The common domain is \( [2,\infty\)\cap\(-\infty,5]=[2,5] \). For addition, always take the intersection of domains.
Step 2
Why this answer is correct
The correct answer is A. ( [2,5] ). The common domain is \( [2,\infty\)\cap\(-\infty,5]=[2,5] \). For addition, always take the intersection of domains.
Step 3
Exam Tip
दोनों फलनों के डोमेन का प्रतिच्छेद \( [2,\infty\)\cap\(-\infty,5]=[2,5] \) है। परीक्षा में जोड़ के लिए हमेशा डोमेन का प्रतिच्छेद लें।
For product, the domain is the intersection of original domains, so only (x=3) is excluded. Check the original domain before simplifying.
Step 2
Why this answer is correct
The correct answer is A. \( \mathbb{R}-{3} \). For product, the domain is the intersection of original domains, so only (x=3) is excluded. Check the original domain before simplifying.
Step 3
Exam Tip
गुणन में डोमेन (f) और (g) के डोमेन का प्रतिच्छेद होता है, इसलिए केवल (x=3) हटेगा। सरलीकरण से पहले मूल डोमेन देखना जरूरी है।
We need (g(x)\neq0), and \(x^2-4=0\) gives \(x=\pm2\). In quotient questions, always exclude zeros of the denominator.
Step 2
Why this answer is correct
The correct answer is A. \( \mathbb{R}-{-2,2} \). We need (g(x)\neq0), and \(x^2-4=0\) gives \(x=\pm2\). In quotient questions, always exclude zeros of the denominator.
Step 3
Exam Tip
हर जगह (g(x)\neq0) चाहिए, और \(x^2-4=0\) से \(x=\pm2\) मिलते हैं। भाग वाले प्रश्नों में हर के शून्य हमेशा हटाएं।
For \(\sqrt{x}\), \(x\ge0\), and for \(\frac{1}{x-4}\), \(x\neq4\). Hence the domain is \( [0,\infty\)-{4} ).
Step 2
Why this answer is correct
The correct answer is A. \( [0,\infty\)-{4} ). For \(\sqrt{x}\), \(x\ge0\), and for \(\frac{1}{x-4}\), \(x\neq4\). Hence the domain is \( [0,\infty\)-{4} ).
Step 3
Exam Tip
\(\sqrt{x}\) के लिए \(x\ge0\) और \(\frac{1}{x-4}\) के लिए \(x\neq4\) चाहिए। इसलिए डोमेन \( [0,\infty\)-{4} ) है।
Although (fg=1), the original functions do not allow (x=-2) and (x=1). It is wrong to enlarge the domain after simplification.
Step 2
Why this answer is correct
The correct answer is A. \( \mathbb{R}-{-2,1} \). Although (fg=1), the original functions do not allow (x=-2) and (x=1). It is wrong to enlarge the domain after simplification.
Step 3
Exam Tip
यद्यपि (fg=1) बनता है, मूल फलनों में (x=-2) और (x=1) अनुमत नहीं हैं। सरलीकृत अभिव्यक्ति से डोमेन बढ़ाना गलत है।
\(\frac{x^2-1}{x-1}=x+1\), but (x=1) makes the original denominator zero. After cancellation, remember the removed point.
Step 2
Why this answer is correct
The correct answer is A. \(x+1,\ \mathbb{R}-{1}\). \(\frac{x^2-1}{x-1}=x+1\), but (x=1) makes the original denominator zero. After cancellation, remember the removed point.
Step 3
Exam Tip
\(\frac{x^2-1}{x-1}=x+1\), पर (x=1) मूल हर को शून्य करता है। रद्द करने के बाद भी हटाए गए बिंदु याद रखें।
The first function needs \(x\neq-1\), and the second needs \(x\neq2\). The domain of the sum is the common part of these conditions.
Step 2
Why this answer is correct
The correct answer is A. \( \mathbb{R}-{-1,2} \). The first function needs \(x\neq-1\), and the second needs \(x\neq2\). The domain of the sum is the common part of these conditions.
Step 3
Exam Tip
पहले फलन में \(x\neq-1\) और दूसरे में \(x\neq2\) चाहिए। जोड़ का डोमेन इन दोनों शर्तों का साझा भाग है।
(\frac{(x+2)(x-2)}{x-2-4}=1), but (h(x)\neq0) gives \(x\neq\pm2\). The original denominator restriction remains after simplification.
Step 2
Why this answer is correct
The correct answer is A. \(1,\ \mathbb{R}-{-2,2}\). (\frac{(x+2)(x-2)}{x-2-4}=1), but (h(x)\neq0) gives \(x\neq\pm2\). The original denominator restriction remains after simplification.
Step 3
Exam Tip
(\frac{(x+2)(x-2)}{x-2-4}=1), पर (h(x)\neq0) से \(x\neq\pm2\) है। सरलीकरण के बाद भी मूल हर की रोक बनी रहती है।
For \(\sqrt{9-x^2}\), \(-3\le x\le3\), and for \(\frac{1}{x}\), \(x\neq0\). Hence the domain is ( [-3,3]-{0} ).
Step 2
Why this answer is correct
The correct answer is A. ( [-3,3]-{0} ). For \(\sqrt{9-x^2}\), \(-3\le x\le3\), and for \(\frac{1}{x}\), \(x\neq0\). Hence the domain is ( [-3,3]-{0} ).
Step 3
Exam Tip
\(\sqrt{9-x^2}\) के लिए \(-3\le x\le3\) और \(\frac{1}{x}\) के लिए \(x\neq0\) है। अतः डोमेन ( [-3,3]-{0} ) है।
Zeros of both denominators must be removed, so \(x=\pm1,\pm2\) are excluded. For subtraction also, take the intersection of domains.
Step 2
Why this answer is correct
The correct answer is A. \( \mathbb{R}-{-2,-1,1,2} \). Zeros of both denominators must be removed, so \(x=\pm1,\pm2\) are excluded. For subtraction also, take the intersection of domains.
Step 3
Exam Tip
दोनों हरों के शून्य हटाने होंगे, इसलिए \(x=\pm1,\pm2\) निषिद्ध हैं। घटाव में भी डोमेन का प्रतिच्छेद लिया जाता है।
The expression is \(x^2-|x|\). Put \(t=|x|\ge0\), then \(t^2-t\) has minimum \(-\frac{1}{4}\). For hard modulus questions, using (t) is helpful.
Step 2
Why this answer is correct
The correct answer is A. \(-\frac{1}{4}\). The expression is \(x^2-|x|\). Put \(t=|x|\ge0\), then \(t^2-t\) has minimum \(-\frac{1}{4}\). For hard modulus questions, using (t) is helpful.
Step 3
Exam Tip
मान \(x^2-|x|\) है, \(t=|x|\ge0\) रखने पर \(t^2-t\) का न्यूनतम \(-\frac{1}{4}\) है। कठिन प्रश्नों में मापांक को (t) से बदलना उपयोगी है।
For (g), \(x\ge1\) is needed, and then (f(x)\neq0) is automatically true. Hence the domain is \( [1,\infty\) ).
Step 2
Why this answer is correct
The correct answer is A. \( [1,\infty\) ). For (g), \(x\ge1\) is needed, and then (f(x)\neq0) is automatically true. Hence the domain is \( [1,\infty\) ).
Step 3
Exam Tip
(g) के लिए \(x\ge1\) चाहिए, और तब (f(x)\neq0) अपने आप सत्य है। इसलिए डोमेन \( [1,\infty\) ) है।
The denominator is (g(x)), so \(\sqrt{x-1}\neq0\) and (x>1) are required. In division, the denominator radical cannot be zero.
Step 2
Why this answer is correct
The correct answer is A. ( \(1,\infty\) ). The denominator is (g(x)), so \(\sqrt{x-1}\neq0\) and (x>1) are required. In division, the denominator radical cannot be zero.
Step 3
Exam Tip
हर (g(x)) है, इसलिए \(\sqrt{x-1}\neq0\) और (x>1) चाहिए। भाग में हर वाले मूल को शून्य नहीं होने देना है।
(\frac{x}{x-1}-\frac{x}{x+1}=\frac{x[(x+1)-(x-1)]}{x-2-1}=\frac{2x}{x-2-1}). Keep brackets correct while taking a common denominator.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{2x}{x^2-1} \). (\frac{x}{x-1}-\frac{x}{x+1}=\frac{x[(x+1)-(x-1)]}{x-2-1}=\frac{2x}{x-2-1}). Keep brackets correct while taking a common denominator.
Step 3
Exam Tip
(\frac{x}{x-1}-\frac{x}{x+1}=\frac{x[(x+1)-(x-1)]}{x-2-1}=\frac{2x}{x-2-1})। समान हर बनाते समय कोष्ठक सही रखें।
The equation \(x^2+\frac{1}{x}=2\) gives (x-3-2x+1=0=(x-1)\(x^2+x-1\)). Since \(x\neq0\), count all valid real roots carefully.
Step 2
Why this answer is correct
The correct answer is A. (2). The equation \(x^2+\frac{1}{x}=2\) gives (x-3-2x+1=0=(x-1)\(x^2+x-1\)). Since \(x\neq0\), count all valid real roots carefully.
Step 3
Exam Tip
समीकरण \(x^2+\frac{1}{x}=2\) से (x-3-2x+1=0=(x-1)\(x^2+x-1\)) मिलता है। \(x\neq0\) होने पर तीन में से दो वास्तविक हल अलग-अलग हैं?
((fg)(x)=x-2\cdot\frac{1}{x}=x), but (x=0) is not in the domain. Hence every non-zero real number is a solution.
Step 2
Why this answer is correct
The correct answer is A. \( \mathbb{R}-{0} \). ((fg)(x)=x-2\cdot\frac{1}{x}=x), but (x=0) is not in the domain. Hence every non-zero real number is a solution.
Step 3
Exam Tip
((fg)(x)=x-2\cdot\frac{1}{x}=x), पर (x=0) डोमेन में नहीं है। अतः हर अशून्य वास्तविक संख्या हल है।
For \(\sqrt{x^2-4}\), \(x\le-2\) or \(x\ge2\), and for (g), \(x\neq5\). Both conditions must be applied together.
Step 2
Why this answer is correct
The correct answer is A. ( \(-\infty,-2]\cup[2,\infty\)-{5} ). For \(\sqrt{x^2-4}\), \(x\le-2\) or \(x\ge2\), and for (g), \(x\neq5\). Both conditions must be applied together.
Step 3
Exam Tip
\(\sqrt{x^2-4}\) के लिए \(x\le-2\) या \(x\ge2\), और (g) के लिए \(x\neq5\) है। दोनों शर्तें साथ लगेंगी।
(f(x)=x+3) only for \(x\neq3\), so (f-g=0) on that domain. Domain is essential when comparing algebraic expressions as functions.
Step 2
Why this answer is correct
The correct answer is A. \( \mathbb{R}-{3} \). (f(x)=x+3) only for \(x\neq3\), so (f-g=0) on that domain. Domain is essential when comparing algebraic expressions as functions.
Step 3
Exam Tip
(f(x)=x+3) केवल \(x\neq3\) पर है, इसलिए (f-g=0) उसी डोमेन पर है। बीजगणितीय समानता और फलन की समानता में डोमेन जरूरी है।
The denominator \(x^2+1\) is never (0), so no real number is excluded. Only real zeros of the denominator are removed.
Step 2
Why this answer is correct
The correct answer is A. \( \mathbb{R} \). The denominator \(x^2+1\) is never (0), so no real number is excluded. Only real zeros of the denominator are removed.
Step 3
Exam Tip
हर \(x^2+1\) कभी (0) नहीं होता, इसलिए कोई वास्तविक संख्या हटेगी नहीं। हर को शून्य बनाने वाली वास्तविक संख्या ही हटाई जाती है।
Since \(x^2+1>0\) for every real (x), both (f) and (g) are defined on all \(\mathbb{R}\). Hence (fg=1) on all \(\mathbb{R}\).
Step 2
Why this answer is correct
The correct answer is A. \( \mathbb{R} \). Since \(x^2+1>0\) for every real (x), both (f) and (g) are defined on all \(\mathbb{R}\). Hence (fg=1) on all \(\mathbb{R}\).
Step 3
Exam Tip
क्योंकि \(x^2+1>0\) हर वास्तविक (x) के लिए है, (f) और (g) दोनों पूरे \(\mathbb{R}\) पर परिभाषित हैं। अतः (fg=1) पूरे \(\mathbb{R}\) पर है।
(\frac{1}{x-1}+\frac{1}{x+1}=\frac{(x+1)+(x-1)}{x-2-1}=\frac{2x}{x-2-1}). Add the numerators correctly after taking the common denominator.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{2x}{x^2-1} \). (\frac{1}{x-1}+\frac{1}{x+1}=\frac{(x+1)+(x-1)}{x-2-1}=\frac{2x}{x-2-1}). Add the numerators correctly after taking the common denominator.
Step 3
Exam Tip
(\frac{1}{x-1}+\frac{1}{x+1}=\frac{(x+1)+(x-1)}{x-2-1}=\frac{2x}{x-2-1})। जोड़ में अंशों को सही तरह जोड़ें।
For \(\sqrt{x-2}\), \(x\ge2\), and the denominator needs \(x-5\neq0\). Hence \( [2,\infty\)-{5} ) is correct.
Step 2
Why this answer is correct
The correct answer is A. \( [2,\infty\)-{5} ). For \(\sqrt{x-2}\), \(x\ge2\), and the denominator needs \(x-5\neq0\). Hence \( [2,\infty\)-{5} ) is correct.
Step 3
Exam Tip
\(\sqrt{x-2}\) के लिए \(x\ge2\) और हर \(x-5\neq0\) चाहिए। इसलिए \( [2,\infty\)-{5} ) सही है।
Here \(f-g=\frac{4}{x}\) and (f+g=2), so \(f^2-g^2=\frac{8}{x}\). The identity (a-2-b-2=(a-b)(a+b)) is very useful.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{8}{x} \). Here \(f-g=\frac{4}{x}\) and (f+g=2), so \(f^2-g^2=\frac{8}{x}\). The identity (a-2-b-2=(a-b)(a+b)) is very useful.
Step 3
Exam Tip
\(f-g=\frac{4}{x}\) और (f+g=2), इसलिए \(f^2-g^2=\frac{8}{x}\)। पहचान (a-2-b-2=(a-b)(a+b)) बहुत उपयोगी है।
For (f), (x>0), and for (g), \(x\ge1\). Their intersection is \( [1,\infty\) ). For product also, take the common domain of both functions.
Step 2
Why this answer is correct
The correct answer is A. \( [1,\infty\) ). For (f), (x>0), and for (g), \(x\ge1\). Their intersection is \( [1,\infty\) ). For product also, take the common domain of both functions.
Step 3
Exam Tip
(f) के लिए (x>0) और (g) के लिए \(x\ge1\) चाहिए, अतः प्रतिच्छेद \( [1,\infty\) ) है। गुणन में भी दोनों फलनों का साझा डोमेन लें।
A. समान नियम और समान डोमेन होने पर/when they have same rule and same domain
Step 1
Concept
(f(x)=x-1) only for \(x\neq-1\), while (g) is on all \(\mathbb{R}\). Equal functions must have both the same rule and the same domain.
Step 2
Why this answer is correct
The correct answer is A. समान नियम और समान डोमेन होने पर / when they have same rule and same domain. (f(x)=x-1) only for \(x\neq-1\), while (g) is on all \(\mathbb{R}\). Equal functions must have both the same rule and the same domain.
Step 3
Exam Tip
(f(x)=x-1) केवल \(x\neq-1\) पर है, जबकि (g) पूरे \(\mathbb{R}\) पर है। समान फलन के लिए नियम के साथ डोमेन भी समान होना चाहिए।
In both functions, (x=-1) and (x=1) are not allowed. Hence the domain of (f-g) is \( \mathbb{R}-{-1,1} \).
Step 2
Why this answer is correct
The correct answer is A. \( \mathbb{R}-{-1,1} \). In both functions, (x=-1) and (x=1) are not allowed. Hence the domain of (f-g) is \( \mathbb{R}-{-1,1} \).
Step 3
Exam Tip
दोनों फलनों में (x=-1) और (x=1) अनुमत नहीं हैं। इसलिए (f-g) का डोमेन \( \mathbb{R}-{-1,1} \) है।
(f-g=x-2-2x+1=(x-1)2), which is (0) only at (x=1). A perfect square helps identify the zero quickly.
Step 2
Why this answer is correct
The correct answer is A. केवल (x=1) पर / only at (x=1). (f-g=x-2-2x+1=(x-1)2), which is (0) only at (x=1). A perfect square helps identify the zero quickly.
Step 3
Exam Tip
(f-g=x-2-2x+1=(x-1)2), जो (0) केवल (x=1) पर होता है। पूर्ण वर्ग देखकर शून्य जल्दी मिल जाता है।
