The denominator is (x), so the function is not defined at (x=0). In exams, make sure the denominator is not (0).
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}-{0}\). The denominator is (x), so the function is not defined at (x=0). In exams, make sure the denominator is not (0).
Step 3
Exam Tip
हर में (x) है, इसलिए (x=0) पर फलन परिभाषित नहीं है। परीक्षा में हर को (0) न बनने दें।
The linear function (f(x)=x+3) is defined for every real (x). In exams, the domain of a basic linear function is usually \(\mathbb{R}\).
Step 2
Why this answer is correct
The correct answer is A. सभी वास्तविक संख्याएँ / All real numbers. The linear function (f(x)=x+3) is defined for every real (x). In exams, the domain of a basic linear function is usually \(\mathbb{R}\).
Step 3
Exam Tip
रेखीय फलन (f(x)=x+3) हर वास्तविक (x) के लिए परिभाषित है। परीक्षा में हर रेखीय फलन का प्रांत सामान्यतः \(\mathbb{R}\) लें।
A. (R) (A) से (B) में function है और range (=B) है/(R) is a function from (A) to (B) and range (=B)
Step 1
Concept
Every \(x\in A\) has a unique image in (1,4,9,16), and all codomain elements are images. In this case, range and codomain are equal.
Step 2
Why this answer is correct
The correct answer is A. (R) (A) से (B) में function है और range (=B) है / (R) is a function from (A) to (B) and range (=B). Every \(x\in A\) has a unique image in (1,4,9,16), and all codomain elements are images. In this case, range and codomain are equal.
Step 3
Exam Tip
हर \(x\in A\) की unique image (1,4,9,16) में है और सभी codomain elements images हैं। इस case में range और codomain equal हैं।
A. (x=1) की images (y=-1) और (y=1) दोनों हैं/(x=1) has images (y=-1) and (y=1)
Step 1
Concept
For (x=1), two possible (y)-values exist, so uniqueness fails. Read an absolute value relation with direction carefully.
Step 2
Why this answer is correct
The correct answer is A. (x=1) की images (y=-1) और (y=1) दोनों हैं / (x=1) has images (y=-1) and (y=1). For (x=1), two possible (y)-values exist, so uniqueness fails. Read an absolute value relation with direction carefully.
Step 3
Exam Tip
(x=1) के लिए दो possible (y) values हैं, इसलिए uniqueness टूटती है। absolute value relation को direction के साथ पढ़ें।
A. valid function with graph ({(1,3),(2,2),(3,1)})
Step 1
Concept
For every \(x\in A\), \(4-x\in A\) and the output is unique. Check both closed output and uniqueness.
Step 2
Why this answer is correct
The correct answer is A. valid function with graph ({(1,3),(2,2),(3,1)}). For every \(x\in A\), \(4-x\in A\) and the output is unique. Check both closed output and uniqueness.
Step 3
Exam Tip
हर \(x\in A\) पर \(4-x\in A\) और output unique है। closed output और uniqueness दोनों check करें।
A. (f) फिर भी function हो सकता है/(f) may still be a function
Step 1
Concept
Two different inputs having the same image does not violate the function rule. Violation occurs when one input has two different images.
Step 2
Why this answer is correct
The correct answer is A. (f) फिर भी function हो सकता है / (f) may still be a function. Two different inputs having the same image does not violate the function rule. Violation occurs when one input has two different images.
Step 3
Exam Tip
दो अलग inputs की same image होना function rule का violation नहीं है। violation तभी होता है जब एक input की दो अलग images हों।
A. हाँ, duplicate ((1,2)) same pair है/Yes, duplicate ((1,2)) is the same pair
Step 1
Concept
In sets, a repeated pair is not counted, so each input has a unique image. Understand duplicate pair and different image separately.
Step 2
Why this answer is correct
The correct answer is A. हाँ, duplicate ((1,2)) same pair है / Yes, duplicate ((1,2)) is the same pair. In sets, a repeated pair is not counted, so each input has a unique image. Understand duplicate pair and different image separately.
Step 3
Exam Tip
sets में repeated pair count नहीं होता, इसलिए each input की unique image है। duplicate pair और different image को अलग-अलग समझें।
A. जब repeated first coordinate की image वही same हो और duplicate pair हटाने पर uniqueness रहे/When the repeated first coordinate has the same image and uniqueness remains after removing duplicate pairs
Step 1
Concept
A duplicate ordered pair is not a separate element in a set. If the same input gives the same output, uniqueness is not broken.
Step 2
Why this answer is correct
The correct answer is A. जब repeated first coordinate की image वही same हो और duplicate pair हटाने पर uniqueness रहे / When the repeated first coordinate has the same image and uniqueness remains after removing duplicate pairs. A duplicate ordered pair is not a separate element in a set. If the same input gives the same output, uniqueness is not broken.
Step 3
Exam Tip
set में duplicate ordered pair अलग element नहीं माना जाता। यदि same input same output ही दे रहा है, तो uniqueness नहीं टूटती।
A. (x=3) के लिए \(y=4\notin B\), इसलिए image नहीं बनती/For (x=3), \(y=4\notin B\), so no image is formed
Step 1
Concept
The required output for (x=3) is (4), which is not in the codomain, so every domain element is not mapped. A codomain restriction can change the relation.
Step 2
Why this answer is correct
The correct answer is A. (x=3) के लिए \(y=4\notin B\), इसलिए image नहीं बनती / For (x=3), \(y=4\notin B\), so no image is formed. The required output for (x=3) is (4), which is not in the codomain, so every domain element is not mapped. A codomain restriction can change the relation.
Step 3
Exam Tip
(x=3) का required output (4) codomain में नहीं है, इसलिए domain का हर element mapped नहीं है। codomain restriction relation को बदल सकती है।
\(|A\times B|=6\), so relations are \(2^6\) and functions are \(3^2\). Keep relation and function counting formulas separate.
Step 2
Why this answer is correct
The correct answer is A. \(2^6\) और \(3^2\) / \(2^6\) and \(3^2\). \(|A\times B|=6\), so relations are \(2^6\) and functions are \(3^2\). Keep relation and function counting formulas separate.
Step 3
Exam Tip
\(|A\times B|=6\), इसलिए relations \(2^6\) और functions \(3^2\) हैं। relation और function counting formulas अलग रखें।
Total relations are \(2^{|A\times B|}=2^6=64\), and functions are \(2^3=8\). Hence not functions (=64-8=56).
Step 2
Why this answer is correct
The correct answer is A. \(2^{6}-2^3=56\). Total relations are \(2^{|A\times B|}=2^6=64\), and functions are \(2^3=8\). Hence not functions (=64-8=56).
Step 3
Exam Tip
कुल relations \(2^{|A\times B|}=2^6=64\) हैं और functions \(2^3=8\) हैं। not functions (=64-8=56)।
The actual outputs are (2,3,4,5), while the codomain is the given set (B). The range is always a subset of the codomain.
Step 2
Why this answer is correct
The correct answer is A. range (={2,3,4,5}), codomain (={1,2,3,4,5}). The actual outputs are (2,3,4,5), while the codomain is the given set (B). The range is always a subset of the codomain.
Step 3
Exam Tip
actual outputs (2,3,4,5) हैं, जबकि codomain given (B) है। range हमेशा codomain का subset होती है।
In every pair, the second coordinate is (4) times the first coordinate. To identify a rule, verify it on all given values.
Step 2
Why this answer is correct
The correct answer is A. (f(x)=4x). In every pair, the second coordinate is (4) times the first coordinate. To identify a rule, verify it on all given values.
Step 3
Exam Tip
हर pair में second coordinate first coordinate का (4) times है। rule पहचानने के लिए सभी given values पर verify करें।
The outputs are (2,1,0,1,2), so the distinct range is ({0,1,2}). In a set, order and repetition are not important.
Step 2
Why this answer is correct
The correct answer is A. \({0,1,2}). The outputs are (2,1,0,1,2), so the distinct range is ({0,1,2}). In a set, order and repetition are not important.
Step 3
Exam Tip
outputs (2,1,0,1,2) हैं, इसलिए distinct range ({0,1,2}) है। set में order और repetition important नहीं होते।
A. identity में \(x\mapsto x\), constant में हर \(x\mapsto 1\)/In identity, \(x\mapsto x\); in constant, every \(x\mapsto 1\)
Step 1
Concept
Identity maps each input to itself, while a constant function sends every input to the same value. Both can be valid functions.
Step 2
Why this answer is correct
The correct answer is A. identity में \(x\mapsto x\), constant में हर \(x\mapsto 1\) / In identity, \(x\mapsto x\); in constant, every \(x\mapsto 1\). Identity maps each input to itself, while a constant function sends every input to the same value. Both can be valid functions.
Step 3
Exam Tip
identity input को उसी पर map करती है, जबकि constant function हर input को same value पर भेजता है। दोनों valid functions हो सकते हैं।
A. (2) की दो different images (b) और (c) हैं/(2) has two different images (b) and (c)
Step 1
Concept
One input (2) is related to two outputs, so it is not a function. In function checks, first notice repeated first coordinates.
Step 2
Why this answer is correct
The correct answer is A. (2) की दो different images (b) और (c) हैं / (2) has two different images (b) and (c). One input (2) is related to two outputs, so it is not a function. In function checks, first notice repeated first coordinates.
Step 3
Exam Tip
एक input (2) दो outputs से जुड़ा है, इसलिए यह function नहीं है। function check में repeated first coordinate पर सबसे पहले ध्यान दें।
\(0^2=0\), \(1^2=1\), and \(2^2=4\). In a finite case, the graph of a function is the set of ordered pairs.
Step 2
Why this answer is correct
The correct answer is A. \({(0,0),(1,1),(2,4)}). \(0^2=0\), \(1^2=1\), and \(2^2=4\). In a finite case, the graph of a function is the set of ordered pairs.
Step 3
Exam Tip
\(0^2=0\), \(1^2=1\), और \(2^2=4\)। function का graph finite case में ordered pairs का set है।
A. नहीं, क्योंकि \(f(3)=9\notin{1,2,3,4,5}\)/No, because \(f(3)=9\notin{1,2,3,4,5}\)
Step 1
Concept
The output is unique, but (f(3)=9) is not in the codomain. For a function, every image must also lie in the codomain.
Step 2
Why this answer is correct
The correct answer is A. नहीं, क्योंकि \(f(3)=9\notin{1,2,3,4,5}\) / No, because \(f(3)=9\notin{1,2,3,4,5}\). The output is unique, but (f(3)=9) is not in the codomain. For a function, every image must also lie in the codomain.
Step 3
Exam Tip
output unique तो है, लेकिन (f(3)=9) codomain में नहीं है। function के लिए image codomain के अंदर भी होनी चाहिए।
At (x=2), the denominator becomes (0), so (f(2)) is not defined. In a rational function, the denominator must not be zero.
Step 2
Why this answer is correct
The correct answer is A. (2). At (x=2), the denominator becomes (0), so (f(2)) is not defined. In a rational function, the denominator must not be zero.
Step 3
Exam Tip
(x=2) पर denominator (0) हो जाता है, इसलिए (f(2)) defined नहीं है। rational function में denominator zero न हो, यह जरूरी है।
It is not empty, but (1) has two images and (3) has no image. In hard MCQs, check both conditions separately.
Step 2
Why this answer is correct
The correct answer is A. \(R=\{(1,1),(1,2),(2,3)\}\). It is not empty, but (1) has two images and (3) has no image. In hard MCQs, check both conditions separately.
Step 3
Exam Tip
यह empty नहीं है, लेकिन (1) की दो images हैं और (3) की image missing है। hard MCQ में दोनों conditions अलग-अलग जांचें।
