{"id":17445,"date":"2022-05-25T20:00:46","date_gmt":"2022-05-25T14:30:46","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=17445"},"modified":"2022-05-17T11:06:26","modified_gmt":"2022-05-17T05:36:26","slug":"mcq-questions-for-class-11-maths-chapter-2","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/mcq-questions-for-class-11-maths-chapter-2\/","title":{"rendered":"MCQ Questions for Class 11 Maths Chapter 2 Relations and Functions with Answers"},"content":{"rendered":"

Students can access the NCERT MCQ Questions for Class 11 Maths Chapter 2 Relations and Functions with Answers Pdf free download aids in your exam preparation and you can get a good hold of the chapter. Use MCQ Questions for Class 11 Maths with Answers<\/a> during preparation and score maximum marks in the exam. Students can download the Relations and Functions Class 11 MCQs Questions with Answers from here and test their problem-solving skills. Clear all the fundamentals and prepare thoroughly for the exam taking help from Class 11 Maths Chapter 2 Relations and Functions Objective Questions.<\/p>\n

Relations and Functions Class 11 MCQs Questions with Answers<\/h2>\n

Students are advised to solve the Relations and Functions Multiple Choice Questions of Class 11 Maths to know different concepts. Practicing the MCQ Questions on Relations and Functions Class 11 with answers will boost your confidence thereby helping you score well in the exam.<\/p>\n

Explore numerous MCQ Questions of Relations and Functions Class 11 with answers provided with detailed solutions by looking below.<\/p>\n

Question 1.
\nThe domain of the function 7-x<\/sup>Px-3<\/sub> is
\n(a) {1, 2, 3}
\n(b) {3, 4, 5, 6}
\n(c) {3, 4, 5}
\n(d) {1, 2, 3, 4, 5}<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) {3, 4, 5}
\nThe function f(x) = 7-x<\/sup>Px-3<\/sub> is defined only if x is an integer satisfying the following inequalities:
\n1. 7 – x \u2265 0
\n2. x – 3 \u2265 0
\n3. 7 – x \u2265 x – 3
\nNow, from 1, we get x \u2264 7 ……… 4
\nfrom 2, we get x \u2265 3 ……………. 5
\nand from 2, we get x \u2264 5 ………. 6
\nFrom 4, 5 and 6, we get
\n3 \u2264 x \u2264 5
\nSo, the domain is {3, 4, 5}<\/p>\n<\/details>\n

\n

Question 2.
\nThe domain of tan-1<\/sup> (2x + 1) is
\n(a) R
\n(b) R – {1\/2}
\n(c) R – {-1\/2}
\n(d) None of these<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) R
\nSince tan-1<\/sup> x exists if x \u2208 (-\u221e, \u221e)
\nSo, tan-1<\/sup> (2x + 1) is defined if
\n-\u221e < 2x + 1 < \u221e
\n\u21d2 -\u221e < x < \u221e
\n\u21d2 x \u2208 (-\u221e, \u221e)
\n\u21d2 x \u2208 R
\nSo, domain of tan-1<\/sup> (2x + 1) is R.<\/p>\n<\/details>\n

\n

Question 3.
\nTwo functions f and g are said to be equal if f
\n(a) the domain of f = the domain of g
\n(b) the co-domain of f = the co-domain of g
\n(c) f(x) = g(x) for all x
\n(d) all of above<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) all of above
\nTwo functions f and g are said to be equal if f
\n1. the domain of f = the domain of g
\n2. the co-domain of f = the co-domain of g
\n3. f(x) = g(x) for all x<\/p>\n<\/details>\n

\n

Question 4.
\nIf the function f : R \u2192 R be given by f(x) = x\u00b2 + 2 and g : R \u2192 R is given by g(x) = x\/(x – 1). The value of gof(x) is
\n(a) (x\u00b2 + 2)\/(x\u00b2 + 1)
\n(b) x\u00b2\/(x\u00b2 + 1)
\n(c) x\u00b2\/(x\u00b2 + 2)
\n(d) none of these<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) (x\u00b2 + 2)\/(x\u00b2 + 1)
\nGiven f(x) = x\u00b2 + 2 and g(x) = x\/(x – 1)
\nNow, gof(x) = g(x\u00b2 + 2) = (x\u00b2 + 2)\/(x\u00b2 + 2 – 1) = (x\u00b2 + 2)\/(x\u00b2 + 1)<\/p>\n<\/details>\n

\n

Question 5.
\nGiven g(1) = 1 and g(2) = 3. If g(x) is described by the formula g(x) = ax + b, then the value of a and b is
\n(a) 2, 1
\n(b) -2, 1
\n(c) 2, -1
\n(d) -2, -1<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) 2, -1
\nGiven, g(x) = ax + b
\nAgain, g(1) = 1
\n\u21d2 a \u00d7 1 + b = 1
\n\u21d2 a + b = 1 ……… 1
\nand g(2) = 3
\n\u21d2 a \u00d7 2 + b = 3
\n\u21d2 2a + b = 3 …….. 2
\nSolve equation 1 and 2, we get
\na = 2, b = -1<\/p>\n<\/details>\n

\n

Question 6.
\nLet f : R \u2192 R be a function given by f(x) = x\u00b2 + 1 then the value of f-1<\/sup> (26) is
\n(a) 5
\n(b) -5
\n(c) \u00b15
\n(d) None of these<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) \u00b15
\nLet y = f(x) = x\u00b2 + 1
\n\u21d2 y = x\u00b2 + 1
\n\u21d2 y – 1 = x\u00b2
\n\u21d2 x = \u00b1\u221a(y – 1)
\n\u21d2 f-1<\/sup> (x) = \u00b1\u221a(x – 1)
\nNow, f-1<\/sup> (26) = \u00b1\u221a(26 – 1)
\n\u21d2 f-1<\/sup> (26) = \u00b1\u221a(25)
\n\u21d2 f-1<\/sup> (26) = \u00b15<\/p>\n<\/details>\n

\n

Question 7.
\nthe function f(x) = x – [x] has period of
\n(a) 0
\n(b) 1
\n(c) 2
\n(d) 3<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) 1
\nLet T is a positive real number.
\nLet f(x) is periodic with period T.
\nNow, f(x + T) = f(x), for all x \u2208 R
\n\u21d2 x + T – [x + T] = x – [x] , for all x \u2208 R
\n\u21d2 [x + T] – [x] = T, for all x \u2208 R
\nThus, there exist T > 0 such that f(x + T) = f(x) for all x \u2208 R
\nNow, the smallest value of T satisfying f(x + T) = f(x) for all x \u2208 R is 1
\nSo, f(x) = x – [x] has period 1<\/p>\n<\/details>\n

\n

Question 8.
\nThe function f(x) = sin (\u200e\u03c0x\/2) + cos (\u03c0x\/2) is periodic with period
\n(a) 4
\n(b) 6
\n(c) 12
\n(d) 24<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) 4
\nPeriod of sin (\u200e\u03c0x\/2) = 2\u03c0\/(\u03c0\/2) = 4
\nPeriod of cos (\u03c0x\/2) = 2\u03c0\/(\u03c0\/2) = 4
\nSo, period of f(x) = LCM (4, 4) = 4<\/p>\n<\/details>\n

\n

Question 9.
\nThe domain of the function f(x) = x\/(1 + x\u00b2) is
\n(a) R – {1}
\n(b) R – {-1}
\n(c) R
\n(d) None of these<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) R
\nGiven, function f(x) = x\/(1 + x\u00b2)
\nSince f(x) is defined for all real values of x.
\nSo, domain(f) = R<\/p>\n<\/details>\n

\n

Question 10.
\nIf f : R \u2192 R is defined by f(x) = x\u00b2 – 3x + 2, the f(f(y)) is
\n(a) x4<\/sup> + 6x\u00b3 + 10x\u00b2 + 3x
\n(b) x4<\/sup> – 6x\u00b3 + 10x\u00b2 + 3x
\n(c) x4<\/sup> + 6x\u00b3 + 10x\u00b2 – 3x
\n(d) x4<\/sup> – 6x\u00b3 + 10x\u00b2 – 3x<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) x4<\/sup> – 6x\u00b3 + 10x\u00b2 – 3x
\nGiven, f(x) = x\u00b2 – 3x + 2
\nNow, f(f(y)) = f(x\u00b2 – 3x + 2)
\n= (x\u00b2 – 3x + 2)\u00b2 – 3(x\u00b2 – 3x + 2) + 2
\n= x4<\/sup> – 6x\u00b3 + 10x\u00b2 – 3x<\/p>\n<\/details>\n

\n

Question 11.
\nIf n is the smallest natural number such that n + 2n + 3n + …. + 99n is a perfect square, then the number of digits in square of n is
\n(a) 1
\n(b) 2
\n(c) 3
\n(d) 4<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) 3
\nGiven that
\nn + 2n + 3n + …. + 99n
\n= n \u00d7 (1 + 2 + 3 + …….. + 99)
\n= (n \u00d7 99 \u00d7 100)\/2
\n= n \u00d7 99 \u00d7 50
\n= n \u00d7 9 \u00d7 11 \u00d7 2 \u00d7 25
\nTo make it perfect square we need 2 \u00d7 11
\nSo n = 2 \u00d7 11 = 22
\nNow n\u00b2 = 22 \u00d7 22 = 484
\nSo, the number of digit in n\u00b2 = 3<\/p>\n<\/details>\n

\n

Question 12.
\nLet f : R – R be a function defined by f(x) = cos(5x + 2), then f is
\n(a) injective
\n(b) surjective
\n(c) bijective
\n(d) None of these<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) None of these
\nGiven, f(x) = cos(2x + 5)
\nPeriod of f(x) = 2\u03c0\/5
\nSince f(x) is a periodic function with period 2\u03c0\/5, so it is not injective.
\nThe function f is not surjective also as its range [-1, 1] is a proper subset of its co-domain R<\/p>\n<\/details>\n

\n

Question 13.
\nThe function f(x) = sin (\u200e\u03c0x\/2) + 2cos (\u03c0x\/3) – tan (\u03c0x\/4) is periodic with period
\n(a) 4
\n(b) 6
\n(c) 8
\n(d) 12<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) 12
\nPeriod of sin (\u200e\u03c0x\/2) = 2\u03c0\/(\u03c0\/2) = 4
\nPeriod of cos (\u03c0x\/3) = 2\u03c0\/(\u03c0\/3) = 6
\nPeriod of tan (\u03c0x\/4) = \u03c0\/(\u03c0\/4) = 4
\nSo, period of f(x) = LCM (4, 6, 4) = 12<\/p>\n<\/details>\n

\n

Question 14.
\nIf the function f : R \u2192 R be given by f(x) = x\u00b2 + 2 and g : R \u2192 R is given by g(x) = x\/(x – 1). The value of gof(x) is
\n(a) (x\u00b2 + 2)\/(x\u00b2 + 1)
\n(b) x\u00b2\/(x\u00b2 + 1)
\n(c) x\u00b2\/(x\u00b2 + 2)
\n(d) none of these<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) (x\u00b2 + 2)\/(x\u00b2 + 1)
\nGiven f(x) = x\u00b2 + 2 and g(x) = x\/(x – 1)
\nNow, gof(x) = g(x\u00b2 + 2) = (x\u00b2 + 2)\/(x\u00b2 + 2 – 1) = (x\u00b2 + 2)\/(x\u00b2 + 1)<\/p>\n<\/details>\n

\n

Question 15.
\nThe domain of the function 7-x<\/sup>Px-3<\/sub> is
\n(a) {1, 2, 3}
\n(b) {3, 4, 5, 6}
\n(c) {3, 4, 5}
\n(d) {1, 2, 3, 4, 5}<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) {3, 4, 5}
\nThe function f(x) = 7-x<\/sup>Px-3<\/sub> is defined only if x is an integer satisfying the following inequalities:
\n1. 7 – x \u2265 0
\n2. x – 3 \u2265 0
\n3. 7 – x \u2265 x – 3
\nNow, from 1, we get x \u2264 7 ………4
\nfrom 2, we get x \u2265 3 …………….5
\nand from 2, we get x \u2264 5 ……….6
\nFrom 4, 5 and 6, we get
\n3 \u2264 x \u2264 5
\nSo, the domain is {3, 4, 5}<\/p>\n<\/details>\n

\n

Question 16.
\nIf f(x) = ex<\/sup> and g(x) = loge<\/sub> x then the value of fog(1) is
\n(a) 0
\n(b) 1
\n(c) -1
\n(d) None of these<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) 1
\nGiven, f(x) = ex<\/sup>
\nand g(x) = logx
\nfog(x) = f(g(x))
\n= f (logx)
\n= elog x<\/sup>
\n= x
\nSo, fog(1) = 1<\/p>\n<\/details>\n

\n

Question 17.
\nA relation R is defined from the set of integers to the set of real numbers as (x, y) = R if x\u00b2 + y\u00b2 = 16 then the domain of R is
\n(a) (0, 4, 4)
\n(b) (0, -4, 4)
\n(c) (0, -4, -4)
\n(d) None of these<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) (0, -4, 4)
\nGiven that:
\n(x, y) \u2208 R \u21d4 x\u00b2 + y\u00b2 = 16
\n\u21d4 y = \u00b1\u221a(16 – x\u00b2)
\nwhen x = 0 \u21d2 y = \u00b14
\n(0, 4) \u2208 R and (0, -4) \u2208 R
\nwhen x = \u00b14 \u21d2 y = 0
\n(4, 0) \u2208 R and (-4, 0) \u2208 R
\nNow for other integral values of x, y is not an integer.
\nHence R = {(0, 4), (0, -4), (4, 0), (-4, 0)}
\nSo, Domain(R) = {0, -4, 4}<\/p>\n<\/details>\n

\n

Question 18.
\nThe period of the function f(x) = sin (2\u03c0x\/3) + cos (\u03c0x\/3)
\n(a) 3
\n(b) 4
\n(c) 12
\n(d) None of these<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) 12
\nGiven, function f(x) = sin (2\u03c0x\/3) + cos (\u03c0x\/2)
\nNow, period of sin (2\u03c0x\/3) = 2\u03c0\/{(2\u03c0\/3)} = (2\u03c0 \u00d7 3)\/(2\u03c0) = 3
\nand period of cos (\u03c0x\/2) = 2\u03c0\/{(\u03c0\/2)} = (2\u03c0 \u00d7 2)\/(\u03c0) = 2 \u00d7 2 = 4
\nNow, period of f(x) = LCM(3, 4) = 12
\nHence, period of function f(x) = sin (2\u03c0x\/3) + cos (\u03c0x\/2) is 12<\/p>\n<\/details>\n

\n

Question 19.
\nIf f(x) = ax + b and g(x) = cx + d and f{g(x)} = g{f(x)} then
\n(a) f(a) = g(c)
\n(b) f(b) = g(b)
\n(c) f(d) = g(b)
\n(d) f(c) = g(a)<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) f(d) = g(b)
\nGiven, f(x) = ax + b and g(x) = cx + d and
\nNow, f{g(x)} = g{f(x)}
\n\u21d2 f{cx + d} = g{ax + b}
\n\u21d2 a(cx + d) + b = c(ax + b) + d
\n\u21d2 ad + b = cb + d
\n\u21d2 f(d) = g(b)<\/p>\n<\/details>\n

\n

Question 20.
\nThe domain of the function f (x) = 1\/(2 – cos 3x) is
\n(a) (1\/3, 1)
\n(b) [1\/3, 1)
\n(c) (1\/3, 1]
\n(d) R<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) R
\nGiven
\nfunction is f(x) = 1\/(2 – cos 3x)
\nSince -1 \u2264 cos 3x \u2264 1 for all x \u2208 R
\nSo, -1 \u2264 2 – cos 3x \u2264 1 for all x \u2208 R
\n\u21d2 f(x) is defined for all x \u2208 R
\nSo, domain of f(x) is R<\/p>\n<\/details>\n

\n

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Students can access the NCERT MCQ Questions for Class 11 Maths Chapter 2 Relations and Functions with Answers Pdf free download aids in your exam preparation and you can get a good hold of the chapter. Use MCQ Questions for Class 11 Maths with Answers during preparation and score maximum marks in the exam. Students …<\/p>\n

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