Level 37 • 34/50 questions • 35 seconds per question.
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If (f(x)=x-2+4x) and (g(x)=3x-5), find ((f-g)(x)).
A. \(x^2+x+5\)
((f-g)(x)=f(x)-g(x)), अतः (x-2+4x-(3x-5)=x-2+x+5)। घटाते समय bracket का sign बदलता है। / ((f-g)(x)=f(x)-g(x)), hence (x-2+4x-(3x-5)=x-2+x+5). While subtracting, change the signs inside the bracket.
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If (f(x)=x+2) and (g(x)=x-3), what is ((fg)(x))?
A. \(x^2-x-6\)
((fg)(x)=f(x)g(x)), इसलिए ((x+2)(x-3)=x-2-x-6)। गुणा में दोनों terms को distribute करें। / ((fg)(x)=f(x)g(x)), so ((x+2)(x-3)=x-2-x-6). In multiplication, distribute both terms properly.
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If (f(x)=x-2-9) and (g(x)=x-3), what is the simplified form of (\left\(\frac{f}{g}\right\)(x))?
A. \(x+3,\ x \neq 3\)
(\frac{x-2-9}{x-3}=\frac{(x-3)(x+3)}{x-3}=x+3), पर (x=3) allowed नहीं है। cancel करने के बाद भी मूल denominator याद रखें। / (\frac{x-2-9}{x-3}=\frac{(x-3)(x+3)}{x-3}=x+3), but (x=3) is not allowed. Even after cancellation, remember the original denominator.
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If (f(x)=\sqrt{x-1}) and (g(x)=\sqrt{5-x}), what is the domain of ((f+g)(x))?
A. ([1,5])
दोनों square roots defined होने चाहिए, इसलिए \(x-1\geq 0\) और \(5-x\geq 0\), जिससे \(x\in[1,5]\)। sum के लिए domains का intersection लें। / Both square roots must be defined, so \(x-1\geq 0\) and \(5-x\geq 0\), giving \(x\in[1,5]\). For a sum, take the intersection of domains.
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If (f(x)=\frac{1}{x-2}) and (g(x)=x+4), what is the domain of ((f+g)(x))?
A. \(\mathbb{R}-{2}\)
(f(x)) में denominator (x-2) शून्य नहीं हो सकता, इसलिए \(x\neq 2\)। polynomial (g(x)) सभी real (x) पर defined है। / In (f(x)), the denominator (x-2) cannot be zero, so \(x\neq 2\). The polynomial (g(x)) is defined for all real (x).
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If (f(x)=3x-2) and (g(x)=x-2+1), what is ((g-f)(2))?
A. (1)
(g(2)=5) और (f(2)=4), इसलिए ((g-f)(2)=5-4=1)। order को ध्यान से पढ़ें क्योंकि (g-f) और (f-g) अलग होते हैं। / (g(2)=5) and (f(2)=4), so ((g-f)(2)=5-4=1). Read the order carefully because (g-f) and (f-g) are different.
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If (f(x)=x-2+2x) and (g(x)=x), what is the domain of (\left\(\frac{f}{g}\right\)(x))?
A. \(\mathbb{R}-{0}\)
quotient में denominator (g(x)=x) शून्य नहीं होना चाहिए, इसलिए \(x\neq 0\)। simplify करने से पहले restriction लगाएं। / In a quotient, the denominator (g(x)=x) must not be zero, so \(x\neq 0\). Apply the restriction before simplifying.
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If (f(x)=\frac{x+1}{x-4}) and (g(x)=x-2), what is the domain of ((fg)(x))?
A. \(\mathbb{R}-{4}\)
product का domain दोनों domains का intersection होता है, और (f(x)) में \(x\neq 4\)। (g(x)) polynomial है, इसलिए कोई नई restriction नहीं है। / The domain of a product is the intersection of both domains, and in (f(x)), \(x\neq 4\). Since (g(x)) is a polynomial, it adds no new restriction.
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If (f(x)=x-2-4) and (g(x)=x+2), which statement is correct about (\left\(\frac{f}{g}\right\)(-2))?
A. यह defined नहीं हैIt is not defined
(g(-2)=0), इसलिए quotient (\left\(\frac{f}{g}\right\)(-2)) defined नहीं है। numerator भी (0) हो तो भी denominator (0) allowed नहीं होता। / (g(-2)=0), so the quotient (\left\(\frac{f}{g}\right\)(-2)) is not defined. Even if the numerator is (0), a zero denominator is not allowed.
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If (f(x)=|x|) and (g(x)=x), what is the value of ((f-g)(-3))?
A. (6)
(f(-3)=3) और (g(-3)=-3), इसलिए ((f-g)(-3)=3-(-3)=6)। modulus में negative input का output positive होता है। / (f(-3)=3) and (g(-3)=-3), so ((f-g)(-3)=3-(-3)=6). A modulus gives a non-negative output.
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If (f(x)=2x+5) and (g(x)=5-2x), what is ((f+g)(x)) equal to?
A. (10)
((f+g)(x)=2x+5+5-2x=10), इसलिए यह constant function है। opposite terms cancel हों तो ध्यान से लिखें। / ((f+g)(x)=2x+5+5-2x=10), so it is a constant function. When opposite terms cancel, write the remaining constant carefully.
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If (f(x)=x-2) and (g(x)=-x-2), which function is ((f+g)(x))?
A. zero function (0)
((f+g)(x)=x-2-x-2=0), इसलिए यह zero function है। समान opposite functions का sum zero देता है। / ((f+g)(x)=x-2-x-2=0), so it is the zero function. The sum of equal opposite functions gives zero.
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If (f(x)=x+1) and (g(x)=\frac{1}{x+1}), what is the value of ((fg)(x))?
A. \(1,\ x\neq -1\)
((fg)(x)=(x+1)\cdot\frac{1}{x+1}=1), पर (x=-1) allowed नहीं है। reciprocal वाली function में denominator check करें। / ((fg)(x)=(x+1)\cdot\frac{1}{x+1}=1), but (x=-1) is not allowed. Always check the denominator in a reciprocal function.
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If (f(x)=\sqrt{x+2}) and (g(x)=\frac{1}{x-1}), what is the domain of ((f+g)(x))?
A. \([-2,\infty\)-{1})
\(\sqrt{x+2}\) के लिए \(x\geq -2\) और \(\frac{1}{x-1}\) के लिए \(x\neq 1\)। intersection से domain \([-2,\infty\)-{1}) मिलता है। / For \(\sqrt{x+2}\), \(x\geq -2\), and for \(\frac{1}{x-1}\), \(x\neq 1\). Their intersection gives \([-2,\infty\)-{1}).
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If (f(x)=x-2-2x) and (g(x)=x-2), what is the simplified form of (\left\(\frac{f}{g}\right\)(x))?
A. \(x,\ x\neq 2\)
(\frac{x-2-2x}{x-2}=\frac{x(x-2)}{x-2}=x), पर \(x\neq 2\)। cancelled factor से मूल restriction खत्म नहीं होती। / (\frac{x-2-2x}{x-2}=\frac{x(x-2)}{x-2}=x), but \(x\neq 2\). A cancelled factor does not remove the original restriction.
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If (f(x)=x-2+3) and (g(x)=2x), what is ((fg)(1))?
A. (8)
(f(1)=4) और (g(1)=2), इसलिए ((fg)(1)=4\cdot2=8)। product में values multiply होती हैं। / (f(1)=4) and (g(1)=2), so ((fg)(1)=4\cdot2=8). In a product, the function values are multiplied.
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If (f(x)=x-4) and (g(x)=x-2+2x), what is ((g+f)(x))?
A. \(x^2+3x-4\)
((g+f)(x)=g(x)+f(x)=x-2+2x+x-4=x-2+3x-4)। addition में order बदलने से answer नहीं बदलता। / ((g+f)(x)=g(x)+f(x)=x-2+2x+x-4=x-2+3x-4). In addition, changing the order does not change the answer.
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If (f(x)=\frac{2}{x+3}) and (g(x)=\frac{1}{x-1}), what is the domain of ((f+g)(x))?
A. \(\mathbb{R}-{-3,1}\)
दोनों denominators non-zero होने चाहिए, इसलिए \(x\neq -3\) और \(x\neq 1\)। rational functions में सभी denominators check करें। / Both denominators must be non-zero, so \(x\neq -3\) and \(x\neq 1\). For rational functions, check all denominators.
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If (f(x)=x-2-1) and (g(x)=x-2+1), what is ((fg)(x))?
A. \(x^4-1\)
(\(x^2-1\)\(x^2+1\)=x-4-1) difference of squares से मिलता है। formula पहचानने से calculation तेज होती है। / (\(x^2-1\)\(x^2+1\)=x-4-1) by difference of squares. Recognizing the formula makes calculation faster.
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If (f(x)=x+2) and (g(x)=x-2+4x+4), what is (\left\(\frac{g}{f}\right\)(x))?
A. \(x+2,\ x\neq -2\)
(g(x)=(x+2)2), इसलिए (\frac{g(x)}{f(x)}=x+2), लेकिन \(x\neq -2\)। quotient में denominator की zero value हटानी जरूरी है। / (g(x)=(x+2)2), so (\frac{g(x)}{f(x)}=x+2), but \(x\neq -2\). In a quotient, remove the zero value of the denominator.
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If (f(x)=\sqrt{x}) and (g(x)=x-2), what is the domain of ((fg)(x))?
A. \([0,\infty\))
\(\sqrt{x}\) के लिए \(x\geq 0\) चाहिए और (g(x)) सभी real (x) पर defined है। इसलिए product का domain \([0,\infty\)) है। / \(\sqrt{x}\) needs \(x\geq 0\), and (g(x)) is defined for all real (x). Therefore, the product domain is \([0,\infty\)).
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If (f(x)=x-2+2) and (g(x)=4x), how can ((f+g)(x)) be written in complete square form?
A. ((x+2)2-2)
((f+g)(x)=x-2+4x+2=(x+2)2-2)। complete square में आधे coefficient का square जोड़कर घटाते हैं। / ((f+g)(x)=x-2+4x+2=(x+2)2-2). In completing the square, add and subtract the square of half the coefficient.
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If (f(x)=\frac{1}{x}) and (g(x)=x), what is the domain of ((f+g)(x))?
A. \(\mathbb{R}-{0}\)
\(\frac{1}{x}\) में (x=0) पर denominator शून्य हो जाता है। इसलिए sum का domain \(\mathbb{R}-{0}\) है। / In \(\frac{1}{x}\), the denominator becomes zero at (x=0). Therefore, the domain of the sum is \(\mathbb{R}-{0}\).
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If (f(x)=x+3) and (g(x)=2x-6), what is ((fg)(0))?
A. (-18)
(f(0)=3) और (g(0)=-6), इसलिए ((fg)(0)=3\cdot(-6)=-18)। sign multiplication में गलती न करें। / (f(0)=3) and (g(0)=-6), so ((fg)(0)=3\cdot(-6)=-18). Do not make a sign error in multiplication.
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If (f(x)=x-2+5x+6) and (g(x)=x+2), what is the simplified form of (\left\(\frac{f}{g}\right\)(x))?
A. \(x+3,\ x\neq -2\)
(x-2+5x+6=(x+2)(x+3)), इसलिए quotient (x+3) है, पर \(x\neq -2\)। factorization के साथ restriction लिखना जरूरी है। / (x-2+5x+6=(x+2)(x+3)), so the quotient is (x+3), but \(x\neq -2\). Write the restriction along with the factorization.
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If (f(x)=x-2-4x+4) and (g(x)=x-2), what is the value of ((fg)(x)) at (x=2)?
A. (0)
(f(2)=0) और (g(2)=0), इसलिए ((fg)(2)=0\cdot0=0)। product में कोई factor zero हो तो product zero होता है। / (f(2)=0) and (g(2)=0), so ((fg)(2)=0\cdot0=0). If any factor is zero, the product is zero.
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If (f(x)=x-2) and (g(x)=2x), what is the minimum value of ((f+g)(x))?
A. (-1)
((f+g)(x)=x-2+2x=(x+1)2-1), इसलिए minimum value (-1) है। complete square से quadratic का minimum आसानी से मिलता है। / ((f+g)(x)=x-2+2x=(x+1)2-1), so the minimum value is (-1). Completing the square helps find the minimum of a quadratic.
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If (f(x)=x-2+1) and (g(x)=x-2-3), what is ((f-g)(5))?
A. (4)
((f-g)(x)=\(x^2+1\)-\(x^2-3\)=4), इसलिए (x=5) पर भी value (4) है। पहले simplify करने से calculation कम होती है। / ((f-g)(x)=\(x^2+1\)-\(x^2-3\)=4), so at (x=5) the value is also (4). Simplifying first reduces calculation.
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If (f(x)=\sqrt{x-4}) and (g(x)=\sqrt{x+1}), what is the domain of ((fg)(x))?
A. \([4,\infty\))
\(\sqrt{x-4}\) के लिए \(x\geq 4\) और \(\sqrt{x+1}\) के लिए \(x\geq -1\)। intersection \([4,\infty\)) है। / For \(\sqrt{x-4}\), \(x\geq 4\), and for \(\sqrt{x+1}\), \(x\geq -1\). The intersection is \([4,\infty\)).
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If (f(x)=\frac{1}{x-2-1}) and (g(x)=x+2), which values are excluded from the domain of ((f+g)(x))?
A. (x=-1) और (x=1)(x=-1) and (x=1)
(x-2-1=(x-1)(x+1)), इसलिए denominator zero (x=1) और (x=-1) पर होता है। domain में denominator zero values हटती हैं। / (x-2-1=(x-1)(x+1)), so the denominator is zero at (x=1) and (x=-1). Values making the denominator zero are excluded from the domain.
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If (f(x)=x-2-5x+6) and (g(x)=x-3), what is (\left\(\frac{f}{g}\right\)(2))?
A. (0)
(f(2)=0) और (g(2)=-1), इसलिए quotient (0) है। denominator non-zero हो तो numerator (0) allowed है। / (f(2)=0) and (g(2)=-1), so the quotient is (0). A zero numerator is allowed when the denominator is non-zero.
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If (f(x)=x+1) and (g(x)=x-2+1), what is the degree of ((f+g)(x))?
A. (2)
((f+g)(x)=x-2+x+2), जिसकी highest power (2) है। polynomial की degree highest exponent से मिलती है। / ((f+g)(x)=x-2+x+2), whose highest power is (2). The degree of a polynomial is given by its highest exponent.
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If (f(x)=\frac{x+2}{x-1}) and (g(x)=x-1), what is the simplified form of ((fg)(x))?
A. \(x+2,\ x\neq 1\)
((fg)(x)=\frac{x+2}{x-1}(x-1)=x+2), पर मूल (f(x)) के कारण \(x\neq 1\)। simplification के साथ domain restriction लिखें। / ((fg)(x)=\frac{x+2}{x-1}(x-1)=x+2), but because of the original (f(x)), \(x\neq 1\). Write the domain restriction with the simplification.
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If (f(x)=x-2+2x+1) and (g(x)=x+1), which statement is correct for (\left\(\frac{f}{g}\right\)(-1))?
A. defined नहीं हैnot defined
(g(-1)=0), इसलिए quotient (-1) पर defined नहीं है। cancel करने से पहले original denominator check करें। / (g(-1)=0), so the quotient is not defined at (-1). Check the original denominator before cancellation.
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