{"id":34964,"date":"2022-08-24T17:20:54","date_gmt":"2022-08-24T11:50:54","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=34964"},"modified":"2022-08-24T17:21:01","modified_gmt":"2022-08-24T11:51:01","slug":"inter-1st-year-maths-1a-functions-solutions-ex-1b","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/inter-1st-year-maths-1a-functions-solutions-ex-1b\/","title":{"rendered":"Inter 1st Year Maths 1A Functions Solutions Ex 1(b)"},"content":{"rendered":"

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions<\/a> Inter 1st Year Maths 1A Functions Solutions Exercise 1(b) will help students to clear their doubts quickly.<\/p>\n

Intermediate 1st Year Maths 1A Functions Solutions Exercise 1(b)<\/h2>\n

I.<\/span><\/p>\n

Question 1.
\nIf f(x) = ex<\/sup> and g(x) = loge<\/sub>x, then show that f o g = g o f and find f-1<\/sup> and g-1<\/sup>.
\nSolution:
\nGiven f(x) = ex<\/sup> and g(x) = loge<\/sub>x
\nNow (f o g) (x) = f(g(x))
\n= f(loge<\/sub>x) [\u2235 g(x) = \\(\\log _{e} x\\)]
\n= \\(e^{\\left(\\log _{e} x\\right)}\\)
\n= x
\n\u2234 (fog) (x) = x ………(1)
\nand (g o f) (x) = g(f(x))
\n= g(ex<\/sup>) [\u2235 f(x) = ex<\/sup>]
\n= loge<\/sub> (ex<\/sup>) [\u2235 g(x) = loge<\/sub>x]
\n= x loge<\/sub> (e)
\n= x(1)
\n= x
\n\u2234 (g o f) (x) = x …….(2)
\nFrom (1) and (2)
\nf o g = g o f
\nGiven f(x) = ex<\/sup>
\nLet y = f(x) = ex<\/sup> \u21d2 x = f-1<\/sup>(y)
\nand y = ex<\/sup> \u21d2 x = loge<\/sub> (y)
\n\u2234 f-1<\/sup>(y) = loge<\/sub> (y) \u21d2 f-1<\/sup>(x) = loge<\/sub> (x)
\nLet y = g(x) = loge<\/sub> (x)
\n\u2235 y = g(x) \u21d2 x = g-1<\/sup>(y)
\n\u2235 y = loge<\/sub> (x) \u21d2 x = ey<\/sup>
\n\u2234 g-1<\/sup>(y) = ey<\/sup> \u21d2 g-1<\/sup>(x) = ex<\/sup>
\n\u2234 f-1<\/sup>(x) = loge<\/sub> (x) and g-1<\/sup>(x) = ex<\/sup><\/p>\n

\"Inter<\/p>\n

Question 2.
\nIf f(y) = \\(\\frac{y}{\\sqrt{1-y^{2}}}\\), g(y) = \\(\\frac{y}{\\sqrt{1+y^{2}}}\\) then show that (fog) (y) = y
\nSolution:
\nf(y) = \\(\\frac{y}{\\sqrt{1-y^{2}}}\\) and g(y) = \\(\\frac{y}{\\sqrt{1+y^{2}}}\\)
\nNow, (fog) (y) = f(g(y))
\n\"Inter
\n\u2234 (fog) (y) = y<\/p>\n

Question 3.
\nIf f : R \u2192 R, g : R \u2192 R are defined by f(x) = 2x2<\/sup> + 3 and g(x) = 3x – 2, then find
\n(i) (fog)(x)
\n(ii) (gof) (x)
\n(iii) (fof) (0)
\n(iv) go(fof) (3)
\nSolution:
\nf : R \u2192 R, g : R \u2192 R and f(x) = 2x2<\/sup> + 3; g(x) = 3x – 2
\n(i) (f o g) (x) = f(g(x))
\n= f(3x – 2) [\u2235 g(x) = 3x – 2]
\n= 2(3x- 2)2<\/sup>\u00a0+ 3 [\u2235 f(x) = 2x2<\/sup> + 3]
\n= 2(9x2<\/sup> – 12x + 4) + 3
\n= 18x2<\/sup> – 24x + 8 + 3
\n= 18x2<\/sup> – 24x + 11<\/p>\n

(ii) (gof) (x) = g(f(x))
\n= g(2x2<\/sup> + 3) [\u2235 f(x) = 2x2<\/sup> + 3]
\n= 3(2x2<\/sup> + 3) – 2 [\u2235 g(x) = 3x – 2]
\n= 6x2<\/sup> + 9 – 2
\n= 6x2<\/sup> + 7<\/p>\n

(iii) (fof) (0) = f(f(0))
\n= f(2(0) + 3) [\u2235 f(x) = 2x2<\/sup> + 3]
\n= f(3)
\n= 2(3)2<\/sup> + 3
\n= 18 + 3
\n= 21<\/p>\n

(iv) g o (f o f) (3)
\n= g o (f (f(3)))
\n= g o (f (2(3)2<\/sup> + 3)) [\u2235 f(x) = 2x2<\/sup> + 3]
\n= g o (f(21))
\n= g(f(21))
\n= g(2(21)2<\/sup> + 3)
\n= g(885)
\n= 3(885) – 2 [\u2235 g(x) = 3x – 2]
\n= 2653<\/p>\n

\"Inter<\/p>\n

Question 4.
\nIf f : R \u2192 R, g : R \u2192 R are defined by f(x) = 3x – 1, g(x) = x2<\/sup> + 1, then find
\n(i) (f o f) (x2<\/sup> + 1)
\n(ii) f o g (2)
\n(iii) g o f (2a – 3)
\nSolution:
\nf : R \u2192 R, g : R \u2192 R and f(x) = 3x – 1 ; g(x) = x2<\/sup> + 1
\n(i) (f o f) (x2<\/sup> + 1)
\n= f(f(x2<\/sup> + 1))
\n= f[3(x2<\/sup> + 1) – 1] [\u2235 f(x) = 3x – 1]
\n= f(3x2<\/sup> + 2)
\n= 3(3x2<\/sup> + 2) – 1
\n= 9x2<\/sup> + 5<\/p>\n

(ii) (f o g) (2)
\n= f(g(2))
\n= f(22<\/sup> + 1) [\u2235 g(x) = x2<\/sup> + 1]
\n= f(5)
\n= 3(5) – 1
\n= 14 [\u2235 f(x) = 3x – 1]<\/p>\n

(iii) (g o f) (2a – 3)
\n= g(f(2a – 3))
\n= g[3(2a – 3) – 1] [\u2235 f(x) = 3x – 1]
\n= g(6a – 10)
\n= (6a – 10)2<\/sup> + 1 [\u2235 g(x) = x2<\/sup> + 1]
\n= 36a2<\/sup> – 120a + 100 + 1
\n= 36a2<\/sup> – 120a + 101<\/p>\n

Question 5.
\nIf f(x) = \\(\\frac{1}{x}\\), g(x) = \u221ax for all x \u2208 (0, \u221e) then find (g o f) (x).
\nSolution:
\nf(x) = \\(\\frac{1}{x}\\), g(x) = \u221ax, \u2200 x \u2208 (0, \u221e)
\n(g o f) (x) = g(f(x))
\n= g(\\(\\frac{1}{x}\\)) [\u2235 f(x) = \\(\\frac{1}{x}\\)]
\n= \\(\\sqrt{\\frac{1}{x}}\\)
\n= \\(\\frac{1}{\\sqrt{x}}\\) [\u2235 g(x) = \u221ax]
\n\u2234 (gof) (x) = \\(\\frac{1}{\\sqrt{x}}\\)<\/p>\n

Question 6.
\nf(x) = 2x – 1, g(x) = \\(\\frac{x+1}{2}\\) for all x \u2208 R, find (g o f) (x).
\nSolution:
\nf(x) = 2x – 1, g(x) = \\(\\frac{x+1}{2}\\) \u2200 x \u2208 R
\n(g o f) (x) = g(f(x))
\n= g(2x – 1) [\u2235 f(x) = 2x – 1]
\n= \\(\\frac{(2 x-1)+1}{2}\\)
\n= x [\u2235 g(x) = \\(\\frac{x+1}{2}\\)]
\n\u2234 (g o f) (x) = x<\/p>\n

\"Inter<\/p>\n

Question 7.
\nIf f(x) = 2, g(x) = x2<\/sup>, h(x) = 2x for all x \u2208 R, then find (f o (g o h)) (x).
\nSolution:
\nf(x) = 2, g(x) = x2<\/sup>, h(x) = 2x, \u2200 x \u2208 R
\n[f o (g o h) (x)]
\n= [f o g (h(x))]
\n= f o g (2x) [\u2235 h(x) = 2x]
\n= f[g(2x)]
\n= f((2x)2<\/sup>) [\u2235 g(x) = x2<\/sup>]
\n= f(4x2<\/sup>) = 2 [\u2235 f(x) = 2]
\n\u2234 [f o (g o h) (x)] = 2<\/p>\n

Question 8.
\nFind the inverse of the following functions.
\n(i) a, b \u2208 R, f : R \u2192 R defined by f(x) = ax + b, (a \u2260 0).
\nSolution:
\na, b \u2208 R, f : R \u2192 R and f(x) = ax + b, a \u2260 0
\nLet y = f(x) = ax + b
\n\u21d2 y = f(x)
\n\u21d2 x = f-1<\/sup>(y) ……..(i)
\nand y = ax + b
\n\u21d2 x = \\(\\frac{y-b}{a}\\) ……..(ii)
\nFrom (i) and (ii)
\nf-1<\/sup>(y) = \\(\\frac{y-b}{a}\\)
\n\u21d2 f-1<\/sup>(x) = \\(\\frac{x-b}{a}\\)<\/p>\n

(ii) f : R \u2192 (0, \u221e) defined by f(x) = 5x<\/sup>
\nSolution:
\nf : R \u2192 (0, \u221e) and f(x) = 5x<\/sup>
\nLet y = f (x) = 5x<\/sup>
\ny = f(x) \u21d2 x = f-1<\/sup>(y) ……(i)
\nand y = 5x<\/sup> \u21d2 log5<\/sub> (y) = x ……..(ii)
\nFrom (i) and (ii)
\nf-1<\/sup>(y) = log5<\/sub>(y) \u21d2 f-1<\/sup>(x) = log5 (x)<\/p>\n

(iii) f : (0, \u221e) \u2192 R defined by f(x) = log2<\/sub> (x).
\nSolution:
\nf : (0, \u221e) \u2192 R and f(x) = log2<\/sub> (x)
\nLet y = f(x) = log2<\/sub> (x)
\n\u2235 y = f(x) \u21d2 x = f-1<\/sup>(y) ……..(i)
\nand y = log2<\/sub>(x) \u21d2 x = 2y
\nFrom (i) and (ii)
\nf-1<\/sup>(y) = 2y \u21d2 f-1<\/sup>(x) = 2x<\/p>\n

\"Inter<\/p>\n

Question 9.
\nIf f(x) = 1 + x + x2<\/sup> + …… for |x| < 1 then show that f-1<\/sup>(x) = \\(\\frac{x-1}{x}\\)
\nSolution:
\nf(x) = 1 + x + x2<\/sup> + ……..
\n\"Inter<\/p>\n

Question 10.
\nIf f : [1, \u221e) \u21d2 [1, \u221e) defined by f(x) = \\(2^{x(x-1)}\\) then find f-1<\/sup>(x).
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

II.<\/span><\/p>\n

Question 1.
\nIf f(x) = \\(\\frac{x-1}{x+1}\\), x \u2260 \u00b11, then verify (f o f-1<\/sup>) (x) = x.
\nSolution:
\nGiven f(x) = \\(\\frac{x-1}{x+1}\\), x \u2260 \u00b11
\nLet y = f(x) = \\(\\frac{x-1}{x+1}\\)
\n\u2235 y = f(x) \u21d2 x = f-1<\/sup>(y) ……(i)
\nand y = \\(\\frac{x-1}{x+1}\\)
\n\"Inter<\/p>\n

Question 2.
\nIf A = {1, 2, 3}, B = {\u03b1, \u03b2, \u03b3}, C = {p, q, r} and f : A \u2192 B, g : B \u2192 C are defined by f = {(1, \u03b1), (2, \u03b3), (3, \u03b2)}, g = {(\u03b1, q), (\u03b2, r), (\u03b3, p)}, then show that f and g are bijective functions and (gof)-1<\/sup> = f-1<\/sup> o g-1<\/sup>.
\nSolution:
\nA = {1, 2, 3}, B = {\u03b1, \u03b2, \u03b3},
\nf : A \u2192 B and f = {(1, \u03b1), (2, \u03b3), (3, \u03b2)}
\n\u21d2 f(1) = \u03b1, f(2) = \u03b3, f(3) = \u03b2
\n\u2235 Distinct elements of A have distinct f – images in B, f: A \u2192 B is an injective function.
\nRange of f = {\u03b1, \u03b3, \u03b2} = B(co-domain)
\n\u2234 f : A \u2192 B is a surjective function.
\nHence f : A \u2192 B is a bijective function.
\nB = {\u03b1, \u03b2, \u03b3}, C = {p, q, r}, g : B \u2192 C and g : {(\u03b1, q), (\u03b2, r), (\u03b3, p)}
\n\u21d2 g(\u03b1) = q, g(\u03b2) = r, g(\u03b3) = p
\n\u2234 Distinct elements of B have distinct g – images in C, g : B \u2192 C is an injective function.
\nRange of g = {q, r, p} = C, (co-domain)
\n\u2234 g : B \u2192 C is a surjective function
\nHence g : B \u2192 C is a bijective function
\nNow f = {(1, \u03b1), (2, \u03b3), (3, \u03b2)}
\ng = {(\u03b1, q), (\u03b2, r), (\u03b3, p)}
\ng o f = {(1, q), (2, p), (3, r)}
\n\u2234 (g o f)-1<\/sup> = {(q, 1), (r, 3), (p, 2)} ………(1)
\ng-1<\/sup> = {(q, \u03b1), (r, \u03b2), (p, \u03b3)}
\nf-1<\/sup> = {(\u03b1, 1), (\u03b3, 2),(\u03b2, 3)}
\nNow f-1<\/sup> o g-1<\/sup> = {(q, 1), (r, 3), (p, 2)} …….(2)
\nFrom eq’s (1) and (2)
\n(gof)-1<\/sup> = f-1<\/sup> o g-1<\/sup><\/p>\n

\"Inter<\/p>\n

Question 3.
\nIf f : R \u2192 R, g : R \u2192 R defined by f(x) = 3x – 2, g(x) = x2<\/sup> + 1, then find
\n(i) (g o f-1<\/sup>) (2)
\n(ii) (g o f)(x – 1)
\nSolution:
\nf : R \u2192 R, g : R \u2192 R and f(x) = 3x – 2
\nf is a bijective function \u21d2 its inverse exists
\nLet y = f(x) = 3x – 2
\n\u2235 y = f(x) \u21d2 x = f-1<\/sup>(y) …….(i)
\nand y = 3x – 2
\n\u21d2 x = \\(\\frac{y+2}{3}\\) ……..(ii)
\nFrom (i) and (ii)
\nf-1<\/sup>(y) = \\(\\frac{y+2}{3}\\)
\n\u21d2 f-1<\/sup>(x) = \\(\\frac{x+2}{3}\\)
\nNow (gof-1<\/sup>) (2)
\n= g(f-1<\/sup>(2))
\n\"Inter
\n\u2234 (g o f-1<\/sup>) (2) = \\(\\frac{25}{9}\\)<\/p>\n

(ii) (g o f) (x -1)
\n= g(f(x – 1))
\n= g(3(x – 1) – 2) [\u2235 f(x) = 3x – 2]
\n= g(3x – 5)
\n= (3x – 5)2<\/sup> + 1 [\u2235 g(x) = x2<\/sup> + 1]
\n= 9x2<\/sup> – 30x + 26
\n\u2234 (g o f) (x – 1) = 9x2<\/sup> – 30x + 26<\/p>\n

Question 4.
\nLet 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> o g-1<\/sup>
\nSolution:
\nf = {(1, a), (2, c), (4, d), (3, b)}
\n\u2234 f-1<\/sup> = {(a, 1), (c, 2), (d, 4), (b, 3)}
\ng-1<\/sup> = {(2, a), (4, b), (1, c), (3, d)}
\n\u2234 g = {(a, 2), (b, 4), (c, 1), (d, 3)}
\n(g o f) = {(1, 2), (2, 1), (4, 3), (3, 4)}
\n\u2234 (gof)-1<\/sup> = {(2, 1), (1, 2), (3, 4), (4, 3)} ……….(1)
\nf-1<\/sup> o g-1<\/sup> = {(2, 1), (4, 3), (1, 2), (3, 4)} ……..(2)
\nFrom eq’s (1) and (2), we observe (gof)-1<\/sup> = f-1<\/sup> o g-1<\/sup><\/p>\n

Question 5.
\nLet f : R \u2192 R, g : R \u2192 R be defined by f(x) = 2x – 3, g(x) = x3<\/sup> + 5 then find (f o g)-1<\/sup> (x).
\nSolution:
\nf : R \u2192 R, g : R \u2192 R and f(x) = 2x – 3 and g(x) = x3<\/sup> + 5
\nNow (fog) (x) = f(g(x))
\n= f(x3<\/sup> + 5) [\u2235 g(x) = x2<\/sup> + 5]
\n= 2(x3<\/sup> + 5) – 3 [\u2235 f(x) = 2x – 3]
\n= 2x3<\/sup> + 7
\n\u2234 (f o g) (x) = 2x3<\/sup> + 7
\nLet y = (f o g) (x) = 2x3<\/sup> + 7
\n\u2235 y = (fog)(x)
\n\u21d2 x = (fog)-1<\/sup> (y) …….(1)
\nand y = 2x3<\/sup> + 7
\n\u21d2 x3<\/sup> = \\(\\frac{y-7}{2}\\)
\n\u21d2 x = \\(\\left(\\frac{y-7}{2}\\right)^{\\frac{1}{3}}\\) …..(2)
\nFrom eq’s (1) and (2),
\n(f o g)-1<\/sup> (y) = \\(\\left(\\frac{y-7}{2}\\right)^{\\frac{1}{3}}\\)
\n\u2234 (f o g)-1<\/sup> (x) = \\(\\left(\\frac{x-7}{2}\\right)^{\\frac{1}{3}}\\)<\/p>\n

\"Inter<\/p>\n

Question 6.
\nLet f(x) = x2<\/sup>, g(x) = 2x<\/sup>. Then solve the equation (f o g) (x) = (g o f) (x)
\nSolution:
\nGiven f(x) = x2<\/sup> and g(x) = 2x<\/sup>
\nNow (f o g) (x) = f(g(x))
\n= f(2x<\/sup>) [\u2235 g(x) = 2x<\/sup>]
\n= (2x<\/sup>)2<\/sup>
\n= 22x<\/sup> [\u2235 f(x) = x2<\/sup>]
\n\u2234 (f o g) (x) = 22x<\/sup> ……(1)
\nand (g o f) (x) = g(f(x))
\n= g(x2<\/sup>) [\u2235 f(x) = x2<\/sup>]
\n= \\((2)^{x^{2}}\\) [\u2235 g(x) = 2x<\/sup>]
\n\u2234 (g o f) (x) = \\((2)^{x^{2}}\\)
\n\u2235 (f o g) (x) = (g o f) (x)
\n\u21d2 22x<\/sup> = \\((2)^{x^{2}}\\)
\n\u21d2 2x<\/sup> = x2<\/sup>
\n\u21d2 x2<\/sup> – 2x = 0
\n\u21d2 x(x – 2) = 0
\n\u21d2 x = 0, x = 2
\n\u2234 x = 0, 2<\/p>\n

Question 7.
\nIf f(x) = \\(\\frac{x+1}{x-1}\\), (x \u2260 \u00b11) then find (fofof) (x) and (fofofof) (x).
\nSolution:
\nf(x) = \\(\\frac{x+1}{x-1}\\), (x \u2260 \u00b11)
\n(i) (fofof) (x) = (fof) [f(x)]
\n= (fof) \\(\\left(\\frac{x+1}{x-1}\\right)\\) [\u2235 f(x) = \\(\\left(\\frac{x+1}{x-1}\\right)\\)]
\n\"Inter<\/p>\n

(ii) (fofofof) (x) = f[(f o f o f) (x)]
\n= f [f(x)] {from (1)}
\n\"Inter
\nIn the above problem if a number of f is even its answer is x and if a number of f is odd its answer is f(x).<\/p>\n","protected":false},"excerpt":{"rendered":"

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Functions Solutions Exercise 1(b) will help students to clear their doubts quickly. Intermediate 1st Year Maths 1A Functions Solutions Exercise 1(b) I. Question 1. If f(x) = ex and g(x) = logex, then show that f o g = g o …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[15],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/34964"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=34964"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/34964\/revisions"}],"predecessor-version":[{"id":34965,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/34964\/revisions\/34965"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=34964"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=34964"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=34964"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}