Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
+1 vote
2.3k views
in Relations and Functions by (47.4k points)
closed by

A function f[- 3, 7 ) → R is defined as follows

\( f(x) = \begin{cases} 4x^2-1; & \quad \text{ } -3≤x<2 \text{ }\\ 3x-2;& \quad \text{} 2≤x≤4 \text{ }\\2x-3;& \quad \text{} 4<x<7 \end{cases}\)

Find 

(i) f(5) + f(6)

(ii) f(1) – f(-3)

(iii) f(-2) – f(4)

(iv) (f(3) + f(-1))/(2f(6) - f(1))

1 Answer

+1 vote
by (47.8k points)
selected by
 
Best answer

Given, f(x) = 4x2 – 1; x = {-3, -2, -1, 0, 1}

f(x) = 3x – 2; x = {2,3,4}

f(x) = 2x – 3; x = {5,6}

(i) f(5) + f(6)

f(x) = 2x – 3

f(5) = 2(5) – 3 = 10 – 3 = 7

f(6) = 2(6) – 3 = 12 – 3 = 9

∴ f(5) + f(6) = 7 + 9 = 16

(ii) f(1) – f(-3)

f(x) = 4x2 – 1

f(1) = 4(1)2 – 1 = 4 – 1 = 3

f(-3) = [4(-3)2 – 1]

= 4 (9) – 1

= 36 – 1 = 35

∴ f(1) – f(-3) = 3 – (35) = -32

(iii) f(-2) – f(4)

f(x) = 4x2 – 1

f(-2) = 4(-2)2 – 1 = 4(4) – 1 = 16 – 1 = 15

f(x) = 3x – 2

f(4) = [3(4) – 2] = 12 – 2 = 10

∴ f(-2) – f(4) = 15 – 10 = 5

(iv) (f(3) + f(-1))/(2f(6) - f(1))

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

...