Maths 1A Important Questions<\/a> TS Inter 1st Year Maths 1A Functions Important Questions Long Answer Type to help strengthen their preparations for exams.<\/p>\nTS Inter 1st Year Maths 1A Functions Important Questions Long Answer Type<\/h2>\n
Question 1.
\nIf f : A \u2192 B, g : B \u2192 C are two bijective functions, then prove that gof : A \u2192 C is also a bijective function. [Mar. 18, 16 (AP), 09 ; May 13, 12, 10, 08, 06, 04 00, 96, 92]
\nAnswer:
\nSince f: A \u2192B is a bijective function
\no f: A \u2192 B is both one-one and onto functions.
\nSince f: A \u2192 B is a one-one function
\n\u21d4 a1<\/sub>, a2<\/sub> \u2208 A, f(a1<\/sub>) = f(a2<\/sub>) \u21d2 a1<\/sub> = a2<\/sub>
\nSince f: A \u2192 B is a onto function \u21d4 \u2203 one element a \u2208 A such that f(a) = b, \u2200 b \u2208 B.
\nSince g: B \u2192 C is a bijective function
\n\u21d4 g: B \u2192 C is both one-one and onto functions.
\nSince g: B \u2192 C is a one-one function
\n\u21d4 b1<\/sub>, b2<\/sub> \u2208 B, g (b1<\/sub>) = g (b2<\/sub>) \u21d2 b1<\/sub> = b2<\/sub>
\nSince g: B \u2192 C is an onto function \u21d4 \u2203 one element b e B such that g(b) = c, \u2200 c \u2208 C.
\nIf f: A \u2192 B, g : B \u2192 C \u21d2 gof: A \u2192 C.<\/p>\nTo prove that gof: A \u2192 C is a one-one function:
\nIf gof: A \u2192 C is a one-one function
\n\u21d4 a1<\/sub>, a2<\/sub> \u2208 A, (gof) (a1<\/sub>) = (gof) (a2<\/sub>) \u21d2 a1<\/sub> = a2<\/sub>
\nNow (gof) (a1<\/sub>) = (gof) (a2<\/sub>)
\ng [f(a1<\/sub>)] = g [ f(a2<\/sub>)] [\u2235 g is one-one]
\nf(a1<\/sub>) = f(a2<\/sub>) [\u2235 f is one-one]
\na1<\/sub> = a2<\/sub>
\nHence, gof: A \u2192 C is a one-one function.<\/p>\nTo prove that gof: A \u2192 C is an onto function:
\nLet c \u2208 C
\nIf gof: A \u2192 C is an onto function \u21d4 \u2203 one element a \u2208 A, such that
\n(gof) (a) = c, \u2200 c \u2208 C.
\nNow (gof) (a) = g [f(a)] = g(b) = c
\nThus for any element c \u2208 C, there is an element a \u2208 A such that (gof) (a) = c.
\n\u2234 gof: A \u2192 C is an onto function.
\nSince gof: A \u2192 C is both one-one function and onto function then
\ngof: A \u2192 C is a bijective function.<\/p>\n
<\/p>\n
Question 2.
\nIf f : A \u2192 B, g : B \u2192 C are two bijective functions, then prove that (gof)-1<\/sup> = f-1<\/sup>og-1<\/sup>. [Mar. ’16 (TS), 14, 11, 10, 06, 04, 02, 00, 92; May 15 (AP) 14, 11, 09, 02, 98, 94 Mar. 19 (AP) ]
\nAnswer:
\nSince f: A \u2192 B, g : B \u2192 C are bijections
\n\u21d2 gof: A \u2192 C is a bijection
\n\u21d2 (gof)-1<\/sup>: C \u2192 A is also a bijection
\nSince f: A \u2192 B is a bijective function
\nthen f-1<\/sup>: B \u2192 A is also a bijective function
\nSince g: B \u2192 C is a bijective function
\nthen g-1<\/sup>: C \u2192 B is also a bijective function
\n\u21d2 f-1<\/sup>og-1<\/sup>: C \u2192 A is also a bijection
\nSince the two functions (gof)-1<\/sup>, f-1<\/sup>og-1<\/sup> are from C \u2192 A their domains are same.
\nLet c \u2208 C
\nSince f : A \u2192 B is onto \u21d4 \u2203 one element
\na \u2208 A such that
\nf(a) = b, \u2200 b \u2208 B
\nf(a) = b \u21d2 f-1<\/sup> (b) = a
\nSince g : B \u2192 C is onto \u21d4 \u2203 one element be B such that g(b) = c, \u2200 c \u2208 C
\ng(b) = c \u21d2 g-1<\/sup>(c) = b
\nNow (gof) (a) = g[f(a)] = g (b) = c \u21d2 a = (gof)-1<\/sup>(c)
\n\u21d2 (gof)-1<\/sup>(c) = a ………………. (1)
\nAlso (f-1<\/sup>og-1<\/sup>) (c) = f-1<\/sup> [g-1<\/sup>(c)] = f-1<\/sup>(b) = a ……………… (2)
\n\u2234 From (1) and (2)
\n(gof)-1<\/sup>(c) = (f-1<\/sup>og-1<\/sup>) (c)
\n\u2234 (gof)-1<\/sup> = f-1<\/sup>og-1<\/sup>.<\/p>\nQuestion 3.
\nLet f : A \u2192 B, is a function and IA<\/sub>, IB<\/sub> are identity functions on A and B respectively. Then prove that foIA<\/sub> = f = IB<\/sub>of. [Mar. 18 (TS); Mar. 13, 08, 05; May 92]
\nAnswer:
\nIf f: A \u2192 B is a function.
\nIf IA<\/sub> and IB<\/sub> are identity functions on A and B respectively.
\ni.e., IA<\/sub> : A \u2192 B, IB<\/sub> : B \u2192 B<\/p>\n(i) IA<\/sub>: A \u2192 A, f: A \u2192 B \u21d2 f o IA<\/sub> : A \u2192 B
\nHence, functions f o IA<\/sub> and f are defined on same domain A.
\nLet a \u2208 A
\n(f o IA<\/sub>) (a) = f[IA<\/sub> (a)] = f(a)
\n\u2234 f o IA<\/sub> = f<\/p>\n(ii) f: A \u2192 B, IB<\/sub>: B \u2192 B \u21d2 IB<\/sub> o f: A \u2192 B
\nThe functions (IB<\/sub> o f) and f are defined on the same domain A.
\nLet a \u2208 A
\nNow (IB<\/sub> o f)(a) = IB<\/sub>[f(a)] = f(a)
\n\u2234 IB<\/sub> o f = f ………………. (2)
\nFrom (1) and (2) we get
\nf o IA<\/sub> = IB<\/sub> o f = f<\/p>\nQuestion 4.
\nIf f: A \u2192 B is a bijection, then prove that fof-1<\/sup> = IB<\/sub> and f-1<\/sup> o f = IA<\/sub>. [Mar. 17, 15 (AP); Mar. 12, 07, 03, 02; May 07, 05, 01 Mar. 19 (TS)]
\nAnswer:
\n<\/p>\n(i) Since f: A \u2192 B is a bijection \u21d2 f-1<\/sup>: B \u2192 A is also a bijection
\nIA<\/sub>; f: A \u2192 B, f-1<\/sup>: B \u2192 A \u21d2 f-1<\/sup> o f: A \u2192 A is also bijection
\nClearly IA<\/sub>: A \u2192 A such that IA<\/sub>(a) = a, \u2200 a \u2208 A
\nLet a \u2208 A
\nSince f-1<\/sup>: B \u2192 A is onto function \u21d4 \u2203 one element b \u2208 B,
\nsuch that
\nf-1<\/sup>(b) = a, \u2200 a \u2208 A
\nf-1<\/sup> (b) = a \u21d2 f(a) = b
\nNow (f-1<\/sup>of) (a) = f-1<\/sup>[f(a)] = f-1<\/sup>(b) = a = IA<\/sub>(a)
\n\u2234 f-1<\/sup>of = IA<\/sub><\/p>\n<\/p>\n
(ii) Since f: A \u2192 B is a bijection \u21d2 f-1<\/sup>: B \u2192 A is also bijection
\nIB<\/sub>: f-1<\/sup>:B \u2192 A, f: A \u2192 B = fof-1<\/sup>:B \u2192 B is also a bijection
\nClearly IB<\/sub>: B \u2192 B such that IB<\/sub>(b) = b, \u2200 b \u2208 B
\nLet b \u2208 B
\nSince f-1<\/sup>: B \u2192 A is an onto function \u21d4 \u2203
\none element b \u2208 B such that f-1<\/sup>(b) = a, \u2200 a \u2208 A
\nf-1<\/sup>(b) = a \u21d2 f(a) = b
\nNow (fof-1<\/sup>) (b) = f[f-1<\/sup>(b)] = f(a) = b = IB<\/sub>(b)
\n\u2234 fof-1<\/sup> = IB<\/sub><\/p>\nQuestion 5.
\nIf f:A \u2192 B, g:B \u2192 A are two functions such that gof = IA<\/sub> and fog = IB<\/sub>, then prove that f is a bijection and g = f-1<\/sup>. [May 15 (TS); Mar. 08, 01; May 03]
\nAnswer:
\n
\n(i) To prove that f is one-one
\nLet a1<\/sub>, a2<\/sub> \u2208 A and since f : A \u2192 B, f(a1<\/sub>), f(a2<\/sub>) \u2208 B
\nNow f(a1<\/sub>) = f(a2<\/sub> ) \u21d2 g[f(a1<\/sub>)] = g[f(a2<\/sub>)]
\n\u21d2 (gof) (a1<\/sub>) = (gof) (a2<\/sub>)
\n\u21d2 IA<\/sub>(a1<\/sub>) = IA<\/sub>(a2<\/sub>)
\n\u2234 a1<\/sub> = a2<\/sub>
\n\u2234 f is one-one<\/p>\n(ii) To prove that f is onto
\nLet b be an element of B
\nIB<\/sub> (b) = (fog) (b)
\n\u21d2 b = f[g(b)] \u21d2 f(g(b)) = b
\ni.e., there exists a pre-image g(b) \u2208 A for b, under the mapping f.
\n\u2234 f is onto
\nThus \u2018f\u2019 is one-one and onto hence, f-1<\/sup>: B \u2192 A exists and is also one-one onto.<\/p>\n(iii) To prove g = f-1<\/sup>
\nNow g:B \u2192 A and f-1<\/sup>:B \u2192 A
\nLet a \u2208 A and b be the f – image of a where b \u2208 B
\n\u2234 f(a) = b \u21d2 a = f-1<\/sup> (b)
\nNow g(b) = g[f(a)] (gof) (a) = IA<\/sub>A(a) = a
\n\u21d2 a = f-1<\/sup> (b)
\n\u2234 g = f-1<\/sup><\/p>\nQuestion 6.
\nIf f:A \u2192 B, g: B \u2192 C and h: C \u2192 D are three functions then prove that ho(gof) = (hog) of. That is composition of functions is associative. [May \u201899, \u201895]
\nAnswer:
\nf:A \u2192 B, g:B \u2192 C, h:C \u2192 D be three functions.
\nf:A \u2192 B, and g:B \u2192 C = gof: A \u2192 C
\nNow gof: A \u2192 C and h:C \u2192 D \u21d2 ho(gof): A \u2192 D
\ng:B \u2192 C and h:C \u2192 D = (hog):B \u2192 D
\nNow f: A \u2192 B \u21d2 hog: B\u2192D
\n(hog)of: A \u2192 D
\nThus ho(gof) and (hog)of both exist and have the same domain A and co-domain D.
\nLet a \u2208 A,
\nHence ho(gof) = (hog) of \u2208 A
\nNow [ho(gof)] (a) = h [(gof) (a)] = h[g(f(a))]
\n= (hog) [f(a)] = [(hog) of] (a)
\n\u2234 [ho(gof)] (a) = [(hog) of] (a)<\/p>\n
<\/p>\n
Question 7.
\nIf f: A \u2192 B, g: B \u2192 C be surjections, then show that gof: A \u2192 C is a surjection. [May 98, 97, 96, 94, 93, 91]
\nAnswer:
\nLet c \u2208 C
\nSince f: A \u2192 B is a onto \u21d4 \u2203 one element
\na \u2208 A such that f(a) = b,\u2200 b \u2208 B
\nSince g: B \u2192 C is a onto \u21d4 \u2203 one element
\nb \u2208 B such that g(b) = c, \u2200 c \u2208 C.
\nIf f: A \u2192 B, g:B \u2192 C = gof: A \u2192 C
\nTo prove that gof : A \u2192 C is a onto
\n
\nIf gof : A \u2192 C is a onto \u21d4 \u2203 one element a \u2208 A
\nsuch that
\n(gof) (a) = c, \u2200 c \u2208 C.
\nNow (gof) (a) = g[f(a)] = g(b) = c
\nThus for any element c \u20ac C, there is an
\nelement a \u2208 A such that (gof) (a) = c.
\n\u2234 gof: A \u2192 C is an onto function.<\/p>\n
Question 8.
\nIf f = ((1, a), (2, c), (4, d), (3, b)} and g-1<\/sup> = {(2, a), (4, b), (1, c), (3, d)}, then show that (gof)-1<\/sup> = f-1<\/sup>og-1<\/sup>. {Mar. 15 (TS); May 07, 93}
\nAnswer:
\nGiven
\nf = {(1, a), (2, c), (4, d), (3, b))
\nf-1<\/sup> = ((a, 1), (c, 2), (d, 4), (b, 3))
\ng = ((a, 2), (b, 4), (c, 1), (d, 3))
\ng-1<\/sup> = {(2, a), (4, b), (1, c), (3, d)}<\/p>\ngof:
\n<\/p>\n
\u2234 gof = {(1, 2), (2, 1), (3, 4), (4, 3)}
\n(gof)-1<\/sup> = {(2, 1), (1, 2), (4, 3), (3, 4)}
\nf-1<\/sup>og-1<\/sup>
\n
\nf-1<\/sup>og-1<\/sup> = {(1, 2), (2, 1), (3, 4), (4, 3)}
\n\u2234 (gof)-1<\/sup> = f-1<\/sup>og-1<\/sup><\/p>\n<\/p>\n
Question 9.
\nIf the function f is defined by
\n
\nthen find the values of
\n(i) f(3)
\n(ii) f(0)
\n(iii) f(-1.5)
\n(iv) f(2) + f(- 2)
\n(v) f(- 5).
\nAnswer:
\nGiven
\n
\n(i) For x > 1; f(x) = x + 2; f(3) = 3 + 2 = 5
\n(ii) For – 1 \u2264 x \u2264 1; f(x) = 2, f(0) = 2
\n(iii) For – 3 < x < – 1; f(x) = x – 1 \u2234 f(- 1.5) = – 1.5 – 1 = – 2.5 (iv) For x > 1, f(x) = x + 2
\nf(2) = 2 + 2 = 4
\nFor – 3 < x < – 1, f(x) = x – 1
\n\u2234 f(- 2) = – 2 – 1 = – 3
\nf(2) + f(- 2) = 4 – 3 = 1
\n(v) f(- 5) is not defined.<\/p>\n
Question 10.
\nIf A = {- 2, – 1, 0, 1, 2) and f: A \u2192 B is a surjection defined by f(x) = x2<\/sup> + x + 1 find B.
\nAnswer:
\nGiven, A = {- 2, – 1, 0, 1, 2)
\nf(x) = x2<\/sup> + x + 1
\nSince f : A \u2192 B is a surjection then f(A) = B
\nf(-2) = (- 2)2<\/sup> – 2 + 1 = 4 – 2 + 1 = 3
\nf(-1) = (- 1)2<\/sup> – 1 + 1 = 1 – 1 + 1 = 1
\nf(0) = 02<\/sup> + 0 + 1 = 1
\nf(1) = 12<\/sup> + 1 + 1 = 1 + 1 + 1 = 3
\nf(2) = 22<\/sup> + 2 + 1 = 4 + 2 + 1 = 7
\n\u2234 B = f(A) = {3. 1, 7}<\/p>\nQuestion 11.
\nIf A = {1, 2, 3, 4} and f:A \u2192 R is a function defined by f(x) = \\(\\frac{x^2-x+1}{x+1}\\), then find the range of f.
\nAnswer:
\nGiven A = {1, 2, 3, 4) and f(x) = \\(\\frac{x^2-x+1}{x+1}\\)
\nSince f: A \u2192 R is a function, then
\n<\/p>\n
Question 12.
\nIf f: Q \u2192 Q, is defined by f(x) = 5x + 4 for all x \u2208 Q, find f-1<\/sup>. [Mar. 17 (TS)]
\nAnswer:
\nLet y = f(x) = 5x + 4
\ny = f(x) \u21d2 x = f-1<\/sup>(y) ……………… (1)
\ny = 5x + 4 \u21d2 y – 4 = 5x
\nx = \\(\\frac{y-4}{5}\\) ………………… (2)
\nFrom (1) and (2),
\nf-1<\/sup>(y) = \\(\\frac{y-4}{5}\\) \u21d2 f(x) = \\(\\frac{x-4}{5}\\), \u2200 x \u2208 Q<\/p>\n<\/p>\n
Question 13.
\nIf f(x) = \\(\\frac{x+1}{x-1}\\) (x \u2260 \u00b1 1), then find (fofofof) (x).
\nAnswer:
\nGiven f(x) = \\(\\frac{x+1}{x-1}\\) (x \u2260 \u00b1 1)
\nNow (fofofof) (x) = f[f[f{f(x)}]]
\n<\/p>\n
Question 14.
\nFind the domain of the real valued function f(x) = \\(\\sqrt{16-x^2}\\).
\nAnswer:
\nGiven f(x) = \\(\\sqrt{16-x^2}\\) \u2208 R
\n\u21d2 16 – x2<\/sup> \u2265 0
\n\u21d2 x2<\/sup> – 16 \u2264 0
\n\u21d2 (x + 4) (x – 4) \u2264 4
\n\u21d2 x \u2208 [- 4, 4]
\n\u2234 Domain of ‘f’ is [- 4, 4]<\/p>\nQuestion 15.
\nFind the domain of the real valued function f(x) = \\(\\sqrt{9-x^2}\\).
\nAnswer:
\nGiven f(x) = \\(\\sqrt{9-x^2}\\) \u2208 R
\n\u21d2 9 – x2<\/sup> \u2265 0
\n\u21d2 x2<\/sup> – 9 \u2264 0
\n\u21d2 (x + 3) (x – 3) \u2264 0
\n\u21d2 x \u2208 [- 3 ,3]
\n\u2234 Domain of f\u2019 is [- 3, 3]<\/p>\nQuestion 16.
\nFind the domain of the real valued function f(x) = \\(\\frac{1}{6 x-x^2-5}\\).
\nAnswer:
\nGiven f(x) = \\(\\frac{1}{6 x-x^2-5}\\) \u2208 R
\n\u21d2 6x – x2<\/sup> – 5 \u2260 0
\n\u21d2 x2<\/sup> – 6x + 5 \u2260 0
\n\u21d2 x2<\/sup> – 5x – x + 5 \u2260 0
\n\u21d2 x(x – 5) – 1 (x – 5) \u2260 0
\n\u21d2 x – 1 \u2260 0 or x – 5 \u2260 0
\n\u21d2 x \u2260 1 or x \u2260 5
\n\u2234 x \u2260 1, 5
\n\u2234 Domain of ‘f’ is R – {1, 5}<\/p>\nQuestion 17.
\nFind the domain of the real valued function f(x) = \\(\\frac{2 x^2-5 x+7}{(x-1)(x-2)(x-3)}\\).
\nAnswer:
\nGiven f(x) = \\(\\frac{2 x^2-5 x+7}{(x-1)(x-2)(x-3)}\\) \u2208 R
\n\u21d2 (x – 1) (x – 2) (x – 3) \u2260 0
\n\u21d2 x – 1 \u2260 0, x – 2 \u2260 0, x – 3 \u2260 0
\n\u21d2 x \u2260 1, x \u2260 2, x \u2260 3
\n\u2234 x \u2260 1, 2, 3
\n\u2234 Domain of ‘f’ is R – {1, 2, 3}<\/p>\n
<\/p>\n
Question 18.
\nFind the domain of the real valued function f(x) = \\(\\frac{\\sqrt{2+x}+\\sqrt{2-x}}{x}\\).
\nAnswer:
\nGiven f(x) = \\(\\frac{\\sqrt{2+x}+\\sqrt{2-x}}{x}\\) \u2208 R
\n
\n\u21d2 2 + x \u2265 0, 2 – x \u2265 0 and x \u2260 0
\nx \u2265 – 2, 2 \u2265 x and x \u2260 0
\nx \u2264 2 and x \u2260 0
\n\u21d2 x \u2208 [- 2, 0) \u222a (0, 2]
\n\u2234 Domain of ‘f’ is [- 2, 0) \u222a (0, 2]<\/p>\n
Question 19.
\nIf f = {(1, 2), (2, – 3), (3, – 1)}, then find
\n(i) 2f
\n(ii) 2 + f
\n(iii) f2<\/sup>
\n(iv) \u221af
\nAnswer:
\nGiven f = {(1, 2), (2, – 3), (3, – 2)}
\nDomain of \u2018f\u2019 is A = {1, 2, 3}
\nf(1) = 2f(2) = – 3, f(3) = – 1<\/p>\n(i) (2f) (x) = 2f(x)
\n(2f) (1) = 2f(1) = 2(2) = 4
\n(2f) (2) = 2f(2) = 2(- 3) = – 6
\n(2f) (3) = 2f(3) = 2(- 1) = – 2
\n\u2234 2f = {(1, 4), (2, – 6),(3, – 2)}<\/p>\n
(ii) (2 + f) (x) = 2 + f(x)
\n(2 + f) (1) = 2 + f(1) = 2 + 2 = 4
\n(2 + f) (2) = 2 + f(2) = 2 – 3 = – 1
\n(2 + f) (3) = 2 + f(3) = 2 – 1 = 1
\n\u2234 2 + f = {(1, 4), (2, – 1), (3, 1)}<\/p>\n
(iii) (f2<\/sup>) (x) = [f(x)]2<\/sup>
\n(f2<\/sup>) (1) = [f(1)]2<\/sup> = 22<\/sup> = 4
\n(f2<\/sup>) (2) = [f(2)]2<\/sup> = (- 3)2<\/sup> = 9
\n(f2<\/sup>) (3) = (f(3)]2<\/sup> = (- 1)2<\/sup> = 1
\n\u2234 f2<\/sup> = {(1, 4), (2, 9), (3, 1)}<\/p>\n(iv) (\u221af)(x) = \u221af(x)
\n(\u221af) (1) = \u221af(1) = \u221a2
\n(\u221af) (2) = \u221af(2) = \u221a- 3 (not valid)
\n(\u221af) (3) = \u221af(3) = \u221a- 1 (not valid)
\n\u2234 \u221af = {(1, \u221a2)}<\/p>\n
Some More Maths 1A Functions Important Questions<\/h3>\n
Question 1.
\nIf f(x) = \\(\\frac{\\cos ^2 x+\\sin ^4 x}{\\sin ^2 x+\\cos ^4 x}\\), \u2200 x \u2208 R then show that f(2012) = 1.
\nAnswer:
\n<\/p>\n
Question 2.
\nIf f: R \u2192 R is defined by f(x) = \\(\\frac{1-x^2}{1+x^2}\\), then show that f(tan \u03b8) = cos 2\u03b8.
\nAnswer:
\nGiven f: R \u2192 R, f(x) = \\(\\frac{1-x^2}{1+x^2}\\)
\nLHS = f(tan \u03b8)
\n= \\(\\frac{1-\\tan ^2 \\theta}{1+\\tan ^2 \\theta}\\) = cos 2\u03b8 + RHS
\n\u2234 f(tan \u03b8) = cos 2\u03b8<\/p>\n
<\/p>\n
Question 3.
\nIf f: R – {\u00b11} \u2192 R is defined by f(x) = log \\(\\left|\\frac{1+x}{1-x}\\right|\\), then show that f(\\(\\left(\\frac{2 x}{1+x^2}\\right)\\)) = 2f(x)
\nAnswer:
\nGiven f: R – {\u00b11} \u2192 R
\n<\/p>\n
Question 4.
\nIf A = {x\/ – 1 \u2264 x \u2264 1}, f(x) = x2<\/sup>, g(x) = x3<\/sup> which of the following are surjections?
\n(i) f : A \u2192 A
\n(ii) g: A \u2192 A
\nAnswer:
\n(i) Given A = {x\/ – 1 \u2264 x \u2264 1}
\n\u2234 A = {- 1, 0, 1}
\nf(x) = x2<\/sup>
\nf(- 1) = (- 1)2<\/sup> = 1
\nf(0) = (0)2<\/sup> = 0
\nf(1) = (1)2<\/sup> = 1
\n\u2234 f = (- 1, 1), (0 , 0), (1, 1))<\/p>\nf: A \u2192 A
\n
\nRange of f(A) = {0, 1) \u2260 A (co-domain)
\n\u2234 f : A \u2192 A is not a surjection.<\/p>\n
(ii) Given A = {x\/ – 1 \u2264 x \u2264 1}
\n\u2234 A = {- 1, 0, 1}
\ng(x) = x3<\/sup>
\ng(- 1) = (- 1)3<\/sup> = – 1
\ng(0) = (0)3<\/sup> = 0
\ng(1) = (1)3<\/sup> = 1
\n\u2234 g = {(- 1, -1), (0, 0), (1, 1)}
\ng: A \u2192 A
\n
\nRange of g(A) = {- 1, 0, 1} = A (co-domain)
\n\u2234 g is a surjection.<\/p>\n<\/p>\n
Question 5.
\nIf f(x) = cos (log x) then show that
\n\\(f\\left(\\frac{1}{x}\\right) \\cdot f\\left(\\frac{1}{y}\\right)-\\frac{1}{2}\\left[f\\left(\\frac{x}{y}\\right)+f(x y)\\right]\\) = 0
\nAnswer:
\nGiven f(x) = cos (log x)
\nf\\(\\left(\\frac{1}{x}\\right)\\) = cos\\(\\left(\\log \\frac{1}{x}\\right)\\) = cos (log x-1<\/sup>)
\n= cos (- log x) = cos (log x)
\nSimilarly f\\(\\left(\\frac{1}{y}\\right)\\) = cos (log y)
\nf\\(\\left(\\frac{x}{y}\\right)\\) = cos \\(\\left(\\log \\left(\\frac{x}{y}\\right)\\right)\\)
\n= cos (log x – log y)
\nf(xy) = cos (log xy)
\n= cos (log x + log y)
\nL.H.S: f\\(\\left(\\frac{1}{x}\\right) \\cdot f\\left(\\frac{1}{y}\\right)-\\frac{1}{2}\\left(f\\left(\\frac{x}{y}\\right)+f(x y)\\right)\\)
\n= cos (log x) . cos (log y) – \\(\\frac{1}{2}\\) [cos (log x – log y) + cos (log x + log y)]
\n= cos (log x) . cos (log y)
\n
\n= cos (log x) . cos (log y) – cos (log x) . cos (log y) = 0
\n= R.H.S<\/p>\nQuestion 6.
\nFind the inverse function of f(x) = log2<\/sub>x.
\nAnswer:
\nGiven f: (0, \u221d) \u2192 R, f(x) = log2<\/sub>x
\nLet y = f(x) = log2<\/sub>x
\ny = f(x) = x = f-1<\/sup>(y) ……………. (1)
\ny = log2<\/sub>x \u21d2 x = 2y<\/sup> (2)
\nFrom (1) & (2)
\nf-1<\/sup>(y) = 2y
\n\u21d2 f-1<\/sup>(x) = 2x<\/p>\nQuestion 7.
\nIf f(x) = 1 + x + x2<\/sup> +…….. for |x| < 1, then show that f-1<\/sup>(x) = \\(\\frac{\\mathbf{x}-1}{\\mathbf{x}}\\).
\nAnswer:
\nGiven that f(x) = 1 + x + x2<\/sup> + ………….
\nf(x) = \\(\\frac{1}{1-\\mathrm{x}}\\)
\n<\/p>\nQuestion 8.
\nFind the domain of the real valued function f(x) = \\(\\frac{1}{\\sqrt{\\mathbf{x}^2-a^2}}\\) (a >0). [Mar.15 (AP)]
\nAnswer:
\nGiven f(x) = \\(\\frac{1}{\\sqrt{x^2-a^2}}\\) \u2208 R
\n\u21d2 x2<\/sup> – a2<\/sup>
\n\u21d2 (x + a) (x – a) > 0
\n\u21d2 x < – a or x > a
\n\u21d2 x \u2208 (- \u221d, – a) \u222a (a, \u221d)
\n\u2234 Domain of f\u2019 is (- \u221d, – a) \u222a (a, \u221d)<\/p>\n<\/p>\n
Question 9.
\nFind the domain of the real valued function f(x) = \\(\\sqrt{(\\mathbf{x}-\\alpha)(\\beta-\\mathbf{x})}\\) (0 < \u03b1 < \u03b2).
\nAnswer:
\nGiven f(x) = \\(\\sqrt{(\\mathbf{x}-\\alpha)(\\beta-\\mathbf{x})}\\) \u2208 R
\n\u21d2 (x – \u03b1) (x – \u03b2) \u2265 0
\n\u21d2 (x – \u03b1) (x – \u03b2) \u2264 0
\n\u21d2 \u03b1 \u2264 x \u2264 \u03b2
\n\u21d2 x \u2208 [\u03b1, \u03b2]
\n\u2234 Domain of ‘f’ is [\u03b1, \u03b2]<\/p>\n
Question 10.
\nFind the domain of the real valued function f(x) = \\(\\sqrt{2-x}+\\sqrt{1+x}\\).
\nAnswer:
\nGiven f(x) = \\(\\sqrt{2-x}+\\sqrt{1+x}\\) \u2208 R
\n\u21d2 2 – x \u2265 0 and 1 + x \u2265 0
\n\u21d2 2 \u2265 x and x \u2265 – 1
\n\u21d2 x \u2264 2 and x \u2265 – 1
\n\u21d2 x \u2208 [- 1, 2]
\n\u2234 Domain of ‘f’ is [-1, 2]
\n<\/p>\n
Question 11.
\nFind the domain of the real valued function f(x) = \\(\\sqrt{|\\mathbf{x}|-\\mathbf{x}}\\)
\nAnswer:
\nGiven f(x) = \\(\\sqrt{|\\mathbf{x}|-\\mathbf{x}}\\) \u2208 R
\n\u21d2 |x| – x \u2265 0
\n\u21d2 |x| \u2265 x
\n\u21d2 x \u2208 R
\n\u2234 Domain of \u2018f\u2019 is \u2018R.<\/p>\n
Question 12.
\nFind the domain and range of the real valued function f(x) = \\(\\frac{2+x}{2-x}\\)
\nAnswer:
\nGiven f(x) = \\(\\frac{2+x}{2-x}\\) \u2208 R
\n\u21d2 2 – x \u2260 0 \u21d2 x \u2260 2
\nDomain of T is R – { 2 }.
\nLet y = f(x) = \\(\\frac{2+x}{2-x}\\)
\ny = \\(\\frac{2+x}{2-x}\\)
\n2yx – xy = 2 + x
\n2y – 2 = x + xy
\n2y – 2 = x(1 + y)
\nx \u2208 R – {2}, y + 1 \u2260 0
\ny \u2260 – 1
\n\u2234 Range of ‘f’ is R – {- 1}.<\/p>\n
Question 13.
\nFind the domain and range of the real valued function f(x) = \\(\\sqrt{9-x^2}\\) [Mar. 15 (TS)]
\nAnswer:
\nGiven f(x) = \\(\\sqrt{9-x^2}\\) \u2208 R
\n\u21d2 9 – x2<\/sup> \u2265 0
\n\u21d2 x2<\/sup> – 9 \u2264 0
\n\u21d2 (x + 3) (x -3) \u2264 0
\n\u21d2 x \u2208 [- 3, 3]
\n\u2234 Domain of \u2018f\u2019 is [- 3, 3]
\nLet y = f(x) = \\(\\sqrt{9-x^2}\\)
\ny = \\(\\sqrt{9-x^2}\\)
\ny2<\/sup> = 9 – x2<\/sup>
\nx2<\/sup> = 9 – y2<\/sup>
\nx = \\(\\sqrt{9-y^2}\\) \u2208 R
\n\u21d2 9 – y2<\/sup> \u2265 0
\n\u21d2 y2<\/sup> – 9 \u2264 0
\n\u21d2 (y + 3) (y – 3) \u2264 0
\n\u21d2 y \u2208 [- 3, 3]
\nBut f(x) attains only non-negative values.
\n\u2234 Range of f = [0, 3].<\/p>\n<\/p>\n
Question 14.
\nDetermine whether the function f(x) = x\\(\\left(\\frac{e^x-1}{e^x+1}\\right)\\) is even or odd.
\nAnswer:
\n
\nSince f(- x) = f(x) then f is an even function.<\/p>\n
Question 15.
\nDetermine whether the function f(x) = log(x + \\(\\sqrt{x^2+1}\\)) is even or odd.
\nAnswer:
\n
\nSince f(- x) = – f(x) then f(x) is an odd function.<\/p>\n
Question 16.
\nFind the domain of the real valued function f(x) = log [x – (x)].
\nAnswer:
\nGiven f(x) = log [x – (x)] \u2208 R
\n\u21d2 x – (x)> 0
\n\u21d2 x > (x)
\nThen x is a non – integer.
\n\u2234 Domain of \u2018f\u2019 is R – Z.<\/p>\n
Question 17.
\nFind the domain of the real valued function f(x) = \\(\\frac{1}{\\log (2-x)}\\).
\nAnswer:
\nGiven f(x) = \\(\\frac{1}{\\log (2-x)}\\) \u2208 R
\n\u21d2 log (2 – x) \u2260 0 and 2 – x > 0
\n\u21d2 log (2 – x) \u2260 log 1 and 2 > x
\n\u21d2 2 – x \u2260 1 and x < 2
\n\u21d2 x \u2260 1
\n\u2234 Domain of ‘f’ is (- \u221d, 1) \u222a (1, 2)<\/p>\n
Question 18.
\nFind the domain of the real valued function f(x) = \\(\\sqrt{\\mathbf{x}-[\\mathbf{x}]}\\).
\nAnswer:
\nGiven f(x) = \\(\\sqrt{\\mathbf{x}-[\\mathbf{x}]}\\) \u2208 R
\n\u21d2 [x] – x \u2265 0 \u21d2 x \u2265 [x] \u21d2 x \u2208 R
\n\u2234 Domain of \u2019f is Z.<\/p>\n
<\/p>\n
Question 19.
\nFind the domain of the real valued function f(x) = \\(\\sqrt{[\\mathbf{x}]-\\mathbf{x}}\\).
\nAnswer:
\nGiven f(x) = \\(\\sqrt{[\\mathbf{x}]-\\mathbf{x}}\\) \u2208 R
\n\u21d2 [x] – x \u2265 0 \u21d2 [x] \u2265 x \u21d2 x \u2208 Z
\n\u2234 Domain of ‘f’ is Z.<\/p>\n
Question 20.
\nIf f and g are real valued functions defined by f(x) = 2x – 1 and g(x) = x2<\/sup> then find
\n(i) (3f – 2g)(x)
\n(ii) (fg) (x)
\n(iii) \\(\\left(\\frac{\\sqrt{f}}{g}\\right)\\)(x)
\n(iv) (f + g + 2) (x)
\nAnswer:
\nGiven f(x) = 2x – 1 and g(x) = x2<\/sup>
\nDomain of f = domain of g R
\nHence the domain of all the functions is R.
\n(i) (3f – 2g) (x) = 3f(x) – 2g(x)
\n= 3(2x – 1) – 2(x2<\/sup>)
\n= 6x – 3 – 2x2<\/sup>
\n= – 2x2<\/sup> + 6x – 3<\/p>\n(ii) (fg)(x) f(x) . g(x)
\n= (2x – 1)(x2<\/sup>) = 2x3<\/sup> – x2<\/sup>.<\/p>\n(iii) \\(\\left(\\frac{\\sqrt{f}}{g}\\right)\\) (x) = \\(\\frac{\\sqrt{f(x)}}{g(x)}\\) = \\(\\frac{\\sqrt{2 x-1}}{x^2}\\)<\/p>\n
(iv) (f + g + 2) (x) = f(x) .g(x) + 2
\n= 2x – 1 + x2<\/sup> + 2
\n= x2<\/sup> + 2x + 1 = (x + 1)2<\/sup><\/p>\nQuestion 21.