{"id":2457,"date":"2026-01-01T16:16:32","date_gmt":"2026-01-01T10:46:32","guid":{"rendered":"https:\/\/www.manabadi.co.in\/boards\/?p=2457"},"modified":"2026-01-03T11:34:15","modified_gmt":"2026-01-03T06:04:15","slug":"ts-inter-1st-year-maths-1a-solutions-cp-1-functions-ex-1c","status":"publish","type":"post","link":"https:\/\/www.manabadi.co.in\/boards\/ts-inter-1st-year-maths-1a-solutions-cp-1-functions-ex-1c\/","title":{"rendered":"TS Inter 1st Year Maths 1A Solutions Chapter 1 Functions Ex 1(c)"},"content":{"rendered":"\n

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) = 1\/(x2<\/sup>\u22121)(x+3)
Answer:
Domain of f is the value of all real x for which (x2<\/sup> \u2013 1) (x + 3) \u2260 0
\u21d4 (x + 1) (x \u2013 1) (x + 3) \u2260 0
\u21d4 x \u2260 \u2013 1, 1, -3
\u2234 Domain of f is, R \u2013 {-1, 1, \u2013 3}
(ii) f(x) = 2x2<\/sup>\u22125x+7\/(x\u22121)(x\u22122)(x\u22123)
Answer:
Here (x \u2013 1) (x \u2013 2) (x \u2013 3) + 0
\u21d4 x \u2260 1, x \u2260 2, x \u2260 3.
Domain of f is, R \u2013 {1, 2, 3}
(iii) f(x) = 1\/log(2\u2212x)
Answer:
f(x) = 1\/log(2\u2212x) \u2208 R
\u21d4 log (2 \u2013 x) \u2260 0 and 2 \u2013 x > 0
\u21d4 2 \u2013 x \u2260 1 and 2 > x
\u21d4 x \u2260 1 and x < 2
\u21d4 x \u2208 (-\u221e, 1) U (1, 2)
(or) x \u2208 (-\u221e, 2) \u2013 {1}
Domain of f = x \u2208 {\u221e, 2} \u2013 {1}<\/p>\n\n\n\n

(iv) f(x) = |x \u2013 3|
Answer:
f(x) = |x \u2013 3| \u2208 R
\u21d4 x \u2208 R
\u2234 Domain of f = R
(v) f(x) = \u221a4x\u2212x2<\/sup>. (May 2005)
Answer:
f(x) = \u221a4x\u2212x2<\/sup> \u2208 R
\u21d4 4x \u2013 x2<\/sup> \u2265 0
\u21d4 x(4 \u2013 x) \u2265 0
\u21d4 x \u2208 [0, 4]
\u2234 Domain of f = [0, 4]
(vi) f(x) = 1\/\u221a1\u2212x2<\/sup>
Answer:
f(x) = 1\/\u221a1\u2212x2<\/sup>2 0 \u21d4 (1 \u2013 x)(1 \u2013 x) > 0 \u21d4 x \u2208 (-1, 1) \u2234 Domain of f = {x\/x \u2208 (-1, 1)} (vii) f(x) = 3x<\/sup>\/x+1
Answer:
f(x) = 3x<\/sup>\/x+1 \u2208 R
\u21d4 3x \u2208 R, \u2200 x \u2208 R and x + 1 \u2260 0
\u21d4 x \u2260 \u2013 1
\u2234 Domain of f = R \u2013 {-1}
(viii) f(x) = \u221ax2<\/sup>\u221225 (May 2012)
Answer:
f(x) = \u221ax2<\/sup>\u221225 \u2208 R
\u21d4 x2 \u2013 25 \u2265 0
\u21d4 (x + 5)(x \u2013 5) \u2265 0
\u21d4 x \u2208 (-\u221e, -5] \u222a [5, \u221e)
\u21d4 x \u2208 R \u2013 (-5, 5)
\u2234 Domain of f is R \u2013 (-5, 5)<\/p>\n\n\n\n

(ix) f(x) = \u221ax\u2212[x]
Answer:
f(x) = \u221ax\u2212[x] \u2208 R
\u21d4 x \u2013 [x] \u2265 0
\u21d4 x \u2265 [x]
\u21d4 x \u2208 R
\u2234 Domain of f is R
(x) f(x) = \u221a[x]\u2212x
Answer:
f(x) = \u221a[x]\u2212x \u2208 R
\u21d4 [x] \u2013 x \u2265 0
\u21d4 [x] \u2265 x
\u21d4 x \u2264 [x]
\u21d4 x \u2208 z
\u2234 Domain of f is Z (Set of injection)<\/p>\n\n\n\n

Question 2.
Find the ranges of the following real valued functions,
(i) log |4 \u2013 x2<\/sup>|<\/h3>\n\n\n\n

Answer:
Let y = f(x) = log |4 \u2013 x2<\/sup>| \u2208 R
\u21d4 4 \u2013 x2<\/sup> \u2260 0 \u21d2 x \u2260 \u00b1 2
y = log|4 \u2013 x2<\/sup>|
\u21d2 |4 \u2013 x2<\/sup>| = ey<\/sup>
ey<\/sup> > 0 \u2200 y \u2208 R
\u2234 Range of f is R.
(ii) \u221a[x]\u2212x
Answer:
Let y = f(x) = \u221a[x]\u2212x \u2208 R
\u21d4 [x] \u2013 x > 0
\u21d4 [x] \u2265 x \u21d4 x \u2264 [x]
\u2234 Domain of f is z
Then Range of f is {0}
\u2234 The Range of f = [1, \u221e]
(iii) sin\u03c0[x]\/1+[x]2<\/sup>
Answer:
Let y = f(x) = sin\u03c0[x]\/1+[x]2<\/sup> \u2208 R
\u21d4 x \u2208 R
\u2234 Domain of f is R
For x \u2208 R, [x] is an integer and sin n [x]= 0 \u2200 n \u2208 R Range of f is {0}
(iv) x2<\/sup>\u22124\/x\u22122
Answer:
Let y = f(x) = x2<\/sup>\u22124\/x\u22122 = (x + 2) \u21d4 x \u2013 2 \u2260 0
\u2234 Domain of f is R \u2013 {2}
Then y = x + 2 \u2234 x \u2260 2, we have y \u2260 4
\u2234 Range of f is R \u2013 {4}.
(v) \u221a9+x2<\/sup>
Answer:
Let y = \u221a9+x2<\/sup> f(x) \u2208 R
Domain of f is R.
When x = 0, f (0) = \u221a9 = \u00b1 3, But when f(0) = 3,
For all values of x e R \u2013 {0}, f (x) > 3
Range of f = {3, \u221e).<\/p>\n\n\n\n

Question 3.
If f and g are real valued functions defined by f(x) \u2013 2x \u2013 1 and g(x) = x2<\/sup>, then find
(i) (3f \u2013 2g)(x)<\/h3>\n\n\n\n

Answer:
(3f \u2013 2g) (x) = 3 f(x) \u2013 2 g(x)= 3 (2x \u2013 1) \u2013 2(x2<\/sup>)
= -2x2<\/sup> + 6x \u2013 3<\/p>\n\n\n\n

(ii) (fg) (x)
Answer:
(fg)(x) = f(x) g(x) = (2x \u2013 1)(x2<\/sup>) = 2x3<\/sup> \u2013 x2<\/sup><\/p>\n\n\n\n

(iii) (\u221af\/g)(x)
Answer:
\u221af(x)\/g(x)=\u221a2x\u22121\/x2<\/sup><\/p>\n\n\n\n

(iv) (f + g + 2)(x)
Answer:
(f + g + 2) (x) = f(x) + g(x) + 2
= 2x \u2013 1 + x2<\/sup> + 2
= x2<\/sup> + 2x + 1 = (x + 1)2<\/sup><\/p>\n\n\n\n

Question 4.
If f = {(1, 2), (2, -3), (3, -1)}, then find
(i) 2f
(ii) 2 + f
(iii) f2<\/sup>
(iv) \u221af
[May 2012, May 2008]<\/h3>\n\n\n\n

Answer:
Given f = {(1, 2), (2, -3), (3, -1)} we have f(1) = 2, f(2) = -3 and f(3) = -1
(i) 2f = {(1, 2 x 2), (2, 2 (-3), (3, 2(-1))}
= {(1. 4). (2, \u2013 6). (3, -2)}
(ii) 2 + f = {(1, 2+2), (2, 2+(-3), (3, 2+(-1))}
= {(1, 4), (2, -1), (3. 1)}<\/p>\n\n\n\n

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

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

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

Question 1.
Find the domains of the following real valued functions
(i) f(x)= \u221ax2<\/sup>\u22123x+2<\/h3>\n\n\n\n

Answer:
f(x) = \u221ax2<\/sup>\u22123x+2 \u2208 R
Domain of f is x2<\/sup> -3x + 2 > 0
\u21d2 (x \u2013 2) (x \u2013 1) > 0
\u21d2 x \u2208 [-\u221e, 1] u [2, \u221e]
\u2234 Domain of f = R \u2013 [1, 2]<\/p>\n\n\n\n

(ii) f (x) = log (x2<\/sup> \u2013 4x + 3)
Answer:
f(x) = log (x2<\/sup> \u2013 4x + 3) \u2208 R
\u21d4 x2<\/sup> \u2013 4x + 3 > 0
\u21d4 (x \u2013 3) (x \u2013 1) > 0
x \u2208 (-\u221e, 1) \u222a (3, \u221e)
Domain f = R \u2013 (1, 3)<\/p>\n\n\n\n

(iii) f(x) = \u221a2+x+\u221a2\u2212x\/x
Answer:
f(x) = \u221a2+x+\u221a2\u2212x\/x \u2208 R
\u21d4 2 + x > 0 2 \u2013 x > 0, x \u2260 0
\u21d4 x > -2, x < 2 x \u2260 0 \u21d4 -2 < x < 2, x \u2260 0 Domain of f is [-2, 2] \u2013 {0} (iv) f(x) = 1\/3\u221ax\u22122 log (4\u2212x)10<\/sup> Answer: f(x) = 1\/3\u221ax\u22122log(4\u2212x)10<\/sup> \u2208 R \u21d4 4 \u2013 x > 0, 4 \u2013 x \u2260 1 and x \u2013 2 \u2260 0
\u21d4 x < 4, x \u2260 3, x \u2260 2 Domain of f is [-\u221e, 4] \u2013 {2, 3} (v) f(x) = \u221a4\u2212x2<\/sup>\/[x]+2 Answer: f(x) = \u221a4\u2212x2<\/sup>\/[x]+2 \u2208 R \u21d4 4 \u2013 x > 0, [x] + 2 > 0 or
4 \u2013 x2<\/sup> < 0 and [x] > + 2 < 0 When 4 \u2013 x2<\/sup> > 0, and [x] + 2 > 0
we have (2 \u2013 x) (2 + x) > 0 and [x] > \u2013 2
\u21d4 x \u2208 [-2, 2] and x \u2208 [-1, \u221e)
\u21d4 x \u2208 [-1, 2] \u2026\u2026\u2026\u2026\u2026.(1)
When 4 \u2013 x2 < 0, and [x] + 2 < 0 \u21d4 (2 + x) (2 \u2013 x) < 0 and [x] + 2 < 0 \u21d4 x \u2208 [-\u221e, -2] \u222a [2, \u221e] and [x] < \u2013 2 \u21d4 x \u2208 [- \u221e, -2] \u222a [2, \u221e] and x \u2208 (- \u221e,-2) \u21d4 x \u2208 [-\u221e, -2] \u2026\u2026\u2026\u2026\u2026\u2026(2) \u2234 from (1) and (2) \u2234 Domain of f is [-\u221e, -2] \u222a {-1, 2} (vi) f(x) = \u221alog0.3<\/sub>(x\u2212x2<\/sup>) Answer: f(x) = \u221alog0.3<\/sub>(x\u2212x2<\/sup>) \u2208 R \u21d4 log0.3<\/sub> (x \u2013 x2) > 0 .
\u21d2 x \u2013 x2<\/sup> < (0.3) 0 \u21d2 x \u2013 x2<\/sup> < 1 \u21d2 -x2<\/sup> + x < 1 \u21d2 -x2<\/sup> + x \u2013 1 < 0 \u21d2 x2<\/sup> \u2013 x + 1 > 0
This is true for all x \u2208 R \u2026..(1)
and x \u2013 x2<\/sup> > 0
\u21d2 x2<\/sup> \u2013 x < 0
\u21d2 x (x \u2013 1) < 0
\u21d2 x \u2208 (0, 1) \u2026\u2026\u2026.(2)
\u2234 Domain of f is R n (0, 1) = (0, 1)
\u2234 Domain of f = (0, 1)
(vii) f(x) = 1\/x+|x|
Answer:
f(x) = 1\/x+|x| \u2208 R
\u21d4 x + |x| \u2260 0 \u21d2 x \u2208 (0, \u221e)
(\u2235 |x| = x if x \u2265 0
= -x if x < 0)
\u2234 Domain of f = (0, \u221e)<\/p>\n\n\n\n

Question 2.
Prove that the real valued function f(x) = x\/ex<\/sup>\u22121 + x\/2 + 1 is an even function on R \u2013 {0} \u2013<\/h3>\n\n\n\n

Answer:
f (x) \u2208 R, ex \u2013 1 \u2260 0
\u21d2 ex \u2260 1 \u21d2 x \u2260 0

Since f(-x) = f(x), the function f is even function on R \u2013 {0}.<\/p>\n\n\n\n

Question 3.
Find the domain and range of the following functions.
(i) f(x) = tan\u03c0[x]\/1+sin\u03c0[x]+[x2<\/sup>]<\/h3>\n\n\n\n

Answer:
f(x) = tan\u03c0[x]\/1+sin\u03c0[x]+[x2<\/sup>] \u2208 R
\u21d4 x \u2208 R; since [x] is an integar so that tan \u03c0 [x] and sin \u03c0 [x] are zero. \u2200 x \u2208 R
Domain of f is R and Range = {0}
(ii) f(x) = x\/2\u22123x
Answer:
f(x) = x\/2\u22123x \u2208 R
\u21d4 2 \u2013 3x \u2260 0 \u21d2 x \u2260 2\/3
\u2234 Domain of f = R \u2013 {2\/3}
Let y = f(x) = x\/2\u22123x
\u21d2 2y \u2013 3xy = x
\u21d2 2y = x(1 + 3y)
\u21d2 x = 2y\/1+3y
\u2234 x \u2208 R \u2013 {2\/3}, 1 + 3y \u2260 0
\u21d2 y \u2260 \u22121\/3
\u2234 Range of f = R \u2013 {\u22121\/3}
(iii) f(x) = |x| + |1 + x|
Answer:
f(x) \u2208 R \u21d4 x \u2208 R
Domain of f = R
\u2234 |x| = x if x > 0
= \u2013 x if x < 0 |1 + x| = 1 + x if 1 + x > 0 ie., x > -1
= \u2013 (1 + x) if 1 + x < 0 ie., x < \u2013 1
For x = 0, f(0) = 1,
x= 1, f(1) = |1| + |1 + 1| = 3
x = 2, f(2) = |2| + |1 + 2| = 2 + 3 = 5
x = -2, f(-2) = |-2| + |1 +(-2)| = 2 + 1 = 3
x = -1, f(-1) = |-1| + |1 + (-1)| = 1<\/p>\n","protected":false},"excerpt":{"rendered":"

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) = 1\/(x2\u22121)(x+3)Answer:Domain of f is the value of all real x for which (x2 \u2013 1) (x + 3) \u2260 0\u21d4 (x + 1) (x \u2013 1) (x + 3) \u2260 […]<\/p>\n","protected":false},"author":2,"featured_media":2463,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1544,15,38],"tags":[],"class_list":["post-2457","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-study-material-tg-inter","category-telangana","category-tg-inter"],"_links":{"self":[{"href":"https:\/\/www.manabadi.co.in\/boards\/wp-json\/wp\/v2\/posts\/2457","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.manabadi.co.in\/boards\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.manabadi.co.in\/boards\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.manabadi.co.in\/boards\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/www.manabadi.co.in\/boards\/wp-json\/wp\/v2\/comments?post=2457"}],"version-history":[{"count":2,"href":"https:\/\/www.manabadi.co.in\/boards\/wp-json\/wp\/v2\/posts\/2457\/revisions"}],"predecessor-version":[{"id":2524,"href":"https:\/\/www.manabadi.co.in\/boards\/wp-json\/wp\/v2\/posts\/2457\/revisions\/2524"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.manabadi.co.in\/boards\/wp-json\/wp\/v2\/media\/2463"}],"wp:attachment":[{"href":"https:\/\/www.manabadi.co.in\/boards\/wp-json\/wp\/v2\/media?parent=2457"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.manabadi.co.in\/boards\/wp-json\/wp\/v2\/categories?post=2457"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.manabadi.co.in\/boards\/wp-json\/wp\/v2\/tags?post=2457"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}