{"id":8397,"date":"2024-09-24T10:22:01","date_gmt":"2024-09-24T04:52:01","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=8397"},"modified":"2024-09-24T10:22:01","modified_gmt":"2024-09-24T04:52:01","slug":"ncert-solutions-class-11-maths-chapter-2-relations-and-functions-miscellaneous-exercise","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-11-maths-chapter-2-relations-and-functions-miscellaneous-exercise\/","title":{"rendered":"NCERT Solutions Class 11 Maths Chapter 2 Relations and Functions (Miscellaneous Exercise)"},"content":{"rendered":"

NCERT Solutions Class 11 Maths\u00a0<\/span><\/strong><\/span><\/h2>\n

The NCERT Solutions in English Language for Class 11 Mathematics Chapter – 2 Relations and Functions <\/strong>Exercise – Miscellaneous has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n

Chapter 2 (Relations and Functions)\u00a0<\/span><\/strong><\/span><\/h2>\n\n\n\n
Miscellaneous Exercise – 2<\/span><\/strong><\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

1. The relation\u00a0f<\/em> is defined by \"NCER
\n<\/strong>The relation\u00a0g<\/em> is defined by \"NCER
\n<\/strong>Show that\u00a0f<\/em>\u00a0is a function and\u00a0g\u00a0<\/em>is not a function. <\/strong><\/span><\/p>\n

Solution –\u00a0 <\/strong>The given relation\u00a0f<\/em> is defined as: \"NCER
\n<\/strong>It is seen that for 0 \u2264\u00a0x<\/em> < 3,
\nf<\/em>(x<\/em>) =\u00a0x<\/em>2\u00a0<\/sup>and for 3 <\u00a0x<\/em> \u2264\u00a010,
\nf<\/em>(x<\/em>) = 3x
\n<\/em>Also, at\u00a0x<\/em> = 3\u00a0<\/span><\/p>\n

f<\/em>(x<\/em>) = 32<\/sup>\u00a0= 9 or\u00a0f<\/em>(x<\/em>) = 3 \u00d7 3 = 9
\ni.e., at\u00a0x<\/em>\u00a0= 3,\u00a0f<\/em>(x<\/em>) = 9\u00a0 <\/span><\/p>\n

Hence, for 0\u00a0\u2264\u00a0x<\/em>\u00a0\u2264\u00a010, the images of\u00a0f<\/em>(x<\/em>) are unique.\u00a0 <\/span><\/p>\n

Therefore, the given relation is a function. Now, In the given relation,\u00a0g<\/em> is defined as
\n\"NCER
\n<\/strong>It is seen that, for\u00a0x<\/em> = 2
\ng<\/em>(x<\/em>) = 22<\/sup>\u00a0= 4 and\u00a0g<\/em>(x<\/em>) = 3 \u00d7 2 = 6
\nThus, element 2 of the domain of the relation\u00a0g<\/em> corresponds to two different images, i.e., 4 and 6.
\nTherefore, this relation is not a function.<\/span><\/p>\n

2. If\u00a0f<\/em>(x<\/em>) =\u00a0x<\/em>2<\/sup>, find \"NCER <\/strong><\/span><\/p>\n

Solution –\u00a0<\/strong>Given, f<\/em>(x<\/em>) =\u00a0x<\/em>2
\n<\/sup>Hence, by putting the condition of f(x) in f(1.1) and f(1),\u00a0 we can find the result of the given equation <\/span>
\n\"NCER<\/span><\/p>\n

3. Find the domain of the function\u00a0 \"NCER <\/strong><\/span><\/p>\n

Solution –\u00a0<\/strong>Given function, \"NCER<\/strong><\/span><\/p>\n

\"NCER
\nIt is clearly notified that, the function f is defined for all real numbers except at x = 6 and x = 2 as the denominator becomes zero otherwise.
\nTherefore, the domain of\u00a0f<\/em> is\u00a0R\u00a0\u2013 {2, 6}.\u00a0<\/span><\/p>\n

4. Find the domain and the range of the real function\u00a0f<\/em>\u00a0defined by\u00a0f<\/em>(x) = \u221a(x \u2013 1).
\n<\/strong>Solution –\u00a0<\/strong>Given real function,
\nf(x) = \u221a(x \u2013 1). <\/span>
\nClearly it is notified, \u221a(x \u2013 1) is defined for (x \u2013 1) \u2265 0. <\/span>
\nHence, the function f(x) = \u221a(x \u2013 1) is defined for x \u2265 1. <\/span>
\nSo that, the domain of f is the set of all real numbers greater than or equal to 1. <\/span>
\nDomain of f = [1, \u221e). <\/span>
\nNow, <\/span>
\nAccording to the condition, x \u2265 1 \u21d2 (x \u2013 1) \u2265 0 \u21d2 \u221a(x \u2013 1) \u2265 0 <\/span>
\nThat\u2019s why, the range of f is the set of all real numbers greater than or equal to 0. <\/span>
\nRange of f = [0, \u221e). <\/span>
\nTherefore, the domain of f is R \u2013 {2, 6}. <\/span><\/p>\n

5. Find the domain and the range of the real function\u00a0f<\/em>\u00a0defined by\u00a0f<\/em>\u00a0(x<\/em>) = |x<\/em> \u2013 1|.
\n<\/strong>Solution – <\/strong>Given real function: f(x) = |x \u2013 1| <\/span>
\nClearly it is notified that, the function |x \u2013 1| is defined for all real numbers. <\/span>
\nHence, Domain of f = R <\/span>
\nAlso, according to the condition , for x \u2208 R, |x \u2013 1| assumes all real numbers. <\/span>
\nSo that, the range of f is the set of all non-negative real numbers.<\/span><\/p>\n

6. Let \"NCER<\/strong>be a function from\u00a0R\u00a0into\u00a0R. Determine the range of\u00a0f<\/em>.<\/strong><\/span><\/p>\n

Solution –\u00a0<\/strong>Given function, \"NCER
\n<\/strong>Substituting values and determining the images, we have
\n\"NCER
\nFrom the above equation, the range of f is the set of all second elements. <\/span>
\nIt can be notified that all these elements are greater than or equal to 0 but less than 1. <\/span>
\n[As the denominator is greater than the numerator.] <\/span>
\nOr, We know that, for x \u2208 R, <\/span>
\nx2 <\/sup>\u2265 0 <\/span>
\nThen, <\/span>
\nx2<\/sup>\u00a0+ 1 \u2265 x2
\n<\/sup>1 \u2265 (x2<\/sup>\u00a0\/ (x2<\/sup> + 1)) <\/span>
\nTherefore, the range of f = [0, 1)<\/span><\/p>\n

7. Let\u00a0f<\/em>,\u00a0g<\/em>:\u00a0R\u00a0\u2192\u00a0R\u00a0be defined, respectively by\u00a0f<\/em>(x<\/em>) =\u00a0x\u00a0<\/em>+ 1,\u00a0g<\/em>(x<\/em>) = 2x<\/em>\u00a0\u2013 3. Find\u00a0f<\/em>\u00a0+\u00a0g<\/em>,\u00a0f<\/em>\u00a0\u2013\u00a0g<\/em>\u00a0and\u00a0f\/g<\/em>.
\n<\/strong>Solution –\u00a0<\/strong>According to the question, let us assume, the functions f, g: R \u2192 R is defined as given conditions f(x) = x + 1, g(x) = 2x \u2013 3. <\/span>
\nNow, <\/span>
\nWe find that (f + g) (x) = f(x) + g(x) = (x + 1) + (2x \u2013 3) = 3x \u2013 2 <\/span>
\nSo that, (f + g) (x) = 3x \u2013 2 <\/span>
\nNow, we find that, (f \u2013 g) (x) = f(x) \u2013 g(x) = (x + 1) \u2013 (2x \u2013 3) = x + 1 \u2013 2x + 3 = \u2013 x + 4 <\/span>
\nSo that, (f \u2013 g) (x) = -x + 4 <\/span>
\n(f\/g(x)) = f(x)\/g(x), g(x) \u2260 0, x \u2208 R <\/span>
\n(f\/g(x)) = x + 1\/ 2x \u2013 3, 2x \u2013 3 \u2260 0 <\/span>
\nSo that, (f\/g(x)) = x + 1\/ 2x \u2013 3, x \u2260 3\/2.<\/span><\/p>\n

8. Let\u00a0f\u00a0<\/em>= {(1, 1), (2, 3), (0, \u20131), (\u20131, \u20133)} be a function from\u00a0Z\u00a0to\u00a0Z\u00a0defined by\u00a0f<\/em>(x<\/em>) =\u00a0ax<\/em>\u00a0+\u00a0b<\/em>, for some integers\u00a0a<\/em>,\u00a0b<\/em>. Determine\u00a0a<\/em>,\u00a0b<\/em>.
\n<\/strong>Solution –\u00a0<\/strong>Given,\u00a0f\u00a0<\/em>= {(1, 1), (2, 3), (0, \u20131), (\u20131, \u20133)}
\nAnd the function defined as,\u00a0\u00a0f<\/em>(x<\/em>) =\u00a0ax<\/em>\u00a0+\u00a0b
\n<\/em>For (1, 1)\u00a0\u2208\u00a0f
\n<\/em>We have,\u00a0\u00a0f<\/em>(1) = 1
\nSo,\u00a0a<\/em>\u00a0\u00d7 1 +\u00a0b<\/em> = 1
\na<\/em>\u00a0+\u00a0b<\/em> = 1\u00a0 \u00a0———— (i)
\nAnd for (0, \u20131)\u00a0\u2208\u00a0f
\n<\/em>We have\u00a0f<\/em>(0) = \u20131
\na<\/em>\u00a0\u00d7 0 +\u00a0b<\/em> = \u20131
\nb<\/em> = \u20131
\nOn substituting\u00a0b<\/em> = \u20131 in (i), we get
\na<\/em>\u00a0+ (\u20131) = 1\u00a0\u21d2\u00a0a<\/em> = 1 + 1 = 2.
\nTherefore, the values of\u00a0a<\/em>\u00a0and\u00a0b<\/em>\u00a0are 2 and \u20131, respectively.<\/span><\/p>\n

9. Let R be a relation from\u00a0N\u00a0to\u00a0N\u00a0defined by R = {(a<\/em>,\u00a0b<\/em>):\u00a0a<\/em>,\u00a0b<\/em>\u00a0\u2208\u00a0N\u00a0and\u00a0a<\/em>\u00a0=\u00a0b<\/em>2<\/sup>}. Are the following true?
\n<\/strong>(i) (a<\/em>,\u00a0a<\/em>)\u00a0\u2208\u00a0R, for all\u00a0a\u00a0<\/em>\u2208\u00a0N
\n(ii) (a<\/em>,\u00a0b<\/em>)\u00a0\u2208\u00a0R, implies (b<\/em>,\u00a0a<\/em>) \u2208 R
\n<\/strong>(iii) (a<\/em>,\u00a0b<\/em>)\u00a0\u2208\u00a0R, (b<\/em>,\u00a0c<\/em>)\u00a0\u2208\u00a0R implies (a<\/em>,\u00a0c<\/em>) \u2208 R
\n<\/strong>Justify your answer in each case. <\/strong><\/span><\/p>\n

Solution –\u00a0 <\/strong>Given relation R = {(a<\/em>,\u00a0b<\/em>):\u00a0a<\/em>,\u00a0b<\/em>\u00a0\u2208\u00a0N\u00a0and\u00a0a<\/em>\u00a0=\u00a0b<\/em>2<\/sup>} <\/span><\/p>\n

(i)<\/strong> It can be seen that 2\u00a0\u2208\u00a0N; however, 2\u00a0\u2260\u00a022<\/sup> = 4.
\nThus, the statement \u201c(a<\/em>,\u00a0a<\/em>)\u00a0\u2208\u00a0R, for all\u00a0a\u00a0<\/em>\u2208\u00a0N\u201d\u00a0is not true.<\/span><\/p>\n

(ii)<\/strong> Its clearly seen that (9, 3)\u00a0\u2208\u00a0N\u00a0because 9, 3\u00a0\u2208\u00a0N\u00a0and 9 = 32<\/sup>.
\nNow, 3\u00a0\u2260\u00a092<\/sup> = 81; therefore, (3, 9)\u00a0\u2209\u00a0N
\nThus, the statement \u201c(a<\/em>,\u00a0b<\/em>)\u00a0\u2208\u00a0R, implies (b<\/em>,\u00a0a<\/em>) \u2208 R\u201d is not true.\u00a0<\/span><\/p>\n

(iii)<\/strong> It\u2019s clearly seen that (16, 4) \u2208 R, (4, 2) \u2208 R because 16, 4, 2 \u2208 N and 16 = 42<\/sup>\u00a0and 4 = 22<\/sup>.
\nNow, 16\u00a0\u2260\u00a022<\/sup> = 4; therefore, (16, 2)\u00a0\u2209\u00a0N
\nThus, the statement \u201c(a<\/em>,\u00a0b<\/em>)\u00a0\u2208\u00a0R, (b<\/em>,\u00a0c<\/em>)\u00a0\u2208\u00a0R implies (a<\/em>,\u00a0c<\/em>) \u2208 R\u201d is not true.\u00a0<\/span><\/p>\n

10. Let A = {1, 2, 3, 4}, B = {1, 5, 9, 11, 15, 16} and\u00a0f\u00a0<\/em>= {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}. Are the following true?
\n<\/strong>(i)\u00a0f<\/em>\u00a0is a relation from A to B
\n(ii)\u00a0f<\/em> is a function from A to B
\n<\/strong>Justify your answer in each case. <\/strong><\/span><\/p>\n

Solution –\u00a0<\/strong>Given,
\nA = {1, 2, 3, 4} and B = {1, 5, 9, 11, 15, 16}
\nSo,
\nA \u00d7 B = {(1, 1), (1, 5), (1, 9), (1, 11), (1, 15), (1, 16), (2, 1), (2, 5), (2, 9), (2, 11), (2, 15), (2, 16), (3, 1), (3, 5), (3, 9), (3, 11), (3, 15), (3, 16), (4, 1), (4, 5), (4, 9), (4, 11), (4, 15), (4, 16)}
\nAlso, given that,
\nf\u00a0<\/em>= {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}<\/span><\/p>\n

(i)<\/strong> A relation from a non-empty set A to a non-empty set B is a subset of the Cartesian product A \u00d7 B. It\u2019s clearly seen that\u00a0f<\/em> is a subset of A \u00d7 B.
\nTherefore,\u00a0f<\/em> is a relation from A to B.\u00a0<\/span><\/p>\n

(ii)<\/strong> As the same first element, i.e., 2 corresponds to two different images (9 and 11), relation\u00a0f\u00a0<\/em>is not a function.<\/span><\/p>\n

11. Let\u00a0f<\/em>\u00a0be the subset of\u00a0Z\u00a0\u00d7\u00a0Z\u00a0defined by\u00a0f\u00a0<\/em>= {(ab<\/em>,\u00a0a<\/em>\u00a0+\u00a0b<\/em>):\u00a0a<\/em>,\u00a0b<\/em>\u00a0\u2208\u00a0Z}. Is\u00a0f<\/em> a function from\u00a0Z\u00a0to\u00a0Z: justify your answer.\u00a0<\/strong><\/span><\/p>\n

Solution –\u00a0<\/strong>Given relation,\u00a0f<\/em> is defined as\u00a0f\u00a0<\/em>= {(ab<\/em>,\u00a0a<\/em>\u00a0+\u00a0b<\/em>):\u00a0a<\/em>,\u00a0b<\/em> \u2208\u00a0Z}
\nWe know that a relation\u00a0f<\/em> from a set A to a set B is said to be a function if every element of set A has unique images in set B.
\nAs 2, 6, \u20132, \u20136\u00a0\u2208\u00a0Z, (2 \u00d7 6, 2 + 6), (\u20132 \u00d7 \u20136, \u20132 + (\u20136))\u00a0\u2208\u00a0f
\n<\/em>i.e., (12, 8), (12, \u20138)\u00a0\u2208\u00a0f
\n<\/em>It\u2019s clearly seen that the same first element, 12, corresponds to two different images (8 and \u20138).
\nTherefore, the relation\u00a0f<\/em> is not a function.\u00a0<\/span><\/p>\n

12. Let A = {9, 10, 11, 12, 13} and let\u00a0f<\/em>: A\u00a0\u2192\u00a0N\u00a0be defined by\u00a0f<\/em>(n<\/em>) = the highest prime factor of\u00a0n<\/em>. Find the range of\u00a0f<\/em>.
\n<\/strong>Solution –\u00a0<\/strong>Given, A = {9, 10, 11, 12, 13}
\nNow,\u00a0f<\/em>: A \u2192 N is defined as\u00a0f<\/em>(n<\/em>) = The highest prime factor of\u00a0n
\n<\/em>So,
\nPrime factor of 9 = 3
\nPrime factors of 10 = 2, 5
\nPrime factor of 11 = 11
\nPrime factors of 12 = 2, 3
\nPrime factor of 13 = 13
\nThus, it can be expressed as
\nf<\/em>(9) = The highest prime factor of 9 = 3
\nf<\/em>(10) = The highest prime factor of 10 = 5
\nf<\/em>(11) = The highest prime factor of 11 = 11
\nf<\/em>(12) = The highest prime factor of 12 = 3
\nf<\/em>(13) = The highest prime factor of 13 = 13
\nThe range of\u00a0f<\/em>\u00a0is the set of all\u00a0f<\/em>(n<\/em>), where\u00a0n<\/em> \u2208\u00a0A.
\nTherefore,
\nRange of\u00a0f<\/em>\u00a0= {3, 5, 11, 13}<\/span><\/p>\n

Go Back To Chapters<\/i><\/b><\/a><\/span><\/p>\n","protected":false},"excerpt":{"rendered":"

NCERT Solutions Class 11 Maths\u00a0 The NCERT Solutions in English Language for Class 11 Mathematics Chapter – 2 Relations and Functions Exercise – Miscellaneous has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter 2 (Relations and Functions)\u00a0 Miscellaneous Exercise – 2 1. The relation\u00a0f is defined by The […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1674],"tags":[1675,1679,1687,1691,1678,1680,1681],"class_list":["post-8397","post","type-post","status-publish","format-standard","hentry","category-class-11-maths","tag-class-11-ncert-mathematics-solutions","tag-class-11-ncert-solutions","tag-ncert-class-11-mathematics-chapter-2-relations-and-functions-solutions","tag-ncert-class-11-mathematics-miscellaneous-exercise-2-solutions","tag-ncert-class-11-mathematics-solutions","tag-ncert-solutions-class-11-mathematics","tag-ncert-solutions-class-11-maths"],"yoast_head":"\nNCERT Class 11 Maths Solutions - Miscellaneous Exercise 2 | TheExamPillar NCERT<\/title>\n<meta name=\"description\" content=\"NCERT Solutions Class 11 Maths\u00a0The NCERT Solutions in English Language for Class 11 Mathematics Chapter - 2 Relations and Functions Exercise - 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