{"id":21848,"date":"2024-01-04T11:38:17","date_gmt":"2024-01-04T06:08:17","guid":{"rendered":"https:\/\/apboardsolutions.in\/?p=21848"},"modified":"2024-01-05T10:54:31","modified_gmt":"2024-01-05T05:24:31","slug":"inter-1st-year-maths-1a-functions-solutions-ex-1c","status":"publish","type":"post","link":"https:\/\/apboardsolutions.in\/inter-1st-year-maths-1a-functions-solutions-ex-1c\/","title":{"rendered":"Inter 1st Year Maths 1A Functions Solutions Ex 1(c)"},"content":{"rendered":"

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

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

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

Question 1.
\nFind the domains of the following real-valued functions.
\n(i) f(x) = \\(\\frac{1}{\\left(x^{2}-1\\right)(x+3)}\\)
\nSolution:
\nf(x) = \\(\\frac{1}{\\left(x^{2}-1\\right)(x+3)}\\) \u2208 R
\n\u21d4 (x2<\/sup> – 1) (x + 3) \u2260 0
\n\u21d4 (x + 1) (x – 1) (x + 3) \u2260 0
\n\u21d4 x \u2260 -1, 1, -3
\n\u2234 Domain of f is R – {-1, 1, -3}<\/p>\n

(ii) f(x) = \\(\\frac{2 x^{2}-5 x+7}{(x-1)(x-2)(x-3)}\\)
\n\u21d4 (x – 1) (x – 2) (x – 3) \u2260 0
\n\u21d4 x \u2260 1, x \u2260 2, x \u2260 3
\n\u2234 Domain of f is R – {1, 2, 3}<\/p>\n

(iii) f(x) = \\(\\frac{1}{\\log (2-x)}\\)
\nSolution:
\nf(x) = \\(\\frac{1}{\\log (2-x)}\\)
\n\u21d4 log (2 – x) \u2260 0 and 2 – x > 0
\n\u21d4 (2 – x) \u2260 1 and 2 > x
\n\u21d4 x \u2260 1 and x < 2
\nx \u2208 (-\u221e, 1) \u222a (1, 2) (or) x \u2208 (-\u221e, 2) – {1}
\n\u2234 Domain of f is {(-\u221e, 2) – {1}}<\/p>\n

(iv) f(x) = |x – 3|
\nSolution:
\nf(x) = |x – 3| \u2208 R
\n\u21d4 x \u2208 R
\n\u2234 The domain of f is R<\/p>\n

(v) f(x) = \\(\\sqrt{4 x-x^{2}}\\)
\nSolution:
\nf(x) = \\(\\sqrt{4 x-x^{2}}\\) \u2208 R
\n\u21d4 4x – x2<\/sup> \u2265 0
\n\u21d4 x(4 – x) \u2265 0
\n\u21d4 x \u2208 [0, 4]
\n\u2234 Domain of f is [0, 4]<\/p>\n

\"Inter<\/p>\n

(vi) f(x) = \\(\\frac{1}{\\sqrt{1-x^{2}}}\\)
\nSolution:
\nf(x) = \\(\\frac{1}{\\sqrt{1-x^{2}}}\\) \u2208 R
\n\u21d4 1 – x2<\/sup> > 0
\n\u21d4 (1 + x) (1 – x) > 0
\n\u21d4 x \u2208 (-1, 1)
\n\u2234 Domain of f is {x\/x \u2208 (-1, 1)}<\/p>\n

(vii) f(x) = \\(\\frac{3^{x}}{x+1}\\)
\nSolution:
\nf(x) = \\(\\frac{3^{x}}{x+1}\\) \u2208 R
\n\u21d4 3x<\/sup> \u2208 R, \u2200 x \u2208 R and x + 1 \u2260 0
\n\u21d4 x \u2260 -1
\n\u2234 Domain of f is R – {-1}<\/p>\n

(viii) f(x) = \\(\\sqrt{x^{2}-25}\\)
\nSolution:
\nf(x) = \\(\\sqrt{x^{2}-25}\\) \u2208 R
\n\u21d4 x2<\/sup> – 25 \u2265 0
\n\u21d4 (x + 5) (x – 5) \u2265 0
\n\u21d4 x \u2208 (-\u221e, -5] \u222a [5, \u221e)
\n\u21d4 x \u2208 R – (-5, 5)
\n\u2234 Domain of f is R – (- 5, 5)<\/p>\n

(ix) f(x) = \\(\\sqrt{x-[x]}\\)
\nSolution:
\nf(x) = \\(\\sqrt{x-[x]}\\) \u2208 R
\n\u21d4 x – [x] \u2265 0
\n\u21d4 x \u2265 [x]
\n\u21d4 x \u2208 R
\n\u2234 Domain of f is R.<\/p>\n

(x) f(x) = \\(\\sqrt{[x]-x}\\)
\nSolution:
\nf(x) = \\(\\sqrt{[x]-x}\\) \u2208 R
\n\u21d4 [x] – x \u2265 0
\n\u21d4 [x] \u2265 x
\n\u21d4 x \u2264 [x]
\n\u21d4 x \u2208 Z
\n\u2234 The domain of f is z (Where z denotes a set of integers)<\/p>\n

\"Inter<\/p>\n

Question 2.
\nFind the ranges of the following real-valued functions.
\n(i) log|4 – x2<\/sup>|
\nSolution:
\nLet y = f(x) = log|4 – x2<\/sup>|
\nf(x) \u2208 R
\n\u21d4 4 – x2<\/sup> \u2260 0
\n\u21d4 x \u2260 \u00b12
\n\u2235 y = log|4 – x2<\/sup>|
\n\u21d2 |4 – x2<\/sup>| = ey<\/sup>
\n\u2235 ey<\/sup> > 0 \u2200 y \u2208 R
\n\u2234 The range of f is R.<\/p>\n

(ii) \\(\\sqrt{[x]-x}\\)
\nSolution:
\nLet y = f(x) = \\(\\sqrt{[x]-x}\\)
\nf(x) \u2208 R
\n\u21d4 [x] – x \u2265 0
\n\u21d4 x \u2264 [x]
\n\u21d4 x \u2208 z
\n\u2234 Domain of f is z. Then range of f is {0}<\/p>\n

(iii) \\(\\frac{\\sin \\pi[x]}{1+[x]^{2}}\\)
\nSolution:
\nLet f(x) = \\(\\frac{\\sin \\pi[x]}{1+[x]^{2}}\\) \u2208 R
\n\u21d4 x \u2208 R
\n\u2234 The domain of f is R
\nFor x \u2208 R, [x] is an integer,
\nsin \u03c0[x] = 0, \u2200 x \u2208 R [\u2235 sin n\u03c0 = 0, \u2200 n \u2208 z]
\n\u2234 Range of f is {0}<\/p>\n

(iv) \\(\\frac{x^{2}-4}{x-2}\\)
\nSolution:
\nLet y = f(x) = \\(\\frac{x^{2}-4}{x-2}\\) \u2208 R
\n\u21d4 y = \\(\\frac{(x+2)(x-2)}{x-2}\\)
\n\u21d4 x \u2260 2
\n\u2234 The domain of f is R – {2}
\nThen y = x + 2, [\u2235 x \u2260 2 \u21d2 y \u2260 4]
\nThen its range R – {4}<\/p>\n

(v) \\(\\sqrt{9+x^{2}}\\)
\nSolution:
\nLet y = f(x) = \\(\\sqrt{9+x^{2}}\\) \u2208 R
\nThe domain of f is R
\nWhen x = 0, f(0) = \u221a9 = 3
\nFor all values of x \u2208 R – {0}, f(x) > 3
\n\u2234 The range of f is [3, \u221e)<\/p>\n

Question 3.
\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)
\nSolution:
\n(i) (3f – 2g)(x)
\nf(x) = 2x – 1, g(x) = x2<\/sup>
\n(3f – 2g) (x) = 3f(x) – 2g(x)
\n= 3(2x – 1) – 2x2<\/sup>
\n= -2x2<\/sup> + 6x – 3<\/p>\n

(ii) (fg) (x)
\n= f(x) . g(x)
\n= (2x – 1) (x2<\/sup>)
\n= 2x3<\/sup> – x2<\/sup><\/p>\n

(iii) \\(\\left(\\frac{\\sqrt{f}}{g}\\right)(x)\\)
\n\\(\\frac{\\sqrt{f(x)}}{g(x)}=\\frac{\\sqrt{2 x-1}}{x^{2}}\\)<\/p>\n

(iv) (f + g + 2) (x)
\n= f(x) + g(x) + 2
\n= (2x – 1) + x2<\/sup> + 2
\n= x2<\/sup> + 2x + 1
\n= (x + 1)2<\/sup><\/p>\n

\"Inter<\/p>\n

Question 4.
\nIf f = {(1, 2), (2, -3), (3, -1)} then find
\n(i) 2f
\n(ii) 2 + f
\n(iii) f2<\/sup>
\n(iv) \u221af
\nSolution:
\nGiven f = {(1, 2), (2, -3), (3, -1)}
\n(i) 2f = {(1, 2 \u00d7 2), (2, 2(-3), (3, 2(-1))}
\n= {(1, 4), (2, -6), (3, -2)}<\/p>\n

(ii) 2 + f = {(1, 2 + 2), (2, 2 + (-3)), (3, 2 + (-1)}
\n= {(1, 4), (2, -1), (3, 1)}<\/p>\n

(iii) f2<\/sup> = {(1, 22), (2, (-3)2), (3, (-1)2)}
\n= {(1, 4), (2, 9), (3, 1)}<\/p>\n

(iv) \u221af = {(1, \u221a2)}
\n\u2235 \u221a-3 and \u221a-1 are not real<\/p>\n

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

Question 1.
\nFind the domains of the following real-valued functions.
\n(i) f(x) = \\(\\sqrt{x^{2}-3 x+2}\\)
\nSolution:
\nf(x) = \\(\\sqrt{x^{2}-3 x+2}\\) \u2208 R
\n\u21d4 x2<\/sup> – 3x + 2 \u2265 0
\n\u21d4 (x- 1) (x – 2) \u2265 0
\n\u21d4 x \u2208 (-\u221e, 1 ] \u222a [2, \u221e]
\n\u2234 The domain of f is R – (1, 2)<\/p>\n

(ii) f(x) = log(x2<\/sup> – 4x + 3)
\nSolution:
\nf(x) = log(x2<\/sup> – 4x + 3) \u2208 R
\n\u21d4 x2<\/sup> – 4x + 3 > 0
\n\u21d4 (x – 1) (x – 3) > 0
\n\u21d4 x \u2208 (-\u221e, 1) \u222a (3, \u221e)
\n\u2234 Domain of f is R – [1, 3]<\/p>\n

\"Inter<\/p>\n

(iii) f(x) = \\(\\frac{\\sqrt{2+x}+\\sqrt{2-x}}{x}\\)
\nSolution:
\nf(x) = \\(\\frac{\\sqrt{2+x}+\\sqrt{2-x}}{x}\\) \u2208 R
\n\u21d4 2 + x \u2265 0, 2 – x \u2265 0, x \u2260 0
\n\u21d4 x \u2265 -2, x \u2264 2, x \u2260 0
\n\u21d4 -2 \u2264 x \u2264 2, x \u2260 0
\n\u21d4 x \u2208 [-2, 2] – {0}
\nDomain of f is [-2, 2] – {0}<\/p>\n

(iv) f(x) = \\(\\frac{1}{\\sqrt[3]{(x-2)} \\log _{(4-x)} 10}\\)
\nSolution:
\nf(x) = \\(\\frac{1}{\\sqrt[3]{(x-2)} \\log _{(4-x)} 10}\\) \u2208 R
\n\u21d4 4 – x > 0, 4 – x \u2260 1 and x – 2 \u2260 0
\n\u21d4 x < 4, x \u2260 3, x \u2260 2
\n\u2234 Domain of f is (-\u221e, 4) – {2, 3}<\/p>\n

(v) f(x) = \\(\\sqrt{\\frac{4-x^{2}}{[x]+2}}\\)
\nSolution:
\nf(x) = \\(\\sqrt{\\frac{4-x^{2}}{[x]+2}}\\) \u2208 R
\nCase (i) 4 – x2<\/sup> \u2265 0 and [x] + 2 > 0 (or) Case (ii) 4 – x2<\/sup> \u2264 0 and [x] + 2 < 0
\nCase (i): 4 – x2<\/sup> \u2265 0 and [x] + 2 > 0
\n\u21d4 (2 – x) (2 + x) \u2265 0 and [x] > -2
\n\u21d4 x \u2208 [-2, 2] and x \u2208 [-1, \u221e]
\n\u21d4 x \u2208 [-1, 2] ……..(1)
\nCase (ii): 4 – x2<\/sup> \u2264 0 and [x] + 2 < 0
\n\u21d4 (2 + x) (2 – x) \u2264 0 and [x] < – 2
\n\u21d4 x \u2208 (-\u221e, -2] \u222a [2, \u221e] and x \u2208 (-\u221e, -2)
\n\u21d4 x \u2208 (-\u221e, -2) ……(2)
\nfrom (1) and (2),
\nDomain of f is (-\u221e, -2) \u222a [-1, 2]<\/p>\n

(vi) f(x) = \\(\\sqrt{\\log _{0.3}\\left(x-x^{2}\\right)}\\)
\nSolution:
\nf(x) = \\(\\sqrt{\\log _{0.3}\\left(x-x^{2}\\right)}\\) \u2208 R
\nThen log0.3<\/sub>(x – x2<\/sup>) \u2265 0
\n\u21d2 x – x2<\/sup> \u2264 (0.3)0<\/sup>
\n\u21d2 x – x2<\/sup> \u2264 1
\n\u21d2 -x2<\/sup> + x – 1 \u2264 0
\n\u21d2 x2<\/sup> – x + 1 \u2265 0
\nThis is true for all x \u2208 R ……..(1)
\nand x – x2<\/sup> \u2265 0
\n\u21d2 x2<\/sup> – x \u2264 0
\n\u21d2 x(x – 1) \u2264 0
\n\u21d2 x \u2208 (0, 1) …….(2)
\nFrom (1) and (2)
\nDomain of f is R ^ (0, 1) = (0, 1)
\n\u2234 The domain of f is (0, 1)<\/p>\n

(vii) f(x) = \\(\\frac{1}{x+|x|}\\)
\nSolution:
\nf(x) = \\(\\frac{1}{x+|x|}\\) \u2208 R
\n\u21d4 x + |x| \u2260 0
\n\u21d4 x \u2208 (0, \u221e)
\n\u2235 |x| = x, if x \u2265 0
\n|x| = -x, if x < 0
\n\u2234 The domain of f is (0, \u221e)<\/p>\n

\"Inter<\/p>\n

Question 2.
\nProve that the real valued function f(x) = \\(\\frac{x}{e^{x}-1}+\\frac{x}{2}+1\\) is an even function on R \\ {0}.
\nSolution:
\nf(x) \u2208 R, ex<\/sup> – 1 \u2260 0
\n\u21d2 ex<\/sup> \u2260 1
\n\u21d2 x \u2260 0
\n\"Inter
\n\u21d2 f(x) is an even function on R – {0}<\/p>\n

Question 3.
\nFind the domain and range of the following functions.
\n(i) f(x) = \\(\\frac{\\tan \\pi[x]}{1+\\sin \\pi[x]+\\left[x^{2}\\right]}\\)
\nSolution:
\nf(x) = \\(\\frac{\\tan \\pi[x]}{1+\\sin \\pi[x]+\\left[x^{2}\\right]}\\) \u2208 R
\n\u21d4 x \u2208 R, since [x] is an integer tan \u03c0[x] and sin \u03c0[x] each is zero for \u2200 x \u2208 R and f(x) \u2208 R
\nDomain of f is R
\nIts range = {0}<\/p>\n

(ii) f(x) = \\(\\frac{x}{2-3 x}\\)
\nSolution:
\n\"Inter<\/p>\n

(iii) f(x) = |x| + |1 + x|
\nSolution:
\nf(x) = |x| + |1 + x| \u2208 R
\n\u21d4 x \u2208 R
\n\u2234 Domain of f is R
\n\u2235 |x| = x, if x \u2265 0
\n= -x, if x < 0
\n|1 + x| = 1 + x, if x \u2265 -1
\n= -(1 + x) if x < -1
\nFor x = 0, f(0) = |0| + |1 + 0| = 1
\nx = 1, f(1) = |1| + |1 + 1| = 1 + 2 = 3
\nx = 2, f(2) = |2| + |1 + 2| = 2 + 3 = 5
\nx = -2, f(-2) = |-2| + |1 + (-2)| = 2 + 1 = 3
\nx = -1, f(-1) = |-1| + |1 +(-1)| = 1 + 0 = 1
\n\u2234 The range of f is [1, \u221e]<\/p>\n","protected":false},"excerpt":{"rendered":"

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Functions Solutions Exercise 1(c) will help students to clear their doubts quickly. Intermediate 1st Year Maths 1A Functions Solutions Exercise 1(c) I. Question 1. Find the domains of the following real-valued functions. (i) f(x) = Solution: f(x) = \u2208 R \u21d4 … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[14],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/21848"}],"collection":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/comments?post=21848"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/21848\/revisions"}],"predecessor-version":[{"id":21849,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/21848\/revisions\/21849"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/media?parent=21848"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/categories?post=21848"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/tags?post=21848"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}