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Every \(x\in A\) has exactly one image in (B). In exams, first check presence and uniqueness for each domain element.
Step 2
Why this answer is correct
The correct answer is A. \(R=\{(1,4),(2,4),(3,5)\}\). Every \(x\in A\) has exactly one image in (B). In exams, first check presence and uniqueness for each domain element.
Step 3
Exam Tip
हर \(x\in A\) की ठीक एक छवि (B) में है। परीक्षा में पहले domain के हर element की उपस्थिति और uniqueness जांचें।
Each of the (4) elements of (A) has (2) choices in (B), so total functions are \(2^4=16\). Remember the formula: functions from (A) to (B) \(=|B|^{|A|}\).
Step 2
Why this answer is correct
The correct answer is A. \(2^4=16\). Each of the (4) elements of (A) has (2) choices in (B), so total functions are \(2^4=16\). Remember the formula: functions from (A) to (B) \(=|B|^{|A|}\).
Step 3
Exam Tip
(A) के हर (4) element के लिए (B) में (2) choices हैं, इसलिए कुल \(2^4=16\) फलन हैं। सूत्र याद रखें: (A) से (B) तक फलन \(=|B|^{|A|}\)।
A function needs exactly one output for each input; different inputs may have the same output. Common mistake: do not treat same image as a violation of function rule.
Step 2
Why this answer is correct
The correct answer is A. (R) फलन है / (R) is a function. A function needs exactly one output for each input; different inputs may have the same output. Common mistake: do not treat same image as a violation of function rule.
Step 3
Exam Tip
एक input की ठीक एक output होना जरूरी है; अलग inputs की same output हो सकती है। common mistake: same image को function rule का violation न मानें।
Here \(2\in A\) has two images (q) and (r). In a function, one input cannot have two different outputs.
Step 2
Why this answer is correct
The correct answer is C. \(R=\{(1,p),(2,q),(2,r),(3,p)\}\). Here \(2\in A\) has two images (q) and (r). In a function, one input cannot have two different outputs.
Step 3
Exam Tip
यहां \(2\in A\) की दो images (q) और (r) हैं। फलन में एक input की दो अलग outputs नहीं हो सकतीं।
Putting (x=0,1,2,3) gives outputs (1,3,5,7). The range is the set of actual images, not necessarily the whole codomain.
Step 2
Why this answer is correct
The correct answer is A. \({1,3,5,7}). Putting (x=0,1,2,3) gives outputs (1,3,5,7). The range is the set of actual images, not necessarily the whole codomain.
Step 3
Exam Tip
(x=0,1,2,3) रखने पर outputs (1,3,5,7) मिलते हैं। range हमेशा actual images का set होता है, पूरा codomain जरूरी नहीं।
A. हर \(a\in A\) से (B) के ठीक एक element तक arrow है/Every \(a\in A\) has exactly one arrow to an element of (B)
Step 1
Concept
In a function, exactly one arrow must start from each domain element. In diagrams, check both arrow direction and arrow count.
Step 2
Why this answer is correct
The correct answer is A. हर \(a\in A\) से (B) के ठीक एक element तक arrow है / Every \(a\in A\) has exactly one arrow to an element of (B). In a function, exactly one arrow must start from each domain element. In diagrams, check both arrow direction and arrow count.
Step 3
Exam Tip
फलन में domain के प्रत्येक element से ठीक एक arrow निकलना चाहिए। diagram में arrows की direction और count दोनों देखें।
A. \(4\in A\) की कोई image नहीं है/\(4\in A\) has no image
Step 1
Concept
Every element of (A) must have an image, but (4) is missing. An unused element of the codomain does not make a function invalid.
Step 2
Why this answer is correct
The correct answer is A. \(4\in A\) की कोई image नहीं है / \(4\in A\) has no image. Every element of (A) must have an image, but (4) is missing. An unused element of the codomain does not make a function invalid.
Step 3
Exam Tip
(A) के हर element की image होनी चाहिए, लेकिन (4) missing है। codomain का कोई unused element होना function को गलत नहीं बनाता।
A. (x=3) के लिए (y=0,1,2) possible हैं/For (x=3), (y=0,1,2) are possible
Step 1
Concept
For (x=3), there are many images, so uniqueness fails. In inequality relations, count possible outputs for one input.
Step 2
Why this answer is correct
The correct answer is A. (x=3) के लिए (y=0,1,2) possible हैं / For (x=3), (y=0,1,2) are possible. For (x=3), there are many images, so uniqueness fails. In inequality relations, count possible outputs for one input.
Step 3
Exam Tip
(x=3) की कई images हैं, इसलिए uniqueness टूटती है। inequality वाले relations में एक input पर possible outputs गिनें।
A. हर \(a\in A\) के लिए ठीक एक \(b\in B\) हो ताकि \((a,b)\in R\)/For every \(a\in A\), there is exactly one \(b\in B\) such that \((a,b)\in R\)
Step 1
Concept
The condition for a function is on domain elements, not codomain elements. In definition questions, the words exactly one are most important.
Step 2
Why this answer is correct
The correct answer is A. हर \(a\in A\) के लिए ठीक एक \(b\in B\) हो ताकि \((a,b)\in R\) / For every \(a\in A\), there is exactly one \(b\in B\) such that \((a,b)\in R\). The condition for a function is on domain elements, not codomain elements. In definition questions, the words exactly one are most important.
Step 3
Exam Tip
फलन की शर्त domain elements पर लगती है, codomain elements पर नहीं। definition questions में exactly one शब्द सबसे जरूरी है।
In any function, the number of ordered pairs equals the number of elements in the domain. Here (|A|=3), so there are (3) pairs.
Step 2
Why this answer is correct
The correct answer is A. (3). In any function, the number of ordered pairs equals the number of elements in the domain. Here (|A|=3), so there are (3) pairs.
Step 3
Exam Tip
किसी भी फलन में ordered pairs की संख्या domain के elements की संख्या के बराबर होती है। यहां (|A|=3), इसलिए pairs (3) हैं।
A. यह constant function है/It is a constant function
Step 1
Concept
Every input has image (2), so it is a constant function. Repeated output does not make a function invalid.
Step 2
Why this answer is correct
The correct answer is A. यह constant function है / It is a constant function. Every input has image (2), so it is a constant function. Repeated output does not make a function invalid.
Step 3
Exam Tip
हर input की image (2) है, इसलिए यह constant function है। एक ही output बार-बार आना function को invalid नहीं बनाता।
A. हर \(a\in A\) के लिए \(f(a)\in B\) unique होता है/For every \(a\in A\), \(f(a)\in B\) is unique
Step 1
Concept
In a function, every domain element has a unique image in the codomain. The whole codomain need not become the range.
Step 2
Why this answer is correct
The correct answer is A. हर \(a\in A\) के लिए \(f(a)\in B\) unique होता है / For every \(a\in A\), \(f(a)\in B\) is unique. In a function, every domain element has a unique image in the codomain. The whole codomain need not become the range.
Step 3
Exam Tip
function में हर domain element की unique image codomain में होती है। यह जरूरी नहीं कि पूरा codomain range बन जाए।
The outputs are (5,0,-3,-4,-3), so distinct values ({-4,-3,0,5}) form the range. Repeated values are not written in a set.
Step 2
Why this answer is correct
The correct answer is A. \({-4,-3,0,5}). The outputs are (5,0,-3,-4,-3), so distinct values ({-4,-3,0,5}) form the range. Repeated values are not written in a set.
Step 3
Exam Tip
outputs (5,0,-3,-4,-3) हैं, इसलिए distinct values ({-4,-3,0,5}) range हैं। set में repeated values नहीं लिखते।
Each of the (m) domain elements has (n) choices in the codomain, so total functions are \(n^m\). The exponent is always the domain size.
Step 2
Why this answer is correct
The correct answer is A. \(n^m\). Each of the (m) domain elements has (n) choices in the codomain, so total functions are \(n^m\). The exponent is always the domain size.
Step 3
Exam Tip
domain के हर (m) element के लिए codomain में (n) choices हैं, इसलिए कुल \(n^m\) फलन हैं। exponent हमेशा domain size पर होता है।
Here (|A|=2) and (|B|=3), so functions \(=3^2=9\). In counting, keep codomain size as base and domain size as exponent.
Step 2
Why this answer is correct
The correct answer is A. \(3^2=9\). Here (|A|=2) and (|B|=3), so functions \(=3^2=9\). In counting, keep codomain size as base and domain size as exponent.
Step 3
Exam Tip
यहां (|A|=2) और (|B|=3), इसलिए functions \(=3^2=9\)। counting में base codomain size और exponent domain size रखें।
In \(y^2=x\), for (x=4), both (y=2) and (y=-2) are possible. Two (y)-values for one (x) do not define a function.
Step 2
Why this answer is correct
The correct answer is B. \(y^2=x\) with (x>0). In \(y^2=x\), for (x=4), both (y=2) and (y=-2) are possible. Two (y)-values for one (x) do not define a function.
Step 3
Exam Tip
\(y^2=x\) में (x=4) पर (y=2) और (y=-2) दोनों मिलते हैं। एक (x) की दो (y) values function नहीं बनातीं।
For absolute value, (|-1|=1), (|0|=0), (|1|=1), and (|2|=2). In an ordered pair, the first coordinate remains the input.
Step 2
Why this answer is correct
The correct answer is A. \({(-1,1),(0,0),(1,1),(2,2)}). For absolute value, (|-1|=1), (|0|=0), (|1|=1), and (|2|=2). In an ordered pair, the first coordinate remains the input.
Step 3
Exam Tip
absolute value में (|-1|=1), (|0|=0), (|1|=1), (|2|=2)। ordered pair में first coordinate input ही रहता है।
For every \(x\in A\), (y=5-x) is unique and lies in (A). For an equation relation, check uniqueness to decide function status.
Step 2
Why this answer is correct
The correct answer is A. यह फलन है / It is a function. For every \(x\in A\), (y=5-x) is unique and lies in (A). For an equation relation, check uniqueness to decide function status.
Step 3
Exam Tip
हर \(x\in A\) के लिए (y=5-x) unique और (A) में है। equation relation को function बनाने के लिए uniqueness check करें।
A. (x=2) के लिए (y=1) और (y=2) दोनों possible हैं/For (x=2), both (y=1) and (y=2) are possible
Step 1
Concept
(x=2) gets two images, so uniqueness fails. In divisibility relations, check all divisors in the codomain for one input.
Step 2
Why this answer is correct
The correct answer is A. (x=2) के लिए (y=1) और (y=2) दोनों possible हैं / For (x=2), both (y=1) and (y=2) are possible. (x=2) gets two images, so uniqueness fails. In divisibility relations, check all divisors in the codomain for one input.
Step 3
Exam Tip
(x=2) की दो images बनती हैं, इसलिए uniqueness नहीं रहती। divisibility relations में एक input के सभी divisors in codomain देखें।
(f(1)=3), (f(2)=4), and (f(3)=5). When forming a relation from a rule, write the image of every domain element.
Step 2
Why this answer is correct
The correct answer is A. \({(1,3),(2,4),(3,5)}). (f(1)=3), (f(2)=4), and (f(3)=5). When forming a relation from a rule, write the image of every domain element.
Step 3
Exam Tip
(f(1)=3), (f(2)=4), और (f(3)=5)। rule से relation बनाते समय हर domain element की image लिखें।
The outputs are (2,4,6), so the smallest codomain can be the set of these images. The codomain must contain all possible outputs.
Step 2
Why this answer is correct
The correct answer is A. \({2,4,6}). The outputs are (2,4,6), so the smallest codomain can be the set of these images. The codomain must contain all possible outputs.
Step 3
Exam Tip
outputs (2,4,6) हैं, इसलिए smallest codomain इन्हीं images का set हो सकता है। codomain में सभी possible outputs होने चाहिए।
A. हाँ, क्योंकि हर (x) की exactly one parity है/Yes, because every (x) has exactly one parity
Step 1
Concept
Every number is either even or odd, not both. Many inputs having the same parity is allowed.
Step 2
Why this answer is correct
The correct answer is A. हाँ, क्योंकि हर (x) की exactly one parity है / Yes, because every (x) has exactly one parity. Every number is either even or odd, not both. Many inputs having the same parity is allowed.
Step 3
Exam Tip
हर number या तो even है या odd, दोनों नहीं। many inputs की same parity होना allowed है।
From the ordered pairs, (f(3)=7) and (f(4)=9), so the sum is (16). While reading function values, the input is the first coordinate.
Step 2
Why this answer is correct
The correct answer is A. (16). From the ordered pairs, (f(3)=7) and (f(4)=9), so the sum is (16). While reading function values, the input is the first coordinate.
Step 3
Exam Tip
ordered pairs से (f(3)=7) और (f(4)=9), इसलिए sum (16) है। function value पढ़ते समय input first coordinate होता 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।
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।
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।
\(0^2=0\), \(1^2=1\), 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. (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।
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।
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।
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 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।
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)।
\(|A\times B|=6\), 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।
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. 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. 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. (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. 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. (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. (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।
