हिंदी
तमिलनाडु बोर्ड ऑफ सेकेंडरी एज्युकेशनएसएसएलसी (अंग्रेजी माध्यम) कक्षा १०

A function f: [– 5, 9] → R is defined as follow : f(x) = {6x+1;-5≤x<25x2-1;2≤x<63x-4;6≤x≤9 Find f(– 3) + f(2) - Mathematics

Advertisements
Advertisements

प्रश्न

A function f: [– 5, 9] → R is defined as follow :

f(x) = `{{:(6x + 1";", -5 ≤ x < 2),(5x^2 - 1";", 2 ≤ x < 6),(3x - 4";", 6 ≤ x ≤ 9):}` Find f(– 3) + f(2)

योग

उत्तर

f(x) = 6x + 1; x = {– 5, – 4, – 3, – 2, – 1, 0, 1}

f(x) = 5x2 – 1; x = {2, 3, 4, 5}

f(x) = 3x – 4; x = {6, 7, 8, 9}

f(– 3) + f(2)

f(x) = 6x + 1

f(– 3) = 6(– 3) + 1 = – 18 + 1 = – 17

f(x) = 5x2 – 1

f(2) = 5(2)2 – 1 = 20 – 1 = 19

f(– 3) + f(2) = – 17 + 19 = 2

shaalaa.com
Special Cases of Functions
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 1: Relations and Functions - Exercise 1.4 [पृष्ठ २५]

APPEARS IN

सामाचीर कलवी Mathematics [English] Class 10 SSLC TN Board
अध्याय 1 Relations and Functions
Exercise 1.4 | Q 10. (i) | पृष्ठ २५
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×