Level 38 • 50/50 questions • 35 seconds per question.
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If (f(x)=x-2+1) and (g(x)=2x-3), what is ((f+g)(x))?
A. \(x^2+2x-2\)
((f+g)(x)=f(x)+g(x)=x-2+1+2x-3=x-2+2x-2)। परीक्षा में समान पदों को ध्यान से जोड़ें। / ((f+g)(x)=f(x)+g(x)=x-2+1+2x-3=x-2+2x-2). In exams, combine like terms carefully.
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If (f(x)=3x+5) and (g(x)=x-2-4), find ((g-f)(x)).
A. \(x^2-3x-9\)
((g-f)(x)=g(x)-f(x)=x-2-4-(3x+5)=x-2-3x-9)। घटाते समय पूरे (f(x)) पर ऋण लगाएँ। / ((g-f)(x)=g(x)-f(x)=x-2-4-(3x+5)=x-2-3x-9). While subtracting, apply the minus sign to the whole (f(x)).
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If (f(x)=x+2) and (g(x)=x-5), what is ((fg)(x))?
A. \(x^2-3x-10\)
((fg)(x)=f(x)g(x)=(x+2)(x-5)=x-2-3x-10)। गुणा में विस्तार के बाद समान पद मिलाएँ। / ((fg)(x)=f(x)g(x)=(x+2)(x-5)=x-2-3x-10). After expanding, combine like terms.
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If (f(x)=x-2-9) and (g(x)=x-3), what is (\left\(\frac{f}{g}\right\)(x)) for \(x \neq 3\)?
A. (x+3)
(\left\(\frac{f}{g}\right\)(x)=\frac{x-2-9}{x-3}=\frac{(x-3)(x+3)}{x-3}=x+3), जहाँ \(x \neq 3\)। हर शून्य न हो, यह हमेशा जाँचें। / (\left\(\frac{f}{g}\right\)(x)=\frac{x-2-9}{x-3}=\frac{(x-3)(x+3)}{x-3}=x+3), where \(x \neq 3\). Always check that the denominator is not zero.
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If (f(x)=\frac{1}{x-2}) and (g(x)=\frac{1}{x+4}), what is the domain of ((f+g)(x))?
A. \(x \in \mathbb{R}, x \neq 2, x \neq -4\)
हरों (x-2) और (x+4) को शून्य नहीं होना चाहिए, इसलिए \(x \neq 2\) और \(x \neq -4\)। परिमेय फलनों में हर पहले जाँचें। / The denominators (x-2) and (x+4) must not be zero, so \(x \neq 2\) and \(x \neq -4\). In rational functions, check denominators first.
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If (f(x)=2x-2) and (g(x)=x+1), what is the value of ((f-g)(-2))?
A. (9)
((f-g)(x)=2x-2-(x+1)), इसलिए ((f-g)(-2)=8-(-1)=9)। पहले फलन बनाकर फिर मान रखें। / ((f-g)(x)=2x-2-(x+1)), so ((f-g)(-2)=8-(-1)=9). First form the function, then substitute the value.
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If (f(x)=x-2+2x) and (g(x)=x-2-2x), what is ((f+g)(3))?
A. (18)
((f+g)(x)=2x-2), अतः ((f+g)(3)=2\cdot 32=18)। विरोधी पद कट सकते हैं, इसलिए सरल रूप देखें। / ((f+g)(x)=2x-2), hence ((f+g)(3)=2\cdot 32=18). Opposite terms may cancel, so simplify first.
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If (f(x)=x-2-1) and (g(x)=2x+1), find ((fg)(2)).
A. (15)
((fg)(2)=f(2)g(2)=(4-1)(5)=15)। उत्पाद में दोनों फलनों का मान अलग-अलग निकालना आसान है। / ((fg)(2)=f(2)g(2)=(4-1)(5)=15). In a product, finding both values separately is often easier.
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If (f(x)=4x-1) and (g(x)=x-2+2), which statement about ((g+f)(x)) and ((f+g)(x)) is correct?
A. दोनों बराबर हैंBoth are equal
फलनों का योग क्रमविनिमेय है, इसलिए ((f+g)(x)=(g+f)(x))। योग में क्रम बदलने से मान नहीं बदलता। / Addition of functions is commutative, so ((f+g)(x)=(g+f)(x)). Changing order in addition does not change the value.
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If (f(x)=x+6) and (g(x)=x-6), what is the simplified form of ((fg)(x))?
A. \(x^2-36\)
((fg)(x)=(x+6)(x-6)=x-2-36)। यह \(a^2-b^2\) का पैटर्न है। / ((fg)(x)=(x+6)(x-6)=x-2-36). This follows the \(a^2-b^2\) pattern.
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If (f(x)=\frac{x+1}{x-1}) and (g(x)=x-1), what is the domain of ((fg)(x))?
A. \(x \in \mathbb{R}, x \neq 1\)
मूल (f(x)) में \(x-1 \neq 0\), इसलिए \(x \neq 1\), भले ही गुणा के बाद सरलीकरण हो। प्रांत मूल अभिव्यक्तियों से तय करें। / In the original (f(x)), \(x-1 \neq 0\), so \(x \neq 1\), even if simplification occurs after multiplication. Decide the domain from original expressions.
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If (f(x)=\frac{2}{x}) and (g(x)=x-2), what is ((fg)(x)) and its domain?
A. \(2x, x \neq 0\)
((fg)(x)=\frac{2}{x}\cdot x-2=2x), पर मूल हर के कारण \(x \neq 0\)। सरलीकरण के बाद भी मूल प्रतिबंध रखें। / ((fg)(x)=\frac{2}{x}\cdot x-2=2x), but \(x \neq 0\) because of the original denominator. Keep original restrictions after simplification.
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If (f(x)=x-2+4x+4) and (g(x)=x+2), what is (\left\(\frac{f}{g}\right\)(x)) for \(x \neq -2\)?
A. (x+2)
(\frac{x-2+4x+4}{x+2}=\frac{(x+2)2}{x+2}=x+2), जहाँ \(x \neq -2\)। हर का शून्य मान अलग से लिखें। / (\frac{x-2+4x+4}{x+2}=\frac{(x+2)2}{x+2}=x+2), where \(x \neq -2\). Write the zero of the denominator separately.
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If (f(x)=|x|) and (g(x)=x), what is the value of ((f-g)(-5))?
A. (10)
((f-g)(-5)=|-5|-(-5)=5+5=10)। निरपेक्ष मान हमेशा गैर-ऋणात्मक होता है। / ((f-g)(-5)=|-5|-(-5)=5+5=10). Absolute value is always non-negative.
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If (f(x)=x-3) and (g(x)=2x-3-1), what is ((2f-g)(x))?
A. (1)
((2f-g)(x)=2x-3-\(2x^3-1\)=1)। कोष्ठक खोलते समय चिह्न बदलना न भूलें। / ((2f-g)(x)=2x-3-\(2x^3-1\)=1). Do not forget sign changes while opening brackets.
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If (f(x)=x-2) and (g(x)=\sqrt{x}), what is the domain of ((f+g)(x))?
A. \(x \geq 0\)
\(x^2\) सभी वास्तविक (x) के लिए परिभाषित है, लेकिन \(\sqrt{x}\) के लिए \(x \geq 0\) चाहिए। योग का प्रांत प्रतिच्छेद है। / \(x^2\) is defined for all real (x), but \(\sqrt{x}\) needs \(x \geq 0\). The domain of the sum is the intersection.
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If (f(x)=\frac{1}{x}) and (g(x)=\sqrt{x-2}), what is the domain of ((fg)(x))?
A. \(x \geq 2\)
\(\sqrt{x-2}\) के लिए \(x \geq 2\) और \(\frac{1}{x}\) के लिए \(x \neq 0\); प्रतिच्छेद \(x \geq 2\) है। सबसे कड़ा प्रतिबंध पहचानें। / \(\sqrt{x-2}\) requires \(x \geq 2\) and \(\frac{1}{x}\) requires \(x \neq 0\); the intersection is \(x \geq 2\). Identify the strongest restriction.
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If (f(x)=x-1) and (g(x)=x+1), what is the value of (\left\(\frac{f}{g}\right\)(0))?
A. (-1)
(\left\(\frac{f}{g}\right\)(0)=\frac{f(0)}{g(0)}=\frac{-1}{1}=-1)। भाग में हर का मान शून्य है या नहीं, पहले देखें। / (\left\(\frac{f}{g}\right\)(0)=\frac{f(0)}{g(0)}=\frac{-1}{1}=-1). In division, first check whether the denominator value is zero.
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If (f(x)=x-2-4) and (g(x)=x-2+4), what is ((g-f)(x))?
A. (8)
((g-f)(x)=\(x^2+4\)-\(x^2-4\)=8)। समान \(x^2\) पद कट जाते हैं। / ((g-f)(x)=\(x^2+4\)-\(x^2-4\)=8). The like \(x^2\) terms cancel.
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If (f(x)=2x+3), (g(x)=x-4), and (h(x)=x-2), what is ((f+g-h)(x))?
A. \(-x^2+3x-1\)
((f+g-h)(x)=2x+3+x-4-x-2=-x-2+3x-1)। कई फलनों में पदों को क्रम से मिलाएँ। / ((f+g-h)(x)=2x+3+x-4-x-2=-x-2+3x-1). With many functions, combine terms in order.
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If (f(x)=x-2+x) and (g(x)=x), what is (\left\(\frac{f}{g}\right\)(x)) for \(x \neq 0\)?
A. (x+1)
(\frac{x-2+x}{x}=\frac{x(x+1)}{x}=x+1), जहाँ \(x \neq 0\)। साझा गुणनखंड काटते समय प्रतिबंध न भूलें। / (\frac{x-2+x}{x}=\frac{x(x+1)}{x}=x+1), where \(x \neq 0\). Do not forget the restriction while cancelling a common factor.
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If (f(x)=\frac{x}{x+2}) and (g(x)=\frac{2}{x+2}), what is the simplified form of ((f+g)(x))?
A. (1), \(x \neq -2\)
((f+g)(x)=\frac{x+2}{x+2}=1), लेकिन \(x+2 \neq 0\)। सरल उत्तर के साथ प्रांत का प्रतिबंध लिखना जरूरी है। / ((f+g)(x)=\frac{x+2}{x+2}=1), but \(x+2 \neq 0\). It is important to write the domain restriction with the simplified answer.
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If (f(x)=3) and (g(x)=x-2-7x), what is ((fg)(x))?
A. \(3x^2-21x\)
((fg)(x)=3\(x^2-7x\)=3x-2-21x)। स्थिर फलन से गुणा करते समय हर पद पर गुणा करें। / ((fg)(x)=3\(x^2-7x\)=3x-2-21x). When multiplying by a constant function, multiply every term.
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If (f(x)=x-2+1), what is ((f+f)(x))?
A. \(2x^2+2\)
((f+f)(x)=2f(x)=2\(x^2+1\)=2x-2+2)। समान फलन जोड़ने पर (2f(x)) बनता है। / ((f+f)(x)=2f(x)=2\(x^2+1\)=2x-2+2). Adding the same function gives (2f(x)).
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If (f(x)=x-2), what is ((ff)(x))?
A. \(x^2-4x+4\)
((ff)(x)=f(x)\cdot f(x)=(x-2)2=x-2-4x+4)। ((ff)(x)) को (f(f(x))) न समझें। / ((ff)(x)=f(x)\cdot f(x)=(x-2)2=x-2-4x+4). Do not confuse ((ff)(x)) with (f(f(x))).
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If (f(x)=\sin x) and (g(x)=\cos x), what is (\(f^2+g^2\)(x))?
A. (1)
(\(f^2+g^2\)(x)=\sin-2 x+\cos-2 x=1)। त्रिकोणमितीय पहचान याद रखें। / (\(f^2+g^2\)(x)=\sin-2 x+\cos-2 x=1). Remember the trigonometric identity.
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If (f(x)=x-2) and (g(x)=2x), what is the value of ((f+g)(a))?
A. \(a^2+2a\)
((f+g)(a)=f(a)+g(a)=a-2+2a)। चर बदलने पर हर (x) की जगह वही मान रखें। / ((f+g)(a)=f(a)+g(a)=a-2+2a). When the variable changes, replace every (x) by that value.
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If (f(x)=x-2-2x) and (g(x)=x-2+2x), what is ((fg)(0))?
A. (0)
(f(0)=0) और (g(0)=0), इसलिए ((fg)(0)=0)। उत्पाद में कोई एक मान शून्य हो तो उत्पाद शून्य होता है। / (f(0)=0) and (g(0)=0), so ((fg)(0)=0). If one value in a product is zero, the product is zero.
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If (f(x)=\frac{1}{x-1}) and (g(x)=x+3), what is the domain of (\left\(\frac{g}{f}\right\)(x))?
A. \(x \in \mathbb{R}, x \neq 1\)
(f(x)) के लिए \(x \neq 1\) चाहिए और (f(x)=\frac{1}{x-1}) कभी शून्य नहीं होता। इसलिए केवल \(x \neq 1\) प्रतिबंध है। / (f(x)) needs \(x \neq 1\), and (f(x)=\frac{1}{x-1}) is never zero. So the only restriction is \(x \neq 1\).
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If (f(x)=x-2-16) and (g(x)=x+4), what is (\left\(\frac{f}{g}\right\)(-1)) for \(x \neq -4\)?
A. (3)
(\left\(\frac{f}{g}\right\)(x)=\frac{(x-4)(x+4)}{x+4}=x-4), इसलिए (x=-1) पर मान (-5) होना चाहिए; पर सही जाँच में \(\frac{-15}{3}=-5\)। / (\left\(\frac{f}{g}\right\)(x)=\frac{(x-4)(x+4)}{x+4}=x-4), so at (x=-1) the value is (-5). Check directly as \(\frac{-15}{3}=-5\).
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A. (-5)
(\frac{x-2-16}{x+4}=\frac{(x-4)(x+4)}{x+4}=x-4), अतः (-1-4=-5)। सरलीकरण के बाद सही मान रखें। / (\frac{x-2-16}{x+4}=\frac{(x-4)(x+4)}{x+4}=x-4), hence (-1-4=-5). Substitute carefully after simplification.
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If (f(x)=\sqrt{9-x-2}) and (g(x)=x+1), what is the domain of ((f+g)(x))?
A. \(-3 \leq x \leq 3\)
वर्गमूल के लिए \(9-x^2 \geq 0\), इसलिए \(-3 \leq x \leq 3\)। रैखिक फलन कोई अतिरिक्त प्रतिबंध नहीं देता। / For the square root, \(9-x^2 \geq 0\), so \(-3 \leq x \leq 3\). The linear function gives no extra restriction.
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If (f(x)=2x-5) and (g(x)=5-2x), what is ((f+g)(x))?
A. (0)
((f+g)(x)=2x-5+5-2x=0)। विपरीत पदों को पहचानना तेज समाधान देता है। / ((f+g)(x)=2x-5+5-2x=0). Recognising opposite terms gives a quick solution.
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If (f(x)=x-2+3x+2) and (g(x)=x+1), what is (\left\(\frac{f}{g}\right\)(2)) for \(x \neq -1\)?
A. (4)
(\frac{x-2+3x+2}{x+1}=\frac{(x+1)(x+2)}{x+1}=x+2), इसलिए (x=2) पर मान (4) है। गुणनखंड बनाकर समय बचाएँ। / (\frac{x-2+3x+2}{x+1}=\frac{(x+1)(x+2)}{x+1}=x+2), so at (x=2) the value is (4). Factorisation saves time.
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If (f(x)=x-2) and (g(x)=x+2), what is the value of ((f+g)(x)-(fg)(x)) at (x=1)?
A. (1)
(f(1)=1) और (g(1)=3), इसलिए ((f+g)(1)-(fg)(1)=4-3=1)। ऐसे प्रश्नों में पहले अलग-अलग मान निकालें। / (f(1)=1) and (g(1)=3), so ((f+g)(1)-(fg)(1)=4-3=1). In such questions, find individual values first.
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If (f(x)=x-2+2) and (g(x)=2x), what are the solutions of ((f-g)(x)=0)?
A. कोई वास्तविक हल नहींNo real solution
((f-g)(x)=x-2-2x+2=(x-1)2+1), जो कभी (0) नहीं होता। वर्ग पूर्ण करने से वास्तविक हल जल्दी दिखता है। / ((f-g)(x)=x-2-2x+2=(x-1)2+1), which is never (0). Completing the square quickly shows real solutions.
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If (f(x)=x-2-1) and (g(x)=x-2+1), what is the range of ((g-f)(x))?
A. ({2})
((g-f)(x)=2), इसलिए इसका परिसर केवल ({2}) है। स्थिर फलन का परिसर एक-सदस्यीय होता है। / ((g-f)(x)=2), so its range is only ({2}). The range of a constant function is a singleton set.
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If (f(x)=x+1) and (g(x)=x-2), what is the degree of ((fg)(x))?
A. (3)
((fg)(x)=(x+1)x-2=x-3+x-2), इसलिए घात (3) है। बहुपद गुणा में उच्चतम घात देखें। / ((fg)(x)=(x+1)x-2=x-3+x-2), so the degree is (3). In polynomial products, check the highest power.
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If (f(x)=x-2-5x) and (g(x)=3x), what are the zeroes of ((f+g)(x))?
A. (0) और (2)(0) and (2)
((f+g)(x)=x-2-2x=x(x-2)), इसलिए शून्यक (0) और (2) हैं। शून्यक के लिए फलन को (0) के बराबर रखें। / ((f+g)(x)=x-2-2x=x(x-2)), so the zeroes are (0) and (2). For zeroes, equate the function to (0).
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If (f(x)=\frac{x-2}{x+1}) and (g(x)=\frac{x+3}{x+1}), what is ((f-g)(x))?
A. \(\frac{-5}{x+1}, x \neq -1\)
समान हर होने पर अंश घटाएँ: (\frac{x-2-(x+3)}{x+1}=\frac{-5}{x+1})। हर (x+1) के कारण \(x \neq -1\)। / With the same denominator, subtract numerators: (\frac{x-2-(x+3)}{x+1}=\frac{-5}{x+1}). Because of denominator (x+1), \(x \neq -1\).
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If (f(x)=2x+1) and (g(x)=x-2), what is ((3f+2g)(x))?
A. (8x-1)
((3f+2g)(x)=3(2x+1)+2(x-2)=8x-1)। गुणांक को पूरे फलन से गुणा करें। / ((3f+2g)(x)=3(2x+1)+2(x-2)=8x-1). Multiply the coefficient by the whole function.
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If (f(x)=x-2+6x+9) and (g(x)=x+3), what is (\left\(\frac{f}{g}\right\)(0)) for \(x \neq -3\)?
A. (3)
(\frac{x-2+6x+9}{x+3}=\frac{(x+3)2}{x+3}=x+3), अतः (x=0) पर मान (3) है। पूर्ण वर्ग पहचान उपयोगी है। / (\frac{x-2+6x+9}{x+3}=\frac{(x+3)2}{x+3}=x+3), hence the value at (x=0) is (3). The perfect square identity is useful.
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If (f(x)=\frac{1}{\sqrt{x-1}}) and (g(x)=x), what is the domain of ((f+g)(x))?
A. (x>1)
हर में \(\sqrt{x-1}\) है, इसलिए (x-1>0) होना चाहिए। अतः प्रांत (x>1) है। / The denominator has \(\sqrt{x-1}\), so (x-1>0) is required. Hence the domain is (x>1).
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If (f(x)=x-2+1) and (g(x)=x-2-1), what is ((f-g)(5))?
A. (2)
((f-g)(x)=\(x^2+1\)-\(x^2-1\)=2), इसलिए किसी भी (x) पर मान (2) है। पहले सरल करने से गणना कम होती है। / ((f-g)(x)=\(x^2+1\)-\(x^2-1\)=2), so at any (x) the value is (2). Simplifying first reduces calculation.
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If (f(x)=2x) and (g(x)=\frac{1}{2}x), what is ((fg)(x))?
A. \(x^2\)
((fg)(x)=2x\cdot \frac{1}{2}x=x-2)। संख्यात्मक गुणांकों को पहले गुणा करें। / ((fg)(x)=2x\cdot \frac{1}{2}x=x-2). Multiply numerical coefficients first.
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If (f(x)=x-2+x+1) and (g(x)=x-2-x+1), what is ((f-g)(x))?
A. (2x)
((f-g)(x)=x-2+x+1-\(x^2-x+1\)=2x)। कोष्ठक खोलकर चिह्न सावधानी से बदलें। / ((f-g)(x)=x-2+x+1-\(x^2-x+1\)=2x). Open brackets and change signs carefully.
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If (f(x)=\frac{x+2}{x-3}) and (g(x)=x-3), what is ((fg)(4))?
A. (6)
((fg)(x)=\frac{x+2}{x-3}(x-3)=x+2), जहाँ \(x \neq 3\)। (x=4) पर मान (6) है। / ((fg)(x)=\frac{x+2}{x-3}(x-3)=x+2), where \(x \neq 3\). At (x=4), the value is (6).
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If (f(x)=\sqrt{x}) and (g(x)=\sqrt{x}), what are ((fg)(x)) and its domain?
A. \(x, x \geq 0\)
((fg)(x)=\sqrt{x}\cdot\sqrt{x}=x), लेकिन दोनों वर्गमूलों के लिए \(x \geq 0\)। सरल रूप से प्रांत नहीं बदलता। / ((fg)(x)=\sqrt{x}\cdot\sqrt{x}=x), but both square roots require \(x \geq 0\). The simplified form does not change the original domain.
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If (f(x)=x-2-3x+2) and (g(x)=x-1), what is (\left\(\frac{f}{g}\right\)(3)) for \(x \neq 1\)?
A. (1)
(\frac{x-2-3x+2}{x-1}=\frac{(x-1)(x-2)}{x-1}=x-2), अतः (x=3) पर मान (1) है। गुणनखंडन से भाग सरल होता है। / (\frac{x-2-3x+2}{x-1}=\frac{(x-1)(x-2)}{x-1}=x-2), so at (x=3) the value is (1). Factorisation makes division simple.
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If (f(x)=x+4) and (g(x)=4-x), what is ((fg)(x)) equal to?
A. \(16-x^2\)
((fg)(x)=(x+4)(4-x)=16-x-2)। इसे ((4+x)(4-x)) के रूप में देखें। / ((fg)(x)=(x+4)(4-x)=16-x-2). View it as ((4+x)(4-x)).
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