algebra-real-functions se related questions ko ek jagah revise karein. Har question me bilingual content, answer feedback aur explanation available hai.
306 questions tagged with algebra-real-functions.
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)=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|) 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)=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)=\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+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)=2x-2-1), what is ((3f)(x))?
A. \(6x^2-3\)
((3f)(x)=3f(x)=3\(2x^2-1\)=6x-2-3)। scalar को पूरी function से multiply करें। / ((3f)(x)=3f(x)=3\(2x^2-1\)=6x-2-3). Multiply the scalar by the whole function.
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If (f(x)=x+1), (g(x)=x-1), and (h(x)=x-2), what is ((f+g+h)(x))?
A. \(x^2+2x\)
((f+g+h)(x)=(x+1)+(x-1)+x-2=x-2+2x)। constants (1) और (-1) cancel हो जाते हैं। / ((f+g+h)(x)=(x+1)+(x-1)+x-2=x-2+2x). The constants (1) and (-1) cancel out.
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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) and (g(x)=2x+1), what is the value of ((f+g)(-1))?
A. (0)
((f+g)(-1)=f(-1)+g(-1)=1+(-1)=0)। पहले functions में value रखें, फिर जोड़ें। / ((f+g)(-1)=f(-1)+g(-1)=1+(-1)=0). First substitute the value in each function, then add.
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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)=\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)=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)=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+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)=2x+3) and (g(x)=x-2-1), what is ((f+g)(x))?
A. \(x^2+2x+2\)
((f+g)(x)=f(x)+g(x)), इसलिए \(2x+3+x^2-1=x^2+2x+2\)। परीक्षा में like terms को साथ जोड़ें। / ((f+g)(x)=f(x)+g(x)), so \(2x+3+x^2-1=x^2+2x+2\). In exams, combine like terms carefully.
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If (f(x)=x-2-1) and (g(x)=x-2+1), what is ((g-f)(x))?
A. (2)
((g-f)(x)=\(x^2+1\)-\(x^2-1\)=2)। घटाते समय दूसरे कोष्ठक के चिह्न बदलते हैं। / ((g-f)(x)=\(x^2+1\)-\(x^2-1\)=2). While subtracting, the signs inside the second bracket change.
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If (f(x)=\sqrt{x+2}) and (g(x)=\frac{1}{x-1}), what is the real domain of ((f+g)(x))?
A. \([-2,\infty\)-{1})
\(\sqrt{x+2}\) के लिए \(x\ge -2\) और \(\frac{1}{x-1}\) के लिए \(x\ne 1\) चाहिए। दोनों शर्तों को साथ लगाएं। / For \(\sqrt{x+2}\), \(x\ge -2\), and for \(\frac{1}{x-1}\), \(x\ne 1\). Apply both conditions together.
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If (f(x)=x+2) and (g(x)=x-2-2x), what is ((fg)(2))?
A. (0)
(f(2)=4) और (g(2)=0), इसलिए ((fg)(2)=4\cdot 0=0)। गुणन में किसी एक मान के (0) होने पर उत्पाद (0) होता है। / (f(2)=4) and (g(2)=0), so ((fg)(2)=4\cdot 0=0). In multiplication, if one value is (0), the product is (0).
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If (f(x)=5x+2) and (g(x)=2x-7), what is the value of ((f-g)(1))?
A. (12)
(f(1)=7) और (g(1)=-5), इसलिए ((f-g)(1)=7-(-5)=12)। ऋणात्मक मान घटाते समय सावधानी रखें। / (f(1)=7) and (g(1)=-5), so ((f-g)(1)=7-(-5)=12). Be careful while subtracting a negative value.
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If (f(x)=x-2+4x) and (g(x)=2x-2-x), what is ((f+g)(x))?
A. \(3x^2+3x\)
((f+g)(x)=x-2+4x+2x-2-x=3x-2+3x)। समान घात वाले पदों को ही जोड़ें। / ((f+g)(x)=x-2+4x+2x-2-x=3x-2+3x). Add only like terms with the same power.
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If (f(x)=x-2+1) and (g(x)=x-2+1), then (\left\(\frac{f}{g}\right\)(x)) equals which expression?
A. (1)
क्योंकि (f(x)=g(x)), इसलिए (\frac{f(x)}{g(x)}=1)। यहाँ \(x^2+1\) कभी (0) नहीं होता, इसलिए सभी वास्तविक (x) मान्य हैं। / Since (f(x)=g(x)), (\frac{f(x)}{g(x)}=1). Here \(x^2+1\) is never (0), so all real (x) are valid.
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If (f(x)=2x+3) and (g(x)=2x-3), what is ((fg)(x))?
A. \(4x^2-9\)
((2x+3)(2x-3)=(2x)2-32=4x-2-9)। यह अंतर वर्ग का सीधा प्रयोग है। / ((2x+3)(2x-3)=(2x)2-32=4x-2-9). This is a direct use of difference of squares.
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If (f(x)=x+1) and (g(x)=x-2+1), what is the domain of (\left\(\frac{g}{f}\right\)(x))?
A. \(\mathbb{R}-{-1}\)
\(\frac{g}{f}\) में हर (f(x)=x+1) है, इसलिए (x=-1) हटेगा। भागफल में हर वाले फलन को ध्यान से देखें। / In \(\frac{g}{f}\), the denominator is (f(x)=x+1), so (x=-1) is excluded. Identify the denominator function carefully.
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If (f(x)=\frac{1}{x+4}) and (g(x)=x-2), what is the domain of ((f-g)(x))?
A. \(\mathbb{R}-{-4}\)
\(\frac{1}{x+4}\) में (x=-4) पर हर (0) है। अंतर का डोमेन भी \(\mathbb{R}-{-4}\) होगा। / In \(\frac{1}{x+4}\), the denominator is (0) at (x=-4). The domain of the difference is also \(\mathbb{R}-{-4}\).
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If (f(x)=x-2-2x) and (g(x)=x), what is ((f+3g)(x))?
A. \(x^2+x\)
((f+3g)(x)=x-2-2x+3x=x-2+x)। रैखिक संयोजन में गुणक सही लगाएं। / ((f+3g)(x)=x-2-2x+3x=x-2+x). Apply the multiplier correctly in a linear combination.
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If (f(x)=0) and (g(x)=x+7), what is ((f+g)(x))?
A. (x+7)
शून्य फलन जोड़ने से दूसरा फलन वैसा ही रहता है। इसलिए ((f+g)(x)=x+7)। / Adding the zero function leaves the other function unchanged. Hence ((f+g)(x)=x+7).
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If (f(x)=1) and (g(x)=x-2+2x), what is ((fg)(x))?
A. \(x^2+2x\)
(1) से गुणा करने पर फलन नहीं बदलता। इसलिए ((fg)(x)=x-2+2x) है। / Multiplying by (1) does not change the function. Therefore ((fg)(x)=x-2+2x).
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If (f(x)=x-2) and (g(x)=x-2), what is ((f-g)(x))?
A. (0)
दोनों फलन समान हैं, इसलिए उनका अंतर हर (x) पर (0) है। समान फलनों का अंतर शून्य फलन होता है। / The two functions are identical, so their difference is (0) for every (x). The difference of identical functions is the zero function.
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If (f(x)=x-2+2) and (g(x)=2x), which completed-square-like form represents ((f+g)(x))?
A. ((x+1)2+1)
((f+g)(x)=x-2+2x+2=(x+1)2+1)। पूर्ण वर्ग पहचानना ग्राफ और रेंज में भी मदद करता है। / ((f+g)(x)=x-2+2x+2=(x+1)2+1). Recognizing a completed square also helps in graphs and range.
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