{"id":2454,"date":"2025-12-31T12:29:12","date_gmt":"2025-12-31T06:59:12","guid":{"rendered":"https:\/\/www.manabadi.co.in\/boards\/?p=2454"},"modified":"2026-01-03T11:34:50","modified_gmt":"2026-01-03T06:04:50","slug":"ts-inter-1st-year-maths-1a-solutions-cp-1-functions-ex-1b","status":"publish","type":"post","link":"https:\/\/www.manabadi.co.in\/boards\/ts-inter-1st-year-maths-1a-solutions-cp-1-functions-ex-1b\/","title":{"rendered":"TS Inter 1st Year Maths 1A Solutions Chapter 1 Functions Ex 1(b)"},"content":{"rendered":"\n

TS Inter 1st Year Maths 1A Functions Solutions Exercise 1(b)<\/h3>\n\n\n\n

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

Question 1.
If f(x) = ex<\/sup>, and g(x) = loge<\/sub>x, then show that fog = gof and find f-1<\/sup> and g-1<\/sup>.<\/h3>\n\n\n\n

Answer:
Given f(x) = ex and g(x) = loge<\/sub>x
Now (fog) (x) = f[g(x)] = f [loge<\/sub>x]
= elogex<\/sup> = x
(gof) (x) = g [f(x)] = g [ex<\/sup>] = loge<\/sub>ex<\/sup> = x
fog = gof
given f(x) = ex<\/sup> = y
then x = f-1<\/sup> (y) and y = ex \u21d2 x = loge<\/sub>y
f-1(y) = logey \u21d2 f-1<\/sup> (x) = loge<\/sub>x
similarly y = g(x) = logex<\/sup>
then x = g-1<\/sup> (y) and y = loge<\/sub>x
\u21d2 x = ey<\/sup>
g-1<\/sup> (y) = ey<\/sup> \u21d2 g-1<\/sup>(x) = ex<\/p>\n\n\n\n

Question 2.
If f(y) = y\/\u221a1\u2212y2<\/sup>, g(y) = y\/\u221a1+y2<\/sup> then show that (fog)(y) = y.<\/h3>\n\n\n\n

Answer:

\u2234 (fog) (y) = y<\/p>\n\n\n\n

Question 3. If R \u2192 R; g : R \u2192 R are defined by . f(x) = 2x2<\/sup> + 3 and g(x) = 3x \u2013 2, then find
(i) (fog) (x)
(ii) (gof) (x)
(iii) (fof)(0)
(iv) go (fof) (3)<\/h3>\n\n\n\n

Answer:
f; R \u2192 R; g : R \u2192 R and
f(x) = 2x2<\/sup> + 3, g(x) = 3x \u2013 2 then
(i) (fog) (x) = f [g (x)] = f (3x \u2013 2)
= 2 [(3x \u2013 2)2<\/sup>] + 3 (\u2235 f (x) = 2x2<\/sup> + 3)
= 2 [9x2<\/sup> \u2013 12x + 4] + 3
= 18x2<\/sup> \u2013 24x + 11
(ii) (gof) (x) = g [f (x)] = g (2x2<\/sup> + 3)
= 3 (2x2<\/sup> + 3) -2 = 6x2<\/sup> + 7<\/p>\n\n\n\n

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

iv) go (fof) (3)
= go [f (f (3))] (v f (x) = 2x2<\/sup> + 3)
= go [f (2(3)2<\/sup> + 3)]
= go [f (21)]
= g [2 (21)2<\/sup> + 3]
= g [2 (441) + 3]
= g [885]
= 3 (885) \u2013 2 = 2653 (\u2235 g(x) = 3x \u2013 2)<\/p>\n\n\n\n

Question 4.
If f:R \u2192 R, g:R \u2192 R are defined by f(x) = 3x \u2013 1, g(x) = x2<\/sup> + 1, then find
(i) (fof) (x2<\/sup> + 1)
(ii) (fog) (2) (March 2012)
(iii) (gof)(2a \u2013 3)<\/h3>\n\n\n\n

Answer:
Given f: R \u2192 R and g : R \u2192 R defined by f (x) = 3x \u2013 1, g (x) = x2<\/sup> + 1
(i)(fof) (x2<\/sup> + 1 ) = f [f (x2<\/sup> + 1)]
= f [3 (x2<\/sup> + 1) \u2013 1]
\u21d2 f [3x2<\/sup> + 2] (\u2235 f (x) = 3x \u2013 1)
= 3 (3\u00d72<\/sup> + 2) \u2013 1 = 9\u00d72<\/sup> + 5<\/p>\n\n\n\n

(ii) (fog) (2) = f [g (2)] = f [22<\/sup> + 1] = f [5]
= 3(5) \u2013 1 = 14<\/p>\n\n\n\n

(iii) (gof ) (2a \u2013 3)
=g[f(2a \u2013 3)]
= g[3(2a \u2013 3) \u2013 1] (\u2235 f(x) = 3x- 1)
= g [6a \u2013 10]
= (6a \u2013 10)2<\/sup> + 1 (\u2235 g(x)=x2<\/sup> + 1)
= 36a2<\/sup> \u2013 120a + 101<\/p>\n\n\n\n

Question 5.
If f(x) = 1\/x, g(x) = \u221ax \u2200 x \u2208 (0, \u221e) then find (gof)(x).<\/h3>\n\n\n\n

Answer:
(gof)(x) = g[f(x)] = g[1\/x]
= 1\/\u221ax (\u2235 g(x) = x)<\/p>\n\n\n\n

Question 6.
f(x) = 2x \u2013 1, g(x) = x+1\/2 \u2200 x \u2208 R, find (gof)(x).<\/h3>\n\n\n\n

Answer:
(gof)(x) = g[f(x)] = g(2x \u2013 1)
= 2x\u22121+1\/2 = x (\u2235 g(x) = 2x\u22121+1\/2)<\/p>\n\n\n\n

Question 7.
If f(x) = 2, g(x) = x2<\/sup>, h(x) = 2x \u2200 x \u2208 R, then find [fo(goh) (x)].<\/h3>\n\n\n\n

Answer:
fo(goh)<\/a>= fog [h(x)]
= fog [2x]
= f [g(2x)]
= f [ (2x)2<\/sup> ] = f (4x2<\/sup>) = 2
\u2234 fo(goh)<\/a> = 2<\/p>\n\n\n\n

Question 8.
Find the inverse of the following functions.
(i) a, b \u2208 R, f: R \u2192 R, defined by f(x) = ax + b, (a \u2260 0).<\/h3>\n\n\n\n

Answer:
a, b \u2208 R, f : R \u2192 R and f(x) = ax + b
\u21d2 y = ax + b = f(x)
\u21d2 x = f-1<\/sup>(y)
= y\u2212b\/a
\u2234 f-1(x) = x\u2212b\/a
(ii) f: R \u2192 (0, \u221e) defined by 5x<\/sup> (March 2011)
Answer:
f: R\u2192 (0, \u221e) and f(x) = 5x<\/sup>
Let y = f (x) = 5x<\/sup> \u21d2 x = f-1<\/sup>(y)
and x = log5<\/sub>y
\u2234 f-1<\/sup>(y) = log5<\/sub>y \u21d2 f-1<\/sup>(x) = log5<\/sub>x<\/p>\n\n\n\n

(iii) f : (0, \u221e) \u2192 R defined by f(x) = log2<\/sub>x
Answer:
Gii\u2019en f: (0, \u221e) \u2192 R defined by f(x) = log2<\/sub>x
Let y = f (x) = log2<\/sub>x then x = f1 (y)
y = log2<\/sub>x \u21d2 x = 2y<\/sup>
\u2234 f-1<\/sup>(y) = 2y \u21d2 f-1<\/sup>(x) = 2x<\/sup><\/p>\n\n\n\n

Question 9.
If f(x) = 1 + x + x2<\/sup> + \u2026\u2026\u2026\u2026.. for |x| < 1, then show that f-1(x) = x\u22121\/x<\/h3>\n\n\n\n

Answer:
Given f(x) = 1 + x + x2<\/sup> + \u2026\u2026\u2026. for |x| < 1
<\/p>\n\n\n\n

Question 10.
If f : [1, \u221e] \u2192 [1, \u221e] defined by f(x) = 2x(x \u2013 1)<\/sup>, then find f-1<\/sup>(x)<\/h3>\n\n\n\n

Answer:
Given f : [1, \u221e] \u2192 [1, \u221e] defined by f(x) = 2x(x \u2013 1)<\/sup>
Let y = f(x) then x = f-1(y)
Also y = 2x(x \u2013 1)<\/sup> \u21d2 x(x \u2013 1) = log2<\/sub>y
\u21d2 x2<\/sup> \u2013 x \u2013 log2<\/sub>y = 0

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

Question 1.
If f(x) = x\u22121\/x+1, x \u2260 \u00b11, then verify (fof-1<\/sup>)(x) = x<\/h3>\n\n\n\n

Answer:
Given f(x) = x\u22121\/x+1, (x \u2260 \u00b11)
and Let y = f(x) \u21d2 x = f-1<\/sup>(x)
<\/p>\n\n\n\n

Question 2.
If 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-1og-1<\/sup>.<\/h3>\n\n\n\n

Answer:
Given A = {1, 2, 3}, B = (\u03b1, \u03b2, \u03b3), c = {p, q, r) and f : A \u2192 B, g : B \u2192 C defined by f ={(1, \u03b1) (2, \u03b3), (3, \u03b2)}and g = {(a, q), (\u03b2, r), (\u03b3, p)}
From the definitions of f and g f (1) = \u03b1, f (2) = \u03b3, f (3) = \u03b2 and g (\u03b1) = q, g (\u03b2) = r, g (\u03b3) = p
Distinct elements of A have distinct imagine in B. Hence f is an Injection. Also, range of f = (a, y, P) and f is a surjection.
\u2234 f is abijection = B similarly distinct elements of B have distinct images in c and g is an Injection.
Also range of \u2018g\u2019 = {q, \u03b3, p} = C;
\u2234 g is a surjection.
Hence g is a bijection.
\u2234 f and g are bijective functions.
Also gof = {(1, q), (2, r), (3, p)}
and (gof-1) = {(q, 1), (r, 2), (p, 3)} \u2026\u2026\u2026\u2026\u2026.(1)
f-1 = {(\u03b1, 1), (\u03b3, 2), (\u03b2, 3)}
and g-1 = {(q, \u03b1), (r, \u03b2), (p, \u03b3)}
\u2234 f-1og-1 ={(q, 1), (r, 2), (p, 3)} \u2026\u2026\u2026\u2026\u2026\u2026(2)
\u2234 From (1) and (2), (gof-1<\/sup>) = f-1og-1<\/sup><\/p>\n\n\n\n

Question 3.
If f:R \u2192 R; g:R \u2192 R defined by f(x) = 3x \u2013 2, g(x) = x2<\/sup> + 1, then find
(i) (gof-1<\/sup>) (2)
(ii)(gof)(x \u2013 1) (March 2008, May 2006)<\/h3>\n\n\n\n

Answer:
Given f: R \u2192 R, g : R \u2192 R defined by f(x) = 3x \u2013 2, g(x) = x2<\/sup> + 1
et y = f (x) then x = f-1<\/sup> (y)
y = 3x \u2013 2 \u21d2 3x = y + 2
\u21d2 x = y+2\/3
\u2234 f-1<\/sup>(y) = 3+2\/3 \u21d2 f-1(x) = x+2\/3
\u2234 (i)(gof-1<\/sup>) (2) = g[f-1<\/sup>(2)] = g[4\/3]
= (4\/3)2<\/sup> + 1 = 16\/9 + 1 = 25\/9
(ii)(gof) (x \u2013 1) = g [f (x \u2013 1)
= g [3 (x \u2013 1) \u2013 2] = g [3x \u2013 5]
= (3x \u2013 5)2<\/sup> + 1
= 9x2<\/sup> \u2013 30x + 26
(\u2235 g(x) = x2<\/sup> + 1)<\/p>\n\n\n\n

Question 4.
Let f = {(1, a), (2, c), (4, d), (3, b)} and g-1 = {(2, a) (4, b), (1, c), (3, d)} then show that (gof)-1<\/sup> = f-1<\/sup>o g-1<\/sup><\/h3>\n\n\n\n

Answer:
Given f = {(1, a), (2, c), (4, d), (3, b)} and g-1 = {(2, a), (4, b), (1, c), (3, d)}
\u2234 g = {(a, 2), (b, 4), (c, 1), (d, 3)} gof = {(1, 2), (2, 1), (4, 3), (3, 4)}
\u2234 (gof)-1 = {(2, 1), (1, 2), (3, 4), (4, 3)}
f-1 = {(a, 1) (c, 2), (d, 4), (b, 3)}
g-1 = {(2, a), (4, b), (1, c), (3, d)}
f(x) = 3x \u2013 2, g(x) = x2<\/sup> + 1
Let y = f (x) then x = f\u201d (y)
\u2234 f-1o<\/sup> g-1<\/sup> = {(2, 1), (1, 2), (4, 3), (3, 4)}
(gof)-1 = f-1o g-1<\/p>\n\n\n\n

Question 5.
Let f:R \u2192 R; g:R \u2192 R be defined by f(x) = 2x \u2013 3, g(x) = x3<\/sup> + 5 then find (fog)-1(x)<\/h3>\n\n\n\n

Answer:
We have from the formula
(fog)-1<\/sup>(x) = (g-1<\/sup>of-1<\/sup>) \u2026\u2026\u2026\u2026..(1)
where f: R \u2192 R and g : R \u2192 R are defined by
f(x) = 2x \u2013 3 and g(x) = x3<\/sup> + 5
Let y = f(x) = 2x \u2013 3 : Then x = f-1<\/sup>(y)
and 2x \u2013 3 = y \u21d2 x = y+3\/2
f-1<\/sup>(x)x+3\/2 \u2026\u2026\u2026..(2)<\/p>\n\n\n\n

Let y = g(x) = x3<\/sup> + 5. Then x = g-1<\/sup>(y) and x3<\/sup> + 5 = y
\u21d2 x = (y \u2013 5)1\/3<\/sup>
g-1<\/sup>(y) = (y \u2013 5)1\/3<\/sup>
g-1<\/sup>(x) = (x \u2013 5)1\/3<\/sup> \u2026\u2026\u2026.(3)<\/p>\n\n\n\n

From (1), (g-1<\/sup>of-1<\/sup>)(x)
<\/p>\n\n\n\n

Question 6.
Let f(x) = x2<\/sup>,g(z) = 2x<\/sup>. Then solve the equation (fog) (x) = (gof) (x)<\/h3>\n\n\n\n

Answer:
Given f(x) = x2<\/sup> and g(x) = 2x<\/sup>
(fog) (x) = f [g(x)] = f [2x<\/sup>] = (2x<\/sup>)2<\/sup> = 22x<\/sup> \u2026\u2026\u2026\u2026\u2026(1)
and (gof)(x) = g[f(x)] = g[x2<\/sup>] = 2x2<\/sup>
\u2234 from (1) and (2), 22x<\/sup> = 2x<\/sup>2<\/sup>
\u21d2 x2<\/sup> \u2013 2x = 0
\u21d2 x(x \u2013 2) =0
\u21d2 x = 0, 2<\/p>\n\n\n\n

Question 7.
If f(x) = x+1\/x\u22121,(x \u2260 \u00b11),then find(fofof)(x) and (fofofof) (z)<\/h3>\n\n\n\n

Answer:
Given f(x) = x+1\/x\u22121, (x \u2260 \u00b1 1)
then (fofof) (x) = fof(f(x)]
<\/p>\n","protected":false},"excerpt":{"rendered":"

TS Inter 1st Year Maths 1A Functions Solutions Exercise 1(b) I. Question 1.If f(x) = ex, and g(x) = logex, then show that fog = gof and find f-1 and g-1. Answer:Given f(x) = ex and g(x) = logexNow (fog) (x) = f[g(x)] = f [logex]= elogex = x(gof) (x) = g [f(x)] = g […]<\/p>\n","protected":false},"author":2,"featured_media":2464,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1544,15,38],"tags":[],"class_list":["post-2454","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\/2454","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=2454"}],"version-history":[{"count":1,"href":"https:\/\/www.manabadi.co.in\/boards\/wp-json\/wp\/v2\/posts\/2454\/revisions"}],"predecessor-version":[{"id":2455,"href":"https:\/\/www.manabadi.co.in\/boards\/wp-json\/wp\/v2\/posts\/2454\/revisions\/2455"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.manabadi.co.in\/boards\/wp-json\/wp\/v2\/media\/2464"}],"wp:attachment":[{"href":"https:\/\/www.manabadi.co.in\/boards\/wp-json\/wp\/v2\/media?parent=2454"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.manabadi.co.in\/boards\/wp-json\/wp\/v2\/categories?post=2454"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.manabadi.co.in\/boards\/wp-json\/wp\/v2\/tags?post=2454"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}