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।
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।
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।
(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।
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।
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।
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।
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।
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।
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।
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।
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।
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।
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. 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।
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. \(4\in A\) 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. 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।
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।
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।
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।
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 \(=|B|^{|A|}\)।
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।
