Use app×
QUIZARD
QUIZARD
JEE MAIN 2026 Crash Course
NEET 2026 Crash Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
733 views
in Sets, Relations and Functions by (52.1k points)
retagged by

Find gof and fog when f: R → R and g : R → R is defined by 

(i) f(x) = 2x + 3 and g(x) = x2 + 5.

(ii) f(x) = 2x + x2 and g(x) = x3

(iii) f (x) = x2 + 8 and g(x) = 3x3 + 1

1 Answer

+1 vote
by (51.0k points)
selected by
 
Best answer

(i) Given, f: R → R and g: R → R

Therefore, gof: R → R and fog: R → R

Also given that f(x) = 2x + 3 and g(x) = x2 + 5

Now, (gof)(x) = g(f(x))

= g(2x +3)

= (2x + 3)2 + 5

= 4x+ 9 + 12x + 5

=4x+ 12x + 14

Now, (fog)(x) = f(g(x))

= f(x2 + 5)

= 2(x2 + 5) + 3

= 2x+ 10 + 3

= 2x2 + 13

(ii) Given as, f: R → R and g: R → R

Therefore, gof: R → R and fog: R → R

f(x) = 2x + x2 and g(x) = x3

(gof)(x)= g(f(x))

= g(2x + x2)

= (2x + x2)3

Now, (fog)(x) = f(g(x))

= f(x3)

= 2(x3) + (x3)2

= 2x+ x6

(iii) Given as, f: R → R and g: R → R

Therefore, gof: R → R and fog: R → R

f(x) = x2 + 8  and g(x) = 3x3 + 1

(gof)(x) = g(f(x))

= g(x2 + 8)

= 3(x2+8)3 + 1

Now, (fog) (x) = f (g (x))

= f(3x3 + 1)

= (3x3+1)2 + 8

= 9x6 + 6x+ 1 + 8

= 9x+ 6x+ 9

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...