In a quotient, the denominator (g(x)) must not be zero, so (x=2) is excluded. In exams, check the domain before simplifying.
Step 2
Why this answer is correct
The correct answer is B. \(\mathbb{R}\setminus{2}\). In a quotient, the denominator (g(x)) must not be zero, so (x=2) is excluded. In exams, check the domain before simplifying.
Step 3
Exam Tip
भागफल में हर (g(x)) शून्य नहीं होना चाहिए, इसलिए (x=2) हटेगा। परीक्षा में सरलीकरण से पहले प्रांत जांचें।
A. वे \(x\ne 4\) पर समान हैं/They are equal for \(x\ne 4\)
Step 1
Concept
(f(x)) simplifies to (x+4), but (f) is undefined at (x=4). For equal functions, check both rule and domain.
Step 2
Why this answer is correct
The correct answer is A. वे \(x\ne 4\) पर समान हैं / They are equal for \(x\ne 4\). (f(x)) simplifies to (x+4), but (f) is undefined at (x=4). For equal functions, check both rule and domain.
Step 3
Exam Tip
(f(x)) सरल होकर (x+4) बनता है, लेकिन (x=4) पर (f) अपरिभाषित है। समान फलनों में नियम के साथ प्रांत भी देखें।
Both square roots require \(2x-1\ge 0\) and \(7-3x\ge 0\). Hence the domain is \(\left[\frac{1}{2},\frac{7}{3}\right]\).
Step 2
Why this answer is correct
The correct answer is A. \(\left[\frac{1}{2},\frac{7}{3}\right]\). Both square roots require \(2x-1\ge 0\) and \(7-3x\ge 0\). Hence the domain is \(\left[\frac{1}{2},\frac{7}{3}\right]\).
Step 3
Exam Tip
दोनों वर्गमूलों के लिए \(2x-1\ge 0\) और \(7-3x\ge 0\) चाहिए। इसलिए प्रांत \(\left[\frac{1}{2},\frac{7}{3}\right]\) है।
For subtraction, both functions must be defined, so (x=-3) and (x=4) are excluded. Always remove values making any denominator zero.
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}\setminus{-3,4}\). For subtraction, both functions must be defined, so (x=-3) and (x=4) are excluded. Always remove values making any denominator zero.
Step 3
Exam Tip
घटाव के लिए दोनों फलन परिभाषित होने चाहिए, इसलिए (x=-3) और (x=4) हटेंगे। हर शून्य कराने वाले मान हमेशा हटाएं।
The denominator of (g(x)) is (x-3), so at (x=3), (g) and (f+g) are undefined. In a sum, both functions must be defined.
Step 2
Why this answer is correct
The correct answer is A. अपरिभाषित / Undefined. The denominator of (g(x)) is (x-3), so at (x=3), (g) and (f+g) are undefined. In a sum, both functions must be defined.
Step 3
Exam Tip
(g(x)) में हर (x-3) है, इसलिए (x=3) पर (g) और (f+g) अपरिभाषित हैं। योग में दोनों फलन परिभाषित होने चाहिए।
The correct answer is A. \(x^2-\frac{4}{x^2}\), \(x\ne 0\). ((fg)(x)=\left\(x+\frac{2}{x}\right\)\left\(x-\frac{2}{x}\right\)=x-2-\frac{4}{x-2}). Apply ((a+b)(a-b)=a-2-b-2).
Even if the product simplifies to (1), the original functions forbid (x=1) and (x=-1). The domain should not be changed after simplification.
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}\setminus{-1,1}\). Even if the product simplifies to (1), the original functions forbid (x=1) and (x=-1). The domain should not be changed after simplification.
Step 3
Exam Tip
भले ही गुणन सरल होकर (1) दिखे, मूल फलनों में (x=1) और (x=-1) निषिद्ध हैं। प्रांत सरलीकरण के बाद नहीं बदलना चाहिए।
(|x-1|=|x+1|) means (x) is equally distant from (1) and (-1), so (x=0). The distance idea is useful in modulus questions.
Step 2
Why this answer is correct
The correct answer is A. (x=0). (|x-1|=|x+1|) means (x) is equally distant from (1) and (-1), so (x=0). The distance idea is useful in modulus questions.
Step 3
Exam Tip
(|x-1|=|x+1|) का अर्थ है कि (x) दोनों बिंदुओं (1) और (-1) से समान दूरी पर है, इसलिए (x=0)। मापांक में दूरी का विचार उपयोगी है।
For (f), \(x^2-9\ne 0\), so \(x\ne \pm3\), and (f(x)) is never (0). Hence only (-3) and (3) are excluded.
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}\setminus{-3,3}\). For (f), \(x^2-9\ne 0\), so \(x\ne \pm3\), and (f(x)) is never (0). Hence only (-3) and (3) are excluded.
Step 3
Exam Tip
(f) के लिए \(x^2-9\ne 0\), इसलिए \(x\ne \pm3\), और (f(x)) कभी (0) नहीं होता। अतः केवल (-3) और (3) हटेंगे।
The square root needs \(x\ge -2\), and the denominator needs \(x\ne 1\). In a product, both conditions apply together.
Step 2
Why this answer is correct
The correct answer is A. \([-2,\infty\)\setminus{1}). The square root needs \(x\ge -2\), and the denominator needs \(x\ne 1\). In a product, both conditions apply together.
Step 3
Exam Tip
वर्गमूल के लिए \(x\ge -2\) और हर के लिए \(x\ne 1\) चाहिए। गुणन में दोनों शर्तें साथ लगती हैं।
\(\frac{f}{g}=\frac{x^2}{4}\) and \(x^2\ge 0\), so the range is \([0,\infty\)). Division by a positive constant does not change sign.
Step 2
Why this answer is correct
The correct answer is A. \([0,\infty\)). \(\frac{f}{g}=\frac{x^2}{4}\) and \(x^2\ge 0\), so the range is \([0,\infty\)). Division by a positive constant does not change sign.
Step 3
Exam Tip
\(\frac{f}{g}=\frac{x^2}{4}\) और \(x^2\ge 0\), इसलिए परास \([0,\infty\)) है। धन स्थिर से भाग देने पर चिन्ह नहीं बदलता।
((f-g)(x)=\frac{x-2}{x-2}=1), but (x=2) is outside the original domain. Keep restrictions even after simplification.
Step 2
Why this answer is correct
The correct answer is A. (1), \(x\ne 2\). ((f-g)(x)=\frac{x-2}{x-2}=1), but (x=2) is outside the original domain. Keep restrictions even after simplification.
Step 3
Exam Tip
((f-g)(x)=\frac{x-2}{x-2}=1), पर (x=2) मूल प्रांत से बाहर है। सरलीकरण के बाद भी प्रतिबंध याद रखें।
Since \(x^2+1>0\) and \(x^2-1<x^2+1\), the ratio is always less than (1). With a positive denominator, the inequality direction does not change.
Step 2
Why this answer is correct
The correct answer is A. सभी \(x\in\mathbb{R}\) / all \(x\in\mathbb{R}\). Since \(x^2+1>0\) and \(x^2-1<x^2+1\), the ratio is always less than (1). With a positive denominator, the inequality direction does not change.
Step 3
Exam Tip
क्योंकि \(x^2+1>0\) और \(x^2-1<x^2+1\), अनुपात हमेशा (1) से छोटा है। धन हर होने पर असमता की दिशा नहीं बदलती।
The domain of (f) is \(x\ne 0\), and \(fg=\frac{1}{x}\cdot x=1\). The product may simplify, but the domain comes from the original functions.
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}\setminus{0}\), (1). The domain of (f) is \(x\ne 0\), and \(fg=\frac{1}{x}\cdot x=1\). The product may simplify, but the domain comes from the original functions.
Step 3
Exam Tip
(f) का प्रांत \(x\ne 0\) है और \(fg=\frac{1}{x}\cdot x=1\)। उत्पाद सरल हो सकता है, पर प्रांत मूल फलन से तय होगा।
(f=(x-2)2), so \(\frac{f}{g}=x-2\), but (x=2) is excluded. Cancelled factors still leave domain restrictions.
Step 2
Why this answer is correct
The correct answer is A. (x-2), \(x\ne 2\). (f=(x-2)2), so \(\frac{f}{g}=x-2\), but (x=2) is excluded. Cancelled factors still leave domain restrictions.
Step 3
Exam Tip
(f=(x-2)2), इसलिए \(\frac{f}{g}=x-2\), पर (x=2) हटेगा। रद्द हुए कारक भी प्रांत प्रतिबंध छोड़ते हैं।
\(\lfloor x\rfloor+{x}=\lfloor x\rfloor+x-\lfloor x\rfloor=x\). Substituting the definition directly makes the question simple.
Step 2
Why this answer is correct
The correct answer is A. (x). \(\lfloor x\rfloor+{x}=\lfloor x\rfloor+x-\lfloor x\rfloor=x\). Substituting the definition directly makes the question simple.
Step 3
Exam Tip
\(\lfloor x\rfloor+{x}=\lfloor x\rfloor+x-\lfloor x\rfloor=x\)। परिभाषा को सीधे रखने से प्रश्न सरल हो जाता है।
\(\sqrt{x}\) needs \(x\ge 0\), and the denominator needs \(x-2\ne 0\). In a quotient, both square-root and denominator restrictions apply.
Step 2
Why this answer is correct
The correct answer is A. \([0,\infty\)\setminus{2}). \(\sqrt{x}\) needs \(x\ge 0\), and the denominator needs \(x-2\ne 0\). In a quotient, both square-root and denominator restrictions apply.
Step 3
Exam Tip
\(\sqrt{x}\) के लिए \(x\ge 0\) और हर \(x-2\ne 0\) चाहिए। भागफल में वर्गमूल और हर दोनों प्रतिबंध लागू होते हैं।
(x-2-5x+6=(x-2)(x-3)), so the form is (x-2), but (x=3) is excluded. Never forget the cancelled denominator.
Step 2
Why this answer is correct
The correct answer is A. (x-2), \(x\ne 3\). (x-2-5x+6=(x-2)(x-3)), so the form is (x-2), but (x=3) is excluded. Never forget the cancelled denominator.
Step 3
Exam Tip
(x-2-5x+6=(x-2)(x-3)), इसलिए रूप (x-2) है पर (x=3) हटेगा। रद्द किए गए हर को कभी न भूलें।
The denominator has \(\sqrt{x-2}\), so (x-2>0), meaning (x>2). A square root in the denominator cannot be zero.
Step 2
Why this answer is correct
The correct answer is A. (\(2,\infty\)). The denominator has \(\sqrt{x-2}\), so (x-2>0), meaning (x>2). A square root in the denominator cannot be zero.
Step 3
Exam Tip
हर में \(\sqrt{x-2}\) है, इसलिए (x-2>0), यानी (x>2)। हर वाले वर्गमूल में शून्य भी निषिद्ध होता है।
The \(\frac{1}{x}\) terms cancel, but (x=0) remains outside the domain. Keep original restrictions even after simplification.
Step 2
Why this answer is correct
The correct answer is A. (2x), \(x\ne 0\). The \(\frac{1}{x}\) terms cancel, but (x=0) remains outside the domain. Keep original restrictions even after simplification.
Step 3
Exam Tip
\(\frac{1}{x}\) वाले पद कटते हैं, पर (x=0) प्रांत से बाहर रहता है। अभिव्यक्ति सरल होने पर भी मूल प्रतिबंध रखें।
A. वे \(x\ne 1\) पर समान हैं/They are equal for \(x\ne 1\)
Step 1
Concept
(f(x)) simplifies to (x+1), but (f) is not defined at (x=1). Equal functions need both the same rule and the same domain.
Step 2
Why this answer is correct
The correct answer is A. वे \(x\ne 1\) पर समान हैं / They are equal for \(x\ne 1\). (f(x)) simplifies to (x+1), but (f) is not defined at (x=1). Equal functions need both the same rule and the same domain.
Step 3
Exam Tip
(f(x)=x+1) सरलीकृत होता है, लेकिन (f) पर (x=1) परिभाषित नहीं है। समान नियम और समान प्रांत दोनों जरूरी हैं।
(x-2-4=(x-2)(x+2)), so (x=2) and (x=-2) are forbidden. For a sum, take the intersection of the domains.
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}\setminus{-2,2}\). (x-2-4=(x-2)(x+2)), so (x=2) and (x=-2) are forbidden. For a sum, take the intersection of the domains.
Step 3
Exam Tip
(x-2-4=(x-2)(x+2)), इसलिए (x=2) और (x=-2) निषिद्ध हैं। योग में दोनों फलनों के प्रांत का प्रतिच्छेद लें।
A. (1) के निकट लेकिन बराबर नहीं/approaches (1) but not equal
Step 1
Concept
\(\frac{x^2+1}{x^2+2}=1-\frac{1}{x^2+2}\), so the value stays below (1) and approaches (1). The actual maximum is not attained here.
Step 2
Why this answer is correct
The correct answer is A. (1) के निकट लेकिन बराबर नहीं / approaches (1) but not equal. \(\frac{x^2+1}{x^2+2}=1-\frac{1}{x^2+2}\), so the value stays below (1) and approaches (1). The actual maximum is not attained here.
Step 3
Exam Tip
\(\frac{x^2+1}{x^2+2}=1-\frac{1}{x^2+2}\), इसलिए मान (1) से छोटा रहकर (1) के निकट जाता है। यहां वास्तविक अधिकतम प्राप्त नहीं होता।
(\frac{1}{x-1}-\frac{1}{x+1}=\frac{x+1-(x-1)}{x-2-1}=\frac{2}{x-2-1}). Watch numerator signs while taking a common denominator.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{2}{x^2-1}\). (\frac{1}{x-1}-\frac{1}{x+1}=\frac{x+1-(x-1)}{x-2-1}=\frac{2}{x-2-1}). Watch numerator signs while taking a common denominator.
Step 3
Exam Tip
(\frac{1}{x-1}-\frac{1}{x+1}=\frac{x+1-(x-1)}{x-2-1}=\frac{2}{x-2-1})। समान हर बनाते समय अंश के चिन्ह देखें।
((f-g)(x)=x-2-4x+4=(x-2)2), which is always non-negative. A perfect square is never negative.
Step 2
Why this answer is correct
The correct answer is A. सभी \(x\in\mathbb{R}\) / all \(x\in\mathbb{R}\). ((f-g)(x)=x-2-4x+4=(x-2)2), which is always non-negative. A perfect square is never negative.
Step 3
Exam Tip
((f-g)(x)=x-2-4x+4=(x-2)2), जो हमेशा अऋण है। पूर्ण वर्ग कभी ऋणात्मक नहीं होता।
With a common denominator, (f-g=\frac{x-2+1-\(x^2-1\)}{x}=\frac{2}{x}), where \(x\ne 0\). In subtraction, both signs of the second numerator change.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{2}{x}\), \(x\ne 0\). With a common denominator, (f-g=\frac{x-2+1-\(x^2-1\)}{x}=\frac{2}{x}), where \(x\ne 0\). In subtraction, both signs of the second numerator change.
Step 3
Exam Tip
समान हर से (f-g=\frac{x-2+1-\(x^2-1\)}{x}=\frac{2}{x}), जहां \(x\ne 0\)। घटाव में दूसरे अंश के दोनों चिन्ह बदलते हैं।
In \(\frac{f}{g}\), both (f) and (g) must be defined and (g(x)\ne 0); this excludes \(x=\pm2\). In a quotient, zero of the second function is also forbidden.
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}\setminus{-2,2}\). In \(\frac{f}{g}\), both (f) and (g) must be defined and (g(x)\ne 0); this excludes \(x=\pm2\). In a quotient, zero of the second function is also forbidden.
Step 3
Exam Tip
\(\frac{f}{g}\) में (f) और (g) दोनों परिभाषित हों और (g(x)\ne 0); इससे \(x=\pm2\) हटते हैं। भागफल में दूसरे फलन का शून्य भी निषिद्ध होता है।
