{"id":5590,"date":"2024-01-07T12:52:55","date_gmt":"2024-01-07T07:22:55","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=5590"},"modified":"2024-01-10T15:59:26","modified_gmt":"2024-01-10T10:29:26","slug":"ts-inter-1st-year-maths-1a-solutions-chapter-1-ex-1c","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-inter-1st-year-maths-1a-solutions-chapter-1-ex-1c\/","title":{"rendered":"TS Inter 1st Year Maths 1A Solutions Chapter 1 Functions Ex 1(c)"},"content":{"rendered":"

Students must practice these\u00a0TS Intermediate Maths 1A Solutions<\/a> Chapter 1 Functions Ex 1(c) to find a better approach to solving the problems.<\/p>\n

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

I.
\nQuestion 1.
\nFind the domains of the following real valued functions. (May 2014, Mar. 14)
\n(i) f(x) = \\(\\frac{1}{\\left(x^2-1\\right)(x+3)}\\)
\nAnswer:
\nDomain of f is the value of all real x for which (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)}\\)
\nAnswer:
\nHere (x – 1) (x – 2) (x – 3) + 0
\n\u21d4 x \u2260 1, x \u2260 2, x \u2260 3.
\nDomain of f is, R – {1, 2, 3}<\/p>\n

(iii) f(x) = \\(\\frac{1}{\\log (2-x)}\\)
\nAnswer:
\nf(x) = \\(\\frac{1}{\\log (2-x)}\\) \u2208 R
\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
\n\u21d4 x \u2208 (-\u221e, 1) U (1, 2)
\n(or) x \u2208 (-\u221e, 2) – {1}
\nDomain of f = x \u2208 {\u221e, 2} – {1}<\/p>\n

(iv) f(x) = |x – 3|
\nAnswer:
\nf(x) = |x – 3| \u2208 R
\n\u21d4 x \u2208 R
\n\u2234 Domain of f = R<\/p>\n

(v) f(x) = \\(\\sqrt{4 x-x^2}\\). (May 2005)
\nAnswer:
\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 = [0, 4]<\/p>\n

(vi) f(x) = \\(\\frac{1}{\\sqrt{1-x^2}}\\)
\nAnswer:
\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 = {x\/x \u2208 (-1, 1)}<\/p>\n

(vii) f(x) = \\(\\frac{3^x}{x+1}\\)
\nAnswer:
\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 = R – {-1}<\/p>\n

(viii) f(x) = \\(\\sqrt{x^2-25}\\) (May 2012)
\nAnswer:
\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-[\\mathrm{x}]}\\)
\nAnswer:
\nf(x) = \\(\\sqrt{x-[\\mathrm{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{[\\mathbf{x}]-\\mathbf{x}}\\)
\nAnswer:
\nf(x) = \\(\\sqrt{[\\mathbf{x}]-\\mathbf{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 Domain of f is Z (Set of injection)<\/p>\n

Question 2.
\nFind the ranges of the following real valued functions,
\n(i) log |4 – x2<\/sup>|
\nAnswer:
\nLet y = f(x) = log |4 – x2<\/sup>| \u2208 R
\n\u21d4 4 – x2<\/sup> \u2260 0 \u21d2 x \u2260 \u00b1 2
\ny = log|4 – x2<\/sup>|
\n\u21d2 |4 – x2<\/sup>| = ey<\/sup>
\ney<\/sup> > 0 \u2200 y \u2208 R
\n\u2234 Range of f is R.<\/p>\n

(ii) \\(\\sqrt{[\\mathbf{x}]-\\mathbf{x}}\\)
\nAnswer:
\nLet y = f(x) = \\(\\sqrt{[\\mathbf{x}]-\\mathbf{x}}\\) \u2208 R
\n\u21d4 [x] – x > 0
\n\u21d4 [x] \u2265 x \u21d4 x \u2264 [x]
\n\u2234 Domain of f is z
\nThen Range of f is {0}
\n\u2234 The Range of f = [1, \u221e]<\/p>\n

(iii) \\(\\frac{\\sin \\pi[x]}{1+[x]^2}\\)
\nAnswer:
\nLet y = f(x) = \\(\\frac{\\sin \\pi[x]}{1+[x]^2}\\) \u2208 R
\n\u21d4 x \u2208 R
\n\u2234 Domain of f is R
\nFor x \u2208 R, [x] is an integer and sin n [x]= 0 \u2200 n \u2208 R Range of f is {0}<\/p>\n

(iv) \\(\\frac{x^2-4}{x-2}\\)
\nAnswer:
\nLet y = f(x) = \\(\\frac{x^2-4}{x-2}\\) = (x + 2) \u21d4 x – 2 \u2260 0
\n\u2234 Domain of f is R – {2}
\nThen y = x + 2 \u2234 x \u2260 2, we have y \u2260 4
\n\u2234 Range of f is R – {4}.<\/p>\n

(v) \\(\\sqrt{9+x^2}\\)
\nAnswer:
\nLet y = \\(\\sqrt{9+x^2}\\) f(x) \u2208 R
\nDomain of f is R.
\nWhen x = 0, f (0) = \u221a9 = \u00b1 3, But when f(0) = 3,
\nFor all values of x e R – {0}, f (x) > 3
\nRange of f = {3, \u221e).<\/p>\n

\"TS<\/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)
\nAnswer:
\n(3f – 2g) (x) = 3 f(x) – 2 g(x)= 3 (2x – 1) – 2(x2<\/sup>)
\n= -2x2<\/sup> + 6x – 3<\/p>\n

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

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

(iv) (f + g + 2)(x)
\nAnswer:
\n(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 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
\n[May 2012, May 2008]
\nAnswer:
\nGiven f = {(1, 2), (2, -3), (3, -1)} we have f(1) = 2, f(2) = -3 and f(3) = -1
\n(i) 2f = {(1, 2 x 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)| (\u2235 \u221a-3 and \u221a-1 are not real)<\/p>\n

II.
\nQuestion 1.
\nFind the domains of the following real valued functions
\n(i) f(x)= \\(\\sqrt{x^2-3 x+2}\\)
\nAnswer:
\nf(x) = \\(\\sqrt{x^2-3 x+2}\\) \u2208 R
\nDomain of f is x2<\/sup> -3x + 2 > 0
\n\u21d2 (x – 2) (x – 1) > 0
\n\u21d2 x \u2208 [-\u221e, 1] u [2, \u221e]
\n\u2234 Domain of f = R – [1, 2]<\/p>\n

(ii) f (x) = log (x2<\/sup> – 4x + 3)
\nAnswer:
\nf(x) = log (x2<\/sup> – 4x + 3) \u2208 R
\n\u21d4 x2<\/sup> – 4x + 3 > 0
\n\u21d4 (x – 3) (x – 1) > 0
\nx \u2208 (-\u221e, 1) \u222a (3, \u221e)
\nDomain f = R – (1, 3)<\/p>\n

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

(iv) f(x) = \\(\\frac{1}{\\sqrt[3]{x-2} \\log (4-x)^{10}}\\)
\nAnswer:
\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
\nDomain of f is [-\u221e, 4] – {2, 3}<\/p>\n

(v) f(x) = \\(\\sqrt{\\frac{4-x^2}{[x]+2}}\\)
\nAnswer:
\nf(x) = \\(\\sqrt{\\frac{4-x^2}{[x]+2}}\\) \u2208 R
\n\u21d4 4 – x > 0, [x] + 2 > 0 or
\n4 – x2<\/sup> < 0 and [x] > + 2 < 0
\nWhen 4 – x2<\/sup> > 0, and [x] + 2 > 0
\nwe have (2 – x) (2 + x) > 0 and [x] > – 2
\n\u21d4 x \u2208 [-2, 2] and x \u2208 [-1, \u221e)
\n\u21d4 x \u2208 [-1, 2] …………….(1)
\nWhen 4 – x2<\/sup> < 0, and [x] + 2 < 0
\n\u21d4 (2 + x) (2 – x) < 0 and [x] + 2 < 0
\n\u21d4 x \u2208 [-\u221e, -2] \u222a [2, \u221e] and [x] < – 2
\n\u21d4 x \u2208 [- \u221e, -2] \u222a [2, \u221e] and x \u2208 (- \u221e,-2)
\n\u21d4 x \u2208 [-\u221e, -2] ………………(2)
\n\u2234 from (1) and (2)
\n\u2234 Domain of f is [-\u221e, -2] \u222a {-1, 2}<\/p>\n

(vi) f(x) = \\(\\sqrt{\\log _{0.3}\\left(x-x^2\\right)}\\)
\nAnswer:
\nf(x) = \\(\\sqrt{\\log _{0.3}\\left(x-x^2\\right)}\\) \u2208 R
\n\u21d4 log0.3<\/sub> (x – x2<\/sup>) > 0 .
\n\u21d2 x – x2<\/sup> < (0.3) 0
\n\u21d2 x – x2<\/sup> < 1
\n\u21d2 -x2<\/sup> + x < 1
\n\u21d2 -x2<\/sup> + x – 1 < 0
\n\u21d2 x2<\/sup> – x + 1 > 0
\nThis is true for all x \u2208 R …..(1)
\nand x – x2<\/sup> > 0
\n\u21d2 x2<\/sup> – x < 0
\n\u21d2 x (x – 1) < 0
\n\u21d2 x \u2208 (0, 1) ……….(2)
\n\u2234 Domain of f is R n (0, 1) = (0, 1)
\n\u2234 Domain of f = (0, 1)<\/p>\n

(vii) f(x) = \\(\\frac{1}{x+|\\mathrm{x}|}\\)
\nAnswer:
\nf(x) = \\(\\frac{1}{x+|\\mathrm{x}|}\\) \u2208 R
\n\u21d4 x + |x| \u2260 0 \u21d2 x \u2208 (0, \u221e)
\n(\u2235 |x| = x if x \u2265 0
\n= -x if x < 0)
\n\u2234 Domain of f = (0, \u221e)<\/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} –
\nAnswer:
\nf (x) \u2208 R, ex<\/sup> – 1 \u2260 0
\n\u21d2 ex<\/sup> \u2260 1 \u21d2 x \u2260 0
\n\"TS-Inter-1st-Year-Maths-1A-Solutions-Chapter-1-Functions-Ex-1c-1\"
\nSince f(-x) = f(x), the function f is even function on R – {0}.<\/p>\n

\"TS<\/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]}\\)
\nAnswer:
\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 integar so that tan \u03c0 [x] and sin \u03c0 [x] are zero. \u2200 x \u2208 R
\nDomain of f is R and Range = {0}<\/p>\n

(ii) f(x) = \\(\\frac{x}{2-3 x}\\)
\nAnswer:
\nf(x) = \\(\\frac{x}{2-3 x}\\) \u2208 R
\n\u21d4 2 – 3x \u2260 0 \u21d2 x \u2260 \\(\\frac{2}{3}\\)
\n\u2234 Domain of f = R – {\\(\\frac{2}{3}\\)}<\/p>\n

Let y = f(x) = \\(\\frac{x}{2-3 x}\\)
\n\u21d2 2y – 3xy = x
\n\u21d2 2y = x(1 + 3y)
\n\u21d2 x = \\(\\frac{2 \\mathrm{y}}{1+3 \\mathrm{y}}\\)
\n\u2234 x \u2208 R – {\\(\\frac{2}{3}\\)}, 1 + 3y \u2260 0
\n\u21d2 y \u2260 \\(\\frac{-1}{3}\\)
\n\u2234 Range of f = R – {\\(\\frac{-1}{3}\\)}<\/p>\n

(iii) f(x) = |x| + |1 + x|
\nAnswer:
\nf(x) \u2208 R \u21d4 x \u2208 R
\nDomain of f = R
\n\u2234 |x| = x if x > 0
\n= – x if x < 0 |1 + x| = 1 + x if 1 + x > 0 ie., x > -1
\n= – (1 + x) if 1 + x < 0 ie., x < – 1
\nFor x = 0, f(0) = 1,
\nx= 1, f(1) = |1| + |1 + 1| = 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<\/p>\n","protected":false},"excerpt":{"rendered":"

Students must practice these\u00a0TS Intermediate Maths 1A Solutions Chapter 1 Functions Ex 1(c) to find a better approach to solving the problems. TS Inter 1st Year Maths 1A Functions Solutions Exercise 1(c) I. Question 1. Find the domains of the following real valued functions. (May 2014, Mar. 14) (i) f(x) = Answer: Domain of f … Read more<\/a><\/p>\n","protected":false},"author":2,"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\/5590"}],"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\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/comments?post=5590"}],"version-history":[{"count":2,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/5590\/revisions"}],"predecessor-version":[{"id":5593,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/5590\/revisions\/5593"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=5590"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=5590"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=5590"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}