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 421-430 of 600 questions.
If (f(x)=|x|+x) and (g(x)=|x|-x), what is ((fg)(x))?
A. (0)
((|x|+x)(|x|-x)=|x|2-x-2=0)। पहचान ((a+b)(a-b)=a-2-b-2) उपयोगी रहती है। / ((|x|+x)(|x|-x)=|x|2-x-2=0). The identity ((a+b)(a-b)=a-2-b-2) is useful.
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If (f(x)=\frac{1}{x-2-4}) and (g(x)=\frac{1}{x-2}), what is the domain of ((f-g)(x))?
A. \( \mathbb{R}\setminus{-2,2} \)
पहले फलन में \(x^2-4\ne 0\) से \(x\ne \pm2\), और दूसरे में \(x\ne 2\)। संयुक्त प्रांत \( \mathbb{R}\setminus{-2,2} \) है। / The first function gives \(x^2-4\ne 0\), so \(x\ne \pm2\), and the second gives \(x\ne 2\). The common domain is \( \mathbb{R}\setminus{-2,2} \).
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If (f(x)=x-3) and (g(x)=x-2), what are the real solutions of ((f-g)(x)=0)?
A. (x=0,1)
(x-3-x-2=x-2(x-1)=0), इसलिए (x=0) या (x=1)। गुणनखंड करने से हल तेजी से मिलते हैं। / (x-3-x-2=x-2(x-1)=0), so (x=0) or (x=1). Factoring gives solutions quickly.
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If (f(x)=\frac{x+2}{x-2}) and (g(x)=\frac{x-2}{x+2}), what is the simplified form of ((f+g)(x))?
A. \(\frac{2x^2+8}{x^2-4},\ x\ne \pm2\)
अंश ((x+2)2+(x-2)2=2x-2+8) है। समान हर बनाते समय पूरे वर्ग ध्यान से खोलें। / The numerator is ((x+2)2+(x-2)2=2x-2+8). Expand both squares carefully while taking a common denominator.
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If (f(x)=x-2+ax) and (g(x)=bx+1), and ((f-g)(x)=x-2-5x-1), what is ((a,b))?
A. ((a,b)=(2,7))
((f-g)(x)=x-2+(a-b)x-1), इसलिए (a-b=-5)। दिए विकल्पों में केवल ((2,7)) सही है। / ((f-g)(x)=x-2+(a-b)x-1), so (a-b=-5). Among the options, only ((2,7)) works.
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A. \(\frac{x+1}{x},\ x\ne 0,-1\)
भाग देने पर \(\frac{1}{x}\div\frac{1}{x+1}=\frac{x+1}{x}\) मिलता है। मूल (g) के लिए \(x\ne -1\) भी जरूरी है। / Dividing gives \(\frac{1}{x}\div\frac{1}{x+1}=\frac{x+1}{x}\). The original (g) also requires \(x\ne -1\).
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If (f(x)=x-2) and (g(x)=|x|), when is ((f-g)(x)) equal to (0)?
A. \(x\in{-1,0,1}\)
\(x^2=|x|\) में \(t=|x|\ge 0\) रखने पर \(t^2=t\), इसलिए (t=0) या (t=1)। अतः (x=-1,0,1) हैं। / In \(x^2=|x|\), put \(t=|x|\ge 0\), so \(t^2=t\). Thus (t=0) or (t=1), giving (x=-1,0,1).
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If (f(x)=\sqrt{x+1}) and (g(x)=\sqrt{2-x}), what is the maximum value of ((fg)(x))?
A. \(\frac{3}{2}\)
((fg)(x)=\sqrt{(x+1)(2-x)}) और अंदर का द्विघात अधिकतम \(\frac{9}{4}\) है। इसलिए अधिकतम \(\frac{3}{2}\) है। / ((fg)(x)=\sqrt{(x+1)(2-x)}), and the quadratic inside has maximum \(\frac{9}{4}\). Therefore the maximum is \(\frac{3}{2}\).
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If (f(x)=x-2-6x+10) and (g(x)=x-1), what is the minimum value of ((f-g)(x))?
A. \(-\frac{9}{4}\)
((f-g)(x)=x-2-7x+11=\(x-\frac{7}{2}\)2-\frac{5}{4}) नहीं, सही स्थिर मान से न्यूनतम \(-\frac{5}{4}\) है। विकल्पों में त्रुटि जांचें। / ((f-g)(x)=x-2-7x+11=\(x-\frac{7}{2}\)2-\frac{5}{4}), so the minimum is \(-\frac{5}{4}\). Check options carefully.
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If (f(x)=\frac{x-2-1}{x+1}) and (g(x)=x-1), when do (f) and (g) give equal values?
A. हर \(x\ne -1\) परFor every \(x\ne -1\)
(f(x)=x-1) पर (x=-1) परिभाषित नहीं है। इसलिए समान मान केवल (f) के प्रांत में मिलते हैं। / (f(x)=x-1), but (x=-1) is not defined. Hence equal values occur only in the domain of (f).
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