{"id":7514,"date":"2024-01-21T02:26:28","date_gmt":"2024-01-20T20:56:28","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=7514"},"modified":"2024-01-23T09:39:00","modified_gmt":"2024-01-23T04:09:00","slug":"ts-inter-1st-year-maths-1a-functions-important-questions","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-inter-1st-year-maths-1a-functions-important-questions\/","title":{"rendered":"TS Inter 1st Year Maths 1A Functions Important Questions"},"content":{"rendered":"

Students must practice these\u00a0TS Inter 1st Year Maths 1A Important Questions<\/a> Chapter 1 Functions\u00a0to help strengthen their preparations for exams.<\/p>\n

TS Inter 1st Year Maths 1A Functions Important Questions<\/h2>\n

Very Short Answer Questions<\/span><\/p>\n

Question 1.
\nIf \\(A=\\left\\{0, \\frac{\\pi}{6}, \\frac{\\pi}{4}, \\frac{\\pi}{3}, \\frac{\\pi}{2}\\right\\}\\) surjection defined by f(x) = cos x, then find B
\nSolution:
\nf: A \u2192 B is a surjection
\n\u21d2 Codomain of B = Range f(A)
\nGiven f(x) = cos x
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 2.
\nfind the domain of the real-valued function f(x)=\\(\\frac{1}{\\log (2-x)}\\)
\nSolution:
\nf(x)=\\(\\frac{1}{\\log (2-x)}\\) is defined for 2 – x > 0 and 2 – x \u2260 1
\n\u21d2 x-2<0 and 2-x
\n\u21d2 x<2 and x \u2260 1 \u21d2 x \u2208(-\u221e, 2)- {1}
\n\u2234 Domain of f = {x \/ x \u2208 (-\u221e, 2)-{1}}<\/p>\n

Question 3.
\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.
\nSolution:
\ni) Given f, g are bijections, f, g are both one one and onto.
\nTo prove that gof : A \u2192 C is one one:
\n\"TS
\n(\u2235 f : A \u2192 B is one one, and
\ng : B \u2192 C are one one)
\n\u2234 gof : A \u2192 C is one one<\/p>\n

ii) To Prove that gof = A \u2192 C is onto:
\nLet C\u2208C; since g : B \u2192 C is on to \u2200 c \u2208 C \u2203
\nb E B such that g(b) = c …………………….. (1)
\nAlso f: A \u2192 B is on to for b\u2208B \u2203 a\u2208A such that f (a) = b …………………….. (2)
\n\u2234 bc = g(b) = g[f(a)]
\n= (gof) (a)
\nHence for c \u2208 C, \u2203 a \u2208 A such that
\ngof : A \u2192 C is onto
\nHence from the above two results
\ngof : A \u2192 C is a Bijection.<\/p>\n

Question 4.
\nIf f : A \u2192 B is a function and lA,IB are identity functions on A, B respectively then prove that folA<\/sub> lB<\/sub>of = f
\nSolution:
\ni) To prove that folA<\/sub> = f
\nSince f : A \u2192 B and \u2018A : A \u2192 A, we have folA<\/sub>:
\nA \u2192 B defined on the same domain A
\n\"TS<\/p>\n

ii) To prove that lB<\/sub>of = f
\nSince f : A \u2192 B and lB<\/sub> : B \u2192 B we have
\nlB<\/sub>of : A \u2192 B defined In the same domain A
\n\"TS<\/p>\n

Question 5.
\nIf f : A \u2192 B is a bijective function, then prove that (i) fof-1<\/sup> = IB<\/sub> (ii) f-1<\/sup>of = IA<\/sub>
\nSolution:
\nTo prove that fof-1<\/sup> = IB<\/sub>
\nGiven f: A \u2192 B is a bijection then we have
\nf-1 <\/sup>: B \u2192 A is also a bijection
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 6.
\nlf f : A \u2192 B, g : B\u2192C are two bijective functions then prove that (gof)-1<\/sup> = f-1 <\/sup>og-1<\/sup>
\nSolution:
\nGiven that f : A\u2192 B and g: B \u2192 C are bijections
\nwe have gof = A \u2192 C is a bijection.
\n\u2234 (gof)-1<\/sup> : C \u2192 A is also a bijection.
\nAlso since f : A\u2192B and g : B \u2192 C are bijections then f-1<\/sup>: B \u2192 A and g-1<\/sup>: c \u2192 B are bijections and hence f-1<\/sup>og-1<\/sup>; c \u2192 A is also a bijection.
\n(gof)-1<\/sup> and f-1<\/sup>og-1<\/sup> are two functions defined or the same domain C.
\nlet c E C and g : B \u2192 C is a bijection \u2203 unique b E B. Such that g(b)=c \u21d2 b=g-1<\/sup> (c)
\nAlso b \u2208 B and f: A \u2192 B is a bijection, \u2203 a unique a \u2208 A such that f (a) b \u21d2 a = f-1<\/sup>(b)
\n\"TS<\/p>\n

Question 7.
\nIf f : A \u2192 B and g : B\u2192 A are two functions such that gof = IA<\/sub> and fog = IB <\/sub>then g = f-1<\/sup>
\nSolution:
\ni) To prove that f is one one.
\n\"TS
\nSo these exists a reimage g(b) A for \u2018b\u2019 Under \u2018f\u2019
\n\u2234 f is onto.
\nHence \u2018f\u2019 is one one, onto and hence a bijection.
\n\u2234 f : B \u2192 A exists and is also one one onto<\/p>\n

\"TS<\/p>\n

iii) To prove g = f-1<\/sup>
\nNow g : B \u2192 A and f-1<\/sup>: B \u2192 A
\nWe have g and f-1 <\/sup>are del med in the same domain B.
\nLet a \u2208 A and b be the f image of \u2018a\u2019 where b\u2208B.
\n\"TS<\/p>\n

Question 8.
\nIf f : A\u2192 B, g : B\u2192C and h : C\u2192 Dare functions then ho (gof) = (hog) of
\nSolution :
\nGiven f : A \u2192 B and g : B \u2192 C we have
\ngof : A\u2192 C
\nNow gof: A \u2192 C and h: C \u2192 D we have
\nho(gof) : A\u2192 D, Also hog : B\u2192D and f : A \u2192 B
\nWe have (hog) of : A \u2192 D
\nHence (hog) of and ho(gof) are defined in the same domain A.
\nLet a \u2208 A then (ho(gof)] (a) = h [(gof) (a)]
\n= h [g [f(a)]
\n= (hog) [f(a)] = [(hog) of] (a)
\n\u2234 ho (gof) = (hog) of.<\/p>\n

Question 9.
\nOn what domain the functions f(x) = x2<\/sup>– 2x and g(x) = – x+6 are equal?
\nSolution:
\nf(x) = g(x)
\nx2<\/sup>– 2x = – x+6
\n= x2<\/sup>-x-6-0
\n= (x-3) (x+2) = 0 = x = -2,3
\n\u2234 f(x) and g(x) are equal on the domain { -2,3}<\/p>\n

Question 10.
\nFind the inverse of the function f(x) = 5x<\/sup>
\nSolution:
\nLet y = 5x<\/sup> = f(x) then x = f-1<\/sup>(y)
\nAlso x = log5<\/sub>y
\n\u2234 f1<\/sup>(y) = log5<\/sub>(y)
\n\u2234 f1<\/sup>(y) log5<\/sub>y \u21d2 f-1<\/sup>(x) = log5<\/sub>x<\/p>\n

Question 11.
\nIf f : R – {o} \u2192 R is deflued by f(x) = x+\\(\\frac{1}{x}\\)!,then prove that [f(x)]2<\/sup> = f(x2<\/sup>) + f(1)
\nSolution:
\n\"TS<\/p>\n

Question 12.
\nIf the function of defined by
\n\"TS
\nthen find the values If exist of f(4); f(2.5), f(-2), f(-4), f(0), f(-7)
\nSolution :
\ni) Since f(x)=3x-2 for x>3
\nf(4) = 3(4) – 2 = 10
\nDomain of f is
\n(- \u221e, – 3)\u222a[-2,2] u(3, \u221e)<\/p>\n

\"TS<\/p>\n

ii) f(2.5) does not exist since 2.5 does not belong to the domain of f.<\/p>\n

iii) f(x) = x2<\/sup>– 2 for x [-2,2]
\nWe have f(-2) = (2) -2 = 2<\/p>\n

iv) f(x) = 2x + I for x < – 3
\nf( – 4)=2(- 4)+ 1= – 7<\/p>\n

v)f(x)=x2 <\/sup>– 2 for x\u2208[-2,2] and f(0) = – 2.<\/p>\n

vi) f(x) = 2x + 1, for x < – 3
\nf(-7) = 2(-7)+ 1 = – 14+ 1 = – 13<\/p>\n

Question 13.
\nDetermine whether the function f: R \u2192 R defined by
\n\"TS
\nis an injection or a surjection or a bijection?
\nSolution :
\nBy definition of the function f(3) = 3
\nand f(1 )= 5(1) – 2 =3
\n\u2234 1 and 3 have same f image
\nHence f is not an injection.
\nLet y \u2208 R then y>2 or y\u22642
\nif y>2 take x = y \u2208 R so that 1(x) = x = y
\n\"TS
\n\u2234 f is a surjection.
\n\u2234 Since f is not an injection it is not a bijection.<\/p>\n

Question 14.
\nFind the domain of definition of the function y(x), given by the equation 2x<\/sup> +2y <\/sup>= 2.
\nSolution :
\n\"TS<\/p>\n

Question 15.
\nIf f : R\u2192 R defined as f(x+y)=f(x)+f(y)\u2200x, y \u2208 R and f(1) = 7, then find \\(\\sum_{r=1}^n f(r)\\)
\nSolution :
\nConsider
\nf(2)=f(1+1) = f(1)+f(1)=2f(1)
\nf(3) = 1(2 + 1) f(2) + f(1) = 2f(1) + f(1) = 3f(1)
\nS\u00ecmilarly f(r) = r f(1)
\n\"TS<\/p>\n

Question 16.
\nIf \\(f(x)=\\frac{\\cos ^2 x+\\sin ^4 x}{\\sin ^2 x+\\cos ^4 x}, \\forall x \\in R\\) then show that f(2012) = 1.
\nSolution :
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 17.
\nIf f : R\u2192 R, g: R \u2192 R defined by f(x) = 4x -1 and g(x)= x2<\/sup>+2 then find
\n(i) (gof) (x)
\n(ii) (gof) \\(\\left(\\frac{a+1}{4}\\right)\\)
\n(iii) (fof) (x)
\n(iv) go (fof) (0)
\nSolution :
\nGiven f(x) = 4x – 1 and g(x) = x2<\/sup>+2
\nWhere f: R \u2192 R and g: R \u2192 R then<\/p>\n

\u00a1) (gof)(x) = g[f(x)] = g[4x – 1]
\n=(4x – 1 )2+2= 16 x 2- 8x+3<\/p>\n

\"TS
\niii) (fof) (x) = f [f(x)] = f[4x -1]
\n= 4 (4x – 1) -1 = 16x – 5<\/p>\n

iv) [go (fof)] (0) = go [f(f(0)]
\n= go [f (-1)] = g[f(-1)]
\n= g[-5] = 25 + 2 = 27<\/p>\n

Question 18.
\nIf f : [0, 3] – [0, 3] is defined by
\n\"TS
\nthen show that f [0, 3] \u2286 [0, 3] and find fof
\nSolution:
\n\"TS<\/p>\n

Question 19.
\nIf f, g : R\u2192 R are defined by
\n\"TS
\nthen find (fog)\u03c0 + (gof) (e)
\nSolution:
\nWe have g(\u03c0) = 0, and f(e) = 1
\n\u2234 (fog) \u03c0 = f [g(\u03c0)] = f(0) = 0
\n(gof)(e)=g[f(e)]2g(1) = – 1
\n\u2234 (fog)(\u03c0)+(gof)(e) = 0 – 1 = – 1<\/p>\n

\"TS<\/p>\n

Question 20.
\nLet A = {1, 2,3} ,B = {a,b,c}, C = {p,q,r}.
\nIf f : A+B, g : B \u2192 C are defined by
\nf= ((1, a), (2, c), (3, b))
\ng = ((a, q), (b, r), (c, p)) then show that f-1<\/sup>og-1<\/sup> = (gof)-1
\n<\/sup>Solution:
\nGiven f : A \u2192 Band g: B \u2192 C we have
\nf-1 <\/sup>{(a, 1), (c, 2), (b, 3)}
\nand g-1<\/sup> = {(q, a), (r, b), (p, c)}
\nf-1<\/sup>og-1<\/sup> {(q, 1), (r, 3), (p, 2)}
\ngof = {(1, q), (2, p), (3, r)}
\n(gof)-1<\/sup> = ((q, 1), (p, 2), (r, 3))
\n\u2234 (gof)-1<\/sup>= f-1<\/sup>og-1<\/sup><\/p>\n

Question 21.
\nIf f : Q \u2192 Q defined by f(x) = 5x + 4 \u2200 x\u2208Q show that f is a bijection and find f-1<\/sup>.
\nSolution:
\nLet x1<\/sub>, x2 <\/sub>\u2208 Q then f(x1<\/sub>) = f(x2<\/sub>)
\n\u21d2 5x1<\/sub> + 4 = 5x2<\/sub> + 4
\n\u21d2 5x1<\/sub> = 5x2<\/sub> \u21d2 x1<\/sub> = x2<\/sub>
\n\u2234 f is an injection.
\n\"TS<\/p>\n

Question 22.
\nFind the domains of the following real-valued functions.<\/p>\n

(i) \\(f(x)=\\frac{1}{6 x-x^2-5}\\)
\nSolution:
\n\"TS<\/p>\n

\"TS<\/p>\n

(ii) \\(f(x)=\\frac{1}{\\sqrt{x^2-a^2}},(a>0)\\)
\nSolution:
\n\"TS<\/p>\n

(iii) \\(f(x)=\\sqrt{(x+2)(x-3)}\\)
\nSolution:
\n\\(f(x)=\\sqrt{(x+2)(x-3)} \\in R\\)
\n\"TS<\/p>\n

(iv) \\(\\mathbf{f}(\\mathbf{x})=\\sqrt{(\\mathbf{x}-\\alpha)(\\beta-\\mathbf{x})}, \\quad(0<\\alpha<\\beta)\\)
\nSolution:
\n\"TS<\/p>\n

(v) \\(f(x)=\\sqrt{2-x}+\\sqrt{1+x}\\)
\nSolution:
\n\"TS<\/p>\n

(vi) \\(f(x)=\\sqrt{x^2-1}+\\frac{1}{\\sqrt{x^2-3 x+2}}\\)
\nSolution:
\n\"TS<\/p>\n

\"TS<\/p>\n

(vii) \\(f(\\mathbf{x})=\\frac{1}{\\sqrt{|\\mathbf{x}|-\\mathbf{x}}}\\)
\nSolution:
\n\"TS<\/p>\n

(viii) \\(\\mathbf{f}(\\mathbf{x})=\\sqrt{|\\mathbf{x}|-\\mathbf{x}}\\)
\nSolution:
\n\"TS<\/p>\n

Question 23.
\nIf f = {(4, 5), (5, 6), (6, – 4) and g = ((4, -4), (6, 5), (8, 5)} then find
\ni) f+g
\nii) f – g
\niii) 2f + 4g
\niv) f+4
\nv) fg
\nvi) \\(\\frac{f}{g}\\)
\nvii) \\(|\\mathbf{f}|\\)
\nviii) \\(\\sqrt{f}\\)
\nix) f2<\/sup>
\nx) f3
\n<\/sup>Solution:
\nGiven f {(4, 5), (5, 6), (6, – 4)] and g = ((4,-4), (6, 5), (8, 5)) then domain of f = {4, 5, 6) and Range of f = {4, 6, 8)
\nDomain of f \u00b1 g = A B = (4, 6)
\n= (domain of f) \u2229 (domain of g)<\/p>\n

i) f+g={(4,5,-4)(6,-4+5))
\n= {(4, 1), (6, 1)}<\/p>\n

ii) f-g= {(4,5+4),(6,-4-5)}
\n= {(4,9), (6,-9)}<\/p>\n

iii) Domain of 2f = {4, 5, 6}
\nDomain of 4g (4, 6, 8)
\nDomain of 2f + 4g = (4, 6)
\n\u22342f = {(4, 10), (5, 12), (6, -8)}
\n4g = {(4, – 16), (6, 20), (8, 20)}
\n\u22342f + 4g = {(4,-6),(6,12)}<\/p>\n

\"TS<\/p>\n

iv) Domain of f + 4 = {4,5, 6}
\nf + 4 = {(4, 9), (5, 10), (6, 0)}<\/p>\n

v) Domain of fg = (domain of f) n (domain of g)
\nA\u2229B = {4, 6}
\n= {(4, (5) (-4), (6, (- 4), (5)}
\n= {(4, – 20), (6, – 20)}<\/p>\n

\"TS<\/p>\n

ix) Domain of f2<\/sup> = (Domain of f(x)] (4, 5, 6)
\n\u2234 f2<\/sup> = ((4, 25), (5, 36), (6, 16))<\/p>\n

x) Domain of f3<\/sup> = (4, 5, 6)
\n\u2234 f3<\/sup> = ((4, 125), (5, 216), (6, -64))<\/p>\n

\"TS<\/p>\n

Question 24.
\nFind the domains and ranges of the following real valued functions.
\n(i) \\(f(x)=\\frac{2+x}{2-x}\\)
\n(ii) \\(f(x)=\\frac{x}{1+x^2}\\)
\n(iii)\\(f(x)=\\sqrt{9-x^2}\\)
\nSolution:
\n\"TS
\n\"TS<\/p>\n

(ii) \\(f(x)=\\frac{x}{1+x^2}\\)
\nSolution:
\n\"TS<\/p>\n

\"TS<\/p>\n

(iii) \\(f(x)=\\sqrt{9-x^2}\\)
\nSolution:
\n\"TS
\n\"TS
\nBut f(x) posses only non negative values Range of f = [0, 3]<\/p>\n

Question 25.
\nIf f(x) = x2<\/sup> and g(x) = I x find the following functions.
\ni) f+g
\nii) f- g
\niii) fg
\niv) 2f
\nv) f2<\/sup>
\nvi) f+3
\nSolution:
\n\"TS<\/p>\n

Question 26.
\nDetermine whether the following functions are even or odd<\/p>\n

(i) f(x) = ax<\/sup> a -x<\/sup> + sin x
\nSolution :
\nA function f is said to be even if [(-x) = f(x) and odd If f(-x) = – f(x)
\nf(x) = ax<\/sup> a -x<\/sup> – sinx
\n= (ax<\/sup> a -x<\/sup> + sin x) = – f(x)
\n\u2234 f is an odd function.<\/p>\n

\"TS<\/p>\n

(ii) \\(f(x)=x\\left(\\frac{e^x-1}{e^x+1}\\right)\\)
\nSolution :
\nGiven \\(f(x)=x\\left(\\frac{e^x-1}{e^x+1}\\right)\\)
\n\"TS<\/p>\n

(iii) f(x) = log \\(\\left(x+\\sqrt{x^2+1}\\right)\\)
\nSolution :
\n\"TS<\/p>\n

Question 27.
\nFind the domains of the following real-valued functions.<\/p>\n

(i) \\(f(\\mathbf{x})=\\frac{1}{\\sqrt{[\\mathbf{x}]^2-[\\mathbf{x}]-2}}\\)
\nSolution:
\n\"TS
\n\"TS<\/p>\n

(ii) f(x) = log (x – [x])
\nSolution:
\nf(x) \u2208 R
\n\u21d4 x – [x] >0 \u21d4 x>[x]
\n\u21d4 x is not an integer.
\n\u2234 Domain of f is R – Z<\/p>\n

\"TS<\/p>\n

(iii) \\(f(x)=\\sqrt{\\log _{10}\\left(\\frac{3-x}{x}\\right)}\\)
\nSolution:
\n\"TS<\/p>\n

(iv) \\(f(x)=\\sqrt{x+2}+\\frac{1}{\\log _{10}(1-x)}\\)
\nSolution:
\n\"TS<\/p>\n

(v) \\(f(x)=\\frac{\\sqrt{3+x}+\\sqrt{3-x}}{x}\\)
\nSolution:
\n\"TS<\/p>\n","protected":false},"excerpt":{"rendered":"

Students must practice these\u00a0TS Inter 1st Year Maths 1A Important Questions Chapter 1 Functions\u00a0to help strengthen their preparations for exams. TS Inter 1st Year Maths 1A Functions Important Questions Very Short Answer Questions Question 1. If surjection defined by f(x) = cos x, then find B Solution: f: A \u2192 B is a surjection \u21d2 … Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/7514"}],"collection":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/comments?post=7514"}],"version-history":[{"count":12,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/7514\/revisions"}],"predecessor-version":[{"id":7781,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/7514\/revisions\/7781"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=7514"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=7514"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=7514"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}