(\left\(\frac{f}{g}\right\)(x)=\frac{x-2-4}{x-2}=x+2) but \(x\ne 2\). Even after simplifying do not forget the original denominator restriction.
Step 2
Why this answer is correct
The correct answer is A. (x+2), \(x\ne 2\). (\left\(\frac{f}{g}\right\)(x)=\frac{x-2-4}{x-2}=x+2) but \(x\ne 2\). Even after simplifying do not forget the original denominator restriction.
Step 3
Exam Tip
(\left\(\frac{f}{g}\right\)(x)=\frac{x-2-4}{x-2}=x+2) पर \(x\ne 2\)। सरल करने के बाद भी मूल हर का प्रतिबंध न भूलें।
Both denominators cannot be zero so \(x\ne 1\) and \(x\ne -1\). For addition the common domain is the intersection of both domains.
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}-{-1,1}). Both denominators cannot be zero so \(x\ne 1\) and \(x\ne -1\). For addition the common domain is the intersection of both domains.
Step 3
Exam Tip
दोनों हर शून्य नहीं हो सकते इसलिए \(x\ne 1\) और \(x\ne -1\)। जोड़ में सामान्य प्रांत दोनों प्रांतों का प्रतिच्छेद होता है।
For square roots we need \(x-2\ge 0\) and \(5-x\ge 0\) so \(2\le x\le 5\). In such questions apply both conditions together.
Step 2
Why this answer is correct
The correct answer is A. \([2,5]). For square roots we need \(x-2\ge 0\) and \(5-x\ge 0\) so \(2\le x\le 5\). In such questions apply both conditions together.
Step 3
Exam Tip
वर्गमूल के लिए \(x-2\ge 0\) और \(5-x\ge 0\) चाहिए इसलिए \(2\le x\le 5\)। ऐसे प्रश्नों में दोनों शर्तें साथ लगाएँ।
Multiplication gives ((fg)(x)=1) but original denominators require \(x\ne 1,-1\). Simplification does not remove restrictions.
Step 2
Why this answer is correct
The correct answer is A. (1), \(x\ne 1,-1\). Multiplication gives ((fg)(x)=1) but original denominators require \(x\ne 1,-1\). Simplification does not remove restrictions.
Step 3
Exam Tip
गुणन से ((fg)(x)=1) मिलता है लेकिन मूल हरों से \(x\ne 1,-1\)। सरलीकरण प्रतिबंधों को मिटाता नहीं है।
If \(x\ge 0\) then (|x|-x=0) and if (x<0) then (|x|-x=-x-x=-2x). For modulus use cases according to sign.
Step 2
Why this answer is correct
The correct answer is A. (0) जब \(x\ge 0\) और (-2x) जब (x<0). If \(x\ge 0\) then (|x|-x=0) and if (x<0) then (|x|-x=-x-x=-2x). For modulus use cases according to sign.
Step 3
Exam Tip
यदि \(x\ge 0\) तो (|x|-x=0) और यदि (x<0) तो (|x|-x=-x-x=-2x)। मापांक में चिन्ह के अनुसार केस बनाएं।
A. दोनों \(x^2\) के बराबर हैं/Both are equal to \(x^2\)
Step 1
Concept
(\(f\circ g\)(x)=|x|2=x-2) and (\(g\circ f\)(x)=|x-2|=x-2). Sometimes changing composition order still gives the same result.
Step 2
Why this answer is correct
The correct answer is A. दोनों \(x^2\) के बराबर हैं / Both are equal to \(x^2\). (\(f\circ g\)(x)=|x|2=x-2) and (\(g\circ f\)(x)=|x-2|=x-2). Sometimes changing composition order still gives the same result.
Step 3
Exam Tip
(\(f\circ g\)(x)=|x|2=x-2) और (\(g\circ f\)(x)=|x-2|=x-2)। कभी-कभी संयोजन क्रम बदलने पर भी परिणाम समान हो सकता है।
In \(\frac{f}{g}\) the denominator is (g(-3)=0) so the value is undefined. Check restrictions before simplifying \(\frac{x^2-9}{x+3}\).
Step 2
Why this answer is correct
The correct answer is A. यह अपरिभाषित है / It is undefined. In \(\frac{f}{g}\) the denominator is (g(-3)=0) so the value is undefined. Check restrictions before simplifying \(\frac{x^2-9}{x+3}\).
Step 3
Exam Tip
\(\frac{f}{g}\) में हर (g(-3)=0) है इसलिए मान अपरिभाषित है। \(\frac{x^2-9}{x+3}\) को सरल करने से पहले प्रतिबंध देखें।
The square root gives \(x\ge -4\) and the denominator gives \(x\ne 1\). The domain of a product is the intersection of both domains.
Step 2
Why this answer is correct
The correct answer is A. \([-4,\infty)-{1}). The square root gives \(x\ge -4\) and the denominator gives \(x\ne 1\). The domain of a product is the intersection of both domains.
Step 3
Exam Tip
वर्गमूल से \(x\ge -4\) और हर से \(x\ne 1\)। गुणन का प्रांत दोनों प्रांतों का प्रतिच्छेद होता है।
((f+g)(x)=2x-1+5-x=x+4), so it is a linear function. Simplify the terms before identifying the type.
Step 2
Why this answer is correct
The correct answer is A. रेखीय फलन / Linear function. ((f+g)(x)=2x-1+5-x=x+4), so it is a linear function. Simplify the terms before identifying the type.
Step 3
Exam Tip
((f+g)(x)=2x-1+5-x=x+4), इसलिए यह रेखीय फलन है। पदों को सरल करके प्रकार पहचानें।
((f-g)(x)=\frac{2x+1}{3}-\frac{x-4}{2}=\frac{4x+2-3x+12}{6}=\frac{x+14}{6}). Use the least common denominator for fractions.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{x+14}{6}). ((f-g)(x)=\frac{2x+1}{3}-\frac{x-4}{2}=\frac{4x+2-3x+12}{6}=\frac{x+14}{6}). Use the least common denominator for fractions.
Step 3
Exam Tip
((f-g)(x)=\frac{2x+1}{3}-\frac{x-4}{2}=\frac{4x+2-3x+12}{6}=\frac{x+14}{6})। भिन्नों में लघुत्तम समापवर्त्य लें।
(\(f\circ g\)(x)=f\(x^2\)=\frac{1}{x-2}) so \(x\ne 0\). In composition also check the domain of the outer function.
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}-{0}). (\(f\circ g\)(x)=f\(x^2\)=\frac{1}{x-2}) so \(x\ne 0\). In composition also check the domain of the outer function.
Step 3
Exam Tip
(\(f\circ g\)(x)=f\(x^2\)=\frac{1}{x-2}) इसलिए \(x\ne 0\)। संयोजन में बाहरी फलन का प्रांत भी जाँचें।
The correct calculation is ((x+1)2-\(x^2+1\)=2x), so option (B) is correct. Always expand both compositions before subtracting.
Step 2
Why this answer is correct
The correct answer is A. (x). The correct calculation is ((x+1)2-\(x^2+1\)=2x), so option (B) is correct. Always expand both compositions before subtracting.
Step 3
Exam Tip
(\(f\circ g\)(x)=(x+1)2) और (\(g\circ f\)(x)=x-2+1), अंतर (2x) नहीं बल्कि (2x)?
(\(f\circ g\)(x)=\frac{3(5x+2)-2}{5}=\frac{15x+4}{5}=3x+\frac{4}{5}). Substitute the whole (g(x)) in place of (x).
Step 2
Why this answer is correct
The correct answer is A. \(3x+\frac{4}{5}\). (\(f\circ g\)(x)=\frac{3(5x+2)-2}{5}=\frac{15x+4}{5}=3x+\frac{4}{5}). Substitute the whole (g(x)) in place of (x).
Step 3
Exam Tip
(\(f\circ g\)(x)=\frac{3(5x+2)-2}{5}=\frac{15x+4}{5}=3x+\frac{4}{5})। अंदर के पूरे (g(x)) को (x) की जगह रखें।
((g-f)(x)=x-2+1-\(x^2-1\)=2), so it is a constant function. Equal highest-degree terms may cancel.
Step 2
Why this answer is correct
The correct answer is A. स्थिर फलन / Constant function. ((g-f)(x)=x-2+1-\(x^2-1\)=2), so it is a constant function. Equal highest-degree terms may cancel.
Step 3
Exam Tip
((g-f)(x)=x-2+1-\(x^2-1\)=2), इसलिए यह स्थिर फलन है। समान उच्च घात पद कट सकते हैं।
(\frac{x-2+3x+2}{x+1}=\frac{(x+1)(x+2)}{x+1}=x+2), but \(x\ne -1\). It is necessary to exclude the zero of the denominator.
Step 2
Why this answer is correct
The correct answer is A. (x+2), \(x\ne -1\). (\frac{x-2+3x+2}{x+1}=\frac{(x+1)(x+2)}{x+1}=x+2), but \(x\ne -1\). It is necessary to exclude the zero of the denominator.
Step 3
Exam Tip
(\frac{x-2+3x+2}{x+1}=\frac{(x+1)(x+2)}{x+1}=x+2), पर \(x\ne -1\)। हर के शून्य मान को हटाना जरूरी है।
For both square roots \(x+1\ge 0\) and \(x-3\ge 0\), hence \(x\ge 3\). Even in subtraction both functions must be defined.
Step 2
Why this answer is correct
The correct answer is A. \([3,\infty)). For both square roots \(x+1\ge 0\) and \(x-3\ge 0\), hence \(x\ge 3\). Even in subtraction both functions must be defined.
Step 3
Exam Tip
दोनों वर्गमूलों के लिए \(x+1\ge 0\) और \(x-3\ge 0\), इसलिए \(x\ge 3\)। घटाव में भी दोनों फलन परिभाषित होने चाहिए।
(\frac{f}{g}=\frac{(x-2)2}{x-2}=x-2) where \(x\ne 2\), so at (x=3) the value is (1). First note the restriction then substitute.
Step 2
Why this answer is correct
The correct answer is A. (1). (\frac{f}{g}=\frac{(x-2)2}{x-2}=x-2) where \(x\ne 2\), so at (x=3) the value is (1). First note the restriction then substitute.
Step 3
Exam Tip
(\frac{f}{g}=\frac{(x-2)2}{x-2}=x-2) जहाँ \(x\ne 2\), इसलिए (x=3) पर मान (1) है। पहले प्रतिबंध फिर मान रखें।
(g(x)=(f+g)(x)-f(x)=x-2+3x+5-(3x+2)=x-2+3). To find the unknown function subtract the known function from the given sum.
Step 2
Why this answer is correct
The correct answer is A. \(x^2+3\). (g(x)=(f+g)(x)-f(x)=x-2+3x+5-(3x+2)=x-2+3). To find the unknown function subtract the known function from the given sum.
Step 3
Exam Tip
(g(x)=(f+g)(x)-f(x)=x-2+3x+5-(3x+2)=x-2+3)। अज्ञात फलन निकालने के लिए दिए गए योग से ज्ञात फलन घटाएँ।
(f(x)=\frac{x-2-1}{x-1}=x+1), but \(x\ne 1\) because (g(x)\ne 0). While finding an unknown factor check the division restriction.
Step 2
Why this answer is correct
The correct answer is A. (x+1), \(x\ne 1\). (f(x)=\frac{x-2-1}{x-1}=x+1), but \(x\ne 1\) because (g(x)\ne 0). While finding an unknown factor check the division restriction.
Step 3
Exam Tip
(f(x)=\frac{x-2-1}{x-1}=x+1), पर (g(x)\ne 0) के लिए \(x\ne 1\)। अज्ञात गुणक निकालते समय विभाजन का प्रतिबंध देखें।
The first function needs \(x\ne -2\) and the second needs \(x\ne 2\). The domain of the sum is obtained by excluding both restrictions.
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}-{-2,2}). The first function needs \(x\ne -2\) and the second needs \(x\ne 2\). The domain of the sum is obtained by excluding both restrictions.
Step 3
Exam Tip
पहले फलन में \(x\ne -2\) और दूसरे में \(x\ne 2\) है। योग का प्रांत दोनों प्रतिबंधों को हटाकर मिलेगा।
(\frac{g}{f}=\frac{x-2-4}{x+2}=\frac{(x-2)(x+2)}{x+2}=x-2), but \(x\ne -2\). Exclude the zero of the cancelled factor too.
Step 2
Why this answer is correct
The correct answer is A. (x-2), \(x\ne -2\). (\frac{g}{f}=\frac{x-2-4}{x+2}=\frac{(x-2)(x+2)}{x+2}=x-2), but \(x\ne -2\). Exclude the zero of the cancelled factor too.
Step 3
Exam Tip
(\frac{g}{f}=\frac{x-2-4}{x+2}=\frac{(x-2)(x+2)}{x+2}=x-2), पर \(x\ne -2\)। रद्द किए गए गुणनखंड का शून्य भी हटाएँ।
The denominator is (g(x)=x+2), so \(x\ne -2\). The numerator is always defined, so only the denominator restriction applies.
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}-{-2}). The denominator is (g(x)=x+2), so \(x\ne -2\). The numerator is always defined, so only the denominator restriction applies.
Step 3
Exam Tip
हर (g(x)=x+2) है, इसलिए \(x\ne -2\)। अंश हमेशा परिभाषित है इसलिए केवल हर का प्रतिबंध लगेगा।
(\(f\circ g\)(x)=\frac{1}{x-1}) needs \(x\ne 1\) and (\(g\circ f\)(x)=\frac{1}{x}-1) needs \(x\ne 0\). In the sum both restrictions combine.
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}-{0,1}). (\(f\circ g\)(x)=\frac{1}{x-1}) needs \(x\ne 1\) and (\(g\circ f\)(x)=\frac{1}{x}-1) needs \(x\ne 0\). In the sum both restrictions combine.
Step 3
Exam Tip
(\(f\circ g\)(x)=\frac{1}{x-1}) के लिए \(x\ne 1\) और (\(g\circ f\)(x)=\frac{1}{x}-1) के लिए \(x\ne 0\)। योग में दोनों प्रतिबंध मिलते हैं।
(f(x)+1=\frac{x+3}{x-3}+1=\frac{2x}{x-3}), where \(x\ne 3\). Make a common denominator when adding a constant.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{2x}{x-3}), \(x\ne 3\). (f(x)+1=\frac{x+3}{x-3}+1=\frac{2x}{x-3}), where \(x\ne 3\). Make a common denominator when adding a constant.
Step 3
Exam Tip
(f(x)+1=\frac{x+3}{x-3}+1=\frac{2x}{x-3}), जहाँ \(x\ne 3\)। स्थिर संख्या जोड़ते समय समान हर बनाएँ।
((f-g)(x)=\frac{1}{x-2}-\frac{1}{x+2}=\frac{4}{x-2-4}), where \(x\ne \pm 2\). Keep restrictions while multiplying denominators.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{4}{x^2-4}), \(x\ne \pm 2\). ((f-g)(x)=\frac{1}{x-2}-\frac{1}{x+2}=\frac{4}{x-2-4}), where \(x\ne \pm 2\). Keep restrictions while multiplying denominators.
Step 3
Exam Tip
((f-g)(x)=\frac{1}{x-2}-\frac{1}{x+2}=\frac{4}{x-2-4}), जहाँ \(x\ne \pm 2\)। हरों का गुणन करते समय प्रतिबंध रखें।
The denominator is (x-2), so (x=2) is excluded even if the numerator has a common factor. Decide restrictions from the original denominator.
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}-{2}). The denominator is (x-2), so (x=2) is excluded even if the numerator has a common factor. Decide restrictions from the original denominator.
Step 3
Exam Tip
हर (x-2) है इसलिए (x=2) हटेगा, भले ही अंश में समान गुणनखंड हो। मूल हर से प्रतिबंध तय करें।
Here (f-2(x)+g-2(x)=\frac{\(x^2-1\)2+(2x)2}{\(x^2+1\)2}=\frac{\(x^2+1\)2}{\(x^2+1\)2}=1). Recognise the identity while adding squares.
Step 2
Why this answer is correct
The correct answer is A. (1). Here (f-2(x)+g-2(x)=\frac{\(x^2-1\)2+(2x)2}{\(x^2+1\)2}=\frac{\(x^2+1\)2}{\(x^2+1\)2}=1). Recognise the identity while adding squares.
Step 3
Exam Tip
यहाँ (f-2(x)+g-2(x)=\frac{\(x^2-1\)2+(2x)2}{\(x^2+1\)2}=\frac{\(x^2+1\)2}{\(x^2+1\)2}=1)। वर्गों को जोड़ते समय सर्वसमिका पहचानें।
