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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।
A. यह (A) से (B) में फलन है/It is a function from (A) to (B)
Step 1
Concept
Each \(x \in A\) has exactly one image, so it is a function. In exams, first check every element of the domain.
Step 2
Why this answer is correct
The correct answer is A. यह (A) से (B) में फलन है / It is a function from (A) to (B). Each \(x \in A\) has exactly one image, so it is a function. In exams, first check every element of the domain.
Step 3
Exam Tip
हर \(x \in A\) की ठीक एक छवि है, इसलिए यह फलन है। परीक्षा में पहले प्रांत के हर अवयव को जांचें।
A. क्योंकि (1) की दो अलग छवियां हैं/Because (1) has two different images
Step 1
Concept
In a function, one first element cannot have two different images. In exams, watch for repeated first elements.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि (1) की दो अलग छवियां हैं / Because (1) has two different images. In a function, one first element cannot have two different images. In exams, watch for repeated first elements.
Step 3
Exam Tip
फलन में किसी एक प्रथम अवयव की दो अलग छवियां नहीं हो सकतीं। परीक्षा में दोहराए गए प्रथम अवयव पर ध्यान दें।
For a function, each element of (A) gets one image in (B), so there are \(3^2\) functions. Do not confuse this with all relations.
Step 2
Why this answer is correct
The correct answer is B. \(3^2\). For a function, each element of (A) gets one image in (B), so there are \(3^2\) functions. Do not confuse this with all relations.
Step 3
Exam Tip
फलन के लिए (A) के हर अवयव को (B) की एक छवि मिलती है, इसलिए \(3^2\) फलन हैं। सभी संबंधों की संख्या से भ्रमित न हों।
B. यह फलन है और यह अनेक-से-एक हो सकता है/It is a function and it can be many-one
Step 1
Concept
Different first elements may have the same image, so a many-one function is valid. The common mistake is treating the same image as an error.
Step 2
Why this answer is correct
The correct answer is B. यह फलन है और यह अनेक-से-एक हो सकता है / It is a function and it can be many-one. Different first elements may have the same image, so a many-one function is valid. The common mistake is treating the same image as an error.
Step 3
Exam Tip
अलग प्रथम अवयवों की समान छवि हो सकती है, इसलिए अनेक-से-एक फलन मान्य है। गलती यह है कि समान छवि को दोष मान लिया जाता है।
For every \(x \in A\), (y=x+1) gives exactly one value in (B). Hence it is a function from (A) to (B).
Step 2
Why this answer is correct
The correct answer is B. (A) से (B) में फलन / A function from (A) to (B). For every \(x \in A\), (y=x+1) gives exactly one value in (B). Hence it is a function from (A) to (B).
Step 3
Exam Tip
हर \(x \in A\) के लिए (y=x+1) का ठीक एक मान (B) में है। इसलिए यह (A) से (B) में फलन है।
A. \(R=\{(1,1),(2,4),(3,9)\}\) और यह फलन है/\(R=\{(1,1),(2,4),(3,9)\}\) and it is a function
Step 1
Concept
Using \(y=x^2\), each (x) gets exactly one image. Even in formula-based relations, check the whole domain.
Step 2
Why this answer is correct
The correct answer is A. \(R=\{(1,1),(2,4),(3,9)\}\) और यह फलन है / \(R=\{(1,1),(2,4),(3,9)\}\) and it is a function. Using \(y=x^2\), each (x) gets exactly one image. Even in formula-based relations, check the whole domain.
Step 3
Exam Tip
\(y=x^2\) रखने पर हर (x) की ठीक एक छवि मिलती है। सूत्र आधारित संबंध में भी प्रांत पूरा जांचना जरूरी है।
यदि \(f:{1,2,3,4}\to{0,1}\) और (f(x)) को (x) के सम-विषम होने से परिभाषित किया गया है, (f(x)=0) जब (x) सम है और (f(x)=1) जब (x) विषम है, तो परिसर क्या है?
B. यह फलन नहीं है क्योंकि (2) की दो छवियां हैं/It is not a function because (2) has two images
Step 1
Concept
The element (2) is linked to both (3) and (4), so the single-image condition fails. Having all first elements is not enough.
Step 2
Why this answer is correct
The correct answer is B. यह फलन नहीं है क्योंकि (2) की दो छवियां हैं / It is not a function because (2) has two images. The element (2) is linked to both (3) and (4), so the single-image condition fails. Having all first elements is not enough.
Step 3
Exam Tip
(2) से (3) और (4) दोनों जुड़े हैं, इसलिए एकल छवि की शर्त टूटती है। केवल सभी प्रथम अवयव आना पर्याप्त नहीं है।
A. हर \(a \in A\) के लिए ठीक एक ((a,b)) वाला उपसमुच्चय/A subset with exactly one ((a,b)) for every \(a \in A\)
Step 1
Concept
A function is a special relation where each domain element has exactly one image. This is the key point in definition-based questions.
Step 2
Why this answer is correct
The correct answer is A. हर \(a \in A\) के लिए ठीक एक ((a,b)) वाला उपसमुच्चय / A subset with exactly one ((a,b)) for every \(a \in A\). A function is a special relation where each domain element has exactly one image. This is the key point in definition-based questions.
Step 3
Exam Tip
फलन संबंध का विशेष रूप है जिसमें हर प्रांत अवयव की ठीक एक छवि होती है। यह परिभाषा आधारित प्रश्नों में सबसे महत्वपूर्ण बिंदु है।
The number of functions is \(2^3=8\) and the number of relations is \(2^{3\cdot2}=64\). Keep the order of the ratio in mind.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{8}{64}\). The number of functions is \(2^3=8\) and the number of relations is \(2^{3\cdot2}=64\). Keep the order of the ratio in mind.
Step 3
Exam Tip
फलनों की संख्या \(2^3=8\) और संबंधों की संख्या \(2^{3\cdot2}=64\) है। अनुपात लेते समय क्रम को ध्यान रखें।
B. यह फलन है क्योंकि हर (x) की ठीक एक छवि है/It is a function because every (x) has exactly one image
Step 1
Concept
Two different (x)-values having the same image does not violate being a function. The condition is only one image for each (x).
Step 2
Why this answer is correct
The correct answer is B. यह फलन है क्योंकि हर (x) की ठीक एक छवि है / It is a function because every (x) has exactly one image. Two different (x)-values having the same image does not violate being a function. The condition is only one image for each (x).
Step 3
Exam Tip
दो अलग (x) की समान छवि होना फलन के विरुद्ध नहीं है। शर्त केवल हर (x) की एक छवि है।
A. क्योंकि \(f(3)=6\notin{1,2,3,4,5}\)/Because \(f(3)=6\notin{1,2,3,4,5}\)
Step 1
Concept
Every image of a function must lie in the codomain. Here the image of (3) is (6), outside the codomain.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि \(f(3)=6\notin{1,2,3,4,5}\) / Because \(f(3)=6\notin{1,2,3,4,5}\). Every image of a function must lie in the codomain. Here the image of (3) is (6), outside the codomain.
Step 3
Exam Tip
फलन की हर छवि सहप्रांत में होनी चाहिए। यहां (3) की छवि (6) सहप्रांत से बाहर है।
In option (B), each of (1,2,3) appears exactly once as a first element. This identifies a valid function.
Step 2
Why this answer is correct
The correct answer is B. \({(1,x),(2,y),(3,z)}\). In option (B), each of (1,2,3) appears exactly once as a first element. This identifies a valid function.
Step 3
Exam Tip
विकल्प (B) में (1,2,3) प्रत्येक प्रथम अवयव के रूप में ठीक एक बार आया है। यही वैध फलन की पहचान है।
A. क्योंकि (x=4) पर (y=2) और (y=-2) दोनों संभव हैं/Because for (x=4), both (y=2) and (y=-2) are possible
Step 1
Concept
For the same (x), two different (y)-values occur, so it is not a function. In square relations, check positive and negative roots.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि (x=4) पर (y=2) और (y=-2) दोनों संभव हैं / Because for (x=4), both (y=2) and (y=-2) are possible. For the same (x), two different (y)-values occur, so it is not a function. In square relations, check positive and negative roots.
Step 3
Exam Tip
एक ही (x) के लिए दो अलग (y) मिल रहे हैं, इसलिए यह फलन नहीं है। वर्ग वाले संबंधों में धन और ऋण मूल जांचें।
B. यह फलन है और इसका परिसर (B) है/It is a function and its range is (B)
Step 1
Concept
Every domain element has one image and both (a,b) occur. Hence it is a function and its range is (B).
Step 2
Why this answer is correct
The correct answer is B. यह फलन है और इसका परिसर (B) है / It is a function and its range is (B). Every domain element has one image and both (a,b) occur. Hence it is a function and its range is (B).
Step 3
Exam Tip
हर प्रांत अवयव की एक छवि है और दोनों (a,b) प्राप्त होते हैं। अतः यह फलन है और परिसर (B) है।
A. क्योंकि (x=2) पर मान अपरिभाषित है/Because it is undefined at (x=2)
Step 1
Concept
At (x=2), the denominator becomes zero, so the value is undefined. A function must have a value at every domain element.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि (x=2) पर मान अपरिभाषित है / Because it is undefined at (x=2). At (x=2), the denominator becomes zero, so the value is undefined. A function must have a value at every domain element.
Step 3
Exam Tip
(x=2) रखने पर हर शून्य हो जाता है, इसलिए मान परिभाषित नहीं है। फलन के लिए हर प्रांत अवयव पर मान होना चाहिए।
B. हर फलन संबंध होता है, पर हर संबंध फलन नहीं होता/Every function is a relation, but every relation is not a function
Step 1
Concept
A function is a special type of relation with the single-image condition. This basic definition is asked often in exams.
Step 2
Why this answer is correct
The correct answer is B. हर फलन संबंध होता है, पर हर संबंध फलन नहीं होता / Every function is a relation, but every relation is not a function. A function is a special type of relation with the single-image condition. This basic definition is asked often in exams.
Step 3
Exam Tip
फलन संबंध का विशेष प्रकार है जिसमें एकल छवि की शर्त होती है। यह मूल परिभाषा परीक्षा में बार-बार पूछी जाती है।
A. सहप्रांत \(B=\{2,4,6,8\}\), परिसर ({2,4,6})/Codomain \(B=\{2,4,6,8\}\), range ({2,4,6})
Step 1
Concept
The codomain is the given set (B), while the range is the set of actual images. Here (8) is in the codomain but not in the range.
Step 2
Why this answer is correct
The correct answer is A. सहप्रांत \(B=\{2,4,6,8\}\), परिसर ({2,4,6}) / Codomain \(B=\{2,4,6,8\}\), range ({2,4,6}). The codomain is the given set (B), while the range is the set of actual images. Here (8) is in the codomain but not in the range.
Step 3
Exam Tip
सहप्रांत दिया हुआ (B) है, जबकि परिसर वास्तविक प्राप्त छवियां हैं। यहां (8) सहप्रांत में है पर परिसर में नहीं।
Two functions are equal when their domain, codomain, and values at every element are the same. The given condition gives (f=g).
Step 2
Why this answer is correct
The correct answer is A. (f=g). Two functions are equal when their domain, codomain, and values at every element are the same. The given condition gives (f=g).
Step 3
Exam Tip
दो फलन समान होते हैं जब उनका प्रांत, सहप्रांत और हर अवयव पर मान समान हो। यहां दी गई शर्त (f=g) देती है।
(y=3x-5) gives exactly one real (y) for every real (x). The other options give zero or multiple (y)-values for some (x).
Step 2
Why this answer is correct
The correct answer is C. \(y=3x-5\). (y=3x-5) gives exactly one real (y) for every real (x). The other options give zero or multiple (y)-values for some (x).
Step 3
Exam Tip
(y=3x-5) हर वास्तविक (x) के लिए ठीक एक वास्तविक (y) देता है। बाकी विकल्पों में कुछ (x) पर शून्य या अनेक (y) मिलते हैं।
A. यह \(\mathbb{R}\) से \(\mathbb{R}\) में फलन है/It is a function from \(\mathbb{R}\) to \(\mathbb{R}\)
Step 1
Concept
For every real (x), (|x|) has one real value. Different inputs with the same image do not make a function invalid.
Step 2
Why this answer is correct
The correct answer is A. यह \(\mathbb{R}\) से \(\mathbb{R}\) में फलन है / It is a function from \(\mathbb{R}\) to \(\mathbb{R}\). For every real (x), (|x|) has one real value. Different inputs with the same image do not make a function invalid.
Step 3
Exam Tip
हर वास्तविक (x) के लिए (|x|) का एक ही वास्तविक मान होता है। समान छवि वाले अलग इनपुट फलन को अमान्य नहीं बनाते।
A. किसी \(a\in A\) के लिए कोई छवि नहीं है/Some \(a\in A\) has no image
Step 1
Concept
If any domain element has no image, it is not a function. Same images or extra codomain elements are usually not a problem.
Step 2
Why this answer is correct
The correct answer is A. किसी \(a\in A\) के लिए कोई छवि नहीं है / Some \(a\in A\) has no image. If any domain element has no image, it is not a function. Same images or extra codomain elements are usually not a problem.
Step 3
Exam Tip
प्रांत का कोई अवयव बिना छवि के रह जाए तो फलन नहीं बनता। समान छवि या अतिरिक्त सहप्रांत अवयव सामान्यतः समस्या नहीं हैं।
A. क्योंकि \(f(4)=4\notin B\)/Because \(f(4)=4\notin B\)
Step 1
Concept
The codomain (B) does not contain (4), so the image of (4) goes outside. In a function, each image must lie in the codomain.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि \(f(4)=4\notin B\) / Because \(f(4)=4\notin B\). The codomain (B) does not contain (4), so the image of (4) goes outside. In a function, each image must lie in the codomain.
Step 3
Exam Tip
सहप्रांत (B) में (4) नहीं है, इसलिए (4) की छवि बाहर चली जाती है। फलन में छवि सहप्रांत के अंदर होनी चाहिए।
A. हाँ, क्योंकि हर (x) के लिए \(y=\sqrt[3]{x}\) अद्वितीय है/Yes, because for every (x), \(y=\sqrt[3]{x}\) is unique
Step 1
Concept
Every real (x) has a unique real cube root. Hence this relation is a function from \(\mathbb{R}\) to \(\mathbb{R}\).
Step 2
Why this answer is correct
The correct answer is A. हाँ, क्योंकि हर (x) के लिए \(y=\sqrt[3]{x}\) अद्वितीय है / Yes, because for every (x), \(y=\sqrt[3]{x}\) is unique. Every real (x) has a unique real cube root. Hence this relation is a function from \(\mathbb{R}\) to \(\mathbb{R}\).
Step 3
Exam Tip
हर वास्तविक (x) का वास्तविक घनमूल अद्वितीय होता है। इसलिए यह संबंध \(\mathbb{R}\) से \(\mathbb{R}\) में फलन है।
There are \(2^3=8\) total functions, and two constant functions have range only ({0}) or ({1}). Hence (8-2=6) functions remain.
Step 2
Why this answer is correct
The correct answer is A. (6). There are \(2^3=8\) total functions, and two constant functions have range only ({0}) or ({1}). Hence (8-2=6) functions remain.
Step 3
Exam Tip
कुल फलन \(2^3=8\) हैं और दो स्थिर फलनों में परिसर केवल ({0}) या ({1}) है। इसलिए (8-2=6) फलन बचते हैं।
Two arrows from one domain element break the single-image condition. In arrow diagrams, this is the fastest check.
Step 2
Why this answer is correct
The correct answer is A. यह फलन नहीं है / It is not a function. Two arrows from one domain element break the single-image condition. In arrow diagrams, this is the fastest check.
Step 3
Exam Tip
एक प्रांत अवयव से दो तीर एकल छवि की शर्त तोड़ते हैं। तीर आरेख में यह सबसे तेज जांच है।
B. नहीं, क्योंकि सहप्रांत अलग हैं/No, because the codomains are different
Step 1
Concept
Equality of functions also requires the codomain to be the same. The same formula alone is not sufficient.
Step 2
Why this answer is correct
The correct answer is B. नहीं, क्योंकि सहप्रांत अलग हैं / No, because the codomains are different. Equality of functions also requires the codomain to be the same. The same formula alone is not sufficient.
Step 3
Exam Tip
फलन की समानता में सहप्रांत भी समान होना चाहिए। केवल समान सूत्र पर्याप्त नहीं है।
In option (A), every domain element has exactly one image. In the others, an image is missing or two images occur.
Step 2
Why this answer is correct
The correct answer is A. \({(-1,1),(0,0),(1,1)}\). In option (A), every domain element has exactly one image. In the others, an image is missing or two images occur.
Step 3
Exam Tip
विकल्प (A) में प्रत्येक प्रांत अवयव की ठीक एक छवि है। अन्य विकल्पों में या तो छवि छूटी है या दो छवियां हैं।
A. यह फलन है क्योंकि \(x+5\in\mathbb{Z}\) हर \(x\in\mathbb{Z}\) के लिए/It is a function because \(x+5\in\mathbb{Z}\) for every \(x\in\mathbb{Z}\)
Step 1
Concept
Adding (5) to an integer gives an integer and the value is unique. An infinite domain is not a problem for being a function.
Step 2
Why this answer is correct
The correct answer is A. यह फलन है क्योंकि \(x+5\in\mathbb{Z}\) हर \(x\in\mathbb{Z}\) के लिए / It is a function because \(x+5\in\mathbb{Z}\) for every \(x\in\mathbb{Z}\). Adding (5) to an integer gives an integer and the value is unique. An infinite domain is not a problem for being a function.
Step 3
Exam Tip
पूर्णांक में (5) जोड़ने पर फिर पूर्णांक मिलता है और मान अद्वितीय होता है। अनंत प्रांत फलन होने में बाधा नहीं है।
There are (3) choices for the common image of the first three elements and (3) choices for the image of (4). Total functions are \(3\cdot3=9\).
Step 2
Why this answer is correct
The correct answer is B. (9). There are (3) choices for the common image of the first three elements and (3) choices for the image of (4). Total functions are \(3\cdot3=9\).
Step 3
Exam Tip
पहले तीन अवयवों की समान छवि के लिए (3) विकल्प हैं और (4) की छवि के लिए (3) विकल्प हैं। कुल \(3\cdot3=9\) फलन हैं।
A. क्योंकि (x<0) के लिए \(\sqrt{x}\) वास्तविक नहीं है/Because for (x<0), \(\sqrt{x}\) is not real
Step 1
Concept
With the whole \(\mathbb{R}\) as domain, negative (x)-values have no real image. Hence it is not a function on the given domain.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि (x<0) के लिए \(\sqrt{x}\) वास्तविक नहीं है / Because for (x<0), \(\sqrt{x}\) is not real. With the whole \(\mathbb{R}\) as domain, negative (x)-values have no real image. Hence it is not a function on the given domain.
Step 3
Exam Tip
पूरे \(\mathbb{R}\) को प्रांत लेने पर ऋणात्मक (x) के लिए वास्तविक मान नहीं मिलता। इसलिए यह दिए गए प्रांत पर फलन नहीं है।
A. हर \(x\in A\) के लिए (f(x)=c) होता है, जहां \(c\in B\)/For every \(x\in A\), (f(x)=c), where \(c\in B\)
Step 1
Concept
In a constant function, all domain elements have the same codomain element as image. It is a simple valid many-one function.
Step 2
Why this answer is correct
The correct answer is A. हर \(x\in A\) के लिए (f(x)=c) होता है, जहां \(c\in B\) / For every \(x\in A\), (f(x)=c), where \(c\in B\). In a constant function, all domain elements have the same codomain element as image. It is a simple valid many-one function.
Step 3
Exam Tip
स्थिर फलन में सभी प्रांत अवयवों की छवि एक ही सहप्रांत अवयव होती है। यह वैध अनेक-से-एक फलन का सरल उदाहरण है।
For a constant function, the common image is chosen from (B). Since (B) has (3) choices, there are (3) constant functions.
Step 2
Why this answer is correct
The correct answer is B. (3). For a constant function, the common image is chosen from (B). Since (B) has (3) choices, there are (3) constant functions.
Step 3
Exam Tip
स्थिर फलन के लिए सभी अवयवों की समान छवि (B) से चुनी जाती है। (B) में (3) विकल्प हैं, इसलिए (3) स्थिर फलन हैं।
B. यह फलन नहीं है क्योंकि (2) की दो छवियां हैं/It is not a function because (2) has two images
Step 1
Concept
In a function every element of (A) must have exactly one image. In exams first check whether any input has two images.
Step 2
Why this answer is correct
The correct answer is B. यह फलन नहीं है क्योंकि (2) की दो छवियां हैं / It is not a function because (2) has two images. In a function every element of (A) must have exactly one image. In exams first check whether any input has two images.
Step 3
Exam Tip
फलन में (A) के हर अवयव की ठीक एक छवि होनी चाहिए। परीक्षा में पहले किसी इनपुट की दो छवियां जांचें।
B. क्योंकि (4) की दो छवियां हैं/Because (4) has two images
Step 1
Concept
For (4), both (y=2) and (y=-2) occur. Two images for one input reject a function.
Step 2
Why this answer is correct
The correct answer is B. क्योंकि (4) की दो छवियां हैं / Because (4) has two images. For (4), both (y=2) and (y=-2) occur. Two images for one input reject a function.
Step 3
Exam Tip
(4) के लिए (y=2) और (y=-2) दोनों मिलते हैं। एक इनपुट की दो छवियां फलन को अस्वीकार कर देती हैं।
A. डोमेन को \([2,\infty\)) कर दें/Change the domain to \([2,\infty\))
Step 1
Concept
The value \(\sqrt{x-2}\) is real only when \(x\ge 2\). In such questions check domain validity before the formula.
Step 2
Why this answer is correct
The correct answer is A. डोमेन को \([2,\infty\)) कर दें / Change the domain to \([2,\infty\)). The value \(\sqrt{x-2}\) is real only when \(x\ge 2\). In such questions check domain validity before the formula.
Step 3
Exam Tip
\(\sqrt{x-2}\) वास्तविक तभी है जब \(x\ge 2\) हो। ऐसे प्रश्नों में सूत्र से पहले डोमेन की वैधता जांचें।
The values of (p) and (q) are fixed, so (r,s) have \(2^2=4\) choices. Remove fixed inputs and apply the formula to the rest.
Step 2
Why this answer is correct
The correct answer is B. (4). The values of (p) and (q) are fixed, so (r,s) have \(2^2=4\) choices. Remove fixed inputs and apply the formula to the rest.
Step 3
Exam Tip
(p) और (q) तय हैं, इसलिए (r,s) के लिए \(2^2=4\) विकल्प हैं। तय इनपुट हटाकर बाकी पर सूत्र लगाएं।
For every \(x\in X\), exactly one (y) is obtained. Here the reverse square relation becomes a function on the given finite sets.
Step 2
Why this answer is correct
The correct answer is A. यह फलन है / It is a function. For every \(x\in X\), exactly one (y) is obtained. Here the reverse square relation becomes a function on the given finite sets.
Step 3
Exam Tip
हर \(x\in X\) के लिए ठीक एक (y) मिलता है। यहां उल्टा वर्ग संबंध भी दिए गए सीमित समुच्चयों में फलन बन रहा है।
In option (C), every element of (A) appears exactly once as the first component. Many inputs may have the same image in a function.
Step 2
Why this answer is correct
The correct answer is C. ({(1,6),(2,6),(3,6)}). In option (C), every element of (A) appears exactly once as the first component. Many inputs may have the same image in a function.
Step 3
Exam Tip
विकल्प (C) में (A) का हर अवयव ठीक एक बार प्रथम घटक के रूप में आता है। फलन में कई इनपुटों की एक ही छवि हो सकती है।
B. यह फलन है और परिसर ({3,4,6,12}) है/It is a function and range is ({3,4,6,12})
Step 1
Concept
For each given (x), \(\frac{12}{x}\) is a natural number. The obtained values are ({12,6,4,3}).
Step 2
Why this answer is correct
The correct answer is B. यह फलन है और परिसर ({3,4,6,12}) है / It is a function and range is ({3,4,6,12}). For each given (x), \(\frac{12}{x}\) is a natural number. The obtained values are ({12,6,4,3}).
Step 3
Exam Tip
प्रत्येक दिए गए (x) के लिए \(\frac{12}{x}\) प्राकृतिक संख्या है। प्राप्त मान ({12,6,4,3}) हैं।
The denominator needs \(x^2-9\ne0\), so \(x\ne -3,3\). In rational functions never allow the denominator to be zero.
Step 2
Why this answer is correct
The correct answer is B. \(\mathbb{R}-{-3,3}\). The denominator needs \(x^2-9\ne0\), so \(x\ne -3,3\). In rational functions never allow the denominator to be zero.
Step 3
Exam Tip
हर में \(x^2-9\ne0\) चाहिए, इसलिए \(x\ne -3,3\)। भिन्न वाले फलनों में हर को शून्य न होने दें।
B. यह फलन है और परिसर ({0,1}) है/It is a function and range is ({0,1})
Step 1
Concept
Odd numbers have image (1) and the even number has image (0). Having the same image is not a problem for a function.
Step 2
Why this answer is correct
The correct answer is B. यह फलन है और परिसर ({0,1}) है / It is a function and range is ({0,1}). Odd numbers have image (1) and the even number has image (0). Having the same image is not a problem for a function.
Step 3
Exam Tip
विषम संख्याओं की छवि (1) और सम संख्या की छवि (0) है। समान छवि होना फलन के लिए बाधा नहीं है।
The expression (ax+6) stays non-negative for all real (x) only when (a=0). For a square-root function on all \(\mathbb{R}\), the linear part must be constant non-negative.
Step 2
Why this answer is correct
The correct answer is A. (\ a=0). The expression (ax+6) stays non-negative for all real (x) only when (a=0). For a square-root function on all \(\mathbb{R}\), the linear part must be constant non-negative.
Step 3
Exam Tip
(ax+6) सभी वास्तविक (x) के लिए अऋण तभी रहेगा जब (a=0) हो। पूरे \(\mathbb{R}\) पर मूल फलन में रैखिक भाग स्थिर अऋण होना चाहिए।
For (2,3,5), the value is (1), and for (1,4), the value is (0). The range is the part of the codomain that is actually obtained.
Step 2
Why this answer is correct
The correct answer is C. ({0,1}). For (2,3,5), the value is (1), and for (1,4), the value is (0). The range is the part of the codomain that is actually obtained.
Step 3
Exam Tip
(2,3,5) के लिए मान (1) और (1,4) के लिए मान (0) है। परिसर सहप्रांत का वह भाग है जो सच में प्राप्त होता है।
A. फलन है और परिसर ({1,2,3,4}) है/It is a function and range is ({1,2,3,4})
Step 1
Concept
For every (x), (y=5-x) is unique and lies in (B). In finite sets check each input image separately.
Step 2
Why this answer is correct
The correct answer is A. फलन है और परिसर ({1,2,3,4}) है / It is a function and range is ({1,2,3,4}). For every (x), (y=5-x) is unique and lies in (B). In finite sets check each input image separately.
Step 3
Exam Tip
हर (x) के लिए (y=5-x) अद्वितीय और (B) में है। सीमित समुच्चय में हर इनपुट की छवि अलग से जांचें।
A. प्रत्येक \(a\in A\) के लिए ठीक एक \(b\in B\) ऐसा है कि \((a,b)\in f\)/For each \(a\in A\), exactly one \(b\in B\) has \((a,b)\in f\)
Step 1
Concept
A function is a special relation where each element of the first set maps exactly once. Every codomain element need not occur.
Step 2
Why this answer is correct
The correct answer is A. प्रत्येक \(a\in A\) के लिए ठीक एक \(b\in B\) ऐसा है कि \((a,b)\in f\) / For each \(a\in A\), exactly one \(b\in B\) has \((a,b)\in f\). A function is a special relation where each element of the first set maps exactly once. Every codomain element need not occur.
Step 3
Exam Tip
फलन संबंध का विशेष रूप है जिसमें प्रथम समुच्चय का हर अवयव ठीक एक बार मैप होता है। सहप्रांत का हर अवयव आना आवश्यक नहीं है।
A. यह (A) से (B) में फलन है और परिसर ({2,3,4}) है/It is a function from (A) to (B) and range is ({2,3,4})
Step 1
Concept
The images of (1,2,3) are (2,3,4), all lying in (B). Remember the difference between codomain and range.
Step 2
Why this answer is correct
The correct answer is A. यह (A) से (B) में फलन है और परिसर ({2,3,4}) है / It is a function from (A) to (B) and range is ({2,3,4}). The images of (1,2,3) are (2,3,4), all lying in (B). Remember the difference between codomain and range.
Step 3
Exam Tip
(1,2,3) की छवियां क्रमशः (2,3,4) हैं और सभी (B) में हैं। सहप्रांत और परिसर में अंतर याद रखें।
A. ((2,3)) या ((2,4)) में से एक हटाएं/Remove one of ((2,3)) or ((2,4))
Step 1
Concept
Only (2) has two images, so removing one of them is enough. To make a function, give each first component exactly one image.
Step 2
Why this answer is correct
The correct answer is A. ((2,3)) या ((2,4)) में से एक हटाएं / Remove one of ((2,3)) or ((2,4)). Only (2) has two images, so removing one of them is enough. To make a function, give each first component exactly one image.
Step 3
Exam Tip
केवल (2) की दो छवियां हैं, इसलिए उनमें से एक हटाना पर्याप्त है। फलन बनाने के लिए हर प्रथम घटक को ठीक एक छवि दें।
For (f(x)=\frac{1}{x}), the value is undefined at (x=0). A function from \(\mathbb{R}\) must have a value for every real (x).
Step 2
Why this answer is correct
The correct answer is B. (f(x)=\frac{1}{x}). For (f(x)=\frac{1}{x}), the value is undefined at (x=0). A function from \(\mathbb{R}\) must have a value for every real (x).
Step 3
Exam Tip
(f(x)=\frac{1}{x}) में (x=0) पर मान परिभाषित नहीं है। \(\mathbb{R}\) से फलन के लिए हर वास्तविक (x) पर मान चाहिए।
There are (3) choices for the common value of (f(1)=f(2)) and \(3^2\) choices for (3,4). Total functions are \(3\cdot3^2=27\).
Step 2
Why this answer is correct
The correct answer is B. (27). There are (3) choices for the common value of (f(1)=f(2)) and \(3^2\) choices for (3,4). Total functions are \(3\cdot3^2=27\).
Step 3
Exam Tip
(f(1)=f(2)) के लिए (3) विकल्प हैं और (3,4) के लिए \(3^2\) विकल्प हैं। कुल \(3\cdot3^2=27\) फलन हैं।
A. यह फलन है और परिसर ({0,1,2}) है/It is a function and range is ({0,1,2})
Step 1
Concept
Each input has only one absolute value, even if two inputs share an image. The obtained values are (0,1,2).
Step 2
Why this answer is correct
The correct answer is A. यह फलन है और परिसर ({0,1,2}) है / It is a function and range is ({0,1,2}). Each input has only one absolute value, even if two inputs share an image. The obtained values are (0,1,2).
Step 3
Exam Tip
हर इनपुट का केवल एक मापांक मान है, भले दो इनपुटों की छवि समान हो। प्राप्त मान (0,1,2) हैं।
B. क्योंकि (x=0) पर मान \(\frac{1}{2}\notin\mathbb{Z}\) है/Because at (x=0), the value is \(\frac{1}{2}\notin\mathbb{Z}\)
Step 1
Concept
Putting (x=0) gives \(\frac{1}{2}\), which is not in the codomain \(\mathbb{Z}\). Every value of a function must lie in the codomain.
Step 2
Why this answer is correct
The correct answer is B. क्योंकि (x=0) पर मान \(\frac{1}{2}\notin\mathbb{Z}\) है / Because at (x=0), the value is \(\frac{1}{2}\notin\mathbb{Z}\). Putting (x=0) gives \(\frac{1}{2}\), which is not in the codomain \(\mathbb{Z}\). Every value of a function must lie in the codomain.
Step 3
Exam Tip
(x=0) देने पर मान \(\frac{1}{2}\) आता है, जो सहप्रांत \(\mathbb{Z}\) में नहीं है। फलन में हर मान सहप्रांत में होना चाहिए।
B. यह फलन है और परिसर ({1,4,9}) है/It is a function and range is ({1,4,9})
Step 1
Concept
Every \(x\in A\) has one image \(x^2\in B\). The element (16) is in the codomain but not in the range.
Step 2
Why this answer is correct
The correct answer is B. यह फलन है और परिसर ({1,4,9}) है / It is a function and range is ({1,4,9}). Every \(x\in A\) has one image \(x^2\in B\). The element (16) is in the codomain but not in the range.
Step 3
Exam Tip
हर \(x\in A\) की एक छवि \(x^2\in B\) है। (16) सहप्रांत में है पर परिसर में नहीं आता।
B. नहीं, क्योंकि \(f(4)=8\notin{1,2,3,4}\)/No, because \(f(4)=8\notin{1,2,3,4}\)
Step 1
Concept
Here (f(4)=16-12+4=8), which is not in the codomain. For finite domains, checking all values is the safe method.
Step 2
Why this answer is correct
The correct answer is B. नहीं, क्योंकि \(f(4)=8\notin{1,2,3,4}\) / No, because \(f(4)=8\notin{1,2,3,4}\). Here (f(4)=16-12+4=8), which is not in the codomain. For finite domains, checking all values is the safe method.
Step 3
Exam Tip
(f(4)=16-12+4=8) है, जो सहप्रांत में नहीं है। सीमित प्रांत में सभी मानों की जांच करना सुरक्षित तरीका है।
In option (B), both (0) and (1) have image (1), and every input has exactly one image. A common image does not prevent a function.
Step 2
Why this answer is correct
The correct answer is B. ({(0,1),(1,1),(2,3)}). In option (B), both (0) and (1) have image (1), and every input has exactly one image. A common image does not prevent a function.
Step 3
Exam Tip
विकल्प (B) में (0) और (1) दोनों की छवि (1) है और हर इनपुट की एक ही छवि है। समान छवि फलन को नहीं रोकती।
A. क्योंकि (x=2) पर सूत्र अपरिभाषित है/Because the formula is undefined at (x=2)
Step 1
Concept
Even though simplification looks like (x+2), the original formula is undefined at (x=2). When deciding domain, check the original denominator.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि (x=2) पर सूत्र अपरिभाषित है / Because the formula is undefined at (x=2). Even though simplification looks like (x+2), the original formula is undefined at (x=2). When deciding domain, check the original denominator.
Step 3
Exam Tip
भले सरलीकरण (x+2) जैसा दिखे, मूल सूत्र (x=2) पर परिभाषित नहीं है। प्रांत तय करते समय मूल हर को देखें।
Each (x) has a unique real cube root lying in the given (Y). A cube-root relation does not give two values like a square-root relation.
Step 2
Why this answer is correct
The correct answer is A. यह फलन है / It is a function. Each (x) has a unique real cube root lying in the given (Y). A cube-root relation does not give two values like a square-root relation.
Step 3
Exam Tip
हर (x) का वास्तविक घनमूल अद्वितीय है और दिए गए (Y) में है। घनमूल संबंध वर्गमूल वाले संबंध जैसा दो मान नहीं देता।
In option (C), (4) has two images (a) and (b). Having all first components is not enough; uniqueness is also required.
Step 2
Why this answer is correct
The correct answer is C. ({(1,a),(2,b),(3,c),(4,a),(4,b)}). In option (C), (4) has two images (a) and (b). Having all first components is not enough; uniqueness is also required.
Step 3
Exam Tip
विकल्प (C) में (4) की दो छवियां (a) और (b) हैं। केवल सभी प्रथम घटकों का होना काफी नहीं, अद्वितीयता भी चाहिए।
A. क्योंकि (1) की पूर्वछवियां (1) और (-1) दोनों हैं/Because (1) has preimages (1) and (-1)
Step 1
Concept
In the inverse relation, (1) is related to both (1) and (-1). Two images for one input make the inverse relation not a function.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि (1) की पूर्वछवियां (1) और (-1) दोनों हैं / Because (1) has preimages (1) and (-1). In the inverse relation, (1) is related to both (1) and (-1). Two images for one input make the inverse relation not a function.
Step 3
Exam Tip
उल्टे संबंध में (1) से (1) और (-1) दोनों जुड़ते हैं। एक इनपुट की दो छवियां होने से उल्टा संबंध फलन नहीं रहता।
The value (0) is produced by even inputs, namely (2) and (4). A preimage contains elements of the domain, not of the codomain.
Step 2
Why this answer is correct
The correct answer is B. ({2,4}). The value (0) is produced by even inputs, namely (2) and (4). A preimage contains elements of the domain, not of the codomain.
Step 3
Exam Tip
(0) वे इनपुट देते हैं जो सम हैं, यानी (2) और (4)। पूर्वछवि में डोमेन के अवयव आते हैं, सहप्रांत के नहीं।
A. क्योंकि \(f(2)=4\notin B\)/Because \(f(2)=4\notin B\)
Step 1
Concept
The image of (2) is (4), which is not in the codomain (B). For a function, every image must lie in the codomain.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि \(f(2)=4\notin B\) / Because \(f(2)=4\notin B\). The image of (2) is (4), which is not in the codomain (B). For a function, every image must lie in the codomain.
Step 3
Exam Tip
(2) की छवि (4) है जो सहप्रांत (B) में नहीं है। फलन के लिए हर छवि सहप्रांत में होनी चाहिए।
For negative (x), the value is (-1), and for positive (x), it is (1). Since (x=0) is not in the domain, there is no issue.
Step 2
Why this answer is correct
The correct answer is A. ({-1,1}). For negative (x), the value is (-1), and for positive (x), it is (1). Since (x=0) is not in the domain, there is no issue.
Step 3
Exam Tip
ऋणात्मक (x) के लिए मान (-1) और धनात्मक (x) के लिए (1) है। (x=0) प्रांत में नहीं है, इसलिए समस्या नहीं है।
For every natural (n), (2n+1) is a natural number. The other options do not give values in \(\mathbb{N}\) for all (n).
Step 2
Why this answer is correct
The correct answer is B. (g(n)=2n+1). For every natural (n), (2n+1) is a natural number. The other options do not give values in \(\mathbb{N}\) for all (n).
Step 3
Exam Tip
प्रत्येक प्राकृतिक (n) के लिए (2n+1) प्राकृतिक संख्या है। अन्य विकल्प सभी (n) के लिए \(\mathbb{N}\) में मान नहीं देते।
A. यह फलन है और परिसर ([0,4]) है/It is a function and range is ([0,4])
Step 1
Concept
Each (x) has a unique image \(x^2\) lying in ([0,4]). Equal outputs do not prevent a relation from being a function.
Step 2
Why this answer is correct
The correct answer is A. यह फलन है और परिसर ([0,4]) है / It is a function and range is ([0,4]). Each (x) has a unique image \(x^2\) lying in ([0,4]). Equal outputs do not prevent a relation from being a function.
Step 3
Exam Tip
हर (x) की छवि \(x^2\) अद्वितीय है और ([0,4]) में है। समान आउटपुट फलन होने में बाधा नहीं है।
C. (f(x)=\begin{cases}x-2,&x\le2\x+3,&x\ge2\end{cases})
Step 1
Concept
In option (C), (x=2) belongs to both rules and gives values (4) and (5). Two values for one input do not define a function.
Step 2
Why this answer is correct
The correct answer is C. (f(x)=\begin{cases}x-2,&x\le2\x+3,&x\ge2\end{cases}). In option (C), (x=2) belongs to both rules and gives values (4) and (5). Two values for one input do not define a function.
Step 3
Exam Tip
विकल्प (C) में (x=2) दोनों नियमों में आता है और मान (4) तथा (5) मिलते हैं। एक ही इनपुट पर दो मान फलन नहीं बनाते।
A function is a subset of \(A\times B\) where one pair is chosen for each element of (A) from (2) choices. Total is \(2^3=8\).
Step 2
Why this answer is correct
The correct answer is B. (8). A function is a subset of \(A\times B\) where one pair is chosen for each element of (A) from (2) choices. Total is \(2^3=8\).
Step 3
Exam Tip
फलन \(A\times B\) का ऐसा उपसमुच्चय है जिसमें (A) के हर अवयव के लिए (2) में से एक जोड़ी चुनी जाती है। कुल \(2^3=8\) हैं।
Here (f(1)=4), (f(2)=2), (f(3)=2), and (f(4)=4), so the range is ({2,4}). For a piecewise rule list every input value.
Step 2
Why this answer is correct
The correct answer is B. ({2,4}). Here (f(1)=4), (f(2)=2), (f(3)=2), and (f(4)=4), so the range is ({2,4}). For a piecewise rule list every input value.
Step 3
Exam Tip
(f(1)=4), (f(2)=2), (f(3)=2), (f(4)=4), इसलिए परिसर ({2,4}) है। खंडित नियम में हर इनपुट का मान लिखें।
B. फलन नहीं है क्योंकि (1) की छवियां (-1) और (1) हैं/It is not a function because (1) has images (-1) and (1)
Step 1
Concept
For (x=1), both (y=-1) and (y=1) are possible. Two (y)-values for one (x) mean the relation is not a function.
Step 2
Why this answer is correct
The correct answer is B. फलन नहीं है क्योंकि (1) की छवियां (-1) और (1) हैं / It is not a function because (1) has images (-1) and (1). For (x=1), both (y=-1) and (y=1) are possible. Two (y)-values for one (x) mean the relation is not a function.
Step 3
Exam Tip
(x=1) के लिए (y=-1) और (y=1) दोनों संभव हैं। एक (x) के दो (y) होने से संबंध फलन नहीं है।
The remainders of (0,1,2,3) are (0,1,2,0). The remainder is unique, so this is a function.
Step 2
Why this answer is correct
The correct answer is A. परिसर ({0,1,2}) है / The range is ({0,1,2}). The remainders of (0,1,2,3) are (0,1,2,0). The remainder is unique, so this is a function.
Step 3
Exam Tip
(0,1,2,3) के शेषफल क्रमशः (0,1,2,0) हैं। शेषफल अद्वितीय होता है, इसलिए यह फलन है।
A. यह पूरे \(\mathbb{R}\) पर कभी फलन नहीं होगा/It will never be a function on all of \(\mathbb{R}\)
Step 1
Concept
The denominator \(x^2-4\) is zero at \(x=\pm2\), so the original formula is undefined there. A zero denominator cannot be fully removed for all real inputs.
Step 2
Why this answer is correct
The correct answer is A. यह पूरे \(\mathbb{R}\) पर कभी फलन नहीं होगा / It will never be a function on all of \(\mathbb{R}\). The denominator \(x^2-4\) is zero at \(x=\pm2\), so the original formula is undefined there. A zero denominator cannot be fully removed for all real inputs.
Step 3
Exam Tip
हर \(x^2-4\) शून्य होता है जब \(x=\pm2\), इसलिए मूल सूत्र वहां अपरिभाषित है। हर की शून्यता को घटाने से पूरी तरह नहीं हटाया जा सकता।
B. \((R={(x,y):y=0\) यदि \(x<3,\ y=1\) यदि \(x\ge3})\)/\((R={(x,y):y=0\) if \(x<3,\ y=1\) if \(x\ge3})\)
Step 1
Concept
Option (B) assigns exactly one value to every (x). The other options miss some inputs or give multiple values.
Step 2
Why this answer is correct
\(The correct answer is B. (R={(x,y):y=0\) यदि \(x<3,\ y=1\) यदि \(x\ge3}) / (R={(x,y):y=0\) if \(x<3,\ y=1\) if \(x\ge3}). Option (B) assigns exactly one value to every (x). The other options miss some inputs or give multiple values.\)
Step 3
Exam Tip
विकल्प (B) हर (x) को ठीक एक मान देता है। बाकी विकल्पों में कुछ इनपुट छूटते हैं या कई मान मिलते हैं।
A. यह फलन है और परिसर ({1,3,5}) है/It is a function and range is ({1,3,5})
Step 1
Concept
The values obtained are (1,3,5), and all lie in the codomain. Not every element of the codomain needs to occur.
Step 2
Why this answer is correct
The correct answer is A. यह फलन है और परिसर ({1,3,5}) है / It is a function and range is ({1,3,5}). The values obtained are (1,3,5), and all lie in the codomain. Not every element of the codomain needs to occur.
Step 3
Exam Tip
मान (1,3,5) मिलते हैं और सभी सहप्रांत में हैं। सहप्रांत के सभी अवयवों का आना आवश्यक नहीं होता।
B. यह फलन नहीं है क्योंकि (A) के अवयवों की छवि नहीं है/It is not a function because elements of (A) have no images
Step 1
Concept
When \(A\ne\emptyset\), the empty relation gives no image to any element of (A). A function requires an image for every input.
Step 2
Why this answer is correct
The correct answer is B. यह फलन नहीं है क्योंकि (A) के अवयवों की छवि नहीं है / It is not a function because elements of (A) have no images. When \(A\ne\emptyset\), the empty relation gives no image to any element of (A). A function requires an image for every input.
Step 3
Exam Tip
जब \(A\ne\emptyset\) हो, तो खाली संबंध (A) के किसी अवयव को छवि नहीं देता। फलन के लिए हर इनपुट की छवि आवश्यक है।
From the empty domain to any set, there is exactly one empty function. The formula \(|B|^{|A|}=3^0=1\) gives the same result.
Step 2
Why this answer is correct
The correct answer is B. (1). From the empty domain to any set, there is exactly one empty function. The formula \(|B|^{|A|}=3^0=1\) gives the same result.
Step 3
Exam Tip
रिक्त प्रांत से किसी भी समुच्चय में एक ही खाली फलन होता है। सूत्र \(|B|^{|A|}=3^0=1\) भी यही देता है।
