+4\) is (4).
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What is the range of (f(x)=x-2+4)?
A. \([4,\infty\))
क्योंकि \(x^2 \ge 0\), इसलिए \(x^2+4 \ge 4\) है। अतः परिसर \([4,\infty\)) है। / Since \(x^2 \ge 0\), \(x^2+4 \ge 4\). Hence the range is \([4,\infty\)).
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