Practice focused topic-wise MCQs with answers and explanations for quick revision and exam preparation.
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If (f(x)=x-2+ax+1) and (g(x)=x-2+bx+1), and ((f-g)(x)=5x), what is (a-b)?
A. (5)
(f-g=(a-b)x), इसलिए ((a-b)x=5x) से (a-b=5)। समान पदों के गुणांक मिलाएं। / (f-g=(a-b)x), so ((a-b)x=5x) gives (a-b=5). Compare coefficients of like terms.
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If (f(x)=\frac{x+3}{x-3}) and (g(x)=\frac{x-3}{x+3}), what is the domain of ((fg)(x))?
A. \(\mathbb{R}\setminus{-3,3}\)
उत्पाद सरल होकर (1) हो सकता है, पर मूल फलनों में (x=3) और (x=-3) निषिद्ध हैं। प्रांत मूल शर्तों से तय होता है। / The product may simplify to (1), but the original functions forbid (x=3) and (x=-3). The domain is decided by original restrictions.
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If (f(x)=\sqrt{x-1}) and (g(x)=x-5), what is the domain of \(\frac{f}{g}\)?
A. \([1,\infty\)\setminus{5})
वर्गमूल के लिए \(x\ge 1\) और हर के लिए \(x\ne 5\) चाहिए। दोनों शर्तों को साथ लगाएं। / The square root needs \(x\ge 1\), and the denominator needs \(x\ne 5\). Apply both conditions together.
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If (f(x)=x-2-6x+9), (g(x)=x-3), and \(x\ne 3\), what is (\left\(\frac{f}{g}\right\)(x))?
A. (x-3)
(f=(x-3)2), इसलिए \(\frac{f}{g}=x-3\) जब \(x\ne 3\)। कटौती करते समय प्रतिबंध साथ लिखें। / (f=(x-3)2), so \(\frac{f}{g}=x-3\) when \(x\ne 3\). Write the restriction while cancelling.
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If (f(x)=2x-2+1), (g(x)=x-2-3), and (h(x)=x-2+4), what is ((f-g+h)(x))?
A. \(2x^2+8\)
(f-g+h=2x-2+1-\(x^2-3\)+x-2+4=2x-2+8)। घटाव में (g) के सभी चिन्ह बदलें। / (f-g+h=2x-2+1-\(x^2-3\)+x-2+4=2x-2+8). Change all signs of (g) while subtracting.
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If (f(x)=\frac{x-2-1}{x+1}) and (g(x)=x-1), on which domain are (f) and (g) equal?
A. \(x\ne -1\)
(x-2-1=(x-1)(x+1)), इसलिए (f=x-1), पर (x=-1) हटता है। समानता में मूल हर का प्रतिबंध रखें। / (x-2-1=(x-1)(x+1)), so (f=x-1), but (x=-1) is excluded. Keep the original denominator restriction in equality.
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If (f(x)=|x+2|+|x-2|) and (g(x)=4), what is ((f-g)(0))?
A. (0)
(f(0)=|2|+|-2|=4) और (g(0)=4), इसलिए अंतर (0) है। मापांक दूरी की तरह काम करता है। / (f(0)=|2|+|-2|=4) and (g(0)=4), so the difference is (0). Modulus works like distance.
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If (f(x)=\frac{2x+1}{x-2+1}) and (g(x)=\frac{x-3}{x-2+1}), what is ((f+g)(x))?
A. \(\frac{3x-2}{x^2+1}\)
समान हर होने से अंश जुड़ते हैं: (2x+1+x-3=3x-2)। \(x^2+1\) कभी शून्य नहीं होता। / With the same denominator, numerators add: (2x+1+x-3=3x-2). \(x^2+1\) is never zero.
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If (f(x)=x-2-2x+2) and (g(x)=x-2+2x+2), what is ((fg)(0))?
C. (4)
(f(0)=2) और (g(0)=2), इसलिए ((fg)(0)=4)। उत्पाद फलन में पहले अलग-अलग मान निकालें। / (f(0)=2) and (g(0)=2), so ((fg)(0)=4). In a product function, find values separately first.
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If (f(x)=x-2-10x+26) and (g(x)=x-2-10x+30), what is the minimum value of \(\frac{f}{g}\)?
A. \(\frac{1}{5}\)
(\frac{f}{g}=\frac{(x-5)2+1}{(x-5)2+5}), जो ((x-5)2=0) पर \(\frac{1}{5}\) देता है। समान वर्ग पद को \(t\ge 0\) रखकर देखें। / (\frac{f}{g}=\frac{(x-5)2+1}{(x-5)2+5}), which gives \(\frac{1}{5}\) at ((x-5)2=0). Set the common square term as \(t\ge 0\).
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