{"id":38612,"date":"2022-12-05T11:26:49","date_gmt":"2022-12-05T05:56:49","guid":{"rendered":"https:\/\/tsboardsolutions.com\/?p=38612"},"modified":"2023-01-18T16:03:09","modified_gmt":"2023-01-18T10:33:09","slug":"maths-1a-functions-important-questions-long-answer-type","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.com\/maths-1a-functions-important-questions-long-answer-type\/","title":{"rendered":"TS Inter 1st Year Maths 1A Functions Important Questions Long Answer Type"},"content":{"rendered":"

Students must practice these 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>\n

TS 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>\n

To 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>\n

To 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

\"TS<\/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>\n

Question 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>\n

Question 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\"TS<\/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

\"TS<\/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>\n

Question 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\"TS
\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>\n

Question 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

\"TS<\/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\"TS
\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>\n

gof:
\n\"TS<\/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\"TS
\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

\"TS<\/p>\n

Question 9.
\nIf the function f is defined by
\n\"TS
\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\"TS
\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>\n

Question 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\"TS<\/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

\"TS<\/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\"TS<\/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>\n

Question 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>\n

Question 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>\n

Question 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

\"TS<\/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\"TS
\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\"TS<\/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

\"TS<\/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\"TS<\/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>\n

f: A \u2192 A
\n\"TS
\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\"TS
\nRange of g(A) = {- 1, 0, 1} = A (co-domain)
\n\u2234 g is a surjection.<\/p>\n

\"TS<\/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\"TS
\n= cos (log x) . cos (log y) – cos (log x) . cos (log y) = 0
\n= R.H.S<\/p>\n

Question 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>\n

Question 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\"TS<\/p>\n

Question 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

\"TS<\/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\"TS<\/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

\"TS<\/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\"TS
\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\"TS
\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

\"TS<\/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>\n

Question 21.
\nFind the domain of the real valued function f(x) = \\(\\sqrt{x^2-3 x+2}\\).
\nAnswer:
\nGiven f(x)= \\(\\sqrt{x^2-3 x+2}\\) \u2208 R
\n\u21d2 x2<\/sup> – 3x + 2 \u2265 0
\n\u21d2 x2<\/sup> – 2x – x + 2 \u2265 0
\n\u21d2 x(x – 2) – 1(x – 2) \u2265 0
\n\u21d2 (x – 1) (x – 2) \u2265 0
\n\u21d2 x \u2264 1 or x \u2265 2
\n\u21d2 x \u2208 (- \u221d, 1] \u222a [2, \u221d)
\n\u2234 Domain of \u2018f\u2019 is (- \u221d, 1] \u222a [2, \u221d)<\/p>\n

Question 22.
\nf:R \u2192 R defined by f(x) = \\(\\frac{2 x+1}{3}\\), then this function Is injection or not ? Justify. (Mar. 15 (TS)
\nAnswer:
\nGiven that f(x) = \\(\\frac{2 x+1}{3}\\)
\nLet x1<\/sub>, x2<\/sub> \u2208 R.
\nTake f(x1<\/sub>) = f(x2<\/sub>) \u21d2 \\(\\frac{2 x_1+1}{3}=\\frac{2 x_2+1}{3}\\)
\n\u21d2 2x1<\/sub> + 1 = 2x2<\/sub> + 1 \u21d2 2x1<\/sub> = 2x2<\/sub> = x1<\/sub> = x2<\/sub>
\n\u2234 f(x1<\/sub>) = f(x2<\/sub>) \u21d2 x1<\/sub> = x2<\/sub>
\n\u21d2 f is one – one.<\/p>\n

\"TS<\/p>\n

Question 23.
\nIf f = {(4, 5), (5, 6),(6, – 4)} and g = ((4, – 4), (6, 5), (8, 5)) then find f + g and fg. [Mar. \u201817(TS)]
\nAnswer:
\nGiven f = {(4, 5), (5, 6), (6, – 4)} and
\ng = {(4, – 4), (6, 5), (8, 5\u2019)) then domain of f = {4, 5, 6) and Range of f = {4, 6, 8}
\nDomain of f + g = A \u2229 B = {4, 6}
\n= (domain of f) \u2229 (domain of g)
\n(i) f.g={(4, 5, – 4), (6, – 4 + 5)}
\n= {(4, 1), (6, 1)}<\/p>\n

(ii) Domain of fg = (domain of f) \u2229 (domain of g)
\n= A \u2229 B = (4, 6)
\n= {(4, 5 \u00d7 – 4).(6, – 4 \u00d7 5)}.
\n= {(4, – 20), (6, – 20)}<\/p>\n

Question 24.
\nIf f(x) = 2x – 1, g(x) = \\(\\frac{x+1}{2}\\) for all x \u2208 R, are two functions, then find,
\n(i) (gof) (x)
\n(ii) (fog) (x) [Mar. 19(TS)]
\nAnswer:
\nGiven f(x) = 2x – 1, g(x) = \\(\\frac{x+1}{2}\\)<\/p>\n

(i) (gof) (x) = g[ f(x) ]
\n= g[2x – 1] = \\(\\frac{2 x-1+1}{2}\\) = \\(\\frac{2 x}{2}\\) = x<\/p>\n

(ii) (fog) (x) = f [g(x)]
\n= \\(f\\left(\\frac{x+1}{2}\\right)\\) = 2\\(\\left(\\frac{\\mathrm{x}+1}{2}\\right)\\) – 1 = x + 1 – 1 = x<\/p>\n","protected":false},"excerpt":{"rendered":"

Students must practice these Maths 1A Important Questions TS Inter 1st Year Maths 1A Functions Important Questions Long Answer Type to help strengthen their preparations for exams. TS Inter 1st Year Maths 1A Functions Important Questions Long Answer Type Question 1. 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