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FUNCTIONS, PREVIOUS EAMCET BITS, 1., , If f: [2, 3] → IR is defined by f ( x ) = x 3 + 3x − 2 , then the range f(x) is contained in the interval :, [EAMCET 2009], 1) [1,12], , 2) [12,34], , 3) [35,50], , 4) [ −12,12], , Ans: 2, Sol. f ( 2 ) = 12 and f ( 3) = 34, , 1) IR − {0}, , [EAMCET 2009], , 2) IR − {0,1,3}, , 3) IR − {0, −1, −3}, , 1⎫, ⎧, 4) IR − ⎨0, −1, −3, + ⎬, 2⎭, ⎩, , .in, , 2., , ∴ Range = [12, 34], 2x − 1, ⎧, ⎫, ∈ IR ⎬ =, ⎨ x ∈ IR : 3, 2, x + 4x + 3x, ⎩, ⎭, , et, Ba, , di, , Ans: 3, 2x − 1, 2x − 1, Sol., =, 2, x ( x + 4x + 3) x ( x + 1)( x + 3), , is not defined if x ( x + 1)( x + 3) = 0 ⇒ x = −3, −1, 0, 3., , Using mathematical induction, the numbers an ‘s are defined by a 0 = 1, a n +1 = 3n 2 + n + a n ( n ≥ 0 ), , then an =, , w, , The number of subsets of {1, 2, 3, …….9} containing at least one odd number is[EAMCET 2009], 1) 324, 2) 396, 3) 496, 4) 512, Ans: 3, , w, , 4., , 4) n 3 + n 2, , w, , .N, , 1) n 3 + n 2 + 1, 2) n 3 − n 2 + 1, 3) n 3 − n 2, Ans: 2, Sol. a 0 = 1, a1 = 1, a 2 = 3 + 1 + a1 = 5 and so on. Verify (2) is correct, , [EAMCET 2009], , Sol. Number of subsets = 29 − 24 = 512 − 16 = 46, 4., , If, , → C is defined by f ( x ) = e 2ix for x ∈ R , then f is (where C denotes the set of all complex, , numbers), 1) one-one, 2) onto, Ans: 4, Sol. f ( x ) = e2ix = cos 2x + i sin 2x, , [EAMCET 2008], 3) one-one and onto 4) neither one-one nor onto, , f ( 0 ) = f ( π ) = 1 ⇒ f is not one one, , There exists not x ∈ R ∋ f ( x ) = 2 ⇒ f is not onto., 5., , If f : R → R and g : R → R are defined by f ( x ) = x and g ( x ) = [ x − 3] for x ∈ R , then, 8, 8⎫, ⎧, ⎨g ( f ( x ) ) : − < x < ⎬ =, 5, 5⎭, ⎩, 1) [0, 1], 2) [1, 2], , [EAMCET 2008], , 3) {–3, –2}, 1, , 4) {2, 3}, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterma

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Functions, , Ans: 3, 8, 8, 8, 8, Sol. − < x < ⇒ 0 ≤ x < ⇒ −3 ≤ x − 3 < − 3, 5, 5, 5, 5, 7, ⇒ −3 ≤ x − 3 < − ⇒ ⎡⎣ x − 3⎤⎦ = −3or − 2, 5, 8, 8⎫, ⎧, ⇒ ⎨g ( f ( x ) ) : − < x < ⎬ = {−3, −2}, 5, 5⎭, ⎩, 6., If f : [ −6, 6] → R defined by f ( x ) = x 2 − 3 for x ∈ R then, , ( fofof )( −1) + ( fofof )( 0 ) + ( fofof )(1) =, , (, , 1) f 4 2, Sol., , ), , (, , 2) f 3 2, , ), , [EAMCET 2008], , (, , 3) f 2 2, , ), , ( 2), , Ans: 1, ( fofof )( −1) + ( fofof )( 0 ) + ( fofof )(1) = −2 + 33 − 2 = 29, , (, , ), , .in, , f 4 2 = 32 − 3 = 29, , di, , ⎛p⎞, p, If Q denotes the set of all rational numbers and f ⎜ ⎟ = p 2 − q 2 for any ∈ Q, then observe, q, ⎝q⎠, the following statements, [EAMCET 2007], ⎛p⎞, p, I) f ⎜ ⎟ is real for each ∈ Q, q, ⎝q⎠, , et, Ba, , 7., , 4) f, , w, , w, , w, , .N, , ⎛p⎞, p, II) f ⎜ ⎟ is complex number for each ∈ Q, q, ⎝q⎠, Which of the following is correct ?, 1) Both I and II are true, 2) I is true, II is false, 3) I is false, II is true, 4) Both I and II are false, Ans: 3, ⎛1⎞, Sol. f ⎜ ⎟ = 1 − 4 = −3 is an imaginary ⇒ I is false, ⎝2⎠, ⎛p⎞, f ⎜ ⎟ = p 2 − q 2 it is a complex number ⇒ II is true, ⎝q⎠, 1, 8., If f : R → is defined by f ( x ) =, for each x ∈ R , then the range of f is, 2 − cos 3x, [EAMCET 2007], ⎛1 ⎞, ⎡1 ⎤, 2) ⎢ ,1⎥, 3) (1, 2), 4) [1, 2], 1) ⎜ ,1⎟, ⎝3 ⎠, ⎣3 ⎦, Ans: 2, Sol. Max. and Min. values of 2 – cos3x are 3 and 1, ⎡1 ⎤, ∴ Range = ⎢ ,1⎥, ⎣3 ⎦, 9., If f : R → and g : → are defined by f(x) = x – [x] and g(x) = [x] for x ∈ , where [x] is, The greatest integer not exceeding x, then for every x ∈, , , f (g ( x )) =, , [EAMCET 2007], , 2, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterma

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Functions, 1) x, Ans: 2, Sol. f ( g ( x ) ), , 2) 0, , 3) f(x), , 4) g(x), , = g ( x ) − ⎡⎣g ( x ) ⎤⎦, = [x] − [x] = 0, , 1, is defined by f ( x ) = x − [ x ] − for x ∈ , where [x] is the greatest integer not, 2, 1⎫, ⎧, [EAMCET 2006], exceeding x, then ⎨ x ∈ : f ( x ) = ⎬ =……, 2⎭, ⎩, 1) Z, the set of all integers, 2) IN, the set of all natural number, 3) φ, the empty set, 4), Ans: 3, 1, Sol. f ( x ) = x − [ x ] − , x ∈, 2, 1, f (x) =, 2, 1 1, ⇒ x − [x] − =, 2 2, ⇒ x − [x] = 1, , If f =, , →, , et, Ba, , di, , .in, , 10., , ⇒ {x} = 1 which is not possible, where {x} denotes the fractional part, If f = →, is defined by f ( x ) = [ 2x ] − 2 [ x ] for x ∈ . where [x] is the greatest integer not, exceeding x, then the range of f is, [EAMCET 2006], 1) {x ∈ : 0 ≤ x ≤ 1}, 2) {0, 1}, , .N, , 11., , 4) {x ∈, , : x ≤ 0}, , w, , w, , w, , 3) {x ∈ : x > 0}, Ans: 2, Sol. f ( x ) = [ 2x ] − 2 [ x ] , x ∈ = 0, = ∀x ∈ where x = a + f, ∋ 0 < f < 0.5, = 1, ∀x ∈, x = a + f where 0.5 ≤ a < 1, ∴ Range = {0, 1}, 12., , x < −4, ⎧ x + 4 for, ⎪, If f : → is defined by f ( x ) = ⎨3x + 2 for −4 ≤ x < 4 then the correct matching of List I, ⎪ x − 4 for, x≥4, ⎩, from List II is, [EAMCET 2006], List – I, List – II, A) f ( −5 ) + f ( −4 ), i) 14, , (, , B) f f ( −8 ), , ), , C) f ( f ( −7 ) + f ( 3) ), , ii) 4, iii) – 11, , 3, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterma

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Functions, , ((, , )), , D) f f f ( f ( 0 ) ) + 1, , iv) – 1, v) 1, vi) 0, , A, iii, iv, , B, vi, iii, , C, ii, ii, , D, v, i, , 1), 2), 3), 4), Ans: 1, Sol. ( A ) f ( −5 ) + f ( −4 ) = ( −5 + 4 ) + 3 ( −4 ) + 2 = −11, , A, iii, iii, , B, iv, vi, , C, ii, v, , D, v, ii, , ( B ) f ( − 8 + 4 ) = f ( −4 ) = 3 ⇒ f ( 4 ) = 0, ( C ) f ⎡⎣( −3) + 11⎤⎦ = f (8 ) = 4, ( D ) f ( f ( f ( 2 ) ) ) = f ( f (8) ) + 1 = f ( 4 ) + 1 = 0 + 1 = 1, , {x ∈, , }, , di, , 2) φ, the empty set, 4) {x ∈ : x > 0}, , The function f : c → c defined by f ( x ) =, , ax + b, for x ∈ c where bd ≠ 0 reduces to a constant, cx + d, [EAMCET 2005], 3) ad = bc, 4) ab = cd, , w, , w, , w, , .N, , function if, 1) a = c, 2) b = d, Ans: 3, ax + b, Sol. f ( x ) =, cx + d, cx + d ) ax + b ( a / c, , [EAMCET 2005], , et, Ba, , 14., , .in, , : ⎡⎣ x − x ⎤⎦ = 5 =, 1), , the set of all real numbers, 3) {x ∈ : x < 0}, Ans: 2, Sol. x − | x |= 2x, ∀x < 0, = 0, ∀x ≥ 0, ∴ x - |x| ≠ 5, 13., , ax + ad / c, , bc − ad, c, a, bc − ad, = constant bc = ad, f (x) = +, c c ( cx + d ), 15., , 2004, For any integer n ≥ 1 , the number of positive divisors of n is denoted by d(n). Then for a prime P,, , ((, , )), , d d d ( P7 ) =, , 1) 1, Ans: 3, , ((, , [EAMCET 2004], 2) 2, , )), , 3) 3, , (, , 4) P, , ), , Sol. d d d ( p 7 ) = d ( d ( 8 ) ) = d d ( 23 ) = d ( 4 ), = d ( 22 ) = 2 + 1 = 3, , 4, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterma

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Functions, 16., , ⎧ 2 if, ⎪, If f : N → Z is defined by f ( n ) = ⎨10 if, ⎪ 0 if, ⎩, , n = 3k, k ∈ Z, , n = 3k + 1, k ∈ Z then {n ∈ N : f ( n ) > 2} =, n = 3k + 2, k ∈ Z, , [EAMCET 2004], 1) {3, 6, 4}, 2) {1, 4, 7}, Ans: 2, Sol. f ( n ) > 2 ⇒ n = 3k + 1, 17., , ⇒ n = 1; n = 4; n = 7, The function f : →, , 3) {4, 7}, , 4) {7}, , is defined by f ( x ) = 3− x . Observe the following statements of it :, , I. f is one-one, II) f is onto, Out of these, true statements are, 1) only I, II, 2) only II, III, Ans:, Sol. f : R → R;f ( x ) = 3− x, , III) f is a decreasing function, , 4) I, II, III, , .in, , 3) only I, III, , [EAMCET 2004], , w, , ⇒ 4x 2 + 12x + 16 = 8, , w, , .N, , et, Ba, , di, , ∴ f(x) is one-one and it is decreasing function, ⎧ [ x ] if −3 < x ≤ −1, ⎪, [EAMCET 2004], 18. If f ( x ) = ⎨ x, if, 1 < x < 1 , then ( x : f ( x ) ≥ 0 ) =, ⎪ ⎡[ x ]⎤ if, 1≤ x < 3, ⎩⎣ ⎦, 1) (–1, 3), 2) [–1, 3), 3) (–1, 3], 4) [–1, 3], Ans: 1, Sol. Verification, 19. I f : → and g : → are definite by f(x) = 2x + 3 and g(x) = x2 + 7 then the values of x such, that g(f(x)) = 8 are, [EAMCET 2003], 1) 1, 2, 2) –1, 2, 3) –1, –2, 4) 1, –2, Ans: 3, Sol. g ( f ( x ) ) = 4x 2 + 12x + 16, , 20., , w, , ⇒ ( x + 1)( x + 2 ) = 0 ⇒ x = −1, −2, Suppose f : [ −2, 2] →, , {, , ⎧ −1 for − 2 ≤ x ≤ 0, is defined f ( x ) = ⎨, ,, ⎩ x − 1 for 0 ≤ x ≤ 2, , }, , then x ∈ [ −2, 2] : x ≤ 0 and f ( x ) = x = ...., 1) {–1}, , 2) {0}, , [EAMCET 2003], , ⎧ 1⎫, 3) ⎨− ⎬, ⎩ 2⎭, , 4) φ, , Ans: 3, 1, 2, ⎛ 1⎞, 1, ⎛1⎞ 1, ∴f ⎜ − ⎟ = f ⎜ ⎟ = −1 = −, 2, ⎝2⎠ 2, ⎝ 2⎠, Hence f(|x|) = x, , Sol. Now take x = −, , 5, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterma

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Functions, , 21., , ⎧ 1⎫, ∴ Domain of f ( x ) = ⎨− ⎬, ⎩ 2⎭, are given f(x) = |x| and g(x) = [x] for each, If f : → and g : →, , {x ∈, , }, , : g ( f ( x )) ≤ f (g ( x )), , 1) z ∪ ( −∞, 0 ), , [EAMCET 2003], , 2) ( −∞, 0 ), , 3) z, , 4), , Ans: 4, Sol. f ( x ) = x ;g ( x ) = [ x ], g (f ( x )) ≤ f (g ( x )), , g ( f ( x ) ) = g ( x ) = ⎡⎣ x ⎤⎦ = [ x ], f (g ( x )) = f [x] = [x], , [x] ≤ [x], , 1) a, Ans: 2, Sol. f ( x ) = ax +, , a2, , then f ( a ) =, ax, 2) 0, , 3) 1, , di, , If f ( x ) = ax +, , [EAMCET 2002], , 4) – 1, , et, Ba, , 22., , .in, , ∴x ∈, , a2, ax, , 1, −3/ 2 ⎤, ⎡ 1, .a + a 2 ⎢ − ( ax ) a ⎥, 2 ax, ⎣ 2, ⎦, 3 −3, a a .a, f ′(a ) =, −, =0, 2a, 2, cos 2 x + sin 4 x, 23. If f ( x ) =, for x ∈ R , then f(2002)=, sin 2 x + cos 4 x, 1) 1, 2) 2, 3) 3, Ans: 1, cos 2 x + sin 4 x, Sol. f ( x ) =, sin 2 x + cos 4 x, 1, 1 − sin 2 2x, 4, =, =1, 1, 1 − sin 2x, 4, ⇒ f ( 2002 ) = 1, , [EAMCET 2002], 4) 4, , w, , w, , w, , .N, , f ′(x) =, , 24., , The function f : R → R is defined by f ( x ) = cos 2 x + sin 4 x for x ∈ R .Then f(R) =, ⎛3 ⎤, 1) ⎜ ,1⎥, ⎝4 ⎦, Ans: 3, , ⎡3 ⎞, 2) ⎢ ,1⎟, ⎣4 ⎠, , ⎡3 ⎤, 3) ⎢ ,1⎥, ⎣4 ⎦, , [EAMCET 2002], 3, ⎛, ⎞, 4) ⎜ ,1⎟, 4, ⎝, ⎠, , 6, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterma

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Functions, Sol. f ( x ) = cos 2 x + sin 4 x, , = cos 2 x + sin 2 x (1 − cos 2 x ), 1, = 1 − sin 2 2x, 4, 2, sin 2x ∈ [ 0,1], , 1, ( 0) = 1, 4, 1, 3, Minimum of f(x) = 1 − (1) =, 4, 4, ⎡3 ⎤, ∴ Range of f(x) = ⎢ ,1⎥, ⎣4 ⎦, If the functions f and g are defined by f ( x ) = 3x − 4, g ( x ) = 2 + 3x for x ∈, , ∴ Maximum of f(x) = 1 −, , 25., , g −1 ( f − 1 ( 5 ) ) =, , 3) 1/3, , x + 4 −1, x−2, ,g (x) =, 3, 3, , f −1 ( 5 ) = 3 g −1 ( f −1 ( 5 ) ) = g −1 ( 3) =, , If f ( x ) = ( 25 − x 4 ), , 1, 3, , ⎡ ⎛ 1 ⎞⎤, for 0 < x < 5 then f ⎢ f ⎜ ⎟ ⎥ =, ⎣ ⎝ 2 ⎠⎦, −3, 2) 2, 3) 2−2, , 1/ 4, , 1) 2−4, Ans: 4, , Sol. f ( x ) = ( 25 − x 4 ), , 1/ 4, , w, , ⇒ f ( f ( x ) ) = ⎡⎣ 25 − ( 25 − x 4 ) ⎤⎦, , 4) 2−1, , =x, , 1, = 2−1, 2, , w, , 27., , [EAMCET 2001], , w, , 1/ 4, , ∴ f ( f (1/ 2 ) ) =, , 4) 1/4, , .N, , 26., , .in, , 2) 1/2, , di, , Sol. f −1 ( x ) =, , [EAMCET 2002], , et, Ba, , 1) 1, Ans: 3, , respectively, then, , ⎧⎪ x / 2, Let z denote the set of all integers Define f : z → z by f ( x ) = ⎨, ⎪⎩ 0, , ( x is even ), . Then f is =, ( x is odd ), [EAMCET 2001], , 1) On to but not one-one, 3) One-one and onto, Ans: 1, Sol. --28., , 2) One –one but not onto, 4) Neither one-one nor onto, , ( x ≤ −1), , ⎧x + 2, ⎪, Let f : R → R be defined by f ( x ) = ⎨ x 2, ⎪2 − x, ⎩, f(1.5) is, 1) 0, Ans: 3, , ( −1 ≤ x ≤ 1) . Then the value of f(–1.75)+f(0.5) +, ( x ≥ 1), [EAMCET 2001], , 2) 2, , 3) 1, , 4) – 1, , 7, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterma

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Functions, , Sol. f ( −1.75 ) + f ( 0.5 ) + f (1.5 ), = ( −1.75 + 2 ) + ( 0.5 ) + 2 − 1.5 = 1, 2, , 29., , The functions f :, , → ,g :, , →, , are defined as follows:, , [EAMCET 2001], , .in, , ⎧⎪0 ( x rational ), ⎧⎪−1 ( x rational ), f (x) = ⎨, ; g (x) = ⎨, . The (f0g) (π) + (gof) (e) =, ⎪⎩1 ( x irrational ), ⎪⎩ 0 ( x irrational ), 1) –1, 2) 0, 3) 1, 4) 2, Ans: 1, Sol. O, = f ( 0 ) + g (1) ( ∵ π and e are irrationals), =0–1=-1, [EAMCET 2000], 30. If f : R → R is defined by f(x) = 2x + |x|, then f(2x) + f(–x) – f(x) =, 1) 2x, 2) 2|x|, 3) –2x, 4) –2|x|, Ans: 2, Sol. f ( x ) = 2x + x, = 2 ( 2x ) + 2x + 2 ( − x ) + − x − ( 2x + x ) = 2 x, , di, , ∴ f ( 2x ) + f ( − x ) − f ( x ), , If f : R → R and g : R → R are defined by f(x) = 2x + 3 and g(x) = x2+ 7, then the value of x for, which f(g(x)) = 25 are, [EAMCET 2000], 1) ±1, 2) ±2, 3) ±3, 4) ±4, Ans: 2, Sol. f ( g ( x ) ) = 25 ⇒ f ( x 2 + 7 ) = 25, , et, Ba, , 31., , w, , 2) {1, 2}, , [EAMCET 2000], , 3) {–1, –2}, , 4) {1, –2}, , w, , 1) {–1, 2}, Ans: 4, Sol. {1, −2} satisfies, , w, , 32., , ∴ x = ±2, {x ∈ R : x − 2 = x 2 } =, , .N, , ⇒ 2 ( x 2 + 7 ) + 3 = 25, , ™™™, , 8, , PDF Watermark Remover DEMO : Purchase from www.PDFWatermarkRemover.com to remove the waterma