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 511-520 of 600 questions.
If (f(x)=x-2-4x+8) and \(x\in[0,5]\), what is the range of (f)?
B. ( [4,13] )
(f(x)=(x-2)2+4), इसलिए न्यूनतम (4) है। सिरों पर (f(0)=8) और (f(5)=13), इसलिए अधिकतम (13) है। / Since (f(x)=(x-2)2+4), the minimum is (4). At the endpoints, (f(0)=8) and (f(5)=13), so the maximum is (13).
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What is the range of (f(x)=\frac{x-2+3}{x-2+1})?
A. ( (1,3] )
\(t=x^2\ge 0\) रखने पर \(f=\frac{t+3}{t+1}=1+\frac{2}{t+1}\)। इसलिए मान (1) से बड़े और (3) तक हैं। / Put \(t=x^2\ge 0\), then \(f=\frac{t+3}{t+1}=1+\frac{2}{t+1}\). Hence values are greater than (1) and up to (3).
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What is the domain of (f(x)=\sqrt{x-2}+\frac{1}{\sqrt{x-5}})?
B. ( \(5,\infty\) )
\(\sqrt{x-2}\) के लिए \(x\ge 2\) और हर वाले वर्गमूल के लिए (x-5>0) चाहिए। संयुक्त शर्त (x>5) है। / For \(\sqrt{x-2}\), \(x\ge 2\), and for the denominator square root, (x-5>0) is needed. The combined condition is (x>5).
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What is the range of (f(x)=\frac{3x-2+1}{x-2+1})?
A. ( [1,3) )
\(t=x^2\ge 0\) रखने पर \(f=\frac{3t+1}{t+1}=3-\frac{2}{t+1}\)। (t=0) पर (1) मिलता है और (3) केवल सीमा मान है। / Put \(t=x^2\ge 0\), then \(f=\frac{3t+1}{t+1}=3-\frac{2}{t+1}\). At (t=0), the value is (1), and (3) is only a limiting value.
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If (f(x)=\frac{1}{x-2-1}), what is the range of (f)?
A. ( (-\infty,-1]\cup\(0,\infty\) )
\(t=x^2-1\) लेने पर \(t\in[-1,\infty\)) और \(t\ne 0\)। इसलिए \(\frac{1}{t}\) के मान ((-\infty,-1]\cup\(0,\infty\)) हैं। / Taking \(t=x^2-1\), we get \(t\in[-1,\infty\)) and \(t\ne 0\). Hence \(\frac{1}{t}\) has values ((-\infty,-1]\cup\(0,\infty\)).
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What is the domain of (f(x)=\frac{1}{\sqrt{\log_{5}x}})?
C. ( \(1,\infty\) )
हर में वर्गमूल है, इसलिए \(\log_{5}x>0\) चाहिए। आधार (5>1) होने से (x>1) मिलता है। / The square root is in the denominator, so \(\log_{5}x>0\) is required. Since the base (5>1), this gives (x>1).
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What is the range of (f(x)=|x-4|+|x+2|)?
A. \( [6,\infty\) )
दो बिंदुओं (-2) और (4) से दूरी का योग न्यूनतम (6) होता है। इनके बीच किसी भी (x) पर वही न्यूनतम मान मिलता है। / The sum of distances from (-2) and (4) has minimum (6). Any (x) between these points gives the same minimum value.
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What is the range of (f(x)=|x-5|-|x-1|)?
A. ( [-4,4] )
दो निश्चित बिंदुओं (1) और (5) से दूरियों का अंतर (-4) से (4) तक रहता है। अंतरालों पर चिह्न हटाकर सीमा मान जांचें। / The difference of distances from fixed points (1) and (5) lies from (-4) to (4). Remove signs interval-wise and check endpoint values.
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If the domain of (f(x)=\sqrt{a-x}) is (\(-\infty,9]\), what is the value of (a)?
A. (9)
\(\sqrt{a-x}\) के लिए \(a-x\ge 0\), यानी \(x\le a\)। दिए गए प्रांत से (a=9) है। / For \(\sqrt{a-x}\), \(a-x\ge 0\), so \(x\le a\). Comparing with the given domain gives (a=9).
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If the domain of (f(x)=\frac{1}{x-2-a}) is \(\mathbb{R}\setminus{-3,3}\), what is (a)?
C. (9)
हर शून्य तब होगा जब \(x^2=a\)। हटे हुए मान \(\pm 3\) हैं, इसलिए (a=9)। / The denominator is zero when \(x^2=a\). The excluded values are \(\pm 3\), so (a=9).
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